question_slug
stringlengths 3
77
| title
stringlengths 1
183
| slug
stringlengths 12
45
| summary
stringlengths 1
160
⌀ | author
stringlengths 2
30
| certification
stringclasses 2
values | created_at
stringdate 2013-10-25 17:32:12
2025-04-12 09:38:24
| updated_at
stringdate 2013-10-25 17:32:12
2025-04-12 09:38:24
| hit_count
int64 0
10.6M
| has_video
bool 2
classes | content
stringlengths 4
576k
| upvotes
int64 0
11.5k
| downvotes
int64 0
358
| tags
stringlengths 2
193
| comments
int64 0
2.56k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
identify-the-largest-outlier-in-an-array
|
SIMPLEST SOLUTION POSSIBLE || BINARY SEARCH :-
|
simplest-solution-possible-binary-search-2sd6
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\ni first started with sorting the array ,so that i can use binary search .
|
Rxhabh_
|
NORMAL
|
2024-12-01T05:25:41.981193+00:00
|
2024-12-01T05:25:41.981225+00:00
| 1,324 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\ni first started with sorting the array ,so that i can use binary search .\n```\nsort(nums.begin(),nums.end());\n```\nnext , i calculated the sum of the complete array .\n```\nint s = 0; // sum of the complete array\n int n = nums.size();\n int ans = INT_MIN;\n for(int i = 0;i<n;i++){\n s+=nums[i];\n }\n```\nthen , i iterate the complete array and for each element i was looking for outlier which must be equal to the (sum of array - 2* nums[i]). \nthen i applied binary search for the rest parts of the array that means , for previous part(from 0 to i-1) and the next part(from i+1 to n-1) .\n```\nfor(int i = 0;i<n;i++){\n int x = s-(2*nums[i]); // the outlier must be equal to sum of array - (2* nums[i]).\n // applying binary search\n int lo = i+1,hi = n-1;\n while(lo<=hi){\n int mid = lo+(hi-lo)/2;\n if(nums[mid] == x ) {if(x>ans) ans = x; break;}\n else if(nums[mid] > x) hi = mid-1;\n else lo = mid+1;\n }\n lo = 0;hi = i-1;\n while(lo<=hi){\n int mid = lo+(hi-lo)/2;\n if(nums[mid] == x ) {if(x>ans) ans = x; break;}\n else if(nums[mid] > x) hi = mid-1;\n else lo = mid+1;\n }\n }\n```\nhere , ans works as the tracker for the largest outlier possible .\n\nplease upvote , i worked damn hard for it . \n\n# Complexity\n- Time complexity: O(NlogN)\n\n- Space complexity:O(1)\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n sort(nums.begin(),nums.end());\n int s = 0; // sum of the complete array\n int n = nums.size();\n int ans = INT_MIN;\n for(int i = 0;i<n;i++){\n s+=nums[i];\n }\n for(int i = 0;i<n;i++){\n int x = s-(2*nums[i]); // the outlier must be equal to sum of array - (2* nums[i]).\n // applying binary search\n int lo = i+1,hi = n-1;\n while(lo<=hi){\n int mid = lo+(hi-lo)/2;\n if(nums[mid] == x ) {if(x>ans) ans = x; break;}\n else if(nums[mid] > x) hi = mid-1;\n else lo = mid+1;\n }\n lo = 0;hi = i-1;\n while(lo<=hi){\n int mid = lo+(hi-lo)/2;\n if(nums[mid] == x ) {if(x>ans) ans = x; break;}\n else if(nums[mid] > x) hi = mid-1;\n else lo = mid+1;\n }\n }\n return ans;\n }\n};\n```\n\n\n
| 14 | 3 |
['C++']
| 2 |
identify-the-largest-outlier-in-an-array
|
Python3 || 6 lines, sort and check || T/S: 51% / 58%
|
python3-6-lines-sort-and-check-ts-1350-m-dfg3
|
Here's the intuition:
The problem is equivalent to removing each distinct integer in nums, and then determining whether there still exists an integer in that re
|
Spaulding_
|
NORMAL
|
2024-12-01T04:43:29.434426+00:00
|
2025-02-11T21:26:50.618753+00:00
| 2,218 | false |
Here's the intuition:
- The problem is equivalent to removing each distinct integer in `nums`, and then determining whether there still exists an integer in that revised array that is exactly half the sum of that array.
Here's the plan:
1. We determine the sum of `nums`, construct a *Counter* for `nums`, and sort `the keys of the counter`in non-increasing order.
2. We iterate through the sorted keys with respect to the intuition above.
3. When we encounter an integer that meets the criterion above, we return it as the solution.
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
ctr, sm = Counter(nums), sum(nums)
cands = sorted(ctr, reverse = True)
for n in cands:
d, m = divmod(sm - n, 2)
if m: continue
if d in cands and (d != n or ctr[d] > 1): return n
```
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
unordered_map<int, int> ctr;
int sm = 0;
for (int num : nums) {
ctr[num]++;
sm += num;}
vector<int> cands;
for (const auto& [num, count] : ctr) {
cands.push_back(num);}
sort(cands.rbegin(), cands.rend());
for (int n : cands) {
int d = (sm - n) / 2;
int m = (sm - n) % 2;
if (m == 0 && ctr.count(d) && (d != n || ctr[d] > 1)) {
return n;}
}
return -1;}
};
```
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
Map<Integer, Integer> ctr = new HashMap<>();
int sm = 0;
for (int num : nums) {
ctr.put(num, ctr.getOrDefault(num, 0) + 1);
sm += num;
}
List<Integer> cands = new ArrayList<>(ctr.keySet());
cands.sort(Collections.reverseOrder());
for (int n : cands) {
int d = (sm - n) / 2;
int m = (sm - n) % 2;
if (m == 0 && ctr.containsKey(d) && (d != n || ctr.get(d) > 1)) {
return n;
}
}
return -1;
}
}
```
[https://leetcode.com/problems/identify-the-largest-outlier-in-an-array/submissions/1467064288/](https://leetcode.com/problems/identify-the-largest-outlier-in-an-array/submissions/1467064288/)
I could be wrong, but I think that time complexity is *O*(*N* + *M* log *M*) and space complexity is *O*(*M*), in which *N* ~ `len(nums)` and *M* ~ distinct integers in `nums`.
| 13 | 0 |
['C++', 'Java', 'Python3']
| 1 |
identify-the-largest-outlier-in-an-array
|
Counter
|
counter-by-votrubac-12u3
|
We get the sum tot of all elements in the array.\n\nThen, for each number n, we check if (tot - n) // 2 exists in the array (using a counter). \n\nNote the spec
|
votrubac
|
NORMAL
|
2024-12-01T04:32:41.180669+00:00
|
2024-12-01T04:32:41.180690+00:00
| 1,397 | false |
We get the sum `tot` of all elements in the array.\n\nThen, for each number `n`, we check if `(tot - n) // 2` exists in the array (using a counter). \n\nNote the special case when `n == (tot - n) // 2`.\n\n**Python 3**\n```python\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n cnt, tot, res = Counter(nums), sum(nums), -inf\n for n in nums:\n if (tot - n) % 2 == 0 and cnt[(tot - n) // 2] > (n == (tot - n) // 2):\n res = max(res, n)\n return res\n```
| 12 | 3 |
[]
| 2 |
identify-the-largest-outlier-in-an-array
|
🔥Beats 100 % 🔥 | O(n) and O(1) | 2 Solutions | ✔️ Explained & Optimized Solution ✔️ |
|
beats-100-on-and-o1-2-solutions-explaine-kd3u
|
Intuition\nWe first calculate the total sum of the array, then iterate through each number k. For each number, we temporarily remove it and check if the sum of
|
ntrcxst
|
NORMAL
|
2024-12-01T04:50:12.275708+00:00
|
2024-12-01T04:56:24.599675+00:00
| 3,510 | false |
# Intuition\nWe first calculate the total **sum** of the array, then iterate through each number `k`. For each number, we temporarily remove it and check if the sum of the remaining numbers `sum - k` is divisible by `2`. If yes, compute the target value `p = (sum - k) / 2`. Check if `p` exists in the remaining numbers using a **frequency map** or **multiset**. If it does, update the maximum outlier `k`. After the check, restore `k` back to the data structure. Finally, return the largest valid `k`, or a **default** value if no such number exists.\n\n# Approach\n\n`Step 1` **Calculate the Total Sum :**\n- Compute the **sum** of all elements in the array. This will help us evaluate the condition of `divisibility by 2`.\n\n`Step 2` **Use a Multiset for Frequency Tracking :**\n\n- A `multiset` keeps track of the occurrences of each element, allowing **efficient addition, removal, and lookup**.\n\n`Step 3` **Iterate Through the Array :**\n\n- For each element:\n - Temporarily remove it from the `multiset`.\n - Check if sum of the remaining elements `(sum - k)` is **even**.\n - If yes, check if half of **sum** exists in the remaining array.\n\n`Step 4` **Update the Answer :**\n- If all conditions are satisfied, update the maximum **outlier** found so far.\n\n`Step 5` **Restore the Multiset :**\n- After evaluating the current number, restore it to the `multiset`.\n\n`Step 6` **Return the Largest Outlier :**\n- If no valid `outlier` is found, return a **default value**.\n\n# Complexity\n- Time complexity : $$O(n)$$\n- Space complexity : $$O(1)$$\n\n# Code\n```java []\nclass Solution \n{\n public int getLargestOutlier(int[] nums) \n {\n // Step 1: Calculate the total sum of the array\n int sum = 0;\n for (int num : nums)\n {\n sum += num;\n }\n \n // Step 2: Use a TreeMap to mimic multiset functionality\n TreeMap<Integer, Integer> frequencyMap = new TreeMap<>();\n for (int num : nums) \n {\n frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1);\n }\n \n // Step 3: Initialize the answer as the smallest possible value\n int ans = Integer.MIN_VALUE;\n\n // Step 4: Iterate through each number in the array\n for (int num : nums) \n {\n // Check if the sum excluding the current number is even\n if ((sum - num) % 2 == 0) \n {\n int target = (sum - num) / 2;\n\n // Temporarily reduce the count of the current number\n frequencyMap.put(num, frequencyMap.get(num) - 1);\n if (frequencyMap.get(num) == 0) \n {\n frequencyMap.remove(num);\n }\n\n // Check if the target exists in the map\n if (frequencyMap.containsKey(target)) \n {\n // Update the maximum outlier found so far\n ans = Math.max(ans, num);\n }\n\n // Restore the current number to the map\n frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1);\n }\n }\n \n // Step 5: Return the largest outlier, or a default value if none found\n return ans == Integer.MIN_VALUE ? -1 : ans;\n }\n}\n```\n\n```C++ []\nclass Solution\n{\npublic:\n int getLargestOutlier(vector<int>& nums)\n {\n // Step 1: Calculate the total sum of the array\n int sum = 0;\n for (int k : nums)\n {\n sum += k;\n }\n\n // Step 2: Use a multiset to track occurrences of elements\n multiset<int> st(nums.begin(), nums.end());\n\n // Step 3: Initialize the answer as the smallest possible value\n int ans = INT_MIN;\n\n // Step 4: Iterate through each number in the array\n for (int k : nums)\n {\n // Check if the sum excluding the current number is even\n if ((sum - k) % 2 == 0)\n {\n // Temporarily remove the current number from the multiset\n st.erase(st.find(k));\n\n // Calculate the target number\n int p = (sum - k) / 2;\n\n // Check if the target exists in the multiset\n if (st.find(p) != st.end())\n {\n // Update the maximum outlier found so far\n ans = max(ans, k);\n }\n\n // Step 5 : Restore the current number back into the multiset\n st.insert(k);\n }\n }\n\n // Step 6: Return the largest outlier, or a default value if none found\n return ans;\n }\n};\n\n```
| 9 | 1 |
['Array', 'Ordered Map', 'C++', 'Java']
| 3 |
identify-the-largest-outlier-in-an-array
|
O(n) | Beats 100%
|
on-beats-100-by-vamsi_kvk-5ces
|
Intuition\n Describe your first thoughts on how to solve this problem. \n- Calculate total sum\n- Store index of each value\n- Iterate through nums\n - Calcula
|
tylerdurdn
|
NORMAL
|
2024-12-01T04:49:36.170706+00:00
|
2024-12-01T04:49:36.170731+00:00
| 1,978 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- Calculate total sum\n- Store index of each value\n- Iterate through nums\n - Calculate curr_sum = total - val\n - if val is an outlier, then curr_sum should be even and \n d[curr_sum // 2] must be equal to 1\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n s = collections.defaultdict(int)\n for i, v in enumerate(nums):\n s[v] = i\n total = sum(nums)\n res = float(\'-inf\')\n for i,num in enumerate(nums):\n curr_sum = total - num\n if curr_sum%2 == 0: \n val = curr_sum // 2\n if val in s and s[val] != i:\n res = max(res, num)\n return res\n```
| 9 | 0 |
['Python3']
| 1 |
identify-the-largest-outlier-in-an-array
|
💡 Find the Largest Outlier in an Array 🧐
|
find-the-largest-outlier-in-an-array-by-hn2ar
|
\uD83E\uDDE0 Intuition\n1. Key Insight \uD83E\uDDD0:\n The array consists of numbers such that one element represents the sum of special numbers, and another
|
lasheenwael9
|
NORMAL
|
2024-12-02T04:46:35.465150+00:00
|
2024-12-04T09:26:09.489652+00:00
| 1,558 | false |
# \uD83E\uDDE0 Intuition\n1. **Key Insight \uD83E\uDDD0**:\n The array consists of numbers such that one element represents the sum of special numbers, and another is the outlier. By removing the sum element and one other element, the remaining numbers should form a valid pair-wise sum. \n2. **Efficient Strategy**: \n - Sort the array for easy traversal. \n - Check for each candidate outlier and validate if the remaining numbers sum correctly to identify it. \n\n---\n\n# \uD83D\uDEE0\uFE0F Approach\n1. **Sort the Array**: \n This simplifies operations like binary search and helps ensure correctness when verifying the sum condition. \n2. **Iterate Backwards**: \n Check potential outliers starting from the largest element, as the goal is to find the largest outlier. \n3. **Binary Search for Validation**: \n Use a binary search to verify the presence of the sum of special numbers. \n\n---\n\n# \u23F3 Complexity \n- **Time Complexity**: \n $$ O(n \\log n) $$ due to sorting and binary search. \n- **Space Complexity**: \n $$ O(1) $$ additional space apart from the input array. \n\n---\n\n\n# Code \uD83D\uDCBB \n\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n sort(nums.begin(), nums.end()); // Sort the array\n int sum = 0;\n unordered_map<int, int> mp; // Frequency map for validation\n\n for (int i : nums) {\n sum += i; // Calculate the total sum\n mp[i]++;\n }\n\n for (int i = nums.size() - 1; i >= 0; i--) {\n double potentialSum = (sum - nums[i]) / 2.0;\n\n if (binary_search(nums.begin(), nums.end(), potentialSum)) {\n // Validate the sum condition\n if ((potentialSum == nums[i] && mp[nums[i]] >= 2) || potentialSum != nums[i]) {\n return nums[i]; // Return the largest valid outlier\n }\n }\n }\n\n return 0; // Default return value (should not occur)\n }\n};\n```\n\n```java []\nclass Solution {\n public boolean binarySearch(int[] nums, double target) {\n int left = 0, right = nums.length - 1;\n while (left <= right) {\n int mid = left + (right - left) / 2;\n if (nums[mid] == target) {\n return true;\n } else if (nums[mid] < target) {\n left = mid + 1;\n } else {\n right = mid - 1;\n }\n }\n return false;\n }\n\n public int getLargestOutlier(int[] nums) {\n Arrays.sort(nums); // Sort the array\n int totalSum = Arrays.stream(nums).sum();\n Map<Integer, Integer> freq = new HashMap<>();\n\n for (int num : nums) {\n freq.put(num, freq.getOrDefault(num, 0) + 1);\n }\n\n for (int i = nums.length - 1; i >= 0; i--) {\n double potentialSum = (totalSum - nums[i]) / 2.0;\n\n // Use the manual binary search function\n if (binarySearch(nums, potentialSum)) {\n if ((potentialSum == nums[i] && freq.get(nums[i]) >= 2) || potentialSum != nums[i]) {\n return nums[i];\n }\n }\n }\n\n return 0;\n }\n}\n\n```\n\n```python []\nclass Solution:\n def binary_search(self, nums: List[int], target: float) -> bool:\n left, right = 0, len(nums) - 1\n while left <= right:\n mid = (left + right) // 2\n if nums[mid] == target:\n return True\n elif nums[mid] < target:\n left = mid + 1\n else:\n right = mid - 1\n return False\n\n def getLargestOutlier(self, nums: List[int]) -> int:\n nums.sort() # Sort the array\n total_sum = sum(nums)\n freq = Counter(nums)\n\n for i in range(len(nums) - 1, -1, -1):\n potential_sum = (total_sum - nums[i]) / 2.0\n\n # Use the manual binary search function\n if self.binary_search(nums, potential_sum):\n if (potential_sum == nums[i] and freq[nums[i]] >= 2) or potential_sum != nums[i]:\n return nums[i]\n\n return 0\n\n```\n---\n\n# Conclusion \uD83D\uDE80 \n\nThis solution efficiently identifies the largest outlier by leveraging sorting and binary search. The logic ensures correctness and optimal performance even for large input sizes. \n\nIf you found this explanation helpful, don\u2019t forget to **upvote** and share your feedback! \uD83D\uDE0A Happy coding!\n\n\n
| 6 | 0 |
['Hash Table', 'Binary Search', 'C++', 'Java', 'Python3']
| 1 |
identify-the-largest-outlier-in-an-array
|
Easiest C++ Solution
|
easiest-c-solution-by-rajvir_singh-go4n
|
Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\ncpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n
|
rajvir_singh
|
NORMAL
|
2024-12-01T04:31:30.826899+00:00
|
2024-12-01T04:31:30.826924+00:00
| 1,314 | false |
# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int ans = INT_MIN, total = 0;\n unordered_map<int,int> mp;\n for(auto it: nums){\n mp[it]++;\n total += it;\n } \n for(auto it: nums){\n int currSum = total-it;\n mp[it]--;\n if(mp[it] == 0) mp.erase(it);\n if(currSum%2 == 0 && mp.find(currSum/2) != mp.end() && it > ans){\n ans = it;\n }\n mp[it]++;\n }\n return ans;\n }\n};\n```
| 6 | 0 |
['Array', 'Hash Table', 'Math', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Hash Table, Math, Prefix Sum O(n)
|
hash-table-math-prefix-sum-on-by-snigdha-yy7v
|
IntuitionThe problem hints that the array consists of:
n - 2 special numbers
1 sum element, which is the sum of the special numbers
1 outlier
Let’s call the tot
|
Nyx_owl
|
NORMAL
|
2025-03-24T07:14:59.881414+00:00
|
2025-03-24T07:14:59.881414+00:00
| 490 | false |
# Intuition
The problem hints that the array consists of:
- `n - 2` special numbers
- 1 sum element, which is the sum of the special numbers
- 1 outlier
Let’s call the total sum of the array `total_sum`. Let `s` be the sum of the special numbers, and let `x` be the outlier.
Then by the structure of the array:
`total_sum = s + x + s = 2s + x`
Thus, rearranged:
`x = total_sum - 2s`
The key idea is that we iterate through all numbers as potential outliers `x`, subtract `x` from the total sum to get `2s`, and check if `2s` is even. If it is, `s = (total_sum - x) // 2`, and we check if `s` exists in the array.
# Approach
1. Compute the total sum of the array.
2. Use a `Counter` to store the frequency of each number in the array for fast lookups.
3. Iterate through each number as a potential outlier `x`.
- Subtract `x` from `total_sum`, and check if the result is even.
- If it is, check if `s = (total_sum - x) // 2` exists in the frequency map.
- Be careful about the case where `x == s`: we need at least two occurrences in the array.
4. Track the maximum valid outlier.
# Complexity
- Time complexity:
$$O(n)$$ — One pass to compute sum, one pass through unique elements, and constant time dictionary lookups.
- Space complexity:
$$O(n)$$ — For the frequency map.
# Code
```python []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
total_sum = sum(nums)
count = Counter(nums)
max_outlier = float('-inf')
for x in count:
remaining = total_sum - x
if remaining % 2 != 0:
continue
s = remaining // 2
if s in count:
# Ensure s and x are from different indices
if s != x or count[x] > 1:
max_outlier = max(max_outlier, x)
return max_outlier
| 3 | 0 |
['Array', 'Hash Table', 'Counting', 'Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Clean, clear, precise! BEGINNER friendly solution!!! Logic clear!
|
clean-clear-precise-beginner-friendly-so-k1os
|
IntuitionTo solve this problem, we need to find the largest potential outlier in an array where:
Total numbers = n
( n - 2 ) elements are special numbers.
|
amogh97
|
NORMAL
|
2024-12-18T00:56:18.616232+00:00
|
2024-12-18T00:56:18.616232+00:00
| 560 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\nTo solve this problem, we need to find the largest potential outlier in an array where:\n\n- Total numbers = n\n\n- ( n - 2 ) elements are special numbers.\n\n- One element of the remaining two is the "sum of (n - 2) special numbers". (lets call this the sum_element)\n\n- The other remaining element is the outlier, which is distinct from both the sum element and the special numbers.\n\n<u>We know that:</u>\n\ntotal_sum = sum of (n-2) elements + sum_element + outlier\n\n\n*Rearranging the above elements:*\n\noutlier = total_sum - sum of (n-2) elements - sum_element\n\n**Since we know that**\nsum_element = sum of (n-2) elements\n\n**We get the equation as:**\noutlier = total_sum - 2 * sum_element\n\n(We will use this equation in code)\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe question asks for largest outlier element in nums array.\n\nAlso, we know that we will be using this equation:\noutlier = total_sum - 2 * sum_element\n\nSo we need to find total_sum and sum_element to find the outlier\n\n1. To find total_sum, its easy! just sum all the elements of the `nums` array by doing `sum(nums)`\n\n2. We also need to store the frequency of every item in `nums` array in `hash_map` (to make sure `sum_element` and outlier are both distinct)\n\n- `sum_element` is just a single element in `hash_map` (which is the sum of (n-2) elements)\n\n3. We iterate through **every key** in the `hash_map`. **Why?** we are checking every element because one of them HAS to be the **outlier**\n- By iterating through every element in `hash_map` we find the correct pair of `sum_element` and outlier\n\n4. And **HOW** do we find the outlier? using the equation above `potential_outlier = total_sum - 2 * sum_element`\n(renamed outlier to potential_outlier)\n\n5. Once we find this `potential_outlier`, we are not done yet. We need to make sure that its distinct and that its the LARGEST outlier\n\n6. To make sure its distinct => we just check that `potential_outlier != sum_element or hash_map[sum_element] > 1`\n\n7. Then we just find the `largest_outlier` by doing `max(potential_outlier, largest_outlier)`\n\n\n\n# Complexity\nTime complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Calculating the total sum: (O(n)), where (n) is the number of elements in the array.\n- Constructing the hash_map: (O(n)).\n- Iterating over the unique keys in the hash_map: (O(m)), where (m) is the number of unique elements, and (m \\leq n).\n- Therefore, the overall time complexity is (O(n)).\n\nSpace complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is dominated by the space used for the frequency map, which stores the count of each unique number. Hence, the space complexity is (O(m)), where (m) is the number of unique elements in the array.\n\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n \n #total_sum = sum of (n-2) elements + sum_element + outlier\n\n #total_sum - sum of (n-2) elements - sum_element = outlier\n\n # sum_element = sum of (n-2) elements\n # outlier = total_sum - 2 * sum element\n\n total_sum = sum(nums)\n\n hash_map = {}\n for item in nums:\n if item in hash_map.keys():\n hash_map[item] += 1\n else:\n hash_map[item] = 1\n \n\n largest_outlier = float(\'-inf\')\n\n\n\n for sum_element in hash_map.keys():\n potential_outlier = total_sum - 2 * sum_element\n\n if potential_outlier in hash_map.keys():\n if potential_outlier != sum_element or hash_map[sum_element] > 1:\n largest_outlier = max(potential_outlier, largest_outlier)\n\n return largest_outlier\n \n\n```
| 3 | 0 |
['Python', 'Python3']
| 1 |
identify-the-largest-outlier-in-an-array
|
Hash Map and Counter Approach || Python3
|
hash-map-and-counter-approach-python3-by-62mb
|
Hash Map and Counter Approach:\n\n# Intuition\n- The problem involves finding the largest "outlier" number where the remaining sum of elements divided by 2 equa
|
lavanya_immaneni
|
NORMAL
|
2024-12-01T18:01:32.848871+00:00
|
2024-12-01T18:01:32.848897+00:00
| 351 | false |
# Hash Map and Counter Approach:\n\n# Intuition\n- The problem involves finding the largest "outlier" number where the remaining sum of elements divided by 2 equals the number itself.\n\n# Approach\n- Use a `Counter` to keep track of the frequency of each number in `nums`.\n- Calculate the sum of the list (`s`).\n- Iterate over each element in `nums`:\n - Compute the potential "outlier" by subtracting the current element from `s`.\n - If the outlier condition holds and the count of the element is 1, skip it.\n - Otherwise, check if `outlier / 2` exists in the `Counter` and update the maximum value.\n- Return the maximum outlier found, or -1 if no valid outlier is found.\n\n# Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```python\nfrom typing import List\nfrom collections import Counter\n\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n maxi = float(\'-inf\')\n numCount = Counter(nums)\n s = sum(nums)\n \n for i in nums:\n outlier = s - i\n if outlier / 2 == i and numCount[i] == 1:\n continue\n elif numCount[outlier / 2]:\n maxi = max(maxi, i)\n \n return maxi if maxi != float(\'-inf\') else -1\n
| 3 | 1 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
[Python3] enumerate all candidates
|
python3-enumerate-all-candidates-by-ye15-94ts
|
Intuition\nLoop through nums as x and check if x can be the outlier. \n If x is eligible to be an outlier, then the remaining array sums to an even number, say
|
ye15
|
NORMAL
|
2024-12-01T04:42:49.315708+00:00
|
2024-12-01T04:42:58.322439+00:00
| 544 | false |
**Intuition**\nLoop through `nums` as `x` and check if `x` can be the outlier. \n* If `x` is eligible to be an outlier, then the remaining array sums to an even number, say `cand`. And `cand/2` appears in `nums` (and cannot be `x`). \n\nFind the max of such `x`. \n\n**Analysis**\nTime complexity `O(N)`\nSpace complexity `O(N)`\n\n**Implementation**\n```\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n freq = Counter(nums)\n total = sum(nums)\n ans = -inf \n for i, x in enumerate(nums): \n cand = total - x\n if not cand & 1: \n cand //= 2\n if cand in freq and (x != cand or freq[cand] > 1): ans = max(ans, x)\n return ans \n```
| 3 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy Code using Multiset
|
easy-code-using-multiset-by-sapilol-j789
|
Code\ncpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int sum = 0;\n for (int it : nums) {\n sum +=
|
LeadingTheAbyss
|
NORMAL
|
2024-12-01T04:37:52.576805+00:00
|
2024-12-01T04:37:52.576844+00:00
| 63 | false |
# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int sum = 0;\n for (int it : nums) {\n sum += it;\n }\n multiset<int> st(nums.begin(), nums.end());\n int maxi = INT_MIN;\n for (int it : nums) {\n if ((sum - it) % 2 == 0) {\n st.erase(st.find(it));\n int p = (sum - it) / 2;\n if (st.find(p) != st.end()) {\n maxi = max(maxi, it);\n }\n st.insert(it);\n }\n }\n return maxi;\n }\n};\n```
| 3 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy Python O(N) solution
|
easy-python-on-solution-by-zijingguo-qqgr
|
IntuitionFind the criteria to decide if each number can be a outlierApproachCalculate the sum of the array
For each number in the array, if the number is the ou
|
ZijingGuo
|
NORMAL
|
2025-02-08T05:41:23.133290+00:00
|
2025-02-08T05:41:23.133290+00:00
| 240 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Find the criteria to decide if each number can be a outlier
# Approach
<!-- Describe your approach to solving the problem. -->
Calculate the sum of the array
For each number in the array, if the number is the outlier, the remaining sum / 2, which should be another special number should still be in the remaining array.
Use hashmap to record the nums frequency. so that we can just check if the count[potential_sum] value
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(N)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(N)
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
total = sum(nums)
count = Counter(nums)
result = -1001
for num in nums:
# in case the potential_sum == num
count[num] -= 1
# remaining sum
curr_sum = total - num
# test if this number can be a outlier
potential_sum = curr_sum / 2
if potential_sum in count and count[potential_sum] > 0:
result = max(result, num)
count[num] += 1
return result
```
| 2 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easiest and Simplest Solution Ever in O(n)
|
easiest-and-simplest-solution-ever-in-on-3yi0
|
Intuition\nThe problem revolves around identifying the largest outlier in the array. An outlier is defined as a number that is neither part of the n-2 special n
|
aaditya_chauhan
|
NORMAL
|
2024-12-05T14:43:12.584210+00:00
|
2024-12-05T14:43:12.584247+00:00
| 107 | false |
# Intuition\nThe problem revolves around identifying the largest outlier in the array. An outlier is defined as a number that is neither part of the `n-2` special numbers nor their sum. By using the total sum of the array, we can derive the sum of the `n-2` special numbers for each potential outlier and validate it.\n\n# Approach\n1. **Calculate Total Sum**:\n Compute the sum of all numbers in the array. This helps derive the sum of the `n-2` special numbers for any given potential outlier.\n\n2. **Store Frequencies**:\n Use an unordered map to store the frequency of each number for efficient lookups.\n\n3. **Iterate Through the Array**:\n For each number in the array:\n - Assume it is the outlier.\n - Compute the sum of the `n-2` special numbers, x = (total_sum - num) / 2\n\n \n - Ensure \\(x\\) is an integer and exists in the array.\n\n4. **Validate Candidates**:\n Check that:\n - The sum is divisible by 2.\n - \\(x\\) exists in the frequency map.\n - \\(x\\) and the candidate outlier are distinct or \\(x\\) occurs more than once.\n\n5. **Update Maximum Outlier**:\n Track the largest valid outlier during the iteration.\n\n6. **Return Result**:\n After iterating through all numbers, return the maximum outlier found.\n\n# Complexity\n- **Time Complexity**: \n - \\(O(n)\\) to calculate the total sum. \n - \\(O(n)\\) to populate the frequency map. \n - \\(O(n)\\) to validate each candidate. \n Overall: \\(O(n)\\).\n\n- **Space Complexity**: \n - \\(O(n)\\) for the frequency map. \n Overall: \\(O(n)\\).\n\n# Code\n```cpp\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n // Step 1: Calculate the total sum of all elements in the array\n int total_sum = accumulate(nums.begin(), nums.end(), 0);\n\n // Step 2: Store the frequencies of elements in a map\n unordered_map<int, int> freq;\n for (int num : nums) {\n freq[num]++;\n }\n\n int max_outlier = INT_MIN; // Initialize the maximum outlier to the smallest possible value\n\n // Step 3: Iterate through each number in the array\n for (int num : nums) {\n // Calculate the potential sum of the n-2 special numbers\n int x = (total_sum - num) / 2;\n\n // Check if (total_sum - num) is divisible by 2 and x exists in the array\n if ((total_sum - num) % 2 == 0 && freq[x] > 0) {\n // Handle the case where x and num are the same and x appears only once\n if (x == num && freq[x] <= 1) continue;\n\n // Update the maximum outlier if the current num is larger\n max_outlier = max(max_outlier, num);\n }\n }\n\n return max_outlier; // Return the largest valid outlier\n }\n};\n
| 2 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Binary Search Solution to find out special number on every possible outlier
|
binary-search-solution-to-find-out-speci-k0y0
|
Approach\n\n#### Key Insights:\n1. Sorting the Array:\n - Sorting simplifies the process of checking for conditions on the numbers.\n - By sorting, we can i
|
ikmishra2002
|
NORMAL
|
2024-12-01T07:36:44.015818+00:00
|
2024-12-01T09:51:21.721271+00:00
| 590 | false |
### Approach\n\n#### Key Insights:\n1. **Sorting the Array**:\n - Sorting simplifies the process of checking for conditions on the numbers.\n - By sorting, we can iterate from the largest to the smallest, reducing unnecessary computations.\n\n2. **Two-Pointer Search**:\n - For each candidate number (potential outlier), perform a binary search on the remaining numbers.\n - Exclude the candidate number from the search to avoid false positives.\n\n3. **Mathematical Validation**:\n - Calculate the total sum of the array (`sum`).\n - If removing a number (`nums[i]`) makes `sum - nums[i]` equal to one of the other numbers in the array, this is the valid outlier.\n\n---\n\n### Implementation\n\nHere\u2019s the code for the solution:\n\n```cpp []\nclass Solution {\n int check(long long sum, vector<int>& nums, int i) {\n int n = nums.size();\n int l = 0, r = n - 1;\n\n while (l <= r) {\n int mid = (l + r) / 2;\n if (mid == i) {\n if (mid < r) mid++; \n else if (mid > l) mid--; \n else return false;\n }\n\n if (sum - nums[mid] == nums[mid])\n return true;\n if (sum - nums[mid] < nums[mid]) r = mid - 1;\n else l = mid + 1;\n }\n return false;\n }\n\npublic:\n int getLargestOutlier(vector<int>& nums) {\n sort(nums.begin(), nums.end());\n long long sum = accumulate(nums.begin(), nums.end(), 0);\n int n = nums.size();\n\n for (int i = n - 1; i >= 0; i--) {\n if (check(sum - nums[i], nums, i)) return nums[i];\n }\n return -1;\n }\n};\n```\n\n```python []\nclass Solution:\n def check(self, total_sum, nums, index):\n n = len(nums)\n left, right = 0, n - 1\n\n while left <= right:\n mid = (left + right) // 2\n if mid == index:\n if mid < right:\n mid += 1\n elif mid > left:\n mid -= 1\n else:\n return False\n \n if total_sum - nums[mid] == nums[mid]:\n return True\n elif total_sum - nums[mid] < nums[mid]:\n right = mid - 1\n else:\n left = mid + 1\n\n return False\n\n def getLargestOutlier(self, nums):\n nums.sort()\n total_sum = sum(nums)\n n = len(nums)\n\n for i in range(n - 1, -1, -1):\n if self.check(total_sum - nums[i], nums, i):\n return nums[i]\n\n return -1\n```\n\n```java []\nimport java.util.*;\n\nclass Solution {\n private boolean check(long sum, int[] nums, int index) {\n int n = nums.length;\n int left = 0, right = n - 1;\n\n while (left <= right) {\n int mid = (left + right) / 2;\n if (mid == index) {\n if (mid < right) mid++;\n else if (mid > left) mid--;\n else return false;\n }\n\n if (sum - nums[mid] == nums[mid]) return true;\n if (sum - nums[mid] < nums[mid]) right = mid - 1;\n else left = mid + 1;\n }\n return false;\n }\n\n public int getLargestOutlier(int[] nums) {\n Arrays.sort(nums);\n long sum = Arrays.stream(nums).asLongStream().sum();\n\n for (int i = nums.length - 1; i >= 0; i--) {\n if (check(sum - nums[i], nums, i)) return nums[i];\n }\n return -1;\n }\n}\n```\n\n```javascript []\nfunction check(totalSum, nums, index) {\n let left = 0, right = nums.length - 1;\n\n while (left <= right) {\n let mid = Math.floor((left + right) / 2);\n if (mid === index) {\n if (mid < right) mid++;\n else if (mid > left) mid--;\n else return false;\n }\n\n if (totalSum - nums[mid] === nums[mid]) return true;\n if (totalSum - nums[mid] < nums[mid]) right = mid - 1;\n else left = mid + 1;\n }\n return false;\n}\n\nfunction getLargestOutlier(nums) {\n nums.sort((a, b) => a - b);\n let totalSum = nums.reduce((a, b) => a + b, 0);\n\n for (let i = nums.length - 1; i >= 0; i--) {\n if (check(totalSum - nums[i], nums, i)) return nums[i];\n }\n return -1;\n}\n```\n---\n\n\n### Complexity Analysis\n1. **Time Complexity**:\n - Sorting the array: \\(O(nlog n)\\)\n - Binary search for each element: \\(O(nlog n)\\)\n - Total: \\(O(nlog n)\\)\n\n2. **Space Complexity**:\n - Sorting uses \\(O(1)\\) extra space (in-place sorting with a mutable array).\n\n---\n\n### Example Walkthrough\n\n#### Input:\n```plaintext\nnums = [6, 1, 3, 10, 2]\n```\n\n#### Execution:\n1. **Sort the array**: `[1, 2, 3, 6, 10]`\n2. Calculate `sum = 22`.\n3. Iterate from the largest number (`10`):\n - Check if removing `10` satisfies the condition. \n - Result: `10` is the largest outlier.\n\n#### Output:\n```plaintext\n10\n```\n
| 2 | 0 |
['Binary Search', 'Python', 'C++', 'Java', 'JavaScript']
| 2 |
identify-the-largest-outlier-in-an-array
|
✅ Simple Java Solution
|
simple-java-solution-by-harsh__005-k1ju
|
CODE\nJava []\npublic int getLargestOutlier(int[] nums) {\n\tArrays.sort(nums);\n\tint sum = 0;\n\tfor(int num : nums) {\n\t\tsum += num;\n\t}\n\n\tint n = nums
|
Harsh__005
|
NORMAL
|
2024-12-01T05:37:40.936946+00:00
|
2024-12-01T05:37:40.936970+00:00
| 520 | false |
## **CODE**\n```Java []\npublic int getLargestOutlier(int[] nums) {\n\tArrays.sort(nums);\n\tint sum = 0;\n\tfor(int num : nums) {\n\t\tsum += num;\n\t}\n\n\tint n = nums.length;\n\tfor(int i=0; i<n; i++) {\n\t\tint remained = sum-nums[i];\n\t\tint find = remained-nums[i];\n\t\tint findIdx = Arrays.binarySearch(nums, find);\n\t\tif(findIdx > -1) {\n\t\t\tboolean flag = false;\n\t\t\tif(findIdx != i) {\n\t\t\t\tflag = true;\n\t\t\t} else if(findIdx == i && ((findIdx>0 && nums[findIdx-1] == find) || (findIdx<n-1 && nums[findIdx+1] == find))) {\n\t\t\t\tflag = true;\n\t\t\t}\n\t\t\tif(flag) return find;\n\t\t}\n\t}\n\treturn -1;\n}\n```
| 2 | 1 |
['Java']
| 1 |
identify-the-largest-outlier-in-an-array
|
Simple Approach to Finding the Largest Outlier in an Array
|
simple-approach-to-finding-the-largest-o-s864
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to find the largest outlier in the array. An outlier is defined as a number
|
leven925
|
NORMAL
|
2024-12-01T04:45:23.963217+00:00
|
2024-12-01T05:40:57.336186+00:00
| 593 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to find the largest outlier in the array. An outlier is defined as a number that is not part of the set of "special numbers" and is neither the sum of two special numbers. The two elements that are not special numbers are the outlier and the sum of the special numbers.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Sum Calculation: First, calculate the total sum of the array elements.\n2. Frequency Count: Build a frequency table to keep track of the occurrences of each number in the array.\n3. Identify Candidates:\n```\n/* Three kinds of number inside the array. */\n/* Special Numbers, Sum of Special numbers, Outlier */\n/* Sum - Outlier = Special Numbers + Sum of Special numbers */\nsumSubtractItself[i] = total \u2212 nums[i]\ncandidates = sumSubtractItself[i] / 2\n\n```\n4. Return the Largest Outlier: Keep track of the largest outlier and return it.\n# Complexity\n- Time complexity: O( N )\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O( N )\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int len = nums.size(), ans = INT_MIN, total = 0;\n vector<int> sumSubtractItself(len);\n unordered_map<int,int> freq;\n for(int i = 0; i < len; i++){\n total += nums[i];\n freq[nums[i]]++;\n } \n for(int i = 0; i < len; i++){\n sumSubtractItself[i] = total - nums[i];\n if(sumSubtractItself[i] % 2 != 0)\n continue;\n int candidates = sumSubtractItself[i] / 2;\n if(freq.count(candidates) && candidates != nums[i])\n ans = max(ans, nums[i]);\n if(freq.count(candidates) && freq[candidates] > 1)\n ans = max(ans, nums[i]);\n }\n return ans;\n }\n};\n```
| 2 | 0 |
['C++']
| 1 |
identify-the-largest-outlier-in-an-array
|
Simple Solution ✅ beats 💯
|
simple-solution-beats-by-1dx8mmgplf-u98t
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThe function appears to find the largest number (outlier) from the input list nums such
|
1DX8MMgplf
|
NORMAL
|
2024-12-01T04:44:27.188232+00:00
|
2024-12-01T04:44:27.188255+00:00
| 249 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe function appears to find the largest number (outlier) from the input list nums such that, when removed, the remaining numbers can form a valid condition related to their sum. Specifically:\n\nThe remaining sum must be even.\nHalf of this sum must also exist in the list of remaining numbers.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. initail Setup\n\n```\nhashset = set(nums)\ntotal = sum(nums)\nfreq = Counter(nums)\nres12456 = -1001\n\n```\n\nhashset: A set of all unique numbers in nums (used for efficient lookups, though unused in the rest of the code).\ntotal: Sum of all numbers in nums.\nfreq: A Counter object to track the frequency of each number in nums.\nres12456: Stores the largest outlier satisfying the condition. It is initialized to -1001 (assuming no number in nums is smaller than -1000).\n\n2. This loop processes each number in nums to check if it can be an outlier.\n\n```\nfor num in nums:\n```\n\n3. Temporarily Removing a Number:\n\n```\ntotal -= num\nfreq[num] -= 1\nif freq[num] == 0:\n freq.pop(num)\n\n```\n\ntotal -= num: Subtract the current number num from the total.\nfreq[num] -= 1: Reduce the frequency of num in the freq counter.\nif freq[num] == 0: Remove num from freq if its frequency becomes zero.\n\n4. Checking Conditions:\n\n```\nif total % 2 == 0 and (total // 2) in freq:\n res12456 = max(res12456, num)\n\n```\n\ntotal % 2 == 0: Ensures the sum of the remaining numbers is even.\n(total // 2) in freq: Checks if half of the remaining total exists as one of the remaining numbers.\nres12456 = max(res12456, num): Updates res12456 to the maximum of its current value and num if conditions are met.\n\n5. Restoring the State:\n\n```\ntotal += num\nfreq[num] += 1\n\n```\n\nAfter processing num, the total and freq are restored to their original state to process the next number.\n\n# Complexity\n- Time complexity: O( N )\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n- Space complexity: O( N )\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n hashset = set(nums)\n total = sum(nums)\n freq = Counter(nums)\n res12456 = -1001\n \n for num in nums:\n total -= num\n freq[num] -= 1\n\n if freq[num] == 0:\n freq.pop(num)\n\n if total % 2 == 0 and (total // 2) in freq:\n res12456 = max(res12456, num)\n\n total += num\n freq[num] += 1\n\n return res12456\n \n \n \n \n```
| 2 | 1 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Solution using Python, hash map. Not the fastest, but intuitive.
|
solution-using-python-hash-map-not-the-f-k756
| null |
faheem00
|
NORMAL
|
2024-12-12T00:49:27.395062+00:00
|
2024-12-12T00:49:27.395062+00:00
| 230 | false |
# Intuition\nIf we consider a value as an outlier, then there\'s another value, which will be the sum of the rest of the values. So, that value will basically be the half of the sum of the remaining values. Therefore, we can just sum the values other than the outliers, divide by 2, and check if that value exists in the subarray.\n\n# Approach\nSum all the values. Consider each value n in nums as an outlier. Get sum of all the other values. Make sure the new sum is even, which means it\'s divisible by 2. Divide that new sum by 2. If that sum exists in the remaining array, that\'s the outlier.\n\nUse a frequency hashmap to check if a number exists in the subarray\n\n# Complexity\n- Time complexity:\n$$O(n)$$ - Each of the sum operation, freq hash creation operation, and looping through the nums array takes O(n) time. The operations inside the loop takes constant time\n\n- Space complexity:\n$$O(n)$$: The freq hash takes O(n) space for worst case.\n\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n # sum all the values\n # consider a num as outlier. Get sum of all the other values\n # Divide that new sum. If that sum exists in the remaining array,\n # that\'s the outlier\n # Use a frequency hashmap to check if a number exists in the subarray\n\n # sum all the values\n sumVals = sum(nums)\n # frequency hashmap\n freqHash = Counter(nums)\n # track largest outlier\n maxO = float("-inf")\n\n for n in nums:\n # adjust occurance of n in freq hash\n if n in freqHash and freqHash[n] > 0:\n freqHash[n] -= 1\n # get sum of all other vals, check if even, and get divided val\n sumOtherVals = sumVals - n\n if sumOtherVals % 2 == 0: \n dividedVal = sumOtherVals / 2\n if dividedVal in freqHash and freqHash[dividedVal] > 0:\n maxO = max(maxO, n)\n if n in freqHash and freqHash[n] >= 0:\n freqHash[n] += 1\n \n return maxO\n```
| 2 | 0 |
['Hash Table', 'Python3']
| 2 |
identify-the-largest-outlier-in-an-array
|
Beats 100%...Freq Array!!
|
beats-100freq-array-by-srirammente-a9p0
|
IntuitionAt first glance, the problem seems to be about identifying an outlier in the array where one of the numbers is the sum of the special numbers in the ar
|
srirammente
|
NORMAL
|
2025-04-03T17:24:48.037835+00:00
|
2025-04-03T17:24:48.037835+00:00
| 27 | false |
# Intuition
At first glance, the problem seems to be about identifying an outlier in the array where one of the numbers is the sum of the special numbers in the array, and the other is the actual outlier. Our primary challenge is efficiently identifying the largest outlier from the given array, leveraging the constraints that one element is the sum of the special numbers, and the other is an outlier.
# Approach
1. Frequency Array: We will use a frequency array to count how often each number appears in the array. Since the values in the array can be negative, we'll use an offset (1000 in this case) to make sure we can map negative values to non-negative indices.
1. Sum of All Elements: We calculate the total sum of the array. This sum will help in calculating potential candidates for the outlier by subtracting the current number.
1. Iterating Through Each Number:
- For each element in the array, we subtract that element from the total sum. This gives us the sum of the remaining elements, which is potentially the sum of the special numbers.
- If the remaining sum is even, we check if half of the remaining sum exists in the frequency array (since two special numbers must sum up to this remaining value).
1. Validating the Outlier: If the remaining sum is valid, and the corresponding element is found in the array, we conclude that the current element is the outlier.
1. Returning the Largest Outlier: During the iteration, we will track the largest outlier and return it as the result.
# Complexity
- Time complexity:
- The first loop to calculate the sum and populate the frequency array takes O(n), where n is the length of the array.
- The second loop to check for each potential outlier also takes O(n).
- So, the overall time complexity is O(n).
- Space complexity:
- We are using a frequency array of fixed size (2001 elements) to handle the range of values from -1000 to 1000. This array takes up O(1) space, as its size does not depend on the input size.
- The space complexity is therefore O(n) due to the space required for storing the input array.
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int n = nums.length;
int[] freq = new int[2001];
int sum = 0;
for (int num : nums) {
freq[num + 1000]++;
sum += num;
}
int max = -1000;
for (int num : nums) {
int rem = sum - num - num;
if (rem + 1000 >= 0 && rem + 1000 <= 2000 && freq[rem + 1000] > (rem == num ? 1 : 0)) {
max = Math.max(rem, max);
}
}
return max;
}
}
```
| 1 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
easy cpp solution
|
easy-cpp-solution-by-bhowmickmadhurima10-mrq3
|
IntuitionThe goal is to find an outlier in the given array. The outlier should satisfy the equation:outlier=total sum of all elements−2×(some element in the arr
|
bhowmickmadhurima1003
|
NORMAL
|
2025-03-22T09:19:15.273616+00:00
|
2025-03-22T09:19:15.273616+00:00
| 105 | false |
# Intuition
The goal is to find an outlier in the given array. The outlier should satisfy the equation:
outlier=total sum of all elements−2×(some element in the array)
This equation arises because:
There is one number in the array that equals the sum of some subset of elements.By rearranging, we find that the outlier should be the difference between the total sum and twice another element in the array.
To efficiently find this outlier, we:
Compute the total sum of the array.
Iterate over the array to check if sum - 2 * nums[i] exists.
Ensure that the found number is at a different index (so we do not use the same number twice).
Return the largest outlier found.
# Approach
1)Sort the array to ensure that we process numbers in increasing order, which simplifies the search for the largest outlier.
2)Compute the total sum of the array.
3)Use a map (mpp) to store each element’s index, allowing quick lookups.
4)Iterate from the largest number to the smallest (to maximize the outlier):
5)For each number nums[i], check if sum - 2 * nums[i] exists in the map.
6)Ensure that the found element is not at the same index as i (to avoid counting the same number twice).
7)Keep track of the largest valid outlier found so far.
8)Return the largest valid outlier.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
Sorting the array: o(NlogN)
computing the total sum =o(n)
building the map = o(n)
finding the outlier = o(n)
overall=O(NlogN)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
o(n) for storing n elements in the map
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
sort(nums.begin(),nums.end());
int n = nums.size();
int sum = 0;
map<int,int> mpp;
for(int i = 0 ; i< n ; i++){
sum+=nums[i];
mpp[nums[i]]=i;
}
int outlier=INT_MIN ;
int i = n-1;
while(i>=0){
if(mpp.count(sum-2*nums[i]) && mpp[sum-2*nums[i]]!=i){
outlier=max(outlier,sum-2*nums[i]);
}
i--;
}
return outlier;
}
};
```
| 1 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
single traversal + unordered_map stl => O(n)
|
single-traversal-unordered_map-stl-on-by-savg
|
IntuitionApproach[a,b,c,d,e] suppose b is an outlier, for each i we will be checking whether (a+b+c+d+e) - (a+c+d+i) is present in our array or not and handling
|
irfanmohammad
|
NORMAL
|
2025-03-19T19:25:45.212517+00:00
|
2025-03-19T19:25:45.212517+00:00
| 17 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
[a,b,c,d,e] suppose b is an outlier, for each i we will be checking whether (a+b+c+d+e) - (a+c+d+i) is present in our array or not and handling some edge-cases, if present then that is our outlier and i will be our special number.
# Complexity
- Time complexity: O(n)
- Space complexity: O(n)
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
int n = nums.size() ;
int sum = accumulate(nums.begin(), nums.end(), 0) ;
unordered_map<int,int> mp ;
for(auto i : nums) mp[i]++;
int ans = -9000 ;
for(int i=0;i<n;i++){
int x = sum - (nums[i] + nums[i]) ;
if(mp.find(x)!=mp.end()) {
if(x==nums[i] and mp[x]==1) continue ;
ans=max(ans, x);
}
}
if(ans==-9000) return -1 ;
return ans;
}
};
```
| 1 | 0 |
['Hash Table', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Optimized Brute-Force + Counting that beats 95%. Mathematical Intuitive Explanation
|
optimized-brute-force-counting-that-beat-hiyi
|
IntuitionIn this problem, we have array nums sized n with:
an element is outlier
an element is sum of another n-2 elements, excluding itself and outlier.
n-2 "o
|
ddhira123
|
NORMAL
|
2025-03-19T18:06:24.542430+00:00
|
2025-03-19T18:06:24.542430+00:00
| 120 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
In this problem, we have array `nums` sized `n` with:
- an element is outlier
- an element is sum of another `n-2` elements, excluding itself and outlier.
- `n-2` "ordinary" elements.
- Every elements/index can have exactly a role from the 3 roles above.
The highest `n` possible here is $$10^5$$, so if we do brute-force on `nums` directly, it will lead to Time-Limit Exceeded.
On the other hand, the `nums[i]` is bounded at `[-1000, 1000]`. Thus, we can use **hash map / hash table** to store counts and brute-force is safe since the search space is at most `2001`.
# Approach 1 : Brute-Force + Hash Map
<!-- Describe your approach to solving the problem. -->
We use brute-force on unique keys (`nums[i]`) to search for the outlier.
Let's say we have `nums[i] = [a, b, c, d ,e]`, where `e` is outlier and `d` is sum of other 3 elements: `a + b + c`.
Let `sumAll` be the sum of all elements in the `nums`.
```
sumAll = a + b + c + d + e
```
We know that `d = a + b + c`, thus:
```
sumAll = d + d + e
```
In another form:
```
2d = (sumAll - e)
```
Therefore, we can brute-force on `e` that is having possible `d` in the nums.
However, there can be a case when `d` equals `e`. In this case, we need to ensure that they are on different indexes, or its **occurences count** is more than 1.
## Algorithm
1. Create a hashmap to store counts of `nums[i]`.
2. Count occurences for every unique `nums[i]`.
3. Count the sum of all elements in `nums`. Let's call it `sumAll`
4. Let `ans` be our variable to store the answer, set it to any number lower than `-1000` (the lowest `nums[i]` can have)
5. For each `key` in the hashmap (Let `key` be our current outlier guess):
- If current `key` is less than or equal to the current answer, skip it (**continue**).
- If there's exist `d = (sumAll - key) / 2`\* in the hashmap :
- If `d` isn't equal to `key` **or** the occurence count of `d` is more than 1, then update the `ans` to `key`.
6. Return `ans`
> \*) If we create a hashmap of integer, additional check of `2d = sumAll - key` is required to prevent false `d` that leads to wrong answer.
## Complexity
Let $$N$$ be size of `nums` and $$M$$ be number of varying unique elements in `nums`
- Time complexity: $$O(M + N)$$
We loop entire `nums` at least once, and may need to loop entire $$M$$ elements in the hashmap.
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(M)$$
This is for the hashmap storing occurence counts of elements in nums.
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
## Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
unordered_map<int, int> m;
int sumAll = 0;
for(int x : nums){
m[x]++; // Count Occurence
sumAll += x; // Count sumAll
}
int ans = -10000;
// Brute-force on key.
// key is it->first here.
for(auto it=m.begin(); it!=m.end(); it++) {
if(it->first <= ans) continue; // skip
// Find the d
auto it1 = m.find((sumAll - it->first) / 2);
// Additional check on d
if(it1 != m.end() && it1->first * 2 == sumAll - it->first) {
// Final check to qualify as outlier
if(it1->first != it->first || it1->second > 1)
ans = it->first;
}
}
return ans;
}
};
```
| 1 | 0 |
['Greedy', 'Counting', 'C++']
| 1 |
identify-the-largest-outlier-in-an-array
|
quick and dirty for interview
|
quick-and-dirty-for-interview-by-rcw-ck50
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
rcw
|
NORMAL
|
2025-03-17T17:27:11.334017+00:00
|
2025-03-17T17:27:11.334017+00:00
| 74 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
sumnum = sum(nums)
res = float('-inf')
for n in freq.keys():
s = sumnum - 2 * n
if (s !=n and s in freq.keys()) or (s == n and freq[n] > 1):
res = max(res, s)
return res
```
| 1 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
easy and beginer solution
|
easy-and-beginer-solution-by-s_malay-ll3p
|
Code
|
s_malay
|
NORMAL
|
2025-03-10T07:56:29.587997+00:00
|
2025-03-10T07:56:29.587997+00:00
| 181 | false |
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
int n=nums.size();
long long sum=0;
map<int,int>mpp;
for(int i=0;i<n;i++)
{
sum+=nums[i];
mpp[nums[i]]++;
}
int ans=INT_MIN;
for(int i=0;i<n;i++)
{
int rest=(sum-nums[i]);
if(rest%2!=0)continue;
rest=rest/2;
if(mpp.find(rest)!=mpp.end())
{
if(nums[i]==rest && mpp[rest]==1)
continue;
else if (nums[i]>ans)
ans=nums[i];
}
}
return ans;
}
};
```
| 1 | 0 |
['Array', 'Hash Table', 'Counting', 'Enumeration', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Most Intuitive Solution!!! [Easy to Understand]
|
most-intuitive-solution-easy-to-understa-dre3
|
IntuitionThe problem is about finding the largest outlier in an array. An outlier is defined as an element that, when removed, makes the sum of the remaining el
|
nishantkumar13
|
NORMAL
|
2024-12-28T06:55:10.240305+00:00
|
2024-12-28T06:55:10.240305+00:00
| 243 | false |
# Intuition
The problem is about finding the largest outlier in an array. An outlier is defined as an element that, when removed, makes the sum of the remaining elements even. We can solve this problem by iterating through the array and checking each element to see if it meets the criteria.
# Approach
1. **Calculate the total sum**: First, calculate the total sum of all elements in the array.
2. **Use a frequency map**: Create a frequency map to store the count of each element in the array.
3. **Iterate through the array**: For each element in the array:
- Calculate the remaining sum by subtracting the current element from the total sum.
- Check if the remaining sum is even.
- If it is, check if the remaining sum divided by 2 exists in the frequency map.
- If it does, update the largest outlier.
4. **Return the largest outlier**: After iterating through the array, return the largest outlier found.
### Example
Consider the array `a = [1, 2, 3, 4, 5]`:
1. Calculate the total sum: `ts = 1 + 2 + 3 + 4 + 5 = 15`.
2. Create a frequency map: `{1: 1, 2: 1, 3: 1, 4: 1, 5: 1}`.
3. Iterate through the array:
- For element 1: `rs = 15 - 1 = 14` (even), `ele = 14 / 2 = 7` (not in the map).
- For element 2: `rs = 15 - 2 = 13` (odd), skip.
- For element 3: `rs = 15 - 3 = 12` (even), `ele = 12 / 2 = 6` (not in the map).
- For element 4: `rs = 15 - 4 = 11` (odd), skip.
- For element 5: `rs = 15 - 5 = 10` (even), `ele = 10 / 2 = 5` (in the map), update outlier to 5.
4. The largest outlier is 5.
# Complexity
- Time complexity: $$O(n)$$, where $$n$$ is the length of the array, since we iterate through the array twice.
- Space complexity: $$O(n)$$, as we are using a frequency map to store the count of each element.
# Code
```java
class Solution {
public int getLargestOutlier(int[] a) {
int ts = 0;
Map<Integer, Integer> freq = new HashMap<>(a.length);
for (int i : a) {
ts += i;
freq.put(i, freq.getOrDefault(i, 0) + 1);
}
int out = Integer.MIN_VALUE;
for (int i : a) {
int rs = ts - i;
if (rs % 2 != 0) continue;
freq.put(i, freq.getOrDefault(i, 0) - 1);
int ele = rs / 2;
if (freq.getOrDefault(ele, 0) > 0) out = Math.max(out, i);
freq.put(i, freq.get(i) + 1);
}
return out;
}
}
| 1 | 0 |
['Math', 'Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Best Solution | Best Explanation | Most Optimum |
|
best-solution-best-explanation-most-opti-fgj5
|
Intuition\nif T is the total sum of all n elements in the array such that n-2 special elements have sum s ( one of remaining 2 elems) and o is the outlier.\n\nT
|
panseja2gourav5
|
NORMAL
|
2024-12-04T15:37:28.197296+00:00
|
2024-12-04T15:37:28.197361+00:00
| 140 | false |
# Intuition\nif `T` is the total sum of all `n` elements in the array such that `n-2` special elements have sum `s` ( one of remaining 2 elems) and `o` is the outlier.\n\nThen `T = 2*S + O`......................(1)\n\nTherefore, `s = (T - O)/2`..............(2)\n\nSince we need to get the maximum value of outlier, by brute force we will check each element of the array to be the outlier, and get the corresponding value of `s` from it, now if the pertiuclar value of `s` does exist in the array, then it means that the corresponding element is an outlier, Therefore like this for each element we will check if it is valid outlier and then get the maximum outlier by comparision.\n\n## Edge Case\n[6,-31,50,-35,41,37,-42,13], wrongly considering 13 as largest outlier, but correct answer is -35.\n\nThere can be case where value `s` comes out to be same as outlier, `O`, from the formula (2), then if the frequency of outlier in array is 1, then we may wrongly consider an element to be a valid outlier.\nTherefore we also check the frequency of each element, when considering it as outlier, such that, if `s` comes out to be equal to `O`, then frequency of this current element being considered as outlier, must be greater than 1.\n\n\n\n\n# Code\n```java []\nclass Solution {\n public int getLargestOutlier(int[] nums) {\n int n = nums.length;\n HashMap<Integer, Integer> map = new HashMap<>();\n\n int sum = 0; \n\n for(int i = 0; i < n; i++){\n sum += nums[i];\n map.put(nums[i], map.getOrDefault(nums[i], 0)+1);;\n } \n\n int mx = Integer.MIN_VALUE;\n\n for(int i = 0; i < n; i++){\n\n int s = (sum - nums[i]);\n\n if(s % 2 == 0 && map.containsKey(s/2)){\n if(s/2 != nums[i] || map.get(nums[i]) > 1){\n mx = Math.max(mx, nums[i]);\n } \n }\n }\n\n return mx;\n }\n}\n```
| 1 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Two-Pointers Detailed Explanation
|
two-pointers-detailed-explanation-by-shi-ztdx
|
we can use Two Pointers also for this problem.Approach:When outlier is right of 1st remaining element(nums[l], sum of special numbers)
If sum of all special ele
|
shivamjha17
|
NORMAL
|
2024-12-04T08:59:32.431015+00:00
|
2024-12-27T08:01:19.081475+00:00
| 53 | false |
# we can use *Two Pointers* also for this problem.
## Approach:
<!-- Describe your approach to solving the problem. -->
#### When outlier is right of 1st remaining element(nums[l], sum of special numbers)
- If ***sum of all special elements*** { Total sum - 1stElement(nums[l]) - outlier(nums[r]) } ***= 1stElement(nums[l])***,
Then return the outlier(nums[r]) as array is sorted this is the max outlier possible.
- If ***sum of all special elements*** { Total sum - 1stElement(nums[l]) - outlier(nums[r]) } ***> 1stElement(nums[l])***,
Then to make the **sum of all special elements** lower, increase the 1stElement(nums[l]) by incrementing l to l+1 (doing r-- further decreases the outlier(nums[r]) and hence increases **sum of all special elements**, so only way is moving right by l++).
- If ***sum of all special elements*** { Total sum - 1stElement(nums[l]) - outlier(nums[r]) } ***< 1stElement(nums[l])***,
Then to make the **sum of all special elements** higher, decrease the outlier(nums[r]) by decrementing r to r-1 (doing l++ further increases the 1stElement(nums[l]) and hence decreases **sum of all special elements**, so only way is moving left by r--).
#### When outlier is left of 1st remaining element(nums[l], sum of special numbers)
- If ***sum of all special elements*** { Total sum - 1stElement(nums[r]) - outlier(nums[l]) } ***= 1stElement(nums[r])***,
Then update the outlier(nums[l]) to ans as we move right we might get higher outlier.
- If ***sum of all special elements*** { Total sum - 1stElement(nums[r]) - outlier(nums[l]) } ***> 1stElement(nums[r])***,
Then to make the **sum of all special elements** lower, increase the outlier(nums[l]) by incrementing l to l+1 (doing r-- further decreases the 1stElement(nums[r]) and hence increases **sum of all special elements**, so only way is moving right by l++).
- If ***sum of all special elements*** { Total sum - 1stElement(nums[r]) - outlier(nums[l]) } ***< 1stElement(nums[r])***,
Then to make the **sum of all special elements** higher, decrease the 1stElement(nums[r]) by decrementing r to r-1 (doing l++ further increases the outlier(nums[l]) and hence decreases **sum of all special elements**, so only way is moving left by r--).
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
// *** Two-Pointer TC: (NlogN), SC: (K), K for sorting algo
Arrays.sort(nums);
int sum = 0;
for (int e : nums)
sum += e;
int l = 0;
int r = nums.length - 1;
// when outlier is right of 1st remaining element(sum of special numbers)
while (l < r) {
if (sum - nums[l] - nums[r] == nums[l]) {
return nums[r];// highest outliear as array is sorted non-decreasing
} else if (sum - nums[l] - nums[r] > nums[l])
l++;
else
r--;
}
l = 0;
r = nums.length - 1;
int ans = -1;
// when outlier is left of 1st remaining element(sum of special numbers)
while (l < r) {
if (sum - nums[r] - nums[l] == nums[r]) {
ans = nums[l];// every time update for higher outlier as we move to right
l++;
} else if (sum - nums[r] - nums[l] > nums[r])
l++;
else
r--;
}
return ans;
}
}
```
# Complexity
- Time complexity: O(NlogN)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
NlogN for sorting algorithm
At max 2N for two loops
- Space complexity: O(K)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
K space used by sorting algorithm
Let me know if there's anything else I can do to improve this approach.
| 1 | 0 |
['Array', 'Two Pointers', 'Sorting', 'Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Detailed Explanation || O(n) || C++ || Easy and Optimized Solution
|
detailed-explanation-on-c-easy-and-optim-i6bx
|
Intuition\n Describe your first thoughts on how to solve this problem. \nWe are given an array nums and need to find the largest outlier. An outlier in this con
|
Abhishekjha6908
|
NORMAL
|
2024-12-03T09:50:04.381815+00:00
|
2024-12-03T09:50:04.381842+00:00
| 62 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe are given an array `nums` and need to find the largest outlier. An outlier in this context is a number in the array such that if it is removed, the sum of the remaining numbers is even and can be split into two equal halves. The goal is to find the largest such outlier.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Calculate the total sum of the array (`tsum`) and store the frequency of each number in the array using a hash map `freq`.\n2. Loop through each number in the array:\n - For each number, temporarily reduce its frequency.\n - Calculate the sum of the remaining numbers (`reqdoublesum = tsum - num`).\n - If the remaining sum is even, check if half of this sum (`reqsum = reqdoublesum / 2`) exists in the frequency map and has a count greater than zero.\n - If the condition is satisfied, update the result with the maximum of num and the current ans.\n3. Return the largest valid outlier found.\n# Complexity\n- Time complexity: $$O(n)$$.\n - We loop through the array twice: once for calculating the sum and frequencies, and once for checking each element. This results in a linear time complexity.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$.\n - We use a hash map to store the frequencies of the numbers, which requires extra space proportional to the size of the array.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n long tsum = 0;\n unordered_map<int,int> freq;\n for(auto it: nums){\n tsum += it;\n freq[it]++;\n }\n\n int ans = INT_MIN;\n for(auto num: nums){\n freq[num]--;\n long reqdoublesum = tsum-num;\n if(reqdoublesum %2 ==0){\n int reqsum = reqdoublesum/2;\n if(freq[reqsum]>0){\n ans = max(ans,num);\n }\n }\n freq[num]++;\n }\n return ans;\n }\n};\n```
| 1 | 0 |
['Hash Table', 'Counting', 'Enumeration', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Javascript - HashMap
|
javascript-hashmap-by-faustaleonardo-o7dz
|
Code\njavascript []\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar getLargestOutlier = function (nums) {\n let sum = 0;\n let ans = -Infinity;
|
faustaleonardo
|
NORMAL
|
2024-12-03T00:54:06.712313+00:00
|
2024-12-03T00:54:21.760984+00:00
| 156 | false |
# Code\n```javascript []\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar getLargestOutlier = function (nums) {\n let sum = 0;\n let ans = -Infinity;\n const map = new Map();\n for (const num of nums) {\n map.set(num, (map.get(num) || 0) + 1);\n sum += num;\n }\n\n for (let i = 0; i < nums.length; i++) {\n const target = (sum - nums[i]) / 2;\n if (!map.has(target)) continue;\n if (target === nums[i] && map.get(target) === 1) continue;\n\n ans = Math.max(nums[i], ans);\n }\n\n return ans;\n};\n\n```
| 1 | 0 |
['Hash Table', 'JavaScript']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple and Clean Solution || (Using Hash Table) ✅✅
|
simple-and-clean-solution-using-hash-tab-eh9p
|
Code\ncpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int n=nums.size();\n unordered_map<int,int> mp;\n
|
Abhi242
|
NORMAL
|
2024-12-02T16:38:06.265659+00:00
|
2024-12-02T16:40:06.209367+00:00
| 40 | false |
# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int n=nums.size();\n unordered_map<int,int> mp;\n int sum=0;\n for(int i=0;i<n;i++){\n mp[nums[i]]++;\n sum+=nums[i];\n }\n int ans=INT_MIN;\n for(int i=0;i<n;i++){\n if((sum-nums[i])%2==0){\n mp[nums[i]]--;\n if(mp[(sum-nums[i])/2]>0){\n ans=max(ans,nums[i]);\n }\n mp[nums[i]]++;\n }\n }\n if(ans==INT_MIN){\n return -1;\n }\n return ans;\n }\n};\n```
| 1 | 0 |
['Hash Table', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Detailed Solution | HashMap | Prefix Sum | O(n)
|
detailed-solution-hashmap-prefix-sum-on-lnpvx
|
Problem Description\nThese are numbers in the array that collectively make up the majority (n - 2) of the array element.\nAnd from rest 2 one is sum of these sp
|
Paramveer01
|
NORMAL
|
2024-12-02T15:29:43.643598+00:00
|
2024-12-02T18:10:43.508043+00:00
| 260 | false |
# Problem Description\nThese are numbers in the array that collectively make up the majority (n - 2) of the array element.\nAnd from rest 2 one is sum of these special number and one outliner.\nWe have to find out the maximum possible value of this outliner. \n\n\n# Intuition\nFrom problem we can say that nums = [[specail number], special numbers sum , outliner]\nTherefor :- Sum of nums = 2*Special_number_Sum + outliner\n\nNow we can jusy calculate sum and use HashMap to find out the sum.\n# Approach \n1. Create a HashMap to dtore all the values in array.\n2. Store sum of array in a variable.\n3. Now traverse array and substract element from the sum.\n4. Then resultant sum will contain the sum of special elements as we well all special elements so resultant sum will always even.\n i.e temp_sum%2==0 \n5. Now check for temp_sum in HashMap if found store that array element in HashMap and also check if nums[ i ] == temp_sum, then occurance of temp_sum should be >1.\n\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\n```java []\nclass Solution {\n public int getLargestOutlier(int[] nums) {\n int n = nums.length;\n int sum = 0;\n HashMap<Integer,Integer> h = new HashMap<>();\n int max = Integer.MIN_VALUE;\n for(int i : nums) {\n sum += i;\n h.put(i,h.getOrDefault(i,0)+1);\n }\n for(int i : nums){\n int sum_left = (sum-i);\n if(sum_left%2!=0) {\n continue;\n }\n if(h.containsKey(sum_left/2)){\n if(sum_left/2!=i) max = Math.max(max,i);\n if(sum_left/2==i && h.get(i)>1) max = Math.max(max,i);\n \n }\n }\n return max;\n\n }\n}\n```\n```c++ []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int n = nums.size();\n int sum = 0;\n std::unordered_map<int, int> h;\n int max = INT_MIN;\n \n for (int i : nums) {\n sum += i;\n h[i]++;\n }\n\n for (int i : nums) {\n int sum_left = sum - i;\n if (sum_left % 2 != 0) {\n continue;\n }\n int target = sum_left / 2;\n if (h.count(target)) {\n if (target != i) max = std::max(max, i);\n if (target == i && h[i] > 1) max = std::max(max, i);\n }\n }\n return max;\n \n }\n};\n```\n```python []\nclass Solution(object):\n def getLargestOutlier(self, nums):\n n = len(nums)\n total_sum = sum(nums)\n freq = {}\n max_val = float(\'-inf\')\n \n for num in nums:\n freq[num] = freq.get(num, 0) + 1\n \n for num in nums:\n sum_left = total_sum - num\n if sum_left % 2 != 0:\n continue\n target = sum_left // 2\n if target in freq:\n if target != num:\n max_val = max(max_val, num)\n elif target == num and freq[num] > 1:\n max_val = max(max_val, num)\n \n return max_val\n \n```
| 1 | 0 |
['Array', 'Hash Table', 'Math', 'Hash Function', 'Python', 'C++', 'Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
🔥Beats 100 % 🔥 | O(n) and O(1) | ✔️ Explained & Optimized Solution ✔️ |
|
beats-100-on-and-o1-explained-optimized-ybqk0
|
\n\n\n### Intuition:\nThe goal is to find the largest "outlier" in the array. An outlier is defined as a number that is neither one of the original special numb
|
GopalDose21
|
NORMAL
|
2024-12-01T20:01:55.974017+00:00
|
2024-12-02T08:43:02.018601+00:00
| 58 | false |
\n\n\n### Intuition:\nThe goal is to find the largest "outlier" in the array. An outlier is defined as a number that is neither one of the original special numbers nor the element representing the sum of those numbers.\n\nTo solve this:\n1. First, we compute the total sum of the array.\n2. Then, for each number in the array, we check if removing it results in an even sum and if half of that sum is present in the array.\n3. If it is, we consider that number as a valid "outlier" and track the largest such outlier.\n\n---\n\n### Approach:\n1. **Compute the Total Sum**: We first calculate the sum of all elements in the array.\n2. **Track Frequencies**: Use an unordered map (`unordered_map<int, int> freq`) to count the occurrences of each element.\n3. **Iterate Through the Array**: For each element:\n - Calculate the new sum by removing the current element.\n - If this new sum is even, check if half of the new sum exists in the array.\n - If it does, update the result with the largest valid outlier.\n4. **Final Check**: If no valid outlier is found, return `-1` as the answer.\n\n---\n\n### Complexity:\n\n- **Time complexity**:\n - Building the frequency map (`unordered_map`) takes $$O(n)$$ time, where $$n$$ is the size of the array.\n - Iterating through the array and checking conditions for each element also takes $$O(n)$$ time.\n - The overall time complexity is $$O(n)$$.\n\n- **Space complexity**:\n - The space complexity is $$O(n)$$ because we use an unordered map to store the frequency of each number in the array, which could store up to $$n$$ distinct numbers.\n\n---\n\n### Code:\n```cpp\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int sum = 0;\n unordered_map<int, int> freq;\n \n // Calculate the total sum and build the frequency map\n for (int num : nums) {\n sum += num;\n freq[num]++;\n }\n\n int res = INT_MIN; // Initialize result as the smallest possible integer\n\n // Iterate over the array to check each number\n for (int num : nums) {\n int newSum = sum - num; // New sum without the current element\n\n // Check if the new sum is even\n if (newSum % 2 == 0) {\n int target = newSum / 2; // Target value we need to check in the array\n\n // If target exists and it\'s not the same number, or it\'s the same number but appears more than once\n if ((target != num && freq.count(target)) ||\n (target == num && freq[num] > 1)) {\n res = max(res, num); // Update the largest outlier found\n }\n }\n }\n\n return res == INT_MIN ? -1 : res; // If no outlier is found, return -1\n }\n};\n```\n\n---\n\n### Explanation:\n1. **Frequency Map**: We calculate the total sum of the array and simultaneously build a frequency map to keep track of how many times each number appears in the array.\n2. **Iterating Through Each Element**: For each element in the array:\n - We compute the new sum after removing the current element.\n - We then check if this new sum is even. If it is, we check if half of the new sum exists in the array (either different from the current element or multiple occurrences of the current element if it\'s the same).\n3. **Result Update**: If we find a valid outlier, we update the result with the largest such element.\n4. **Edge Case**: If no valid outlier is found, we return `-1`.\n\n---\n\nThis solution efficiently solves the problem by leveraging the frequency map for fast lookups and checking the outlier condition in linear time.
| 1 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy Go Solution | Using Map
|
easy-go-solution-using-map-by-uamalik-mp2c
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
UAMALIK
|
NORMAL
|
2024-12-01T17:21:42.913613+00:00
|
2024-12-01T17:21:42.913670+00:00
| 43 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```golang []\nfunc getLargestOutlier(nums []int) int {\n mp := make(map[int]int) // Frequency map\n\ttotal := 0 // Sum of all numbers\n\tfor _, it := range nums {\n\t\tmp[it]++\n\t\ttotal += it\n\t}\n\n\tans := math.MinInt // Smallest possible integer\n\tfor _, it := range nums {\n\t\t// Temporarily reduce the count of the current number\n\t\tmp[it]--\n\t\tif mp[it] == 0 {\n\t\t\tdelete(mp, it) // Remove the key if count is 0\n\t\t}\n\n\t\tcur := total - it // Current sum excluding `it`\n\n\t\t// Check if `cur` is even and `cur/2` exists in the map\n\t\tif cur%2 == 0 {\n\t\t\tif _, exists := mp[cur/2]; exists {\n\t\t\t\tif it > ans {\n\t\t\t\t\tans = it // Update the answer if `it` is greater\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\n\t\t// Restore the original count of `it`\n\t\tmp[it]++\n\t}\n return ans\n}\n```
| 1 | 0 |
['Go']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple sorting solution. With maths.
|
simple-sorting-solution-with-maths-by-sh-nt9o
|
We have to basically find the outlier. \n\nWe assume every element is outlier and map the array. \n\nThe remaining portion has two things. Sum of the array and
|
SharmaTushar1
|
NORMAL
|
2024-12-01T17:10:52.770260+00:00
|
2024-12-01T17:10:52.770307+00:00
| 65 | false |
We have to basically find the outlier. \n\nWe assume every element is outlier and map the array. \n\nThe remaining portion has two things. Sum of the array and the array itself. So, it\'ll be 2*(the element which is the sum of the array)\n\ntotal_sum = 2*x + y\n\nx is the sum of the array that we are given \ny is the outlier\n\nby this we can say that:\nx = (total_sum-y)/2\n\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n total=sum(nums)\n hMap = defaultdict(int)\n # To get the largest we will just sort in descending.\n nums=sorted(nums, reverse=True)\n for num in nums:\n hMap[num]+=1\n for num in nums:\n if (total-num)%2!=0:\n continue\n hMap[num]-=1\n if (total-num)//2 in hMap and hMap[(total-num)//2]>0:\n return num\n hMap[num]+=1\n return -1\n\n# total - cur = 2*x => x = (total-cur)/2\n```
| 1 | 0 |
['Python3']
| 1 |
identify-the-largest-outlier-in-an-array
|
Using Binary Search
|
using-binary-search-by-boolean_autocrat0-8qiu
|
Intuition\n Describe your first thoughts on how to solve this problem. \n you have to choose two variables one Outlier, one is sum so you have to run two loo
|
Boolean_Autocrat04
|
NORMAL
|
2024-12-01T13:56:48.935444+00:00
|
2024-12-01T13:56:48.935468+00:00
| 27 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n you have to choose two variables one Outlier, one is sum so you have to run two loops.But constraint won\'t let you do that.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSo you run the outer loop from n-1 to 0 and for inner loop you can apply binary search because the array can be sorted.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nNlogN\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int f(int sum,int outlier,int s,int e,vector<int>&nums){\n while(s<=e){\n int m=s+(e-s)/2;\n if((sum-2*nums[m])==0) return m;\n else if((sum-2*nums[m])>0) { s = m+1;}\n else{ e=m-1;}\n }\n return -1;\n }\n int getLargestOutlier(vector<int>& nums) {\n int n=nums.size();\n int sum=accumulate(nums.begin(), nums.end(), 0);\n sort(nums.begin(),nums.end());\n for(int i=n-1;i>=0;i--){\n int outlier=nums[i];\n int sum1=sum-outlier;\n if(i!=n-1){\n int res2=f(sum1,outlier,i+1,n-1,nums);\n if(res2!=-1) return nums[i];\n }\n int res=f(sum1,outlier,0,i-1,nums);\n if(res!=-1) return nums[i];\n }\n return -1;\n }\n};\n\n```
| 1 | 0 |
['Binary Search', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Beats 100% in memory, time, C++, easy to understand
|
beats-100-in-memory-time-c-easy-to-under-kn4g
|
\n\n# Approach\nSort the vector and traverse every element from opposite side. If nums[i] is the answer, then the sum excluding nums[i] should be 1) even 2) (su
|
yellowfence
|
NORMAL
|
2024-12-01T09:05:58.838276+00:00
|
2024-12-01T09:05:58.838307+00:00
| 65 | false |
\n\n# Approach\nSort the vector and traverse every element from opposite side. If nums[i] is the answer, then the sum excluding nums[i] should be 1) even 2) (sum-num[i])/2 should be present in the vector and the index shouldn\'t be equal to i. If all these conditions are met, then the answer is nums[i].\n\n# Complexity\n- Time complexity: O(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n sort(nums.begin(), nums.end());\n int sum = accumulate(nums.begin(), nums.end(), 0);\n int n = nums.size();\n for(int i = n-1;i>=0;i--) {\n sum -= nums[i];\n if(sum%2 == 0) {\n int ind = lower_bound(nums.begin(), nums.end(), sum/2) - nums.begin();\n if(ind < n && ind != i && nums[ind] == sum/2) {\n return nums[i];\n }\n }\n sum += nums[i];\n }\n return -1;\n }\n};\n```
| 1 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
SOLVED C++
|
solved-c-by-mukeshhdhadhariya-4fxb
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
mukeshhdhadhariya
|
NORMAL
|
2024-12-01T06:55:43.143964+00:00
|
2024-12-01T06:55:43.144001+00:00
| 42 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n multiset<int> numSet(nums.begin(), nums.end()); // Insert all elements into the multiset\n int total = 0;\n int max1 = INT_MIN;\n\n // Step 1: Calculate total and find the max value\n for (int a : nums) {\n total += a;\n max1 = max(max1, a); // Track the largest element\n }\n\n int ans = INT_MIN; // Variable to store the largest outlier\n\n // Step 2: Check for outliers\n for (int a : nums) {\n int remainingSum = total - a; // The sum of the special numbers should be (total - a)\n \n // If remaining sum is odd, it can\'t be split into two integer special numbers\n if (remainingSum % 2 != 0) continue;\n\n int p = remainingSum / 2; // The sum of the special numbers\n\n // Step 3: Check if \'p\' exists in the multiset and if \'p\' != \'a\' (distinct numbers)\n if (numSet.count(p) > 0) {\n // If p == a, we need to ensure that there are at least two occurrences of a.\n if (p == a && numSet.count(a) > 1) {\n ans = max(ans, a); // Update largest outlier\n }\n // If p != a, just check the count\n else if (p != a) {\n ans = max(ans, a); // Update largest outlier\n }\n }\n }\n\n // Step 4: If no outlier is found, return the largest element in nums\n return (ans == INT_MIN) ? max1 : ans;\n }\n};\n```
| 1 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy Linear Time Solution
|
easy-linear-time-solution-by-chakuriprar-fyhw
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
chakuriprarthi16
|
NORMAL
|
2024-12-01T06:24:29.359241+00:00
|
2024-12-01T06:24:29.359277+00:00
| 74 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int totalSum = accumulate(begin(nums), end(nums), 0);\n unordered_map<int, int> numCount;\n int answer = INT_MIN;\n\n for (int val : nums)\n numCount[val]++;\n\n for (int val : nums) {\n numCount[val]--;\n int sum = totalSum - val;\n\n if (sum % 2 == 0 and numCount[sum / 2] > 0)\n answer = max(answer, val);\n\n numCount[val]++;\n }\n\n return answer;\n }\n};\n```
| 1 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy HashMap Based Solution By Pui Pui
|
easy-hashmap-based-solution-by-pui-pui-b-exfd
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
pui_pui
|
NORMAL
|
2024-12-01T06:01:21.340401+00:00
|
2024-12-01T06:01:21.340435+00:00
| 366 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n# Code\n```java []\nclass Solution {\n public int getLargestOutlier(int[] nums) {\n\n // So here is the algorithm that I was able to find out.\n // 1. First we keep track of the count of the numbers and store their frequencies in the hashmap\n // 2. Next we, iterate through each number and consider them as the outlier\n // a. If we consider them an outlier, then the remaining sum has to be even\n // b. If the remaining sum is even, the special number is half of the sum. and if that special\n // number exists, then the number at the given index can be considered as an outlier.\n // Likewise we will keep on checking each of the numbers for a potential outlier and finally\n // get the maximum one.\n \n Map<Integer, Integer> countMap = new HashMap<>();\n\n // calculate the total sum here\n int totalSum = 0;\n\n for (int element : nums) {\n countMap.put(element, countMap.getOrDefault(element, 0) + 1);\n\n // we get the total sum of the elements in the array\n totalSum += element;\n }\n\n\n\n int totalNums = nums.length;\n\n \n\n int maxOutlier = Integer.MIN_VALUE;\n\n for (int i = 0; i < totalNums; i++) {\n // I consider the current number as the outlier\n int currentOutlier = nums[i];\n\n // we reduce the count of the current outlier in the map\n // this is taking into the consideration that the current number will not be\n // considered.\n countMap.put(currentOutlier, countMap.get(currentOutlier) - 1);\n\n // now I generate the sum without considering the current number\n int sumWithoutOutlier = totalSum - currentOutlier;\n\n if ((sumWithoutOutlier & 1) == 0) {\n // the sum is even\n // so we take the half of the sum\n int goodNumber = sumWithoutOutlier >> 1;\n if (countMap.getOrDefault(goodNumber, 0) > 0) {\n // this means that the number exists\n // so the current outlier can be a valid candidate\n maxOutlier = Math.max(maxOutlier, currentOutlier);\n }\n }\n\n // we reset the count of the current outlier in the map\n countMap.put(currentOutlier, countMap.get(currentOutlier) + 1);\n }\n\n // we finally return the max outlier.\n return maxOutlier;\n \n }\n}\n```
| 1 | 0 |
['Hash Table', 'Java']
| 1 |
identify-the-largest-outlier-in-an-array
|
Simple Hashmap Soln
|
simple-hashmap-soln-by-catchme999-u450
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
catchme999
|
NORMAL
|
2024-12-01T05:39:09.824786+00:00
|
2024-12-01T05:39:09.824817+00:00
| 181 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, a: List[int]) -> int:\n d=Counter(a)\n s=sum(a)\n ans=-float("inf")\n for i in a:\n t=(s-i)//2 \n d[i]-=1\n # print(t)\n if(d[t]>0 and (s-i)%2==0):\n # print(t)\n # print(d)\n ans=max(ans,i)\n d[i]+=1\n return ans\n \n \n```
| 1 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy intutive solution. O(N) .
|
easy-intutive-solution-on-by-harsh1937-m9v6
|
\n# Complexity\n- Time complexity:O(N)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(N)\n Add your space complexity here, e.g. O(n) \n\n#
|
harsh1937
|
NORMAL
|
2024-12-01T04:42:28.792507+00:00
|
2024-12-01T04:42:28.792563+00:00
| 92 | false |
\n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nimport java.util.HashMap;\n\nclass Solution {\n public int getLargestOutlier(int[] nums) {\n int sum = 0;\n for (int num : nums) {\n sum += num;\n }\n\n HashMap<Integer, Integer> freqMap = new HashMap<>();\n for (int num : nums) {\n freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);\n }\n\n int ans = Integer.MIN_VALUE;\n\n for (int num : nums) {\n int remainingSum = sum - num;\n // System.out.println("Remaining sum: " + remainingSum);\n\n if (remainingSum % 2 == 0) {\n int p = remainingSum / 2;\n // System.out.println("Checking for p: " + p);\n\n if (freqMap.containsKey(p)) {\n if (p == num && freqMap.get(p) < 2) continue;\n\n ans = Math.max(ans, num);\n // System.out.println("Ans: " + ans);\n }\n }\n }\n\n return ans == Integer.MIN_VALUE ? -1 : ans;\n }\n}\n\n```
| 1 | 0 |
['Hash Table', 'Math', 'Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java HashMap O(N) Time and Space
|
java-hashmap-on-time-and-space-by-abascu-nxyu
|
\nclass Solution {\n public int getLargestOutlier(int[] nums) {\n \n int n = nums.length;\n \n Map<Integer, Integer> map = new Ha
|
abascus
|
NORMAL
|
2024-12-01T04:32:54.780760+00:00
|
2024-12-01T04:32:54.780795+00:00
| 75 | false |
```\nclass Solution {\n public int getLargestOutlier(int[] nums) {\n \n int n = nums.length;\n \n Map<Integer, Integer> map = new HashMap<>();\n \n int sum = 0;\n \n for(int i = 0; i < n; i++) {\n \n sum += nums[i];\n \n map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);\n }\n \n int maxOutlier = -1001;\n \n for(int i = 0; i < n; i++) {\n \n int remainingSum = sum - nums[i];\n \n map.put(nums[i], map.get(nums[i]) - 1);\n \n if(remainingSum % 2 == 0 && map.getOrDefault(remainingSum / 2, 0) > 0) {\n \n maxOutlier = Math.max(maxOutlier, nums[i]);\n }\n \n map.put(nums[i], map.get(nums[i]) + 1);\n }\n \n return maxOutlier;\n \n }\n}\n```
| 1 | 0 |
[]
| 0 |
identify-the-largest-outlier-in-an-array
|
Fast C++ Solution
|
fast-c-solution-by-artsor53-0p9f
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
artsor53
|
NORMAL
|
2024-12-01T04:32:28.290652+00:00
|
2024-12-01T04:32:28.290671+00:00
| 99 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int n = nums.size();\n long long sum = 0;\n unordered_map<int, int> f;\n for (int x : nums) {\n sum += x;\n ++f[x];\n }\n int res = -1001;\n for (int i = 0; i < n; ++i) {\n long long newSum = sum - nums[i];\n --f[nums[i]];\n if (newSum % 2 == 0) {\n if (f.find(newSum / 2) != f.end() && f[newSum / 2] > 0) res = max(res, nums[i]); \n }\n ++f[nums[i]];\n }\n return res;\n }\n};\n```
| 1 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
C++ | Traverse and See if the Current Element is the Target or not
|
c-traverse-and-see-if-the-current-elemen-lwhc
|
Intuition\n Describe your first thoughts on how to solve this problem. \nFought Soooo many bugs in this :()\n\n1. Calculate the totalSum.\n2. Also put all the e
|
nishsolvesalgo
|
NORMAL
|
2024-12-01T04:32:21.999295+00:00
|
2024-12-01T04:38:28.793095+00:00
| 16 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFought Soooo many bugs in this :()\n\n1. Calculate the totalSum.\n2. Also put all the elements in a map basically to find out if the outlier is present or not.\n3. Traverse and see if we can make the current element Target.\n\nLemme know if it\'s not optimal, this is what i could think of, off the top of my head. \n\n# Code\n```cpp []\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int totalSum = accumulate(nums.begin(), nums.end(), 0);\n // Try to make every element the target and find if the residual is present or not..\n unordered_map<int, int> umap;\n for(int i=0; i<nums.size(); i++) umap[nums[i]]++;\n int maxOutlier = INT_MIN;\n for(int i=0; i<nums.size(); i++){\n int target = nums[i];\n int temp = totalSum - target;\n if(umap.count(temp - target) && (temp - target != target || umap[temp - target] > 1)) maxOutlier = max(maxOutlier, temp - target);\n }\n return maxOutlier;\n }\n};\n```
| 1 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
C++ Math Check Each Idx For Outlier
|
c-math-check-each-idx-for-outlier-by-bra-o2jh
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
bramar2
|
NORMAL
|
2024-12-01T04:32:16.007502+00:00
|
2024-12-01T04:32:16.007526+00:00
| 31 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\n#include <bits/stdc++.h>\nusing namespace std;\nclass Solution {\npublic:\n int getLargestOutlier(vector<int>& nums) {\n int n = nums.size();\n int sum = accumulate(nums.begin(), nums.end(), 0);\n unordered_map<int, int> freq;\n for(int num : nums) ++freq[num];\n int mx = -1001;\n for(int i = 0; i < n; i++) {\n \tsum -= nums[i]; --freq[nums[i]];\n // One of them is the sum so\n // [a, b, c, .., (a+b+c+..)]\n // sum = 2*(a+b+c+..)\n // check if (a+b+c+..) exists which is sum/2\n \tif(sum % 2 == 0 && freq[sum / 2] > 0) mx = max(mx, nums[i]);\n\n \tsum += nums[i]; ++freq[nums[i]];\n }\n return mx;\n }\n};\n```
| 1 | 0 |
['Math', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Python Simple O(N) Solution with explanation in comments
|
python-simple-on-solution-with-explanati-ol83
|
Code
|
kevincparas
|
NORMAL
|
2024-12-13T02:20:04.877046+00:00
|
2024-12-13T02:20:04.877046+00:00
| 57 | false |
\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n # total_sums - outlier / 2 == sum_of_specials(s)\n # if s in array then s is an outlier\n # edge case where outlier == s, need to make sure there is more than 1 of that element, we will use a hash map to count\n\n count_map = Counter(nums)\n total_sum = sum(nums)\n max_outlier = float(\'-inf\')\n\n for num in nums:\n #num is potential outlier\n sum_of_specials = (total_sum - num) / 2\n if sum_of_specials in count_map:\n if sum_of_specials == num: #edge case scenario\n if count_map[num] > 1: \n max_outlier = max(max_outlier, num)\n else:\n max_outlier = max(max_outlier, num)\n \n return max_outlier\n\n\n\n\n\n\n```
| 1 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Check if Sum of all remaining equal double of one item
|
check-if-sum-of-all-remaining-equal-doub-3u5f
| null |
Sam_Sundar
|
NORMAL
|
2024-12-12T03:33:00.828492+00:00
|
2024-12-12T03:33:00.828492+00:00
| 51 | false |
# Intuition\nThe value of outlier will be equal to the sum of all the elements subtracted by twice of the actual sum of elements\n\n# Complexity\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n def getLargestOutlier(self, nums: List[int]) -> int:\n #nums.sort()\n freq = Counter(nums)\n ans = -1001\n ts = sum(nums)\n n = len(nums)\n for i in range(n):\n outlier = ts - (2*nums[i])\n #print()\n if freq[outlier] - (1 if outlier == nums[i] else 0):\n ans = max(ans,outlier)\n return ans\n\n\n```
| 1 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java Solution
|
java-solution-by-kiranmai_reddy-tmzc
|
Code
|
kiranmai_reddy
|
NORMAL
|
2025-04-09T19:27:14.454304+00:00
|
2025-04-09T19:27:14.454304+00:00
| 1 | false |
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
HashMap<Integer, Integer> map = new HashMap<>();
int total = 0;
for (int n : nums) {
total += n;
map.put(n, map.getOrDefault(n, 0) + 1);
}
int out = Integer.MIN_VALUE;
for (int n : nums) {
int current = n;
int sum = total - current;
map.put(current, map.get(current) - 1);
if ((sum & 1) == 0) {
int even = sum >> 1;
if (map.getOrDefault(even, 0) > 0)
out = Math.max(out, current);
}
map.put(current, map.get(current) + 1);
}
return out;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
beats 95% runtime (java)
|
beats-95-runtime-java-by-dpasala-5oab
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
dpasala
|
NORMAL
|
2025-04-07T01:07:05.412302+00:00
|
2025-04-07T01:07:05.412302+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int total = 0;
Map<Integer, Integer> map = new HashMap<>();
for (int n : nums) {
total += n;
map.put(n, map.getOrDefault(n, 0) + 1);
}
int max = Integer.MIN_VALUE;
for (int k : map.keySet()) {
int out = total - 2 * k;
if (map.containsKey(out)) {
if (out != k || map.get(out) > 1) {
max = Math.max(max, out);
}
}
}
return max;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Intuitive and simple java solution
|
intuitive-and-simple-java-solution-by-pr-oq4z
|
IntuitionTotalSum = specialNumbersSum + 2nd last number + outlier;
we know -> 2nd last number = sum(specialNumbers)i.e. TotalSum = 2 * (2nd last number) + outli
|
pravachan
|
NORMAL
|
2025-04-03T21:30:49.807662+00:00
|
2025-04-03T21:30:49.807662+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
TotalSum = specialNumbersSum + 2nd last number + outlier;
we know -> 2nd last number = sum(specialNumbers)
i.e. TotalSum = 2 * (2nd last number) + outlier;
or outlier = TotalSum - 2 * (2nd last number)
# Approach
<!-- Describe your approach to solving the problem. -->
We can find the total sum easy in O(N)
We don't know what is the 2nd last number and so we can try all the options and see if we can maximize the outlier with below formula.
Outlier = TotalSum - 2 * (nums[i])
Edge case note: Note while confirming the outlier in each iteration, we have to be mindful of 2 conditions
1. The identified outlier needs to be a part of nums. i.e. it should be an element of nums.
2. The outlier itself cannot also be the 2nd elemment. i.e. in case of dupes, have the same index.
So we use a count Map to ensure we have more than one if the Outlier is also equal to the sum(2nd element).
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(N) for iterating over all the elemenets on nums to calculate the total.
Map add operations will be amortizied O(1) and so we will consider it O(1).i.e. O(N) * O(1) = O(N)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
we are using a countmap for all elements in nums. So worst case they are all unique and we store N elements - O(N)
O(N)
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int total = 0;
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for( int i = 0; i < nums.length; i++ ){
total += nums[i];
map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
}
int maxOutlier = Integer.MIN_VALUE;
for( int i = 0; i < nums.length; i++ ){
int sum = 2 * nums[i];
int out = total - sum;
if( map.containsKey(out) && ( nums[i] != out || map.get(out) > 1 ) )
maxOutlier = Math.max(maxOutlier, out);
}
return maxOutlier;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java Solution | Applying Mathematics | O(N)
|
java-solution-applying-mathematics-on-by-geah
|
IntuitionApproach
Check for each number as outlier
Rest values if they have one value as sum it should be divisible by 2
It should contain sum/2 also in rest of
|
rajeevkri
|
NORMAL
|
2025-03-31T09:36:45.948494+00:00
|
2025-03-31T09:36:45.948494+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
- Check for each number as outlier
- Rest values if they have one value as sum it should be divisible by 2
- It should contain sum/2 also in rest of the element
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
$$O(n)$$
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
$$O(n)$$
# Code
```java []
import java.util.HashMap;
import java.util.Map;
class Solution {
public int getLargestOutlier(int[] nums) {
int totalSum = 0;
Map<Integer, Integer> counts = new HashMap<>();
// Calculate total sum and count frequencies
for (int num : nums) {
totalSum += num;
counts.put(num, counts.getOrDefault(num, 0) + 1);
}
int outlier = Integer.MIN_VALUE;
for (int num : nums) {
// Temporarily remove current number
counts.put(num, counts.get(num) - 1);
totalSum -= num;
// Check if remaining sum is even and contains sum/2
if (totalSum % 2 == 0 && counts.getOrDefault(totalSum / 2, 0) > 0) {
outlier = Math.max(outlier, num);
}
// Restore the number
counts.put(num, counts.get(num) + 1);
totalSum += num;
}
return outlier;
}
}
```
| 0 | 0 |
['Array', 'Hash Table', 'Math', 'Counting', 'Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Explain each steps! || HashMap
|
explain-each-steps-hashmap-by-qiyu71-upu4
|
IntuitionThe key solution to this question is to figure out the formula:Total Sum = sum(nums) = a element whose value is the sum of all the special numbers + su
|
Qiyu71
|
NORMAL
|
2025-03-30T21:02:38.782684+00:00
|
2025-03-30T21:02:38.782684+00:00
| 5 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The key solution to this question is to figure out the formula:
Total Sum = sum(nums) = a element whose value is the sum of all the special numbers + sum(sepcial numbers) + outlier
= 2 * the element whose value is the sum of all the special numbers + outlier
Therefore, we can transfer the formula a little to describe the relationship between the ‘sum’ element and outlier:
outlier = Total sum - 2 * the 'sum' element
-------------------
From the question, we could get the other information: special numbers, the sum element, and the outlier must have distinct indices, but may share the same value.
It decides that we should collect the frequency of each elements to make sure that outlier and the sum element are not the same element.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(n)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(n)
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
# initalize the sum and num count dictionary
totalsum = 0
num_count = defaultdict(int)
# Loop through nums to get the total sum and count dictionary
for num in nums:
totalsum += num
num_count[num] += 1
# the lower bound is -1000
result = -1003
# loop through unique elements to guess the outlier and the sum element
for num in num_count.keys():
possi_outlier = totalsum - 2 * num
if possi_outlier in num_count.keys():
# possi_outlier and the sum element are not the same
if possi_outlier != num or num_count[possi_outlier] > 1:
result = max(possi_outlier,result)
return result
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple Python Solution - O(n)
|
simple-python-solution-on-by-jason_chent-xk2c
|
IntuitionApproachHashmap to store the occurance of the numbers and remember to add back the assumed outlier after the checkComplexity
Time complexity:
Space co
|
jason_chent
|
NORMAL
|
2025-03-30T04:35:12.708480+00:00
|
2025-03-30T04:35:12.708480+00:00
| 1 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
Hashmap to store the occurance of the numbers and remember to add back the assumed outlier after the check
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python []
class Solution(object):
def getLargestOutlier(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
# sum of special values = n, one sum = n, one outlier = a, we want to max a
total = sum(nums)
diction = {}
number = set(nums)
for num in nums:
if num not in diction:
diction[num] = 1
else:
diction[num] += 1
res = -1001
for num in nums:
# assume this num is outlier, then check the rest fit into the condition or not.
tempsumtwo = total - num
if tempsumtwo % 2 == 1:
continue
else:
tempsum = tempsumtwo // 2
if tempsum in diction:
diction[num] -= 1
if diction[tempsum] > 0 and res < num:
res = num
diction[num] += 1
else:
continue
print(total, res)
return res
```
| 0 | 0 |
['Python']
| 0 |
identify-the-largest-outlier-in-an-array
|
HashMap and sum
|
hashmap-and-sum-by-awuxiaoqi-dlzl
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
labao
|
NORMAL
|
2025-03-29T18:29:02.684720+00:00
|
2025-03-29T18:29:02.684720+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int sum = 0;
HashMap<Integer, List<Integer>> vToI = new HashMap<>();
int max = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; ++i) {
sum += nums[i];
if(!vToI.containsKey(nums[i])) {
vToI.put(nums[i], new LinkedList<>());
}
vToI.get(nums[i]).add(i);
}
for (int i = 0; i < nums.length; ++i) {
int sumRest = sum - nums[i] * 2;
if (vToI.containsKey(sumRest) && (vToI.get(sumRest).size() > 1 || vToI.get(sumRest).get(0) != i)) {
max = Math.max(sumRest, max);
}
}
return max;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple O(N) Java solution
|
simple-on-java-solution-by-atharvadj97-fjp9
|
IntuitionThe inuition is to try every number as an outlier, and with that thought in mind, do following things -
Understand that for this number to be a valid o
|
atharvadj97
|
NORMAL
|
2025-03-27T00:54:06.581791+00:00
|
2025-03-27T00:54:06.581791+00:00
| 5 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The inuition is to try every number as an outlier, and with that thought in mind, do following things -
* Understand that for this number to be a valid outlier, the remaining sum should be able to split in two parts, wherein one part is sum of x numbers, and the remaining part is a number.
* sumWithoutOutlier = totalSum - outlier
if(sumWithoutOutlier / 2) == 0, means it can be split, now check if it is present in nums, for which you can add all numbers to a hashMap with their index.
* if(sumWithoutOutlier / 2) is present then this is valid outlier, take care of edge case such that (sumWithoutOutlier / 2) is not the current nums[i] itself,
Hence get (sumWithoutOutlier / 2) from hashMap and check if it is not equal to curr idx, if not then it is valid outlier.
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(N)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(N)
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int sum = 0;
Map<Integer, Integer> hMap = new HashMap<Integer, Integer>();
int outlier = Integer.MIN_VALUE;
for(int i = 0; i < nums.length; i++){
sum += nums[i];
hMap.put(nums[i], i);
}
for(int i = 0; i < nums.length; i++){
int curr = sum - nums[i];
if(curr % 2 == 0){
if(hMap.containsKey(curr / 2) && hMap.get(curr / 2) != i){
outlier = Math.max(outlier, nums[i]);
}
}
}
return outlier;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Optimized Counting with Explanation
|
optimized-counting-with-explanation-by-s-26zr
|
IntuitionApproachComplexity
Time complexity: O(n)
Space complexity: O(n)
Code
|
stalebii
|
NORMAL
|
2025-03-26T02:51:39.544359+00:00
|
2025-03-26T02:51:39.544359+00:00
| 9 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
total = sum(nums)
counter = Counter(nums) # stores {num: freq}
res = float('-inf')
for outlier, freq in counter.items():
delta = total - outlier
# stop; should be even
if delta % 2 != 0:
continue
# generate candidate
cand = delta // 2
# alreday seen it
if cand in counter:
# ensure cand and outlier are from different indices
if cand == outlier and freq > 1:
res = max(res, outlier)
# don't need to check index
elif cand != outlier:
res = max(res, outlier)
return res
```
| 0 | 0 |
['Array', 'Hash Table', 'Math', 'Counting', 'Enumeration', 'Python', 'Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy py sol :)
|
easy-py-sol-by-user0490hi-vjhn
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
user0490hI
|
NORMAL
|
2025-03-25T14:22:33.935454+00:00
|
2025-03-25T14:22:33.935454+00:00
| 7 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
tot = 0
dic = {}
for i in range(len(nums)):
tot += nums[i]
if nums[i] not in dic:
dic[nums[i]] = 1
else:
dic[nums[i]] += 1
jvrc = float('-inf')
print(tot)
for i in range(len(nums)):
temp = nums[i]+nums[i]
if tot-temp in dic:
if ((tot-temp)!=nums[i]):
jvrc = max(jvrc, (tot-temp))
elif ((tot-temp)==nums[i] and dic[tot-temp] > 1):
jvrc = max(jvrc, (tot-temp))
return jvrc
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Kotlin | Count Array
|
kotlin-count-array-by-jschnall-a4xd
|
Intuition & ApproachJust read my comment at the topComplexity
Time complexity:
O(n) where n is the size of nums
Space complexity:
O(1)
Code
|
jschnall
|
NORMAL
|
2025-03-25T01:28:07.646397+00:00
|
2025-03-25T01:30:10.112387+00:00
| 7 | false |
# Intuition & Approach
Just read my comment at the top
# Complexity
- Time complexity:
$$O(n)$$ where n is the size of nums
- Space complexity:
$$O(1)$$
# Code
```kotlin []
class Solution {
fun getLargestOutlier(nums: IntArray): Int {
// Plan: Sum all the numbers, try removing each number in turn from largest to smallest.
// See if the remainder is equal to 2 * one of the remaining values in it.
// Change of plan: List isn't sorted, it'd be nlogn to sort descending so we can stop at first found,
// as opposed to n to simply check every item in the list. Let's do that instead
// Also the range for nums is only 2001, so we can keep an array instead of a map
// val counts = nums.asList().groupingBy { it }.eachCount()
val counts = IntArray(2001)
val sum = nums.sum()
var max = Integer.MIN_VALUE
for (num in nums) {
counts[num + 1000]++
}
for (num in nums) {
// Check that it divides evenly
if (abs(sum - num) % 2 == 1) continue
val target = (sum - num) / 2
if (target < -1000 || target > 1000) continue // Prevent ArrayIndexOutOfBounds
val count = counts[target + 1000] // ?: 0
if (target == num) {
if (count > 1) max = maxOf(max, num)
} else {
if (count > 0) max = maxOf(max, num)
}
}
return max
}
}
```
| 0 | 0 |
['Kotlin']
| 0 |
identify-the-largest-outlier-in-an-array
|
Hashing | Explained
|
hashing-explained-by-prashant404-ut43
|
Idea:
Find total sum = T and a frequency map of numbers
Consider each number as outlier (L) one by one
Sum of special numbers = (T - L) / 2
This is an invalid o
|
prashant404
|
NORMAL
|
2025-03-24T21:51:13.931487+00:00
|
2025-03-24T21:51:13.931487+00:00
| 6 | false |
**Idea:**
* Find total sum = T and a frequency map of numbers
* Consider each number as outlier (L) one by one
* Sum of special numbers = (T - L) / 2
* This is an invalid outlier candidate if:
* T - L is odd
* S does not exist in the map
* S exists but only once, which means S is the same as L, but outlier can't be the same number (i.e. the same index) as the sum of special numbers
* Else this is a valid candidate, find max of all such Ls
>T/S: O(n)/O(n), where n = size(nums)
```java []
public int getLargestOutlier(int[] nums) {
var max = Integer.MIN_VALUE;
var map = new HashMap<Integer, Integer>();
var total = getTotal(nums, map);
for (var outlier : nums) {
var diff = total - outlier;
var specialSum = diff >> 1;
if (diff % 2 == 0
&& map.containsKey(specialSum)
&& (specialSum != outlier || map.get(specialSum) > 1))
max = Math.max(max, outlier);
}
return max;
}
private int getTotal(int[] nums, Map<Integer, Integer> map) {
var total = 0;
for (var num : nums) {
total += num;
map.compute(num, (_, v) -> v == null ? 1 : ++v);
}
return total;
}
```
***Please upvote if this helps***
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
O(n) solution...
|
on-solution-by-arogau-l3q4
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
arogau
|
NORMAL
|
2025-03-24T02:38:53.933220+00:00
|
2025-03-24T02:38:53.933220+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
from collections import Counter
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
numshash = Counter(nums)
total = sum(nums)
maxans = float('-inf')
for i in nums:
target = (total - i)/2
if target in numshash:
if target == i:
if numshash[target] >= 2:
maxans = max(maxans, i)
else:
maxans = max(maxans, i)
return maxans
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Hashmap + Math | C++ | O(N)
|
hashmap-math-c-on-by-samdak-6a26
|
IntuitionSet each number as outlier, try to find whether other elements can make sum element and special numbers.Based on constraints, would not be able to brut
|
samdak
|
NORMAL
|
2025-03-22T12:52:15.303936+00:00
|
2025-03-22T12:52:15.303936+00:00
| 9 | false |
# Intuition
Set each number as outlier, try to find whether other elements can make sum element and special numbers.
Based on constraints, would not be able to brute force as it is O(N^2).
A key observation is that if we exclude the outlier, the remaining sum of the numbers in the array is 2 * sum element.
# Approach
1. Create a hashmap to store frequencies of each number in the array.
2. Calculate the sum of all elements in the array, `totalSum`.
3. Initialize a `largest` variable to keep track of the largest outlier.
4. Iterate through `nums`, using each number as a candidate for `outlier`
- At each step, calculate the `remainingSum`, which is `totalSum - outlier`
- `remainingSum` is only valid if it is even, if is not, we can ignore the current outlier candidate
- If it is even, find the `sumElement`, which is `remainingSum / 2`
- From here, there are three cases
- `sumElement` is not in the hashmap, then we can continue
- `sumElement` is in the hashmap and `sumElement != outlier`, meaning the current outlier is a valid candidate
- `sumElement == outlier` but the frequency of sumElement in the hashmap is only 1, it needs to be 2 to be valid, so we ignore this
- If it is a valid outlier, we maximize using `largest`
5. At the end, we return `largest`
# Complexity
- Time complexity: O(N) for two for loops
- Space complexity: O(N) as at most N unique numbers will be stored in the hashmap
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
std::unordered_map<int, int> numFreq;
int totalSum{0};
for (const auto& num: nums) {
numFreq[num]++;
totalSum += num;
}
int largest{INT_MIN};
for (const auto& outlier: nums) {
auto remainingSum{totalSum - outlier};
if (remainingSum % 2 != 0) continue;
auto sumElement{remainingSum / 2};
if (!numFreq.contains(sumElement) || (sumElement == outlier && numFreq[sumElement] == 1)) continue;
largest = std::max(largest, outlier);
}
return largest;
}
};
```
| 0 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java | JavaScript | TypeScript | C++ | C# | Kotlin | Go Solution.
|
java-javascript-typescript-c-c-kotlin-go-3zj5
|
The presented solution applies a frequency array instead of a Hash Map. The problem can be solved very efficiently with a frequency array due to the small input
|
LachezarTsK
|
NORMAL
|
2025-03-22T09:11:54.464127+00:00
|
2025-03-22T09:11:54.464127+00:00
| 11 | false |
The presented solution applies a frequency array instead of a Hash Map. The problem can be solved very efficiently with a frequency array due to the small input range, [-1000, 1000], as of March 2025.
```Java []
public class Solution {
private static final int[] RANGE_OF_VALUES = {-1000, 1000};
private static final int OFFSET_FOR_NEGATIVES = 1000;
public int getLargestOutlier(int[] input) {
int[] frequency = new int[1 + RANGE_OF_VALUES[1] + OFFSET_FOR_NEGATIVES];
int totalSumValues = 0;
for (int value : input) {
++frequency[value + OFFSET_FOR_NEGATIVES];
totalSumValues += value;
}
return findMaxOutlier(totalSumValues, frequency);
}
private int findMaxOutlier(int totalSumValues, int[] frequency) {
int maxOutlier = Integer.MIN_VALUE;
for (int value = RANGE_OF_VALUES[1]; value >= RANGE_OF_VALUES[0]; --value) {
if (frequency[value + OFFSET_FOR_NEGATIVES] == 0) {
continue;
}
double candidate = ((double) totalSumValues - value) / 2;
if (isValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency)) {
maxOutlier = value;
break;
}
}
return maxOutlier;
}
private boolean isValidCandidateForSpecialNumberThatIsNotOutlier(double candidate, int value, int[] frequency) {
return candidate == (int) candidate
&& candidate >= RANGE_OF_VALUES[0]
&& candidate <= RANGE_OF_VALUES[1]
&& frequency[(int) candidate + OFFSET_FOR_NEGATIVES] > 0
&& (value != (int) candidate || frequency[value + OFFSET_FOR_NEGATIVES] > 1);
}
}
```
```JavaScript []
/**
* @param {number[]} input
* @return {number}
*/
var getLargestOutlier = function (input) {
this.RANGE_OF_VALUES = [-1000, 1000];
this.OFFSET_FOR_NEGATIVES = 1000;
const frequency = new Array(1 + this.RANGE_OF_VALUES[1] + this.OFFSET_FOR_NEGATIVES).fill(0);
let totalSumValues = 0;
for (let value of input) {
++frequency[value + this.OFFSET_FOR_NEGATIVES];
totalSumValues += value;
}
return findMaxOutlier(totalSumValues, frequency);
};
/**
* @param {number} totalSumValues
* @param {number[]} frequency
* @return {number}
*/
function findMaxOutlier(totalSumValues, frequency) {
let maxOutlier = Number.MIN_SAFE_INTEGER;
for (let value = this.RANGE_OF_VALUES[1]; value >= this.RANGE_OF_VALUES[0]; --value) {
if (frequency[value + this.OFFSET_FOR_NEGATIVES] === 0) {
continue;
}
const candidate = (totalSumValues - value) / 2;
if (isValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency)) {
maxOutlier = value;
break;
}
}
return maxOutlier;
}
/**
* @param {number} candidate
* @param {number} value
* @param {number[]} frequency
* @return {boolean}
*/
function isValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency) {
return candidate === Math.floor(candidate)
&& candidate >= this.RANGE_OF_VALUES[0]
&& candidate <= this.RANGE_OF_VALUES[1]
&& frequency[candidate + this.OFFSET_FOR_NEGATIVES] > 0
&& (value !== candidate || frequency[value + this.OFFSET_FOR_NEGATIVES] > 1);
}
```
```TypeScript []
function getLargestOutlier(input: number[]): number {
this.RANGE_OF_VALUES = [-1000, 1000];
this.OFFSET_FOR_NEGATIVES = 1000;
const frequency = new Array(1 + this.RANGE_OF_VALUES[1] + this.OFFSET_FOR_NEGATIVES).fill(0);
let totalSumValues = 0;
for (let value of input) {
++frequency[value + this.OFFSET_FOR_NEGATIVES];
totalSumValues += value;
}
return findMaxOutlier(totalSumValues, frequency);
};
function findMaxOutlier(totalSumValues: number, frequency: number[]): number {
let maxOutlier = Number.MIN_SAFE_INTEGER;
for (let value = this.RANGE_OF_VALUES[1]; value >= this.RANGE_OF_VALUES[0]; --value) {
if (frequency[value + this.OFFSET_FOR_NEGATIVES] === 0) {
continue;
}
const candidate = (totalSumValues - value) / 2;
if (isValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency)) {
maxOutlier = value;
break;
}
}
return maxOutlier;
}
function isValidCandidateForSpecialNumberThatIsNotOutlier(candidate: number, value: number, frequency: number[]): boolean {
return candidate === Math.floor(candidate)
&& candidate >= this.RANGE_OF_VALUES[0]
&& candidate <= this.RANGE_OF_VALUES[1]
&& frequency[candidate + this.OFFSET_FOR_NEGATIVES] > 0
&& (value !== candidate || frequency[value + this.OFFSET_FOR_NEGATIVES] > 1);
}
```
```C++ []
#include <span>
#include <limits>
#include <vector>
using namespace std;
class Solution {
static constexpr array<int, 2> RANGE_OF_VALUES = { -1000, 1000 };
static const int OFFSET_FOR_NEGATIVES = 1000;
public:
int getLargestOutlier(vector<int>& input) const {
array<int, 1 + RANGE_OF_VALUES[1] + OFFSET_FOR_NEGATIVES> frequency{};
int totalSumValues = 0;
for (const auto& value : input) {
++frequency[value + OFFSET_FOR_NEGATIVES];
totalSumValues += value;
}
return findMaxOutlier(totalSumValues, frequency);
}
private:
int findMaxOutlier(int totalSumValues, span<const int> frequency) const {
int maxOutlier = numeric_limits<int>::min();
for (int value = RANGE_OF_VALUES[1]; value >= RANGE_OF_VALUES[0]; --value) {
if (frequency[value + OFFSET_FOR_NEGATIVES] == 0) {
continue;
}
double candidate = (static_cast<double>(totalSumValues) - value) / 2;
if (isValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency)) {
maxOutlier = value;
break;
}
}
return maxOutlier;
}
bool isValidCandidateForSpecialNumberThatIsNotOutlier(double candidate, int value, span<const int> frequency) const {
return candidate == (int)candidate
&& candidate >= RANGE_OF_VALUES[0]
&& candidate <= RANGE_OF_VALUES[1]
&& frequency[(int)candidate + OFFSET_FOR_NEGATIVES] > 0
&& (value != (int)candidate || frequency[value + OFFSET_FOR_NEGATIVES] > 1);
}
};
```
```C# []
using System;
public class Solution
{
private static readonly int[] RANGE_OF_VALUES = { -1000, 1000 };
private static readonly int OFFSET_FOR_NEGATIVES = 1000;
public int GetLargestOutlier(int[] input)
{
int[] frequency = new int[1 + RANGE_OF_VALUES[1] + OFFSET_FOR_NEGATIVES];
int totalSumValues = 0;
foreach (int value in input)
{
++frequency[value + OFFSET_FOR_NEGATIVES];
totalSumValues += value;
}
return FindMaxOutlier(totalSumValues, frequency);
}
private int FindMaxOutlier(int totalSumValues, int[] frequency)
{
int maxOutlier = int.MinValue;
for (int value = RANGE_OF_VALUES[1]; value >= RANGE_OF_VALUES[0]; --value)
{
if (frequency[value + OFFSET_FOR_NEGATIVES] == 0)
{
continue;
}
double candidate = ((double)totalSumValues - value) / 2;
if (IsValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency))
{
maxOutlier = value;
break;
}
}
return maxOutlier;
}
private bool IsValidCandidateForSpecialNumberThatIsNotOutlier(double candidate, int value, int[] frequency)
{
return candidate == (int)candidate
&& candidate >= RANGE_OF_VALUES[0]
&& candidate <= RANGE_OF_VALUES[1]
&& frequency[(int)candidate + OFFSET_FOR_NEGATIVES] > 0
&& (value != (int)candidate || frequency[value + OFFSET_FOR_NEGATIVES] > 1);
}
}
```
```Kotlin []
import kotlin.math.floor
class Solution {
private companion object {
val RANGE_OF_VALUES = intArrayOf(-1000, 1000)
const val OFFSET_FOR_NEGATIVES = 1000
}
fun getLargestOutlier(input: IntArray): Int {
val frequency = IntArray(1 + RANGE_OF_VALUES[1] + OFFSET_FOR_NEGATIVES)
var totalSumValues = 0
for (value in input) {
++frequency[value + OFFSET_FOR_NEGATIVES]
totalSumValues += value
}
return findMaxOutlier(totalSumValues, frequency)
}
private fun findMaxOutlier(totalSumValues: Int, frequency: IntArray): Int {
var maxOutlier = Int.MIN_VALUE
for (value in RANGE_OF_VALUES[1] downTo RANGE_OF_VALUES[0]) {
if (frequency[value + OFFSET_FOR_NEGATIVES] == 0) {
continue
}
val candidate: Double = (totalSumValues.toDouble() - value) / 2
if (isValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency)) {
maxOutlier = value
break
}
}
return maxOutlier
}
private fun isValidCandidateForSpecialNumberThatIsNotOutlier(candidate: Double, value: Int, frequency: IntArray): Boolean {
return candidate == floor(candidate)
&& candidate >= RANGE_OF_VALUES[0]
&& candidate <= RANGE_OF_VALUES[1]
&& frequency[candidate.toInt() + OFFSET_FOR_NEGATIVES] > 0
&& (value != candidate.toInt() || frequency[value + OFFSET_FOR_NEGATIVES] > 1)
}
}
```
```Go []
package main
import "math"
var RANGE_OF_VALUES = [2]int{-1000, 1000}
const OFFSET_FOR_NEGATIVES = 1000
func getLargestOutlier(input []int) int {
frequency := make([]int, 1 + RANGE_OF_VALUES[1] + OFFSET_FOR_NEGATIVES)
var totalSumValues = 0
for _, value := range input {
frequency[value + OFFSET_FOR_NEGATIVES]++
totalSumValues += value
}
return findMaxOutlier(totalSumValues, frequency)
}
func findMaxOutlier(totalSumValues int, frequency []int) int {
var maxOutlier = math.MinInt
for value := RANGE_OF_VALUES[1]; value >= RANGE_OF_VALUES[0]; value-- {
if frequency[value + OFFSET_FOR_NEGATIVES] == 0 {
continue
}
candidate := float64(totalSumValues - value) / 2
if isValidCandidateForSpecialNumberThatIsNotOutlier(candidate, value, frequency) {
maxOutlier = value
break
}
}
return maxOutlier
}
func isValidCandidateForSpecialNumberThatIsNotOutlier(candidate float64, value int, frequency []int) bool {
return candidate == math.Floor(candidate) &&
int(candidate) >= RANGE_OF_VALUES[0] &&
int(candidate) <= RANGE_OF_VALUES[1] &&
frequency[int(candidate) + OFFSET_FOR_NEGATIVES] > 0 &&
(value != int(candidate) || frequency[value + OFFSET_FOR_NEGATIVES] > 1)
}
```
| 0 | 0 |
['C++', 'Java', 'Go', 'TypeScript', 'Kotlin', 'JavaScript', 'C#']
| 0 |
identify-the-largest-outlier-in-an-array
|
My first crashout :)
|
my-first-crashout-by-shimonocean-8u84
|
i might swerve bend that corner woaaaooaaahCode
|
shimonocean
|
NORMAL
|
2025-03-21T20:27:41.699120+00:00
|
2025-03-21T20:27:41.699120+00:00
| 3 | false |
i might swerve bend that corner woaaaooaaah
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
# (n-2) numbers are summed
# (1) number is the sum of (n-2)
# (1) number is an outlier
# at every n in nums
# treat n as the outlier
# sum nums without n / 2 == candidate sum of (n-2), call potentialSum
# if potentialSum in nums:
# n is outlier if max outlier seen so far
numsSum = sum(nums)
numsMap = defaultdict(list)
# ansMap = {"outlier":float('-inf'), "diff":float('-inf')}
largestOutlier = float('-inf')
# print(f"SUM IS: {numsSum}")
for i, n in enumerate(nums):
numsMap[n].append(i)
# print(numsMap)
for i in range(len(nums)):
candidateOutlier = nums[i]
potentialSum = numsSum - nums[i]
potentialNum = (potentialSum / 2) # sum of special nums
curDiff = abs(candidateOutlier - potentialNum)
# print(f"{candidateOutlier}, {curDiff}")
if potentialNum in numsMap and i != numsMap[potentialNum][0]:
# if (len(numsMap[candidateOutlier]) == 1 and candidateOutlier != potentialNum) or (candidateOutlier == potentialNum and len(numsMap[candidateOutlier]) == 2):
# print(f"Passed condition: {candidateOutlier}, {curDiff}")
# ansMap["outlier"] = candidateOutlier
# ansMap["diff"] = curDiff
largestOutlier = max(largestOutlier, candidateOutlier)
return largestOutlier
# return ansMap["outlier"]
# 5
# 10
# 5
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
O(n)
|
on-by-user5285zn-kh16
|
After removing a candidate x for the outlier, we need to check if the remaining sum is the double of some value y.
|
user5285Zn
|
NORMAL
|
2025-03-21T09:57:20.774979+00:00
|
2025-03-21T09:57:20.774979+00:00
| 2 | false |
After removing a candidate $x$ for the outlier, we need to check if the remaining sum is the double of some value $y$.
```rust []
use std::collections::{BTreeSet, HashMap};
impl Solution {
pub fn get_largest_outlier(nums: Vec<i32>) -> i32 {
let dbl = nums.iter()
.enumerate()
.fold(HashMap::<i32, Vec<usize>>::new(), |mut dbl, (i, &x)| {
dbl.entry(2*x).or_default().push(i);
dbl
});
let mut res = None;
let total = nums.iter().sum::<i32>();
for (i, x) in nums.into_iter().enumerate() {
let y = total - x;
if dbl.contains_key(&y) && dbl[&y] != vec![i] {
res = res.max(Some(x))
}
}
res.unwrap()
}
}
```
| 0 | 0 |
['Rust']
| 0 |
identify-the-largest-outlier-in-an-array
|
using hash map and total sum
|
using-hash-map-and-total-sum-by-yunxia_7-j6tg
|
IntuitionThe largest outlier need to found. So sort the nums and try the largest number as the outlier. If it is the outlier, then the sum of remaining numbers
|
yunxia_79
|
NORMAL
|
2025-03-20T20:32:57.131380+00:00
|
2025-03-20T20:32:57.131380+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The largest outlier need to found. So sort the nums and try the largest number as the outlier. If it is the outlier, then the sum of remaining numbers should be 2*(one of the remaining number)=> the sum of remaining numbers should be even and sum/2 can be found in the remaining numbers.
So the total sum can be calculated and hashmap can be used to help searching.
# Approach
<!-- Describe your approach to solving the problem. -->
use the hash map to store the num and its count
get the sum of all
try the current number to be outlier to see if the remaining sum is even and remaining sum /2 can be found in the remaining number.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
sort is O(nlogn). Since this problem is only need to know the larget outlier, sort isn't necessary and can use one variable to store the largest outlier and update it when the larger one is found.
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(n) ---use hash map to store the num count
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
unordered_map<int, int> num_count;
int sum = 0;
for(int i = 0; i < nums.size(); i++)
{
sum += nums[i];
num_count[nums[i]]++;
}
sort(nums.begin(), nums.end());
for(int i = nums.size() - 1; i >= 0; i--)
{
int sumL = sum - nums[i];
if(sumL % 2)
{
continue;
}
num_count[nums[i]]--;
if(num_count[sumL/2] > 0)
{
return nums[i];
}
num_count[nums[i]]++;
}
return -1;
}
};
```
| 0 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
python solution
|
python-solution-by-gurdjieff-uenr
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
gurdjieff
|
NORMAL
|
2025-03-18T22:01:12.344572+00:00
|
2025-03-18T22:01:12.344572+00:00
| 7 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
val = sum(nums)
counter = Counter(nums)
res = - inf
for n in nums:
remain = val - n
if remain % 2 == 0 and remain // 2 in counter:
if remain // 2 != n or (remain // 2 == n and counter[n] >= 2):
res = max(res, n)
return res
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
python solution, moving the goalposts and simulate
|
python-solution-moving-the-goalposts-and-6j5w
|
weekly contest 442 resultIntuitionstart off confusingly, but then I have an idea of simulation. I will treat each number as if they are the sum of all special n
|
stephen-moon
|
NORMAL
|
2025-03-17T16:05:53.127307+00:00
|
2025-03-23T04:12:52.650394+00:00
| 76 | false |
# weekly contest 442 result
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
start off confusingly, but then I have an idea of simulation. I will treat each number as if they are the sum of all special numbers `take the shot`, then I can use two steps to derive the outlier in this particular case `move the goalpost to match the shot`. Use a hashMap to determine if the identified outlier is inside the input nums.
now I can use this to extract all the pretential outliers, so return the maximum out of them.
- I imagine the approach to be linear, and then I checked the constraint. linear complexity should be fine👌👍.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
$$O(n)$$
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
$$O(n)$$
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
hashMap = Counter(nums)
sum_ = sum(nums)
res = -float('inf')
for num in nums:
# pretend num is the sum of all special numbers.
curSum = sum_ - num # curSum --> num + outlier
outlier = curSum - num
if outlier in hashMap and (outlier != num or hashMap[outlier] > 1): # if outlier exists inside nums, update the res
res = max(res, outlier)
return res
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simplest | C++ | O(n) | two pass
|
simplest-c-on-two-pass-by-tilotkarsanket-so1e
|
IntuitionSimple MathApproachcompute sum of all numbers.
Check all the numbers one by one,
If a number is outlier then (sum-outlier)/2 should be present in the a
|
tilotkarsanket
|
NORMAL
|
2025-03-16T06:01:23.016974+00:00
|
2025-03-16T06:01:23.016974+00:00
| 5 | false |
# Intuition
Simple Math
# Approach
compute sum of all numbers.
Check all the numbers one by one,
If a number is outlier then (sum-outlier)/2 should be present in the array.
# Complexity
- Time complexity:
$$O(n)$$
Two passes on array
- Space complexity:
$$O(n)$$
To store map.
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums)
{
int sum = 0;
unordered_map<int, int> m;
int ans = INT_MIN;
for(int i=0;i<nums.size();i++)
{
sum = sum + nums[i];
m[nums[i]]++;
}
for(int i=0;i<nums.size();i++){
int outlier = nums[i];
int remaining = sum - outlier;
if(remaining%2==0 && m.find(remaining/2)!=m.end()){
if(remaining/2 != outlier) ans = max(ans, outlier);
if(remaining/2 == outlier && m[outlier]>1) ans = max(ans, outlier);
}
}
return ans;
}
};
```
| 0 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Two pass solution, O(N) time and space
|
two-pass-solution-on-time-and-space-by-p-lwby
|
IntuitionApproachComplexity
Time complexity: O(N)
Space complexity: O(N)
Code
|
PTool
|
NORMAL
|
2025-03-15T04:15:01.476402+00:00
|
2025-03-15T04:15:01.476402+00:00
| 8 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
# exactly n - 2 elements are special numbers. One of the remaining two elements is the sum of these special numbers
# therefore in nums there exist sum(sublist) = s
# for a valid outlier "o" to exist
# sum(nums)-o = 2s
# where s is a number in nums
# calculate the sum of nums
# count the frequency of each digit in nums
# iterate over nums
# consider x as the outlier
# decrease the count of x in the freq_map (we want to exclude the outlier from calculating is s in valid)
# subtract x from the total
# check if total is an even number and that total//2 exists in the frequency map
# if true then update soln to max(soln, x)
# increase the count of x again
# return soln
# T: O(N)
# S: O(N)
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
total = sum(nums)
counts = Counter(nums)
outlier = float('-inf')
for candidate in counts:
curr_sum = total - candidate # curr_sum = 2s if candidate is an outlier
counts[candidate] -= 1
if curr_sum % 2 == 0 and counts[curr_sum//2] > 0:
outlier = max(outlier, candidate)
counts[candidate] += 1
return outlier
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple O(N) T and S Java Solution with explanation.
|
simple-on-t-and-s-java-solution-with-exp-9656
|
Intuition:
We need to find the largest number in the array that, when removed,
results in a sum that is evenly divisible by 2.
We achieve this by checking if (s
|
smit-1z
|
NORMAL
|
2025-03-11T15:32:02.353410+00:00
|
2025-03-11T15:32:02.353410+00:00
| 17 | false |
Intuition:
We need to find the largest number in the array that, when removed,
results in a sum that is evenly divisible by 2.
We achieve this by checking if (sum - num) / 2 exists in the array.
Approach:
1. Calculate the total sum of the array.
2. Use a HashMap to store the frequency of each number.
3. Iterate through each number and check:
- If removing the number makes the sum even.
- If (sum - num) / 2 exists in the HashMap.
- If so, update the largest outlier.
4. Return the largest valid outlier, or -1 if none exists.
Complexity Analysis:
- Time complexity: O(n), since we iterate through the array twice.
- Space complexity: O(n), due to the HashMap storage.
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
HashMap<Integer, Integer> map = new HashMap<>();
int sum = 0;
int largestOutlier = Integer.MIN_VALUE;
for (int num : nums) {
map.put(num, map.getOrDefault(num, 0) + 1);
sum += num;
}
for (int num : nums) {
if ((sum - num) % 2 == 0) {
int target = (sum - num) / 2;
if (map.containsKey(target)) {
map.put(num, map.get(num) - 1);
if (map.get(target) > 0) {
largestOutlier = Math.max(largestOutlier, num);
}
map.put(num, map.get(num) + 1);
}
}
}
return largestOutlier == Integer.MIN_VALUE ? -1 : largestOutlier;
}
}
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple O(n) in Java
|
simple-on-in-java-by-turingcomplete-vhfs
|
Code
|
turingcomplete
|
NORMAL
|
2025-03-10T01:27:05.671178+00:00
|
2025-03-10T01:27:05.671178+00:00
| 16 | false |
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
var total = Arrays.stream(nums).sum();
var seen = new HashSet<Integer>();
var outlier = Integer.MIN_VALUE;
for (var i = 0; i < nums.length; i++) {
// diff = outlier + sumOfSpecials || sum + sumOfSpecials || outlier + sum + (sumOfSpecials - nums[i])
var diff = total - nums[i];
// if nums[i] is the outlier
if (diff % 2 == 0 && seen.contains(diff / 2)) {
outlier = Math.max(outlier, nums[i]);
}
// if num[i] is the sum
if (seen.contains(diff - nums[i])) {
outlier = Math.max(outlier, diff - nums[i]);
}
seen.add(nums[i]);
}
return outlier;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
using map as n(unique numbers) much less than total numbers
|
using-map-as-nunique-numbers-much-less-t-icvf
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
aasaini
|
NORMAL
|
2025-03-07T19:32:47.880034+00:00
|
2025-03-07T19:32:47.880034+00:00
| 11 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
/*
constraints:
nums, n
n-2 => special numbers
one of the remaining 2 => sum(special numbers)
other => outlier
^ distinct indices but could be of same value
atleast one potential outlier exists
return the largest outlier
brute force:
*/
class Solution {
public int getLargestOutlier(int[] nums) {
int sum = 0;
int max_outlier = Integer.MIN_VALUE;
Map<Integer, Integer> map = new HashMap<>();
for(int i =0; i< nums.length; i++)
{
sum+=nums[i];
if(map.containsKey(nums[i]))
map.put(nums[i], map.get(nums[i])+1);
else
map.put(nums[i],1);
}
for(int i=0;i<nums.length;i++){
int potential_outlier = nums[i];
int remaining_sum = sum - nums[i];
if(remaining_sum % 2 == 0 && map.containsKey(remaining_sum/2) &&
((potential_outlier != remaining_sum/2) || map.get(remaining_sum/2)>1)){
max_outlier = Math.max(potential_outlier, max_outlier);
}
}
return max_outlier;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Python3 solution | Binary search
|
python3-solution-binary-search-by-florin-x8ql
|
IntuitionUse binary search.ApproachWe have an array nums with some properties.
nums = [special_numbers, special_numbers_sum, outliner],
where sum(special_number
|
FlorinnC1
|
NORMAL
|
2025-03-07T18:21:53.697440+00:00
|
2025-03-07T18:21:53.697440+00:00
| 17 | false |
# Intuition
Use binary search.
# Approach
We have an array nums with some properties.
- nums = [special_numbers, special_numbers_sum, outliner],
- where sum(special_numbers) == special_numbers_sum.
Sort nums.
So we can check for each number if it is an outliner and if in the remaining array there is another number which is equal to the sum of the remaining numbers (which are all special).
This means we can rewrite it like this:
```
total_sum = sum(nums)
counter = Counter(nums)
sum_w8_outliner = total_sum - outliner
```
if there is an element from nums for this to be true:
```
sum_of_special == nums[mid] and ((nums[mid] == outliner and counter[outliner] >= 2) or nums[mid] != outliner)
```
it means that the chosen element is the sum of the special elements and the outliner is correct and we can pick it.
# Complexity
- Time complexity:
O(m) where m = 2000 for the counter
- Space complexity:
O(n*logn)
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
ans = -math.inf
nums.sort()
def get_number(nums, outliner, total_sum, counter):
sum_w8_outliner = total_sum - outliner
start, end = 0, len(nums)-1
while start <= end:
mid = (start + end) // 2
sum_of_special = sum_w8_outliner - nums[mid]
if sum_of_special == nums[mid] and ((nums[mid] == outliner and counter[outliner] >= 2) or nums[mid] != outliner):
return outliner
elif sum_of_special > nums[mid]:
start = mid + 1
else:
end = mid - 1
return None
total_sum = sum(nums)
counter = Counter(nums)
for i in range(len(nums)):
result = get_number(nums, nums[i], total_sum, counter)
if result is not None:
ans = max(ans, result)
return ans
```
| 0 | 0 |
['Binary Search', 'Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java map solution easier to understand than frequency count
|
java-map-solution-easier-to-understand-t-neh8
|
IntuitionSeeing everyone using a frequency count map, while we could actually use a value to index map, which i think is easier to understand.ApproachAssume tar
|
dorothy_zheng
|
NORMAL
|
2025-03-06T22:02:28.043360+00:00
|
2025-03-06T22:02:28.043360+00:00
| 8 | false |
# Intuition
Seeing everyone using a frequency count map, while we could actually use a value to index map, which i think is easier to understand.
# Approach
Assume target is the number that is the sum of the n - 2 nums, so we have target, outlier and other n - 2 numbers summing up to target in the array. So total sum - outlier == 2 * target. For each outlier, we find if target is in the map and its index should not be the outlier's index. We use a valueToIndex map to find the index.
# Complexity
- Time complexity:
O(n)
- Space complexity:
<O(n)
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int n = nums.length;
Map<Integer, Integer> valueToIndex = new HashMap<>();
int totalSum = 0;
for (int i = 0; i < nums.length; i++) {
valueToIndex.put(nums[i], i);
totalSum += nums[i];
}
int globalMax = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
if ((totalSum - nums[i]) % 2 != 0) {
continue;
}
if (valueToIndex.containsKey((totalSum - nums[i]) / 2) &&
valueToIndex.get((totalSum - nums[i]) / 2) != i) {
globalMax = Math.max(globalMax, nums[i]);
}
}
return globalMax;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Without Sorting O(n) in C++
|
without-sorting-on-in-c-by-ismetdagli0-ja3h
|
IntuitionThe problem requires identifying an outlier in an array where:
n - 2 elements are special numbers.
One element is the sum of those special numbers.
The
|
ismetdagli0
|
NORMAL
|
2025-03-03T03:16:20.342621+00:00
|
2025-03-03T03:16:20.342621+00:00
| 5 | false |
# Intuition
The problem requires identifying an outlier in an array where:
- n - 2 elements are special numbers.
- One element is the sum of those special numbers.
- The remaining element is the outlier, which is not part of the sum.
To solve this:
1. Compute the total sum of the array.
2. Iterate through each number and check if removing it results in the remaining numbers forming a valid sum.
3. Find the largest valid outlier among the candidates.
# Complexity
- Time complexity:
- O(N): First loop computes prefixSum and populates the unordered_set.
O(N): Second loop checks for valid outliers.
Overall: O(N) (since each lookup in unordered_set is O(1) on average).
- Space complexity:
- O(N): We store all elements in unordered_set.
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
int size = nums.size();
int prefixSum = 0;
unordered_set<float> history;
for (int i = 0; i < size; i++){
prefixSum+=nums[i];
history.insert(nums[i]);
}
int largestOutlier = INT_MIN;
for (int i = 0; i < size; i++){
if(history.find((prefixSum-nums[i])/2.0) != history.end() && (prefixSum-nums[i])/2.0 != nums[i] ){
cout << prefixSum << nums[i] << endl ;
largestOutlier = max(largestOutlier, nums[i]);
}
}
if( largestOutlier == INT_MIN){
for (int i = 0; i < size; i++){
if(history.find((prefixSum-nums[i])/2.0) != history.end() ){
cout << prefixSum << nums[i] << endl ;
largestOutlier = max(largestOutlier, nums[i]);
}
}
}
return largestOutlier;
}
};
```
| 0 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Solution is not easy but I tried to make it simple. Please upvote if you like the solution
|
solution-is-not-easy-but-i-tried-to-make-b0jc
|
IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(n)
Code
|
gandhip1361
|
NORMAL
|
2025-03-02T18:48:32.642240+00:00
|
2025-03-02T18:48:32.642240+00:00
| 8 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(n)
- Space complexity:
O(n)
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int totalSum = 0, maxElement = Integer.MIN_VALUE;
// Calculate total sum and find the max element
for (int num : nums) {
totalSum += num;
maxElement = Math.max(maxElement, num);
}
HashMap<Integer, Integer> map = new HashMap<>();
for (int num : nums) {
map.put(num, map.getOrDefault(num, 0) + 1);
}
int outlier = Integer.MIN_VALUE;
for (int num : nums) {
int remainingSum = totalSum - num;
if (remainingSum % 2 == 0) {
int p = remainingSum / 2;
if (map.containsKey(p)) {
if (p == num && map.get(p) < 2) {
continue;
}
outlier = Math.max(outlier, num);
}
}
}
return outlier;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple Python, O(n) solution (99%) (Simply check if array has half of the Sum of remaining items)
|
python-on-on-solution-99-by-anujgrgv-ginl
|
Simply check if half of the sum of remaining items is present in the listComplexity
Time complexity: O(n)
Space complexity: O(n)
Code
|
anujgrgv
|
NORMAL
|
2025-03-01T09:20:09.961654+00:00
|
2025-03-01T09:25:52.030754+00:00
| 14 | false |
Simply check if half of the sum of remaining items is present in the list
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
count = Counter(nums)
total = sum(nums)
pot_outlier = float(-inf)
for key in count:
if key > pot_outlier:
currtotal = total - key
count[key]-=1
if currtotal % 2 == 0 and (currtotal//2 in count) and (count[currtotal//2]>0):
pot_outlier = key
count[key]+=1
return pot_outlier
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Hashmap: O(n) solution
|
hashmap-on-solution-by-washik34-opnq
|
IntuitionApproachComplexity
Time complexity: O(n)
Space complexity: O(n)
Code
|
washik34
|
NORMAL
|
2025-03-01T02:17:20.660457+00:00
|
2025-03-01T02:17:20.660457+00:00
| 13 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: $$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(n)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
hashmap = dict()
for num in nums:
if num not in hashmap:
hashmap[num] = 0
hashmap[num] += 1
total = sum(nums)
mx = -inf
for num in nums:
new_total = total - num
if new_total%2 == 0 and new_total//2 in hashmap:
if new_total//2 != num:
mx = max(mx,num)
elif hashmap[num] >= 2:
mx = max(mx,num)
return mx
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Beat 100% of Solutions: The Most Efficient Way to Find the Largest Outlier
|
beat-100-of-solutions-the-most-efficient-79yd
|
Largest Outlier Problem SolutionApproachKey ObservationIf we assume a number is an outlier, then the remaining numbers should sum to (totalsum - outlier), meani
|
kuangsith
|
NORMAL
|
2025-02-28T21:34:39.801871+00:00
|
2025-02-28T21:34:39.801871+00:00
| 9 | false |
# Largest Outlier Problem Solution
---
## Approach
### Key Observation
If we assume a number is an outlier, then the remaining numbers should sum to `(totalsum - outlier)`, meaning that:`
(totalsum - outlier)/2
` must be a valid special number present in `nums`.
### Edge Case Handling
A special case arises when:
$$
totalsum = 3 \times outlier
$$
In this case, removing `outlier` still leaves `(totalsum - outlier)/2` in the set, potentially causing an incorrect result. To handle this:
1. Check if `totalsum / 3` exists in `nums` **at least twice** (since it must be distinct from the sum element).
2. If true, set it as a candidate for the outlier.
### Efficient Iteration
- Convert `nums` to a **set** for O(1) lookup.
- Iterate through unique numbers in `nums`:
- Check if `(totalsum - num)/2` exists in the set.
- Update the maximum outlier found.
This solution runs in **O(n) time** and beats 100% of solutions in performance.
### Why This Approach is More Efficient
Most other solutions would check for the potential outlier as follows:
```python
if potential_outlier in num_counts:
if potential_outlier != num or num_counts[num] > 1:
largest_outlier = max(largest_outlier, potential_outlier)
```
However, this approach is wasteful because we need to check ```potential_outlier != num``` again every time. Instead, we handle the special case upfront (```totalsum = 3 * outlier```), avoiding redundant checks in the main loop.
---
## Implementation
```python
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
totalsum = sum(nums)
setnum = set(nums)
maxoutlier = float('-inf')
if totalsum/3 in setnum and nums.count(totalsum/3) > 1:
maxoutlier = int(totalsum/3)
for num in [x for x in setnum if x!=totalsum/3]:
if (totalsum-num)/2 in setnum:
#potential outlier
maxoutlier = max(maxoutlier,num)
return maxoutlier
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
[Python3] Good enough
|
python3-good-enough-by-parrotypoisson-7u6z
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
parrotypoisson
|
NORMAL
|
2025-02-27T02:44:36.671062+00:00
|
2025-02-27T02:44:36.671062+00:00
| 5 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
nums.sort()
mapping = {}
summ = 0
for x in nums:
mapping[x] = mapping.get(x, 0) + 1
summ += x
for x in nums[::-1]:
other = (summ-x) / 2
if mapping.get(other, 0) >= 1 + (x==other):
return x
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
python
|
python-by-mykono-7ol7
|
Code
|
Mykono
|
NORMAL
|
2025-02-24T04:51:18.035370+00:00
|
2025-02-24T04:51:18.035370+00:00
| 4 | false |
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
# find smallest special_sum, sum(nums) - 2 * special_sum
# cnt freq
# for each element, check if possible
cnt = Counter(nums)
total_sum = sum(nums)
res = -inf
for k, v in cnt.items():
# outlier k, special_sum = (total - k) / 2
if k < res:
continue
if (total_sum - k) % 2 == 1:
continue
special_sum = (total_sum - k) // 2
if special_sum not in cnt or special_sum == k and cnt[k] < 2:
continue
res = max(res, k)
return res
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Math pattern + hashmap
|
math-pattern-hashmap-by-lhying-ywqk
|
IntuitionApproachComplexity
Time complexity: O(n)
Space complexity: O(m)
Code
|
lhying
|
NORMAL
|
2025-02-22T05:16:34.977738+00:00
|
2025-02-22T05:16:34.977738+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(m)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
# 1 sum, 1 outlier
# outlier = total_sum − 2×sum
# if outlier != sum
# if outlier == sum, there should be 2 identical numbers -> so we need a hashmap
# for each unique number
# 1. treat it as the candidate sum
# 2. get candidate outlier using the candidate sum
total_sum = sum(nums)
num_counts = Counter(nums)
largest_outlier = float('-inf')
for num in num_counts.keys():
potential_sum = num
potential_outlier = total_sum - 2 * potential_sum
if potential_outlier in num_counts:
if potential_outlier != potential_sum:
largest_outlier = max(largest_outlier, potential_outlier)
else: # ==
if num_counts[potential_sum] > 1:
largest_outlier = max(largest_outlier, potential_outlier)
return largest_outlier
# time: O(n)
# - constructing the Counter: n
# - iterating over the unique keys: m
# space: O(m)
# - Counter: unique keys = m
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
swift solution with well explanation
|
swift-solution-with-well-explanation-by-bc1h3
|
IntuitionThe goal is to find the largest outlier in an array. An outlier, in this context, is an element that deviates significantly from the rest of the elemen
|
ChocoletEY
|
NORMAL
|
2025-02-21T04:00:28.742030+00:00
|
2025-02-21T04:00:28.742030+00:00
| 18 | false |
# Intuition
The goal is to find the largest outlier in an array. An outlier, in this context, is an element that deviates significantly from the rest of the elements in the array. Instead of using the mean or median, this method identifies an outlier based on a mathematical relationship derived from the sum of the elements and their counts.
# Approach
1. **Calculate the Total Sum of the Array**: This helps in deriving potential outliers using a mathematical relationship.
2. **Count Occurrences of Each Number**: A dictionary (or hashmap) is used to keep track of the frequency of each element in the array.
3. **Identify Potential Outliers**: For each unique number in the array, compute a potential outlier using the formula: `total_sum - 2 * num`. This formula identifies a number that would balance out the total sum if the number `num` was removed.
4. **Check for Valid Outliers**: Check if the calculated potential outlier exists in the dictionary and ensure that it is a valid outlier (either it is not the same as `num` or it appears more than once).
5. **Track the Largest Outlier**: Keep track of the largest outlier found during the traversal of the dictionary keys.
# Complexity
- Time complexity: O(n), where n is the number of elements in the array. We traverse the array twice: once to count occurrences and once to identify outliers.
- Space complexity: O(n), due to the storage needed for the dictionary.
# Code
```swift []
class Solution {
func getLargestOutlier(_ nums: [Int]) -> Int {
// Step 1: Calculate the total sum of the array
let totalSum = nums.reduce(0, +)
// Step 2: Create a dictionary to count the occurrences of each number
var numCounts = [Int: Int]()
for num in nums {
numCounts[num, default: 0] += 1
}
// Step 3: Initialize the largest outlier to negative infinity
var largestOutlier = Int.min
// Step 4: Traverse the dictionary keys to find the largest outlier
for num in numCounts.keys {
let potentialOutlier = totalSum - 2 * num
if let count = numCounts[potentialOutlier], (potentialOutlier != num || numCounts[num]! > 1) {
largestOutlier = max(largestOutlier, potentialOutlier)
}
}
// Step 5: Return the largest outlier
return largestOutlier
}
}
```
| 0 | 0 |
['Swift']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java O(N) HashMap
|
java-on-hashmap-by-etherwei-7juu
|
IntuitionIf we pick a number off the nums array, how can we in O(1) time determine if it's an outliner? The easiest way is if the sum of the special numbers is
|
etherwei
|
NORMAL
|
2025-02-16T16:19:06.342989+00:00
|
2025-02-16T16:19:06.342989+00:00
| 17 | false |
# Intuition
If we pick a number off the nums array, how can we in O(1) time determine if it's an outliner? The easiest way is if the sum of the special numbers is there, then it has to be `(sum(nums) - outliner) / 2`
# Complexity
- Time complexity:
O(N)
- Space complexity:
O(N)
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
// [-5, -3, -2, 10]
HashMap<Integer, Integer> counter = new HashMap<>();
int sum = 0;
for (int n : nums) {
sum += n;
int current = counter.getOrDefault(n, 0);
counter.put(n, current + 1);
}
int biggestNumber = -10000;
for (int i = nums.length - 1; i >= 0; --i) {
int out = nums[i];
int rest = sum - out;
if (rest % 2 != 0) continue;
int total_sum = rest / 2;
if (out == total_sum) {
if (counter.get(total_sum) >= 2) {
if (out > biggestNumber) {
biggestNumber = out;
}
}
} else {
if (counter.containsKey(total_sum)) {
if (out > biggestNumber) {
biggestNumber = out;
}
}
}
}
return biggestNumber;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Python using Dict
|
python-using-dict-by-ryud54-4o0r
|
IntuitionThere is a special pattern for finding the largest outlier in such an array. By having a quick lookup of the sum of the special numbers, we can iterate
|
ryud54
|
NORMAL
|
2025-02-15T21:21:27.265122+00:00
|
2025-02-15T21:21:27.265122+00:00
| 5 | false |
# Intuition
There is a special pattern for finding the largest outlier in such an array. By having a quick lookup of the sum of the special numbers, we can iterate the array and determine the largest outlier.
# Approach
1. map each number to its indices
2. iterate through the array and for each number, find the corresponding sum of special numbers, which is the difference between the total sum of the numbers in the array and the current number divided by 2.
3. First we have to make sure that the there is such a number that equals this sum of special numbers and also the current number should not be equal to this sum unless there are multiple occurrences.
4. If the current number is larger than that largest outlier, update the largest number as the current number.
# Complexity
- Time complexity:
O(n)
- Space complexity:
O(n)
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
n = len(nums)
array_sum = sum(nums)
largest_outlier = None
outlier_dict = {}
for i in range(n):
curr_num = nums[i]
if curr_num not in outlier_dict.keys():
outlier_dict[curr_num] = [i]
else:
outlier_dict[curr_num].append(i)
for i in range(n):
curr_num = nums[i]
removed_sum = array_sum - curr_num # => 2 * sum of special nums
sum_of_special_nums = removed_sum/2
if sum_of_special_nums in outlier_dict.keys() and (curr_num != sum_of_special_nums or len(outlier_dict[sum_of_special_nums]) > 1):
if not largest_outlier or curr_num > largest_outlier:
largest_outlier = curr_num
return largest_outlier
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
PHP | Hashmap | O(n)
|
php-hashmap-on-by-aryonbe-qnu7
|
Code
|
aryonbe
|
NORMAL
|
2025-02-12T13:48:18.771039+00:00
|
2025-02-12T13:48:18.771039+00:00
| 1 | false |
# Code
```php []
class Solution {
/**
* @param Integer[] $nums
* @return Integer
*/
function getLargestOutlier($nums) {
$T = array_sum($nums);
$count = array_count_values($nums);
$res = PHP_INT_MIN;
foreach($nums as $num){
$outlier = $T - $num -$num;
if($outlier === $num){
if($count[$num] > 1){
$res = max($res, $outlier);
}
}
else if(isset($count[$outlier])){
$res = max($res, $outlier);
}
}
return $res;
}
}
```
| 0 | 0 |
['PHP']
| 0 |
identify-the-largest-outlier-in-an-array
|
Javascript | Hashmap | O(n)
|
javascript-hashmap-on-by-aryonbe-33v3
|
Code
|
aryonbe
|
NORMAL
|
2025-02-12T13:40:01.407371+00:00
|
2025-02-12T13:40:01.407371+00:00
| 19 | false |
# Code
```javascript []
/**
* @param {number[]} nums
* @return {number}
*/
var getLargestOutlier = function(nums) {
let T = 0;
let res = -Infinity;
const map = new Map();
for(const num of nums){
T += num;
map.set(num, (map.get(num) || 0) + 1);
}
for(let i = 0; i < nums.length; i++){
const outlier = T - nums[i] - nums[i];
if(outlier == nums[i]){
if(map.get(nums[i]) > 1){
res = Math.max(res, outlier);
}
}
else if(map.has(outlier)){
res = Math.max(res, outlier);
}
}
return res;
};
```
| 0 | 0 |
['JavaScript']
| 0 |
identify-the-largest-outlier-in-an-array
|
Python | Hashmap | O(n)
|
python-hashmap-on-by-aryonbe-63gj
|
Code
|
aryonbe
|
NORMAL
|
2025-02-12T13:04:55.397308+00:00
|
2025-02-12T13:04:55.397308+00:00
| 3 | false |
# Code
```python3 []
from collections import defaultdict
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
idx = defaultdict(set)
for i,num in enumerate(nums):
idx[num].add(i)
T = sum(nums)
res = float('-inf')
for i,num in enumerate(nums):
R = T-num
if R%2 == 0 and R//2 in idx:
if len(idx[R//2]) > 1:
res = max(res, num)
elif i not in idx[R//2]:
res = max(res, num)
return res
```
| 0 | 0 |
['Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Easy C++
|
easy-c-by-dee_two-q6l2
|
Code
|
dee_two
|
NORMAL
|
2025-02-11T08:11:07.264281+00:00
|
2025-02-11T08:11:07.264281+00:00
| 7 | false |
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
int res=-1001;
int total=accumulate(nums.begin(), nums.end(), 0);
unordered_map<int, int> freq;
for(int &i: nums)
freq[i]++;
for(auto [outlier, v]: freq)
{
int sum=total-outlier;
freq[outlier]--;
//cout<<outlier<<"-----";
for(auto [k, _v]: freq)
{
if(_v)
{
int sum2=sum-k;
//cout<<"["<<sum2<<"]"<<", "<<endl;
if(freq[k]>0 && sum2==k)
{
res=max(res, outlier);
//cout<<k<<", "<<sum2<<endl;
}
}
}
//cout<<endl;
freq[outlier]++;
}
return res;
}
};
```
| 0 | 0 |
['Array', 'Hash Table', 'Counting', 'Enumeration', 'C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java O(N), 85ms solution
|
java-on-85ms-solution-by-mashihur-wyh6
|
IntuitionJava O(N), 85ms solutionApproachComplexity
Time complexity:
O(N)
Space complexity:
O(N)
Code
|
mashihur
|
NORMAL
|
2025-02-02T19:58:44.096961+00:00
|
2025-02-02T19:58:44.096961+00:00
| 11 | false |
# Intuition
Java O(N), 85ms solution
# Approach
# Complexity
- Time complexity:
O(N)
- Space complexity:
O(N)
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int i, sum = 0, outlier = Integer.MIN_VALUE, temp;
HashMap<Integer, Integer> map = new HashMap<>();
for (i = 0; i < nums.length; i++) {
sum += nums[i];
map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
}
for (i = 0; i < nums.length; i++) {
temp = sum - nums[i];
// Need to verify num[i] is outlier and not special number
if (temp % 2 == 0 && map.containsKey(temp/2) && (temp/2 != nums[i] || map.get(temp/2) > 1)) {
outlier = Math.max(outlier, nums[i]);
}
}
return outlier;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java solution using HashMap
|
java-solution-using-hashmap-by-rishabhva-jzww
|
Approach
Calculate the total sum.
Store the values of each number in the array in a hashmap. A HashSet would have worked but we might have a scenario when the o
|
rishabhvallecha
|
NORMAL
|
2025-01-31T08:57:50.190804+00:00
|
2025-01-31T08:58:06.468870+00:00
| 14 | false |
# Approach
<!-- Describe your approach to solving the problem. -->
- Calculate the total sum.
- Store the values of each number in the array in a hashmap. A HashSet would have worked but we might have a scenario when the outlier number chosen is the same as half of the (remainingSum / 2), and that would detect the presence of (remainingSum / 2) in the remaining numbers of the array. So use a hashmap to keep track of the number of occurrences of the numbers.
- Iterate through the array.
- currSum = totalSum - currentNum
- Subtract the current number and check if the currSum is even. Skip if it is odd.
- Check the map if the (currSum / 2) is present. If it is, also check that the current number is not the same as (currSum / 2) or if it is, perform a third check if there exists another occurrence of that particular number, which is different from the outlier chosen.
- Update max if conditions are met.
# Complexity
- Time complexity: $$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(n)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int getLargestOutlier(int[] nums) {
int totalSum = 0;
Map<Integer, Integer> mp = new HashMap<>();
for(int i: nums) {
totalSum += i;
mp.put(i, mp.getOrDefault(i, 0)+1);
}
int max = Integer.MIN_VALUE;
for(int i: nums) {
int currSum = totalSum - i;
if(currSum % 2 != 0) continue;
if(mp.containsKey(currSum / 2)) {
if((currSum / 2 != i) || (currSum / 2 == i && mp.get(currSum / 2) > 1)) {
max = Math.max(max, i);
}
}
}
return max;
}
}
```
| 0 | 0 |
['Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Java solution using Binary Search
|
java-solution-using-binary-search-by-ris-1wcd
|
Approach
Sort the array.
Iterate from the end to find the potentially largest outlier.
Subtract the outlier to find the remaining sum of numbers.
If the remaini
|
rishabhvallecha
|
NORMAL
|
2025-01-31T08:50:25.931659+00:00
|
2025-01-31T08:50:25.931659+00:00
| 8 | false |
# Approach
<!-- Describe your approach to solving the problem. -->
- Sort the array.
- Iterate from the end to find the potentially largest outlier.
- Subtract the outlier to find the remaining sum of numbers.
- If the remaining sum is odd, that outlier is not possible.
- If it is even, binary search the remaining array to check if the (remainingSum / 2) is present in the array. Search space can be restricted by comparing the value of nums[outlierIndex] with (remainingSum / 2).
- If found, return nums[outlierIndex].
# Complexity
- Time complexity: $$O(nlogn)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public boolean binarySearch(int[] nums, int l, int r, int num) {
while(l <= r) {
int mid = l + (r-l)/2;
if(nums[mid] == num) return true;
if(nums[mid] > num) {
r = mid-1;
} else {
l = mid+1;
}
}
return false;
}
public int getLargestOutlier(int[] nums) {
Arrays.sort(nums);
int totalSum = 0;
for(int i: nums) {
totalSum += i;
}
int outlierIndex = nums.length-1;
while(true) {
int currSum = totalSum - nums[outlierIndex];
if(currSum % 2 == 0) {
if(nums[outlierIndex] > currSum / 2) {
if(binarySearch(nums, 0, outlierIndex-1, currSum / 2)) return nums[outlierIndex];
} else {
if(binarySearch(nums, outlierIndex+1, nums.length - 1, currSum / 2)) return nums[outlierIndex];
}
}
outlierIndex--;
}
}
}
```
| 0 | 0 |
['Binary Search', 'Java']
| 0 |
identify-the-largest-outlier-in-an-array
|
Python Easy Solution. O(N) Time Complexity.
|
python-easy-solution-on-time-complexity-3xmpz
|
Intuition
total_sum = sum of (n-2) elements + outlier + special number.
It is given that, sum(n-2) elements = special number.
So, total_sum - outlier = sum(n-2
|
hwaawh
|
NORMAL
|
2025-01-30T22:18:01.441212+00:00
|
2025-01-30T22:18:01.441212+00:00
| 25 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
- **total_sum = sum of (n-2) elements + outlier + special number.**
- It is given that, **sum(n-2) elements = special number.**
- So, **total_sum - outlier = sum(n-2) elements + special number.**
- This implies that, **total_sum - outlier = 2 * special number.**
- Which means that **special number = (total_sum - outlier)/ 2**
# Approach
<!-- Describe your approach to solving the problem. -->
1. We calculate the sum of array and declare max_outlier = -inf.
2. Create a frequency map of all elements.
3. The idea is that, for every ith element of the array, we treat the number as an outlier and check if the above theorem of finding special number is true.
4. If the theorem is proven, then we check if the special number is existing in freq dict or not and then update max_outlier.
5. Whenever considering an element as outlier, make sure to reduce its count before checking for special number in freq dict because, we must not take the currrent outlier into account when checking for special number.
6. Once max_outlier is updated, make sure to put back the current outlier into freq dict, so that it can be used by other elements.
# Complexity
- Time complexity: $$O(n)$$
# Code
```python3 []
class Solution:
def getLargestOutlier(self, nums: List[int]) -> int:
n = len(nums)
total_sum = sum(nums)
max_outlier = -inf
freq_dict = {}
for i in range(n):
key = nums[i]
if nums[i] in freq_dict:
freq_dict[key] += 1
else:
freq_dict[key] = 1
for i in range(n):
current_outlier = nums[i]
required_sum = total_sum - current_outlier
if required_sum % 2 != 0:
continue
freq_dict[current_outlier] -= 1
spl_number = required_sum // 2
if (spl_number) in freq_dict and freq_dict[spl_number] != 0:
max_outlier = max(max_outlier, current_outlier)
freq_dict[current_outlier] += 1
return max_outlier
```
| 0 | 0 |
['Array', 'Hash Table', 'Counting', 'Enumeration', 'Python', 'Python3']
| 0 |
identify-the-largest-outlier-in-an-array
|
Simple C++ solution. Beats >99%.
|
simple-c-solution-beats-99-by-heychiraga-2v5x
|
IntuitionWe know that exactly one number is outlier and the other is a sum of all numbers. Therefore we simply sum all the numbers first and then do a linear sc
|
heychiragarora
|
NORMAL
|
2025-01-30T17:49:04.510069+00:00
|
2025-01-30T17:51:45.867153+00:00
| 8 | false |
# Intuition
We know that exactly one number is outlier and the other is a sum of all numbers. Therefore we simply sum all the numbers first and then do a linear scan to check if the given number can be an _outlier_. The remaining sum will be the double of the "other" non-special number. Therefore we halve the sum and check if the "other" number was present in the original array. There can be multiple solutions, therefore we only take the next possible answer if it's greater than the last found answer.
# Approach
* Create a frequency map `allNums` of all the numbers.
* In the same loop we can also sum all the numbers and store it in `sum`.
* Then we do a linear scan testing if the given number can be an outlier.
* For this, we deduct the current candidate for the `outlier` from the `sum`
* We check if the sum is divisble by 2. If so, we get the other non-special number by halving the deducted `sum`.
* Then we check if this `other` number is present in the dictionary.
* One edge case we need to lookout for is if `other` equals `outlier`, the frequency of this number must be at least 2 to make it a valid solution.
* If the above condition satisfies, we store the `outlier` if it's greater than the previously found answer.
# Complexity
- **Time complexity**: $$O(n)$$ for summing and counting the frequency, and then again $$O(n)$$ for the linear scan. So the total is $$O(n)$$.
- **Space complexity**: $$O(n)$$ for storing the frequency of all the numbers.
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
unordered_map<int, int> allNums;
int sum = 0;
for (int n : nums) {
allNums[n]++;
sum += n;
}
int ans = INT_MIN;
for (int outlier : nums) {
int sumWithoutOutlier = sum - outlier;
if (sumWithoutOutlier % 2 == 0) {
int other = sumWithoutOutlier / 2;
auto it = allNums.find(other);
if (it != allNums.end()) {
if (other == outlier) {
if (it->second >= 2) {
ans = max(ans, outlier);
}
} else {
ans = max(ans, outlier);
}
}
}
}
return ans;
}
};
```
| 0 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Identify the Largest Outlier in an Array
|
identify-the-largest-outlier-in-an-array-00b4
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
sid9857
|
NORMAL
|
2025-01-30T15:04:03.216783+00:00
|
2025-01-30T15:04:03.216783+00:00
| 8 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
int n = nums.size();
long long temp = 0;
map<int, int> m;
int ans = INT_MIN;
for (int i = 0; i < n; i++) {
temp += nums[i];
m[nums[i]] = i;
}
for (int i = 0; i < n; i++) {
if ((temp - nums[i]) % 2 != 0) {
continue;
} else {
int t = (temp - nums[i]) / 2;
if (m.find(t) != m.end()) {
if (m[t] != i) {
// cout<<nums[i]<< " "<<t<<" "<<m[t]<<" "<<i<<endl;
ans = max(ans, nums[i]);
}
}
}
}
return ans;
}
};
```
| 0 | 0 |
['C++']
| 0 |
identify-the-largest-outlier-in-an-array
|
Using basic math and hash map | C++
|
using-basic-math-and-hash-map-c-by-samee-4g6n
|
IntuitionThe total sum of array would be -> sum_element + n-2 special element(basically equal to sum_element) + outlier.
So if we subtract sum_element two times
|
sameerahmed56
|
NORMAL
|
2025-01-29T21:43:53.674837+00:00
|
2025-01-29T21:43:53.674837+00:00
| 11 | false |
# Intuition
The total sum of array would be -> sum_element + n-2 special element(basically equal to sum_element) + outlier.
So if we subtract sum_element two times from total sum we get the outlier.
# Approach
Store all values of the array into a hash map or another array.
Sum all the element of array and use that to get the outlier.
If there are multiple outlier we need to return the max.
**Edge Case:** The outlier and sum_element can be equal but not of same index so make sure of that.
# Complexity
- Time complexity:
O(N) as we are iterating array 2 times.
- Space complexity:
O(N) as we are storing all values in a hash map.
# Code
```cpp []
class Solution {
public:
int getLargestOutlier(vector<int>& nums) {
unordered_map<int,int> um;
int n = nums.size(), allSum = 0;
for(int i=0; i<n; i++){
allSum += nums[i];
um[nums[i]] = i;
}
int res = INT_MIN;
for(int i=0; i<n; i++){
int rem = allSum - (2*nums[i]);
if(um.find(rem) != um.end() && um[rem] != i){
res = max(res, rem);
}
}
return res;
}
};
```
| 0 | 0 |
['C++']
| 0 |
minimum-pair-removal-to-sort-array-i
|
Beginner way to solve this question✅✅✅
|
beginner-way-to-solve-this-question-by-n-6clx
|
IntuitionWe need to make the array non-decreasing by repeatedly merging the smallest consecutive pair until the array is sorted.Approach
Find the smallest cons
|
nilestiwari_7
|
NORMAL
|
2025-04-06T04:04:36.734832+00:00
|
2025-04-06T04:04:36.734832+00:00
| 1,914 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We need to make the array non-decreasing by repeatedly merging the smallest consecutive pair until the array is sorted.
# Approach
<!-- Describe your approach to solving the problem. -->
1. Find the smallest consecutive pair by their sum.
2. Merge this pair and repeat until the array is non-decreasing.
3. Track the number of operations.
# Complexity
- Time complexity:$$O(n^2)$$ due to repeated scans of the array.
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:$$O(n)$$ for storing the modified array.
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
bool isNondecreasing(vector<int>& nums) {
for (int i = 1; i < nums.size(); i++) {
if (nums[i] < nums[i - 1]) {
return false;
}
}
return true;
}
int minimumPairRemoval(vector<int>& nums) {
int ops = 0;
while (!isNondecreasing(nums)) {
int min_sum = INT_MAX;
int index = -1;
//step 1 - Find the smallest consecutive pair by their sum.
for (int i = 0; i < nums.size() - 1; i++) {
int sum = nums[i] + nums[i + 1];
if (sum < min_sum) {
min_sum = sum;
index = i;
}
}
vector<int>temp;
// step 2 - Merge this pair and repeat until the array is non-decreasing.
for(int i = 0;i<nums.size();i++){
if(i==index){
temp.push_back(nums[i]+nums[i+1]);
i++;
}else{
temp.push_back(nums[i]);
}
}
nums = temp;
// step 3 - Track the number of operations.
++ops;
}
return ops;
}
};
```
```python []
from typing import List
class Solution:
def isNondecreasing(self, nums: List[int]) -> bool:
for i in range(1, len(nums)):
if nums[i] < nums[i - 1]:
return False
return True
def minimumPairRemoval(self, nums: List[int]) -> int:
ops = 0
while not self.isNondecreasing(nums):
min_sum = float('inf')
index = -1
# step 1 - Find the smallest consecutive pair by their sum.
for i in range(len(nums) - 1):
sum_pair = nums[i] + nums[i + 1]
if sum_pair < min_sum:
min_sum = sum_pair
index = i
# step 2 - Merge this pair and repeat until the array is non-decreasing.
temp = []
i = 0
while i < len(nums):
if i == index:
temp.append(nums[i] + nums[i + 1])
i += 2
else:
temp.append(nums[i])
i += 1
nums = temp
# step 3 - Track the number of operations.
ops += 1
return ops
```
``` javascript []
/**
* @param {number[]} nums
* @return {number}
*/
var minimumPairRemoval = function(nums) {
// Helper function to check if the array is non-decreasing
const isNondecreasing = (nums) => {
for (let i = 1; i < nums.length; i++) {
if (nums[i] < nums[i - 1]) {
return false;
}
}
return true;
};
let ops = 0;
// Continue until the array is non-decreasing
while (!isNondecreasing(nums)) {
let minSum = Infinity;
let index = -1;
//step 1 - Find the smallest consecutive pair by their sum.
for (let i = 0; i < nums.length - 1; i++) {
let sumPair = nums[i] + nums[i + 1];
if (sumPair < minSum) {
minSum = sumPair;
index = i;
}
}
// step 2 - Merge this pair and repeat until the array is non-decreasing.
let temp = [];
let i = 0;
while (i < nums.length) {
if (i === index) {
temp.push(nums[i] + nums[i + 1]);
i += 2; // Skip next element as it is merged
} else {
temp.push(nums[i]);
i++;
}
}
nums = temp;
// step 3 - Track the number of operations.
ops++;
}
return ops;
};
```
| 7 | 0 |
['Simulation', 'C++', 'Python3', 'JavaScript']
| 2 |
minimum-pair-removal-to-sort-array-i
|
A different Brute-force method in Java with less runtime.
|
a-different-brute-force-method-in-java-w-5lkn
|
IntuitionWe have to find the minimum number of operations needed to make the array non-decreasing.Operation:
Select the adjacent pair with the minimum sum in nu
|
YaswanthKesuboyina
|
NORMAL
|
2025-04-06T06:17:53.106674+00:00
|
2025-04-06T06:17:53.106674+00:00
| 718 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We have to find the minimum number of operations needed to make the array non-decreasing.
**Operation:**
- Select the adjacent pair with the minimum sum in nums. If multiple such pairs exist, choose the leftmost one.
- Replace the pair with their sum.
# Approach
<!-- Describe your approach to solving the problem. -->
***My approach:***
We have to check if the array is non decreasing by checking each element if there is an element in the array which is greater than its next element then we stop and apply operarion and In the operation we find minimum sum and replace the pair with minimum sum and in place of replacing two elements sum as one element we assign sum to first value of the pair and will shift all the remaing elements in the array to left by 1 step. we decrement n by 1 as last element is already shifted. we will again start for checking the array if it is non decreasing and continue the loop until the array is non decreasing. I took count variable to calculate no of times operation is done.
# Code
```java []
class Solution {
public void operation(int[] nums,int n){
int min_index=0;
int minsum=nums[0]+nums[1];
for(int i=0;i<n-1;i++){
if(nums[i]+nums[i+1]<minsum){
min_index=i;
minsum=nums[i]+nums[i+1];
}
}
nums[min_index]=minsum;
for(int i=min_index+1; i<n-1;i++){
nums[i]=nums[i+1];
}
}
public int minimumPairRemoval(int[] nums) {
int n=nums.length;
int[] pairsum=new int[n-1];
int count=0;
for(int i=0; i<n-1;i++){
if(nums[i]>nums[i+1]){
count++;
operation(nums,n);
i=-1;
n--;
}
}
return count;
}
}
```
| 6 | 0 |
['Array', 'Java']
| 1 |
minimum-pair-removal-to-sort-array-i
|
🚀 Best Python Solution | Beginner Friendly | Clean Code + Explanation
|
best-python-solution-beginner-friendly-c-cunc
|
IntuitionThe task is to transform the array into a non-decreasing array by repeatedly removing two adjacent elements and replacing them with their sum. The key
|
PPanwar29
|
NORMAL
|
2025-04-06T04:16:09.599926+00:00
|
2025-04-06T04:54:34.139847+00:00
| 1,050 | false |
# Intuition
The task is to transform the array into a **non-decreasing array** by repeatedly removing two adjacent elements and replacing them with their sum. The key observation is:
To minimize the number of operations, we should always merge the pair with the smallest sum to reduce the chance of creating larger decreasing values.
# 🧭 Approach
- Check if the array is non-decreasing. If yes, return 0.
- Otherwise, find the adjacent pair `(i, i+1)` with the minimum sum.
- Replace both elements with their sum, effectively reducing the array size by 1.
- Count the operation.
Repeat until the array is sorted in non-decreasing order.
# ⏱️ Complexity
**Time Complexity:**
- **Worst case:** `(On*2)`
- Each operation takes `O(n)` to find the minimum sum pair, and there can be up to `O(n)` operations.
**Space Complexity:** `O(n)`
- Due to slicing and copying the array in each operation.
# Code
```python3 []
class Solution:
def minimumPairRemoval(self, nums: List[int]) -> int:
def is_non_decreasing(arr):
return all(arr[i] <= arr[i+1] for i in range(len(arr) - 1))
operations = 0
while not is_non_decreasing(nums):
min_sum = float('inf')
min_index = -1
for i in range(len(nums) - 1):
pair_sum = nums[i] + nums[i+1]
if pair_sum < min_sum:
min_sum = pair_sum
min_index = i
# Replace the pair with their sum
new_num = nums[min_index] + nums[min_index + 1]
nums = nums[:min_index] + [new_num] + nums[min_index + 2:]
operations += 1
return operations
```
| 6 | 0 |
['Python3']
| 5 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.