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stringdate 2013-10-25 17:32:12
2025-04-12 09:38:24
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stringdate 2013-10-25 17:32:12
2025-04-12 09:38:24
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
find-the-highest-altitude
|
[C++] Single Pass Solution Explained, 100% Time, 65% Space
|
c-single-pass-solution-explained-100-tim-u2hh
|
To solve this problem, we will need 2 variables, both initialised to 0:\n res is our accumulator, storing the maximum altitude seen so far;\n curr is the curren
|
ajna
|
NORMAL
|
2021-02-07T08:59:51.238038+00:00
|
2021-02-07T08:59:51.238071+00:00
| 1,519 | false |
To solve this problem, we will need 2 variables, both initialised to `0`:\n* `res` is our accumulator, storing the maximum altitude seen so far;\n* `curr` is the current altitude.\n\nFor each element `n` in `gain`, we will:\n* update `curr` with `n`;\n* updated `res` so that it is going to be the highest value between itself and `curr`.\n\nOnce done, we can return `res` :)\n\nThe code:\n\n```cpp\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n // support variables\n int res = 0, curr = 0;\n for (int n: gain) {\n // updating curr\n curr += n;\n // updating res\n res = max(res, curr);\n }\n return res;\n }\n};\n```
| 7 | 0 |
['C', 'C++']
| 1 |
find-the-highest-altitude
|
The Altitude Challenge: Reach the Peak!
|
the-altitude-challenge-reach-the-peak-by-ur1q
|
IntuitionThe problem at hand is about determining the largest altitude reached after a series of altitude gains and losses, given as a list of integers where ea
|
Akhil333_Vijayakumar
|
NORMAL
|
2025-03-19T11:29:15.237072+00:00
|
2025-04-01T12:55:18.143673+00:00
| 344 | false |
# Intuition
The problem at hand is about determining the largest altitude reached after a series of altitude gains and losses, given as a list of integers where each element represents the change in altitude at each step.
To solve this problem, we need to track the cumulative altitude after each step and identify the highest altitude reached during the entire process.
# Approach
We can approach this problem by iterating through the given list of altitude gains and maintaining a running sum of the altitude as we process each change. Here's how we can approach the problem:
Initialize the initial altitude: We start at an altitude of 0, and we will track the highest altitude reached using a variable.
Iterate through the gains: For each element in the list, we adjust the current altitude and check if it exceeds the highest altitude recorded so far.
Track the maximum altitude: After each update to the current altitude, we keep track of the maximum altitude reached so far.
Return the result: After iterating through all the altitude gains, the result will be the maximum altitude reached.
# Complexity
- Time complexity:
O(n) where n is the length of the gain list. We are iterating through the list once, performing constant time operations for each step.
- Space complexity:
O(n), since we are only maintaining a few variables to track the current altitude and the maximum altitude.
# Code
```python3 []
class Solution:
def largestAltitude(self, gain: List[int]) -> int:
res = [0]
for i in gain:
val = res[-1]+i
res.append(val)
return max(res)
```
| 6 | 0 |
['Python3']
| 2 |
find-the-highest-altitude
|
Easy C++ Solution | Beats 100%
|
easy-c-solution-beats-100-by-krishanbrar-syo7
|
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
|
KrishanBrar
|
NORMAL
|
2023-07-04T06:05:08.743759+00:00
|
2023-07-04T06:05:08.743795+00:00
| 138 | 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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n= gain.size();\n for(int i=1; i<n; i++){\n gain[i]= gain[i-1] + gain[i];\n }\n int ans=0;\n for(int i=0; i<n; i++){\n if(ans<gain[i]){\n ans= gain[i];\n }\n }\n return ans;\n }\n};\n```
| 6 | 0 |
['Array', 'Prefix Sum', 'C++']
| 1 |
find-the-highest-altitude
|
🔥100% Beats🔥||🔥C++🔥||🔥GET MAX PREFIX SUM🔥
|
100-beatscget-max-prefix-sum-by-ganeshku-yu92
|
Code\n\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int ans = 0, s = 0;\n for(auto &i: gain){\n s += i;\n
|
ganeshkumawat8740
|
NORMAL
|
2023-06-19T01:44:59.246230+00:00
|
2023-06-19T01:44:59.246247+00:00
| 1,604 | false |
# Code\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int ans = 0, s = 0;\n for(auto &i: gain){\n s += i;\n ans = max(ans,s);\n }\n return ans;\n }\n};\n```
| 6 | 0 |
['Array', 'Prefix Sum', 'C++']
| 3 |
find-the-highest-altitude
|
Easy python solution
|
easy-python-solution-by-tusharkhanna575-4xu9
|
\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n ans=[0]\n for i in range(len(gain)):\n ans.append(ans[i]+ga
|
tusharkhanna575
|
NORMAL
|
2022-05-17T10:47:41.049515+00:00
|
2022-05-17T10:47:41.049562+00:00
| 844 | false |
```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n ans=[0]\n for i in range(len(gain)):\n ans.append(ans[i]+gain[i]) \n return max(ans)\n```\n\n\nplease upvote if you got the concept\ncomment for any help
| 6 | 0 |
['Array', 'Python', 'Python3']
| 0 |
find-the-highest-altitude
|
Rust One-Liner
|
rust-one-liner-by-eliottcodesonleet-c06v
|
TL;DR give me the answer:\nRust\nimpl Solution {\n pub fn largest_altitude(gain: Vec<i32>) -> i32 {\n gain.into_iter().fold((0, 0), |(c, m), x| (c + x
|
eliottcodesonleet
|
NORMAL
|
2024-06-20T12:17:54.612376+00:00
|
2024-06-20T12:17:54.612398+00:00
| 59 | false |
# TL;DR give me the answer:\n```Rust\nimpl Solution {\n pub fn largest_altitude(gain: Vec<i32>) -> i32 {\n gain.into_iter().fold((0, 0), |(c, m), x| (c + x, m.max(c + x))).1\n }\n}\n```\n\n\n# Approach\n\n## 1. Base intuition\nThis is a "find the maximum" problem which means we can apply dynamic programming principles to it. The final solution is a bottom-up (iterative) approach.\nFor this problem, we need to find the maximum altitude of *any* point in the graph. Mathematically, we are given differences in altitude between points, which might be thought of as a derivative. Thus we should compute the discrete summation and track the maximum value of the sum at each step.\n```Rust\n```\n\n## 2. Refining into one line\nWe can start by recognizing there is no need to store anything but the *running* maximum:\n```Rust\npub fn largest_altitude(gain: Vec<i32>) -> i32 {\n let mut current_altitude = 0;\n let mut maximum_altitude = 0;\n for gain_iterator in 0..(gain.len() - 1) {\n current_altitude += gain[gain_iterator];\n if (current_altitude > maximum_altitude){\n maximum_alititude = current_altitude;\n }\n }\n return maximum_altitude;\n}\n```\nBy utilising iterators on gain instead of `usize` (which is what `gain_iterator` defaults to otherwise):\n```Rust\nfor gain_iterator in gain.into_iter() {\n current_altitude += gain_iterator;\n```\nBut the magic really happens with `fold()`.\nFirst, we can take a hint from `current_altitude` and `maximum_altitude` as our *closure*:\n\n```Rust\n(0, 0) == (current_altitude, maximum_altitude)\n```\n\nThen, we need to update current altitude by the difference given to `fold()` by each iterator: `|(0, 0), gain_iterator|`.\n\nWe know this tuple needs to be updated with the new altitude:\n```Rust\n(current_altitude + gain_iterator, ...\n```\nand the running maximum between that new altitude and the previous maximum:\n```Rust\n... maximum_altitude.max(current_altitude + gain_iterator) )\n```\nSince the variable names don\'t matter, let\'s abbreviate:\n- c = current_altitude\n- m = maximum_altitude\n- x = gain_iterator \n\nThe closure expression is now:\n```Rust\n|(c, m), x| (c + x, m.max(c + x))\n```\n... which we use in `fold()`:\n```Rust\ngain.into_iter().fold((0, 0), |(c, m), x| (c + x, m.max(c + x)))\n```\nAnd we can easily access the maximum element of the tuple using `.1`:\n```Rust\ngain.into_iter().fold((0, 0), |(c, m), x| (c + x, m.max(c + x))).1\n```\nA `return` statement is unnecessary too, because by dropping the `;` at the end of this line, it becomes an evaluation.\n\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nimpl Solution {\n pub fn largest_altitude(gain: Vec<i32>) -> i32 {\n gain.into_iter().fold((0, 0), |(c, m), x| (c + x, m.max(c + x))).1\n }\n}\n```
| 5 | 0 |
['Rust']
| 0 |
find-the-highest-altitude
|
[EASIEST] [JAVA] [SOLUTION] [100%] for [BEGINNERS]
|
easiest-java-solution-100-for-beginners-z56fw
|
Approach\n1. Initialize Variables:\n - Set sum to 0 (cumulative altitude).\n - Set max to 0 (highest altitude).\n\n2. Iterate Through Gain Array:\n - U
|
RajarshiMitra
|
NORMAL
|
2024-05-24T17:16:42.256318+00:00
|
2024-06-16T06:56:44.012594+00:00
| 19 | false |
# Approach\n1. **Initialize Variables**:\n - Set `sum` to 0 (cumulative altitude).\n - Set `max` to 0 (highest altitude).\n\n2. **Iterate Through Gain Array**:\n - Use a `for` loop to iterate through each element in the `gain` array.\n\n3. **Update Cumulative Sum**:\n - Add the current `gain` value to `sum`.\n\n4. **Update Maximum Altitude**:\n - If `sum` is greater than or equal to `max`, update `max` with `sum`.\n\n5. **Return Result**:\n - Return the value of `max`.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int sum=0, max=0;\n for(int i=0; i<gain.length; i++){\n sum += gain[i];\n max = sum>=max ? sum:max;\n }\n return max;\n }\n}\n```\n\n\n
| 5 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
Easiest Approach || beats 100% || STL
|
easiest-approach-beats-100-stl-by-prakha-wwfc
|
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(
|
prakhar-singh
|
NORMAL
|
2023-06-19T09:38:33.837985+00:00
|
2023-06-19T09:38:33.838015+00:00
| 1,077 | false |
# 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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n vector<int>temp = {0};\n for(auto it:gain){\n temp.push_back(it + temp.back());\n } \n return *max_element(temp.begin(), temp.end());\n }\n};\n```
| 5 | 0 |
['C++']
| 1 |
find-the-highest-altitude
|
Java Solution for Find the Highest Altitude Problem
|
java-solution-for-find-the-highest-altit-st51
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to simulate the biker\'s road trip by calculating
|
Aman_Raj_Sinha
|
NORMAL
|
2023-06-19T03:37:51.687339+00:00
|
2023-06-19T03:37:51.687357+00:00
| 749 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to simulate the biker\'s road trip by calculating the altitude at each point. We start with an initial altitude of 0 and update it based on the gain between consecutive points. We keep track of the highest altitude reached during the trip.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize the highest altitude and current altitude variables to 0.\n1. Iterate through the gain array.\n1. For each element in the gain array, update the current altitude by adding the gain.\n1. Use the Math.max function to compare the current altitude with the highest altitude and update the highest altitude if necessary.\n1. After iterating through the entire gain array, return the highest altitude.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of the solution is O(n), where n is the length of the gain array. This is because we iterate through the gain array once to calculate the altitudes.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of the solution is O(1) because we only use a constant amount of extra space to store the highest altitude and current altitude variables. The space usage does not depend on the size of the input.\n\n# Code\n```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int highestAltitude = 0;\n int currentAltitude = 0;\n \n for (int i = 0; i < gain.length; i++) \n {\n currentAltitude += gain[i];\n highestAltitude = Math.max(highestAltitude, currentAltitude);\n }\n \n return highestAltitude;\n }\n}\n```
| 5 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
2 Easy Solutions🔥 || C++🔥 || Fastest🔥🔥🔥
|
2-easy-solutions-c-fastest-by-black_mamb-x2ip
|
\n# Approach : Brute Force\n Describe your approach to solving the problem. \n\n# Code 1\n\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain
|
Black_Mamba01
|
NORMAL
|
2023-06-19T00:45:10.332187+00:00
|
2023-06-19T00:45:10.332205+00:00
| 827 | false |
\n# Approach : Brute Force\n<!-- Describe your approach to solving the problem. -->\n\n# Code 1\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n \n int altitude=0,max=0;\n for(int i=0;i<gain.size();i++){\n altitude=altitude+gain[i];\n\t\t\t\n if(max<altitude)\n max=altitude;\n }\n return max;\n }\n};\n```\n\n# Approach: Using Vector\n<!-- Describe your approach to solving the problem. -->\n\n# Code 2\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n \n vector<int> v;\n v.push_back(0);\n int mx =0;\n for(int i =0; i<gain.size(); i++){\n int s = v[i];\n v.push_back(gain[i]+s);\n if(mx<v[i+1]){\n mx = v[i+1];\n }\n }\n \n for(int i =0; i<v.size(); i++){\n cout<<v[i]<<" ";\n }\n \n cout<<endl;\n \n return mx; \n }\n};\n```
| 5 | 0 |
['C++']
| 1 |
find-the-highest-altitude
|
🚀 Prefix Sum 🚀 Loop🚀 Easy Code 🚀 Python3 🚀
|
prefix-sum-loop-easy-code-python3-by-sid-62ed
|
Please Upvote if you find this useful \uD83D\uDE0A\n\n# Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O
|
siddp6
|
NORMAL
|
2023-01-21T15:28:50.518872+00:00
|
2023-01-21T15:28:50.518915+00:00
| 642 | false |
## Please Upvote if you find this useful \uD83D\uDE0A\n\n# Complexity\n- Time complexity: $$O(n)$$\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```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n sol = 0\n curHeight = 0\n\n for Height in gain:\n curHeight += Height\n sol = max(sol, curHeight)\n \n return sol\n```\n\n## Please Upvote if you find this useful \uD83D\uDE0A\n
| 5 | 0 |
['Array', 'Prefix Sum', 'Python', 'Python3']
| 0 |
find-the-highest-altitude
|
Java - Easy Solution || Understanable 100%
|
java-easy-solution-understanable-100-by-m6wzt
|
\nclass Solution {\n public int largestAltitude(int[] gain) {\n int sum = 0;\n int maxNum = 0;\n\n for(int i = 0; i<gain.length; i++){\
|
deleted_user
|
NORMAL
|
2022-08-30T13:30:50.096229+00:00
|
2022-08-30T13:30:50.096275+00:00
| 629 | false |
```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int sum = 0;\n int maxNum = 0;\n\n for(int i = 0; i<gain.length; i++){\n sum+= gain[i];\n if (sum>maxNum)\n maxNum = sum;\n \n }\n \n return maxNum;\n \n }\n}\n```
| 5 | 0 |
['Java']
| 1 |
find-the-highest-altitude
|
Simple and Easy to Understand with Explanation
|
simple-and-easy-to-understand-with-expla-ur3v
|
Video Explanation\nhttps://youtu.be/tmZ_Ev1kZN8\n\nThere are two ways to solve this:\n1. using arrays to store altitudes then finding max altitude in that array
|
qkrrbtjd90
|
NORMAL
|
2021-03-24T19:06:46.029985+00:00
|
2021-03-24T19:10:35.586666+00:00
| 1,282 | false |
Video Explanation\nhttps://youtu.be/tmZ_Ev1kZN8\n\nThere are two ways to solve this:\n1. using arrays to store altitudes then finding max altitude in that array\n2. store max and current altitude to a variable, then compare current altitude to max altitude. If current altitude is greater simply overwrite max altitude\n\n\n/**\n * 1. store altitudes in an array\n * 2. find max altitute\n * @param {number[]} gain array of net gain/loss in altitude between two points\n * @returns {number} maximum altitude\n * //* Time: O(n)\n * Space: O(n)\n */\n\nRuntime: 80 ms, faster than 61.27% of JavaScript online submissions for Find the Highest Altitude.\nMemory Usage: 38.7 MB, less than 39.62% of JavaScript online submissions for Find the Highest Altitude.\n\n```\n\nconst largestAltitude1 = gain => {\n\tconst altitudes = [0];\n\n\tfor (let i = 0; i < gain.length; i++) {\n\t\taltitudes.push(altitudes[i] + gain[i]);\n\t}\n\n\treturn Math.max(...altitudes);\n};\n\n```\n\nRuntime: 80 ms, faster than 61.27% of JavaScript online submissions for Find the Highest Altitude.\nMemory Usage: 38.7 MB, less than 39.62% of JavaScript online submissions for Find the Highest Altitude.\n\n\n```\nconst largestAltitude2 = gain => {\n\treturn Math.max(\n\t\t...gain.reduce((acc, cv, i) => (acc.push(cv + acc[i]), acc), [0])\n\t);\n};\n```\n\n\n/**\n * 1. store current altitude and maximum altitude\n * 2. compare current altitude and maximum altitude and if current altitude is greater, overwrite max\n * @param {number[]} gain array of net gain/loss in altitude between two points\n * @returns {number} maximum altitude\n * //* Time: O(n)\n * //* Space: O(1)\n */\n\nRuntime: 76 ms, faster than 80.64% of JavaScript online submissions for Find the Highest Altitude.\nMemory Usage: 38.4 MB, less than 70.57% of JavaScript online submissions for Find the Highest Altitude.\n\n\n```\nconst largestAltitude3 = gain => {\n\tlet [max, currentAltitude] = [0, 0];\n\n\tfor (let i = 0; i < gain.length; i++) {\n\t\tcurrentAltitude = gain[i] + currentAltitude;\n\t\tif (currentAltitude > max) max = currentAltitude;\n\t}\n\n\treturn max;\n};\n```\n
| 5 | 0 |
['JavaScript']
| 2 |
find-the-highest-altitude
|
Rust: 0ms - Single Expression Solutions
|
rust-0ms-single-expression-solutions-by-ab9va
|
Using fold():\n\nimpl Solution {\n pub fn largest_altitude(gain: Vec<i32>) -> i32 {\n gain.into_iter()\n .fold((0, 0), |(res, h), g| (res.m
|
seandewar
|
NORMAL
|
2021-03-22T18:33:29.413491+00:00
|
2021-03-22T18:33:29.413525+00:00
| 167 | false |
**Using `fold()`**:\n```\nimpl Solution {\n pub fn largest_altitude(gain: Vec<i32>) -> i32 {\n gain.into_iter()\n .fold((0, 0), |(res, h), g| (res.max(h + g), h + g))\n .0\n }\n}\n```\n---\n\n**Using `scan()`**:\n```\nimpl Solution {\n pub fn largest_altitude(gain: Vec<i32>) -> i32 {\n std::iter::once(0)\n .chain(gain.into_iter())\n .scan(0, |h, g| {\n *h += g;\n Some(*h)\n })\n .max()\n .unwrap()\n }\n}\n```\n---\n\nTime complexity: `O(n)`.\nExtra-space complexity: `O(1)`.\n
| 5 | 0 |
['Rust']
| 0 |
find-the-highest-altitude
|
Beats 100%✅|| Easy Python Code
|
beats-100-easy-python-code-by-jeet-vyas-airy
|
\n\n---\n\n\n\n# Code\npython []\nclass Solution(object):\n def largestAltitude(self, gain):\n """\n :type gain: List[int]\n :rtype: int
|
Jeet-Vyas
|
NORMAL
|
2024-10-22T17:41:25.335530+00:00
|
2024-10-22T17:41:25.335588+00:00
| 478 | false |
\n\n---\n\n\n\n# Code\n```python []\nclass Solution(object):\n def largestAltitude(self, gain):\n """\n :type gain: List[int]\n :rtype: int\n """\n altitude = 0\n max_altitude = 0\n for g in gain:\n altitude += g\n max_altitude = max(max_altitude,altitude)\n\n return max_altitude\n \n```
| 4 | 0 |
['Python']
| 2 |
find-the-highest-altitude
|
🎒 "Climb to the Top! 🏔️ Calculating the Highest Altitude Step-by-Step ⬆️⬇️"
|
climb-to-the-top-calculating-the-highest-mgsk
|
Problem Intuition \uD83E\uDDED:\n\nImagine you\'re hiking \uD83E\uDD7E on a mountain \uD83C\uDF04. You start at sea level (altitude 0) and every step you take e
|
sajidmujawar
|
NORMAL
|
2024-10-09T08:56:54.214895+00:00
|
2024-10-09T08:56:54.214942+00:00
| 481 | false |
# Problem Intuition \uD83E\uDDED:\n\nImagine you\'re hiking \uD83E\uDD7E on a mountain \uD83C\uDF04. You start at sea level (altitude 0) and every step you take either goes up \u2B06\uFE0F or down \u2B07\uFE0F, depending on the gain values in an array. Your goal is to figure out the highest altitude you reach during your hike! Sounds like fun, right? Let\'s dive into how to solve this step by step.\n\n# Approach\n1. Start at Sea Level (altitude = 0) \uD83C\uDF0A:\n <ul><li> Initially, you\'re at 0 altitude.</li>\n <li>The first step might make you go up \u2B06\uFE0F or down \u2B07\uFE0F. We need to keep track of the highest point we reach. Let\'s call this maxi (short for maximum altitude).</li>\n</ul>\n\n2. Cumulative Altitude Calculation \uD83D\uDD22:\n <ul><li>The values in gain[] don\u2019t tell you directly where you are but how much the altitude changes with each step.</li>\n <li>For each step, we update our current altitude based on the previous step: gain[i] = gain[i] + gain[i-1]. This way, we\u2019re calculating the total altitude after each step.</li>\n</ul>\n\n3. Track the Highest Altitude \uD83C\uDFD4\uFE0F:\n <ul><li>With each new altitude, we check if it\'s the highest we\'ve reached so far.</li>\n <li>We do this by comparing the current altitude gain[i] to maxi. If gain[i] is bigger, we update maxi.</li>\n</ul>\n\n4. Return the Maximum Altitude \uD83C\uDFC5:\n <ul><li>After checking all steps, maxi will have the highest altitude we\u2019ve reached during the hike! This is what we return.</li></ul>\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(1)$$ \n\n# Example Walkthrough (\uD83D\uDEB6 + \u26F0\uFE0F = fun!):\n\nLet\'s say we have the gain[] array: [-5, 1, 5, 0, -7].\n\nWe start at sea level:\n<ul><li>\n \uD83C\uDF0A Altitude = 0\n</li>\n</ul>\nNow let\u2019s walk through each step:\n<ul>\n<li>\n Step 1: -5\n \u2B07\uFE0F You go down by 5. Altitude is now 0 + (-5) = -5 \u274C (not higher than 0).\n</li>\n<li>\n Step 2: +1\n \u2B06\uFE0F You go up by 1. Altitude is now -5 + 1 = -4 \u274C (still not higher than 0).\n</li>\n<li>\n Step 3: +5\n \u2B06\uFE0F You climb by 5. Altitude is now -4 + 5 = 1 \uD83C\uDFC5 (new highest point so far!).\n</li>\n<li>\n Step 4: +0\n \u2796 You stay the same. Altitude remains 1. No change \uD83D\uDED1.\n</li>\n<li>\n Step 5: -7\n \u2B07\uFE0F You go down by 7. Altitude is now 1 + (-7) = -6 \u274C (lower than 1).\n</li>\n</ul>\nFinally, the highest altitude you reached was 1, so that\'s what we return! \uD83C\uDF89\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int maxi=max(0,gain[0]);\n for(int i=1;i<gain.size();i++)\n {\n gain[i]+=gain[i-1];\n maxi=max(maxi,gain[i]);\n }\n\n return maxi;\n }\n};\n```\n\n
| 4 | 0 |
['Math', 'Prefix Sum', 'C++']
| 0 |
find-the-highest-altitude
|
✅ BEATS 100% OF SOLUTION ☑️
|
beats-100-of-solution-by-2005115-ugx6
|
PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT\n# CONNECT WITH ME\n### https://www.linkedin.com/in/pratay-nandy-9ba57b229/\n#### https://www.instagram.com/pratay_nand
|
2005115
|
NORMAL
|
2024-01-10T17:18:24.268351+00:00
|
2024-01-10T17:18:24.268385+00:00
| 536 | false |
# **PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT**\n# **CONNECT WITH ME**\n### **[https://www.linkedin.com/in/pratay-nandy-9ba57b229/]()**\n#### **[https://www.instagram.com/pratay_nandy/]()**\n\n# Approach\nExplanation:\n\n1. **Variable Initialization:**\n - `maxi`: It represents the maximum altitude reached during the iteration. Initialized to 0.\n - `sum`: It represents the cumulative altitude gain. Initialized to 0.\n\n2. **Iterate Through the Gain Vector:**\n - The `for` loop iterates through each element in the `gain` vector.\n\n3. **Update Cumulative Altitude Gain:**\n - `sum += gain[i];`: Add the current gain value to the running sum, representing the cumulative altitude gain.\n\n4. **Update Maximum Altitude:**\n - `maxi = max(maxi, sum);`: Update the `maxi` variable with the maximum altitude reached so far. It uses the `max` function to compare the current `sum` with the existing `maxi` and stores the larger of the two.\n\n5. **Return Maximum Altitude:**\n - After the loop completes, the function returns the maximum altitude reached (`maxi`).\n\nThe function essentially calculates the cumulative altitude gain during the iteration through the `gain` vector and keeps track of the maximum altitude reached. The final result is the maximum altitude reached during the entire journey.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:**0(N)**\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:**0(1)**\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int maxi = 0;\n int sum = 0;\n for(int i = 0 ;i < gain.size();i++)\n {\n sum += gain[i];\n maxi = max(maxi,sum);\n }\n return maxi;\n \n }\n};\n```
| 4 | 0 |
['Array', 'Prefix Sum', 'C++']
| 0 |
find-the-highest-altitude
|
BEST SOLUTION BEATING 100%
|
best-solution-beating-100-by-yatiraj_upa-wau7
|
Intuition\n Describe your first thoughts on how to solve this problem. \nIt find the largest altitude reached by accumulating the gains at each step from an ini
|
Yatiraj_Upadhyay
|
NORMAL
|
2023-11-23T07:07:41.495255+00:00
|
2023-11-23T07:07:41.495278+00:00
| 35 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt find the largest altitude reached by accumulating the gains at each step from an initial altitude of zero. The gain array represents the gain or loss in altitude at each step of a journey. The task is to find the maximum altitude reached during the journey.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInitialize an array arr of size n+1, where n is the length of the gain array. This array arr is used to keep track of the altitude after each step.\nIterate through the gain array and calculate the altitude after each step by accumulating the gains/losses. Store these altitudes in the arr array.\nFind the maximum altitude reached by iterating through the arr array and keeping track of the maximum value found.\n\n# Complexity\n**Time Complexity:**\nInitializing the arr array takes O(n) time, where n is the length of the gain array.\nThe first loop to calculate altitudes from gains takes O(n) time since it iterates through the gain array.\nThe second loop to find the maximum altitude also takes O(n) time as it traverses the arr array.\nOverall, the time complexity of the code is O(n).\n\n**Space Complexity:**\nAn additional array arr of size n+1 is used, resulting in O(n) space complexity, where n is the length of the gain array.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int n=gain.length;\n int[] arr=new int[n+1];\n for(int i=1;i<arr.length;i++){\n arr[i]=arr[i-1]+gain[i-1];\n }\n int m=0;\n for(int i=0;i<arr.length;i++){\n if(m<arr[i])\n m=arr[i];\n }\n return m;\n \n \n }\n}\n```
| 4 | 0 |
['Array', 'Prefix Sum', 'Java']
| 0 |
find-the-highest-altitude
|
C++ sol
|
c-sol-by-drspieler-dp3y
|
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
|
drspieler
|
NORMAL
|
2023-07-10T00:35:52.879669+00:00
|
2023-07-10T00:35:52.879688+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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n=gain.size();\n int hi=0, alt=0;\n for(int i=0; i<n; i++){\n alt+=gain[i];\n hi=max(hi, alt);\n }\n return hi;\n }\n};\n```
| 4 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
Best O(N) Solution
|
best-on-solution-by-kumar21ayush03-g1ft
|
Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n
|
kumar21ayush03
|
NORMAL
|
2023-06-23T14:48:35.196827+00:00
|
2023-06-23T14:48:35.196861+00:00
| 8 | false |
# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int ans = 0, cur = 0;\n for (int i = 0; i < gain.size(); i++) {\n cur += gain[i];\n ans = max (ans, cur);\n }\n return ans;\n }\n};\n```
| 4 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
Easy JAVA Code || Beginner Friendly || Beats 100%
|
easy-java-code-beginner-friendly-beats-1-oxwt
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThe given code represents a Java class called Solution with a method largestAltitude. T
|
20250315.ramh9932
|
NORMAL
|
2023-06-19T13:27:06.212297+00:00
|
2023-06-19T13:27:06.212325+00:00
| 657 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe given code represents a Java class called Solution with a method largestAltitude. The method takes an array gain as input, which represents the altitude gains at various points. The goal is to find the highest altitude reached during the journey.\n\nHere\'s how the code works:\n\nIt initializes two variables max and temp to 0.\n\nmax will keep track of the maximum altitude reached so far.\ntemp will store the current altitude as the loop progresses.\nIt then loops through the gain array from the beginning to the end.\n\nAt each iteration, it adds the current gain gain[i] to the temp variable, representing the altitude at that point.\nIt updates the max variable by taking the maximum value between the current max and temp using the Math.max method.\nThis ensures that max always holds the highest altitude reached during the journey up to the current point.\nFinally, the method returns the maximum altitude stored in the max variable.\n\nIn summary, the largestAltitude method calculates the altitude gained at each point and keeps track of the highest altitude reached during the journey. The maximum altitude is then returned as the result.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach used in the given code is a simple iterative approach to find the largest altitude reached during a journey. Here\'s a step-by-step explanation of the approach:\n\nInitialize two variables, max and temp, to 0. These variables will be used to track the maximum altitude reached and the current altitude, respectively.\n\nIterate through the gain array using a for loop. At each iteration:\n\nAdd the current gain gain[i] to the temp variable. This represents the altitude at the current point.\nUpdate the max variable by taking the maximum value between the current max and temp using the Math.max() method. This ensures that max always holds the highest altitude reached during the journey up to the current point.\nAfter iterating through the entire gain array, the max variable will hold the largest altitude reached during the journey.\n\nReturn the value stored in the max variable as the result.\n\nIn summary, the approach calculates the altitude at each point by adding the gains from the gain array. It keeps track of the maximum altitude reached so far and returns that value as the final result.\n\n# Complexity\n- Time complexity:O(n)\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```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int max=0,temp=0;\n for(int i=0;i<gain.length;i++){\n temp+=gain[i];\n max=Math.max(max,temp);\n }\n return max;\n }\n}\n```
| 4 | 0 |
['Java']
| 1 |
find-the-highest-altitude
|
🚴♂️ Find the Highest Altitude | LeetCode Solutions in Python 💻
|
find-the-highest-altitude-leetcode-solut-z8js
|
Intuition\nMy first thought to solve this problem is to use an incremental approach. Given that we start at altitude 0, we can keep adding the \'gain\' at each
|
vanAmsen
|
NORMAL
|
2023-06-19T10:56:55.658119+00:00
|
2023-06-19T11:07:30.905561+00:00
| 379 | false |
# Intuition\nMy first thought to solve this problem is to use an incremental approach. Given that we start at altitude 0, we can keep adding the \'gain\' at each step to get the new altitude and track the maximum altitude reached at any point. This approach makes sense as the altitude at a given point is simply the sum of all the gains till that point.\n\nhttps://www.youtube.com/watch?v=xKg7CkstOaM\n\n# Approach\nWe start by initializing two variables, \'max_altitude\' and \'current_altitude\', both to 0. Then, we loop through the \'gain\' list, adding each value to \'current_altitude\'. After each addition, we compare \'current_altitude\' with \'max_altitude\'. If \'current_altitude\' is greater, we update \'max_altitude\' to the new value. The logic behind this is simple - we are climbing up (or down) a mountain, and we keep track of the highest point we\'ve reached so far. Once we\'ve gone through the entire list, we return \'max_altitude\' as the highest altitude reached.\n\n# Complexity\n- Time complexity:\nThe time complexity of this solution is O(n), where n is the length of the \'gain\' list. This is because we are traversing the \'gain\' list once.\n\n- Space complexity:\nThe space complexity of the solution is O(1). This is because we are using a constant amount of space to store the current and maximum altitudes, regardless of the input size.\n\n# Code\n```python []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n # Initialize the maximum altitude and the current altitude to 0 \n max_altitude = 0 \n current_altitude = 0 \n # Loop through the gain list \n for g in gain: \n # Add the current gain to the current altitude \n current_altitude += g \n # Check if the current altitude is higher than the max altitude \n if current_altitude > max_altitude: \n # If it is, update the max altitude \n max_altitude = current_altitude \n # Return the max altitude \n return max_altitude \n```\n```javascript []\nvar largestAltitude = function(gain) {\n let maxAltitude = 0;\n let currentAltitude = 0;\n \n for(let g of gain) {\n currentAltitude += g;\n if(currentAltitude > maxAltitude)\n maxAltitude = currentAltitude;\n }\n return maxAltitude;\n};\n```\n```rust []\nimpl Solution {\n pub fn largest_altitude(gain: Vec<i32>) -> i32 {\n let mut max_altitude = 0;\n let mut current_altitude = 0;\n\n for &g in gain.iter() {\n current_altitude += g;\n if current_altitude > max_altitude {\n max_altitude = current_altitude;\n }\n }\n max_altitude\n }\n}\n```\n```Go []\nfunc largestAltitude(gain []int) int {\n maxAltitude := 0\n currentAltitude := 0\n \n for _, g := range gain {\n currentAltitude += g\n if currentAltitude > maxAltitude {\n maxAltitude = currentAltitude\n }\n }\n return maxAltitude\n}\n\n```\n```C# []\npublic class Solution {\n public int LargestAltitude(int[] gain) {\n int maxAltitude = 0;\n int currentAltitude = 0;\n \n foreach(int g in gain){\n currentAltitude += g;\n if(currentAltitude > maxAltitude)\n maxAltitude = currentAltitude;\n }\n return maxAltitude;\n }\n}\n```\n
| 4 | 0 |
['Go', 'Python3', 'Rust', 'JavaScript', 'C#']
| 0 |
find-the-highest-altitude
|
C# LINQ one-liner
|
c-linq-one-liner-by-kimsey0-b5om
|
csharp\npublic int LargestAltitude(int[] gain) => gain.Aggregate((Current: 0, Max: 0),\n (t, g) => (t.Current + g, Math.Max(t.Max, t.Current + g)), t => t.Ma
|
kimsey0
|
NORMAL
|
2023-06-19T07:46:48.629155+00:00
|
2023-12-21T11:30:21.771879+00:00
| 270 | false |
```csharp\npublic int LargestAltitude(int[] gain) => gain.Aggregate((Current: 0, Max: 0),\n (t, g) => (t.Current + g, Math.Max(t.Max, t.Current + g)), t => t.Max);\n```
| 4 | 0 |
[]
| 1 |
find-the-highest-altitude
|
C++ Solution || Easy to Understand || Prefix Sum
|
c-solution-easy-to-understand-prefix-sum-5jrj
|
Intuition\nThe description of the problem is not very clear. I recommned you to see the test cases 2-3 times & you will be able to understand the probelem.\n\n#
|
Chetan_Batra2001
|
NORMAL
|
2023-06-19T07:20:24.236515+00:00
|
2023-06-19T07:21:58.667417+00:00
| 408 | false |
# Intuition\nThe description of the problem is not very clear. I recommned you to **see the test cases 2-3 times** & you will be able to understand the probelem.\n\n# Approach\n keep on finding prefix sum\n\n# Complexity\n- Time complexity: O(n)\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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n = gain.size();\n int prefixSum = 0 ;\n int ans = -1e9;\n for(int i =1;i<=n;i++){\n prefixSum += gain[i-1];\n ans = max( ans , prefixSum);\n }\n if(ans < 0) return 0;\n return ans;\n }\n};\n```
| 4 | 0 |
['Array', 'Prefix Sum', 'C++']
| 1 |
find-the-highest-altitude
|
C++ | Linear Scan or Iterative |
|
c-linear-scan-or-iterative-by-ashish_mad-mwos
|
Intuition\n Describe your first thoughts on how to solve this problem. My initial thoughts on solving this problem would be as follows:\n\n1. Initialize two var
|
ashish_madhup
|
NORMAL
|
2023-06-19T07:03:15.348134+00:00
|
2023-06-19T07:03:15.348151+00:00
| 269 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->My initial thoughts on solving this problem would be as follows:\n\n1. Initialize two variables, `altitude` and `max`, to 0. These will be used to keep track of the current altitude and the maximum altitude reached during the journey, respectively.\n\n2. Iterate through the given `gain` vector. At each step, update the `altitude` by adding the gain at the current step.\n\n3. Check if the updated `altitude` is greater than the current maximum altitude (`max`). If it is, update `max` with the new value of `altitude`.\n\n4. After iterating through all the elements in the `gain` vector, `max` will hold the maximum altitude reached during the journey.\n\n5. Return the value of `max` as the result.\n\nThis approach calculates the altitude at each step of the journey and keeps track of the maximum altitude achieved so far. By the end of the journey, it will have the maximum altitude reached.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->This approach calculates the altitude at each step of the journey by summing up the gains. It then updates the maximum altitude if a higher altitude is encountered. By the end of the journey, the maximum altitude reached is returned as the result.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> O(n) \n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --> O(1) \n\n# Code\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain)\n { \n int altitude=0,max=0;\n for(int i=0;i<gain.size();i++)\n {\n altitude=altitude+gain[i]; // Increase the altitude by the gain at each step\t\n if(max<altitude)\n max=altitude; // Update the maximum altitude if the current altitude is higher\n }\n return max; // Return the maximum altitude\n }\n};\n\n```\n\n
| 4 | 0 |
['Array', 'Dynamic Programming', 'Iterator', 'Prefix Sum', 'C++']
| 0 |
find-the-highest-altitude
|
Kotlin | one-liner
|
kotlin-one-liner-by-samoylenkodmitry-hf08
|
\n#### Join me on Telegram\nhttps://t.me/leetcode_daily_unstoppable/250\n#### Problem TLDR\nMax running sum\n#### Intuition\nJust sum all the values and compute
|
SamoylenkoDmitry
|
NORMAL
|
2023-06-19T03:43:59.574721+00:00
|
2023-06-19T03:43:59.574739+00:00
| 214 | false |
\n#### Join me on Telegram\nhttps://t.me/leetcode_daily_unstoppable/250\n#### Problem TLDR\nMax running sum\n#### Intuition\nJust sum all the values and compute the `max`\n\n#### Approach\nLet\'s write Kotlin `fold` one-liner\n#### Complexity\n- Time complexity:\n$$O(n)$$\n- Space complexity:\n$$O(1)$$\n#### Code\n```\n\n fun largestAltitude(gain: IntArray): Int = gain\n .fold(0 to 0) { (max, sum), t -> maxOf(max, sum + t) to (sum + t) }\n .first\n\n```
| 4 | 0 |
['Kotlin']
| 1 |
find-the-highest-altitude
|
Easy C++/Python several different solutions
|
easy-cpython-several-different-solutions-umzu
|
Intuition\n Describe your first thoughts on how to solve this problem. \ngain[i]=altitude[i+1]-altitude[i]\nFind max({altitude[i]})\n# Approach\n Describe your
|
anwendeng
|
NORMAL
|
2023-06-19T01:37:11.177241+00:00
|
2023-06-19T03:05:01.262711+00:00
| 1,653 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ngain[i]=altitude[i+1]-altitude[i]\nFind max({altitude[i]})\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```C++ []\n\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n=gain.size();\n int hi=0, alt=0;\n for(int i=0; i<n; i++){\n alt+=gain[i];\n hi=max(hi, alt);\n }\n return hi;\n }\n};\n```\n```python []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n n=len(gain)\n hi=0\n alt=0\n for x in gain:\n alt+=x\n hi=max(hi, alt)\n return hi\n```\n# 2nd C++ uses STL vector & max_element\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n=gain.size();\n vector<int> alt(n+1, 0);\n alt[0]=0;\n for(int i=1; i<=n; i++){\n alt[i]=gain[i-1]+alt[i-1];\n cout<<alt[i]<<",";\n }\n return *max_element(alt.begin(), alt.end());\n }\n};\n```\n# 3rd C++ solution: No need for extra array alt[]\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n=gain.size();\n for(int i=1; i<n; i++){\n gain[i]+=gain[i-1];//No need for extra array\n }\n gain.push_back(0);//0 for the beginning\n return *max_element(gain.begin(), gain.end());\n }\n};\n```\n
| 4 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
Java Easy Solution |0ms| |100% faster|
|
java-easy-solution-0ms-100-faster-by-ujj-xlqn
|
class Solution {\n public int largestAltitude(int[] gain) {\n\t\n int max = 0;\n int current = 0;\n for(int i=0; i<gain.length; i++) {\n
|
Ujjwall19
|
NORMAL
|
2022-08-29T13:55:01.319119+00:00
|
2022-08-29T13:55:01.319143+00:00
| 310 | false |
class Solution {\n public int largestAltitude(int[] gain) {\n\t\n int max = 0;\n int current = 0;\n for(int i=0; i<gain.length; i++) {\n current += gain[i];\n max = Math.max(current, max);\n }\n return max;\n }\n}
| 4 | 0 |
['Array', 'Java']
| 0 |
find-the-highest-altitude
|
Python prefix sum solution
|
python-prefix-sum-solution-by-leovam-lqht
|
py\n\n\'\'\'\nw: array, prefix sum\nh: this is a prefix-sum like problem, but we only\n need to care about the last element in the list\n so we only need
|
leovam
|
NORMAL
|
2021-01-25T04:44:46.536482+00:00
|
2021-01-25T04:44:46.536623+00:00
| 791 | false |
```py\n\n\'\'\'\nw: array, prefix sum\nh: this is a prefix-sum like problem, but we only\n need to care about the last element in the list\n so we only need to store the last element\n and compare to previous result\n \nTime: O(N)\nspace: O(1)\n\n\'\'\'\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n res = 0\n start = 0\n for i in gain:\n start += i\n res = max(res, start)\n \n return res\n \n\'\'\'\nprefix-sum: a = [0, 1, 2, 3]\nwhat is the sum up to the current index in the array?\n\nres = [a[0]]\nfor i in a[1:]:\n res.append(res[-1]+i)\n \n\'\'\'\n```
| 4 | 0 |
['Prefix Sum', 'Python']
| 1 |
find-the-highest-altitude
|
Easy / 100% Beats / Java
|
easy-100-beats-java-by-praveensuthar3016-0b2e
|
Code
|
praveensuthar3016
|
NORMAL
|
2025-02-03T05:50:15.323244+00:00
|
2025-02-03T05:50:15.323244+00:00
| 353 | false |
# Code
```java []
class Solution {
public int largestAltitude(int[] gain) {
int max= 0;
int value = 0;
for(int i = 0 ; i < gain.length ; i++){
value += gain[i];
max = Math.max(max, value);
}
return max;
}
}
```
| 3 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
[C++] PSA + Sorting (Beats 100%!)
|
c-psa-sorting-beats-100-by-zzzqwq-t433
|
ApproachStandard PSA approach, then finding the maximum element in the array.Code
|
zzzqwq
|
NORMAL
|
2024-12-18T19:50:51.564587+00:00
|
2024-12-18T19:50:51.564587+00:00
| 198 | false |
# Approach\nStandard PSA approach, then finding the maximum element in the array.\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int psa[103];\n int largestAltitude(vector<int>& gain) {\n for (int i = 1; i <= gain.size(); ++i)\n psa[i] = psa[i-1]+gain[i-1];\n return *max_element(psa, psa+103);\n }\n};\n```
| 3 | 0 |
['Sorting', 'Prefix Sum', 'C++']
| 0 |
find-the-highest-altitude
|
Beats 100%✅✅|| 1 Liner Solution🤩
|
beats-100-1-liner-solution-by-jay-vyas-mez3
|
Give me Upvote\u2B06\uFE0F if you find it helpful\uD83D\uDE0A\n\n---\n\n# Code\npython3 []\nfrom itertools import accumulate\n\nclass Solution:\n def largest
|
Jay-Vyas
|
NORMAL
|
2024-10-22T17:38:37.691518+00:00
|
2024-10-22T17:38:37.691552+00:00
| 568 | false |
# **[Give me Upvote\u2B06\uFE0F if you find it helpful\uD83D\uDE0A]()**\n\n---\n\n# Code\n```python3 []\nfrom itertools import accumulate\n\nclass Solution:\n def largestAltitude(self, gain):\n return max(accumulate(gain, initial=0))\n\n```
| 3 | 0 |
['Python', 'Python3']
| 1 |
find-the-highest-altitude
|
Beats 94.41% || Easy Solution 🤓✅
|
beats-9441-easy-solution-by-heikonagl-n6k2
|
\n\n\n# Intuition\nThe problem is asking for the largest altitude reached given a series of altitude gains and losses during a bike ride. My first thought is to
|
nailinger
|
NORMAL
|
2024-10-05T11:51:53.239382+00:00
|
2024-10-05T11:51:53.239413+00:00
| 334 | false |
\n\n\n# Intuition\nThe problem is asking for the largest altitude reached given a series of altitude gains and losses during a bike ride. My first thought is to maintain a running total of the current altitude as we go through each gain/loss, and at each step, check if the current altitude is the highest we have encountered so far. This would allow us to track the largest altitude efficiently in a single pass.\n\n# Approach\n1. **Initialize two variables**: One to keep track of the current altitude (`current`) and another to store the maximum altitude encountered so far (`max_gain`).\n2. **Iterate through the `gain` array**: For each element, update the `current` altitude by adding the current gain/loss value to the previous `current` value.\n3. **Update `max_gain`**: After updating the current altitude, check if it\u2019s the highest altitude encountered. If it is, update `max_gain`.\n4. **Return `max_gain`** after finishing the loop, as it represents the largest altitude reached during the ride.\n\n# Complexity\n- Time complexity:\n The time complexity is $$O(n)$$ where $$n$$ is the number of elements in the `gain` list. We only iterate through the list once.\n\n- Space complexity:\n The space complexity is $$O(1)$$ since we only use two extra variables (`current` and `max_gain`) regardless of the input size.\n\n# Code\n```python\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n current = 0\n max_gain = current\n for alt in gain:\n current += alt\n max_gain = max(max_gain, current)\n return max_gain
| 3 | 0 |
['Prefix Sum', 'Python3']
| 1 |
find-the-highest-altitude
|
SIMPLE PREFIX SUM C++ SOLUTION
|
simple-prefix-sum-c-solution-by-jeffrin2-ys44
|
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
|
Jeffrin2005
|
NORMAL
|
2024-07-29T14:45:49.543366+00:00
|
2024-07-29T14:45:49.543403+00:00
| 1,227 | 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<!-- 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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n = gain.size();\n vector<int>pref(n+1,0);\n pref[0] = gain[0];\n for(int i = 1; i < n; i++){\n pref[i] = pref[i-1] + gain[i];\n }\n return *max_element(pref.begin(), pref.end());\n }\n};\n```
| 3 | 0 |
['C++']
| 1 |
find-the-highest-altitude
|
Find the Highest Altitude | Java
|
find-the-highest-altitude-java-by-_ajit_-1u1k
|
Code\n\nclass Solution {\n public int largestAltitude(int[] gain) {\n int highestAltitude = 0;\n int altitude = 0;\n\n for (int i = 0; i
|
_Ajit_Singh_
|
NORMAL
|
2024-04-11T07:03:30.256615+00:00
|
2024-04-11T07:54:27.921254+00:00
| 1,266 | false |
# Code\n```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int highestAltitude = 0;\n int altitude = 0;\n\n for (int i = 0; i < gain.length; i++) {\n altitude += gain[i]; \n\n if (altitude > highestAltitude)\n highestAltitude = altitude;\n }\n\n return highestAltitude;\n }\n}\n```
| 3 | 0 |
['Array', 'Prefix Sum', 'Java']
| 0 |
find-the-highest-altitude
|
Prefix sum...easy solution with c++
|
prefix-sumeasy-solution-with-c-by-error_-3q9h
|
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
|
error_adp
|
NORMAL
|
2024-03-14T13:10:54.997050+00:00
|
2024-03-14T13:10:54.997078+00:00
| 462 | 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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n=gain.size();\n int ps[n+1];\n ps[0]=gain[0];\n for(int i=1; i<n; i++){\n ps[i] = ps[i-1]+gain[i];\n }\n sort(ps,ps+n);\n return (ps[n-1]<0 ? 0:ps[n-1]);\n }\n};\n```
| 3 | 0 |
['Prefix Sum', 'C++']
| 0 |
find-the-highest-altitude
|
Find the highest altitude using prefix sum approach
|
find-the-highest-altitude-using-prefix-s-0n8l
|
\n# Code\nC++ []\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n = gain.size();\n int sum = 0;\n int highe
|
meurudesu
|
NORMAL
|
2024-01-30T15:50:45.464471+00:00
|
2024-03-05T13:19:39.090570+00:00
| 389 | false |
\n# Code\n```C++ []\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n = gain.size();\n int sum = 0;\n int highest = 0;\n for(int i=0;i<n;i++) {\n sum += gain[i];\n highest = max(highest, sum);\n }\n return highest;\n }\n};\n```\n```Python3 []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n highest = 0\n currHeight = 0\n for i in range(len(gain)):\n currHeight += gain[i]\n highest = max(highest, currHeight)\n return highest\n```
| 3 | 0 |
['Array', 'Prefix Sum', 'C++', 'Python3']
| 0 |
find-the-highest-altitude
|
Easy C++ Solution
|
easy-c-solution-by-deva766825_gupta-zihd
|
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
|
deva766825_gupta
|
NORMAL
|
2024-01-28T13:58:58.799635+00:00
|
2024-01-28T13:58:58.799673+00:00
| 1,364 | 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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int sum=0;\n int max_1=0;\n vector<int> ans;\n for(int i=0;i<gain.size();i++){\n sum=sum+gain[i];\n max_1=max(max_1,sum);\n }\n return max_1;\n\n \n }\n};\n```
| 3 | 0 |
['C++']
| 1 |
find-the-highest-altitude
|
easy solution everyone will understand!! no complexities
|
easy-solution-everyone-will-understand-n-hiw9
|
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
|
Sajal_
|
NORMAL
|
2024-01-19T05:13:18.824004+00:00
|
2024-01-19T05:13:18.824027+00:00
| 666 | 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```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n list1 = [0]\n temp = 0\n for i in gain:\n \n temp = temp + i\n list1.append(temp)\n return max(list1)\n```
| 3 | 0 |
['Python3']
| 0 |
find-the-highest-altitude
|
Simple Solution in both Python and C++ languages😊
|
simple-solution-in-both-python-and-c-lan-auld
|
Solution \nPython3 []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n highest_point = 0\n prev_altitude = 0\n fo
|
k_a_m_o_l
|
NORMAL
|
2023-12-17T21:46:35.977748+00:00
|
2023-12-17T21:46:35.977764+00:00
| 329 | false |
# Solution \n```Python3 []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n highest_point = 0\n prev_altitude = 0\n for i in gain:\n prev_altitude += i\n highest_point = max(highest_point, prev_altitude)\n\n return highest_point\n```\n```C++ []\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int highest_point = 0;\n int prev_point = 0;\n for (int i = 0; i < gain.size(); i++){\n prev_point += gain.at(i);\n highest_point = max(highest_point, prev_point);\n\n }\n\n return highest_point;\n \n }\n};\n```\n\n\n
| 3 | 0 |
['Python', 'C++', 'Python3']
| 0 |
find-the-highest-altitude
|
Python 3 EASY SOLUTION
|
python-3-easy-solution-by-ayush_092005-iyxi
|
Intuition and Approach\nI have created a list for storing all the possible values of altitude and have kept 0 in it beacuse in the question it is giving that it
|
ayush_092005
|
NORMAL
|
2023-11-09T13:55:45.436876+00:00
|
2023-11-09T13:55:45.436899+00:00
| 249 | false |
# Intuition and Approach\nI have created a list for storing all the possible values of altitude and have kept 0 in it beacuse in the question it is giving that it starts with altitude equal to 0. And i have created a variable sum to add the values from the list gain and then appending it to the new list l.\n\n\n# Complexity\n\nTime Complexity: O(N)\nSpace Complexity: O(N)\n# Code\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n l=[0]\n sum=0\n for i in gain:\n sum+=i\n l.append(sum)\n return max(l)\n```
| 3 | 0 |
['Python3']
| 0 |
find-the-highest-altitude
|
Easy to understand python solution
|
easy-to-understand-python-solution-by-dr-t17x
|
Approach\nTo solve this problem, we need to store the original element and move to the next element by adding the difference. Thus, cur_num stores the value fro
|
drboss06
|
NORMAL
|
2023-11-07T09:11:45.185569+00:00
|
2023-11-07T09:11:45.185598+00:00
| 692 | false |
# Approach\nTo solve this problem, we need to store the original element and move to the next element by adding the difference. Thus, cur_num stores the value from the list with heights. At each iteration we look at the maximum value.\n\n# Complexity\n- Time complexity:\nThe complexity of this algorithm is O(n) and it is executed in 28 ms\n\n- Space complexity:\n16.10MB\n\n# Code\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n max_num = 0\n cur_num = 0\n\n for i in range(len(gain)):\n if max_num < cur_num + gain[i]:\n max_num = cur_num + gain[i]\n\n cur_num += gain[i] \n \n return max_num\n \n```
| 3 | 0 |
['Python3']
| 1 |
find-the-highest-altitude
|
❤️C++ Easy Solution || Leet code solution
|
c-easy-solution-leet-code-solution-by-sk-5s3n
|
\n Describe your first thoughts on how to solve this problem. \n\n# Approach : Prefix SUM\n Describe your approach to solving the problem. \n\n# Complexity\n- T
|
skumarsingh
|
NORMAL
|
2023-08-08T06:02:09.545533+00:00
|
2023-08-08T06:02:09.545559+00:00
| 83 | false |
\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach : Prefix SUM\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(n)\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# **Please UpVote if u like**\n\n# Code\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int alt=0;\n int ans=0;\n for(auto x:gain)\n {\n alt+=x;\n ans=max(ans,alt);\n }\n return ans;\n }\n};\n```\n# **Please UpVote if u like**
| 3 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
[Java] Prefix Sum
|
java-prefix-sum-by-xiangcan-ajvv
|
\nclass Solution {\n public int[] findDiagonalOrder(List<List<Integer>> nums) {\n int m = nums.size(), maxSum = 0, size = 0, index = 0;\n List<
|
xiangcan
|
NORMAL
|
2023-06-19T21:49:10.628982+00:00
|
2023-06-19T21:49:10.629000+00:00
| 129 | false |
```\nclass Solution {\n public int[] findDiagonalOrder(List<List<Integer>> nums) {\n int m = nums.size(), maxSum = 0, size = 0, index = 0;\n List<Integer>[] map = new ArrayList[100001];\n for (int i = 0; i < m; i++) {\n size += nums.get(i).size();\n for (int j = 0; j < nums.get(i).size(); j++) {\n int sum = i + j;\n if (map[sum] == null) map[sum] = new ArrayList<>();\n map[sum].add(nums.get(i).get(j));\n maxSum = Math.max(maxSum, sum);\n }\n }\n int[] res = new int[size];\n for (int i = 0; i <= maxSum; i++) {\n List<Integer> cur = map[i];\n for (int j = cur.size() - 1; j >= 0; j--) {\n res[index++] = cur.get(j);\n }\n }\n return res;\n }\n}\n```
| 3 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
✅Easy Java solution||Beginner Friendly🔥
|
easy-java-solutionbeginner-friendly-by-d-5unp
|
Complexity\n- Time complexity: O(n)\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# Co
|
deepVashisth
|
NORMAL
|
2023-06-19T19:36:41.862540+00:00
|
2023-06-19T19:36:41.862559+00:00
| 55 | false |
# Complexity\n- Time complexity: O(n)\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```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int count = 0;\n int max = 0;\n int currSum = 0;\n for(int i = 0; i < gain.length; i ++){\n currSum += gain[i];\n if(currSum > max){\n max = currSum;\n }\n }\n return max;\n }\n}\n```\n**If you really found my solution helpful please upvote it, as it motivates me to post such kind of codes and help the coding community, if you have some queries or some improvements please feel free to comment and share your views.\nHappy Coding**
| 3 | 0 |
['Array', 'Prefix Sum', 'Java']
| 0 |
find-the-highest-altitude
|
Easy level -100% beats in [Java][C++][Python3] -Tc: O(n) Sc: O(1) || Array || Sum||Detailed Approach
|
easy-level-100-beats-in-javacpython3-tc-owt28
|
Intuition\n Describe your first thoughts on how to solve this problem. \nwe aims to find the largest altitude reached given a series of altitude gains. It calcu
|
Anamika_aca
|
NORMAL
|
2023-06-19T15:00:24.279281+00:00
|
2023-06-19T15:00:24.279310+00:00
| 126 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe aims to find the largest altitude reached given a series of altitude gains. It calculates the cumulative altitude by adding the gains at each step and keeps track of the maximum cumulative altitude reached.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIt iterates through the gain vector, representing the altitude gains. It initializes two variables, maxc and cal, to keep track of the maximum cumulative altitude and the current cumulative altitude, respectively.\n\nIn each iteration, it adds the current gain gain[i] to cal. Then, it updates maxc by taking the maximum of the current maxc and cal. This ensures that maxc always holds the highest altitude reached so far.\n\nAfter iterating through all the gains, it returns maxc, which represents the largest altitude reached.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this code is O(n), where n is the number of elements in the gain vector. The code iterates through each gain once, performing constant-time operations for each iteration.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this code is O(1) since it only uses a constant amount of additional space for the variables n, maxc, and cal. The space required does not depend on the size of the input.\n\n\n\n# Code\n```Java []\nclass Solution {\n public int largestAltitude(int[] gain) {\n int n = gain.length;\n int maxc = 0, cal = 0;\n for (int i = 0; i < n; i++) {\n cal += gain[i];\n maxc = Math.max(maxc, cal);\n }\n return maxc;\n }\n}\n```\n```C++ []\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n = gain.size();\n int maxc=0, cal=0;\n \n for(int i =0; i<n; i++){\n cal += gain[i]; \n maxc = max( maxc, cal);\n \n }\n return maxc;\n\n }\n};\n```\n```Python []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n n = len(gain)\n maxc = 0\n cal = 0\n for i in range(n):\n cal += gain[i]\n maxc = max(maxc, cal)\n return maxc\n```
| 3 | 0 |
['Array', 'Prefix Sum', 'C++', 'Java', 'Python3']
| 0 |
find-the-highest-altitude
|
[Kotlin] Solution
|
kotlin-solution-by-devle79-g460
|
Approach\n Describe your approach to solving the problem. \nWell..\n\n# Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Spa
|
devle79
|
NORMAL
|
2023-06-19T10:47:50.381247+00:00
|
2023-06-19T10:47:50.381277+00:00
| 176 | false |
# Approach\n<!-- Describe your approach to solving the problem. -->\nWell..\n\n# Complexity\n- Time complexity: $$O(n)$$\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```\nclass Solution {\n\n fun largestAltitude(gain: IntArray): Int {\n var currentAltitude = 0\n var highestAltitude = 0\n\n gain.forEach {\n currentAltitude += it\n highestAltitude = maxOf(highestAltitude, currentAltitude)\n }\n\n return highestAltitude\n }\n}\n```
| 3 | 0 |
['Kotlin']
| 0 |
find-the-highest-altitude
|
CPP || O(N)
|
cpp-on-by-ayoaditya-u9d9
|
\n\n# Complexity\n- Time complexity: O(N)\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
|
ayoaditya
|
NORMAL
|
2023-06-19T10:28:41.210929+00:00
|
2023-06-19T10:28:41.210955+00:00
| 5 | false |
\n\n# Complexity\n- Time complexity: O(N)\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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) \n {\n int high = 0;\n for(int i=0;i<gain.size();i++) \n {\n if(i==0)\n { \n if(high<gain[i])\n {\n high=gain[i];\n }\n }\n else\n {\n gain[i]+=gain[i-1];\n if(high<gain[i])\n {\n high=gain[i];\n }\n }\n } \n return high;\n }\n};\n```
| 3 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
one liner + detailed solution | Python
|
one-liner-detailed-solution-python-by-tr-17u3
|
Hi!\nPlease comment if you have questions or suggestions regarding this post!\nPlease upvote if you find this post interesting!\n\n# one liner\n\nclass Solution
|
TrickyUnicorn
|
NORMAL
|
2023-06-19T09:30:56.103754+00:00
|
2023-06-19T09:30:56.103772+00:00
| 235 | false |
# Hi!\nPlease **comment** if you have questions or suggestions regarding this post!\nPlease **upvote** if you find this post interesting!\n\n# one liner\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n return max(*accumulate(gain), 0)\n```\n# detailed Solution\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n res = s = 0\n for v in gain:\n s += v\n res = max(res, s)\n return res\n```\n# Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)
| 3 | 0 |
['Array', 'Iterator', 'Python', 'Python3']
| 0 |
find-the-highest-altitude
|
EASIEST CPP SOLUTION
|
easiest-cpp-solution-by-drunkdoctor-vhd1
|
Intuition\nhere we have to find the max gain so we will just use the prefix sum technique.\n\n# Approach\ncreate an altitude pointer and a maxi pointer as soon
|
drunkdoctor
|
NORMAL
|
2023-06-19T06:46:24.478269+00:00
|
2023-06-19T06:46:24.478293+00:00
| 327 | false |
# Intuition\nhere we have to find the max gain so we will just use the prefix sum technique.\n\n# Approach\ncreate an altitude pointer and a maxi pointer as soon as you get max gain update your maxi .\n\n# Complexity\n- Time complexity:\no(n) (only one loop )\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int altitude =0;\n int maxi =0;\n for( int i =0;i<gain.size();i++){\n altitude=altitude+gain[i];\n if( maxi < altitude)\n maxi = altitude;\n \n }\n return maxi;\n \n }\n};\n```
| 3 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
Clear Explanation with Easy Linear Scan Solution
|
clear-explanation-with-easy-linear-scan-x1ajl
|
Intuition\n Describe your first thoughts on how to solve this problem. \nTraverse through the array and keep track of the altitude at the $i$\'th index.\n\n# Ap
|
bigbullboy
|
NORMAL
|
2023-06-19T06:08:57.306863+00:00
|
2023-06-19T06:08:57.306886+00:00
| 138 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTraverse through the array and keep track of the altitude at the $i$\'th index.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nLet us keep track of the heights that the biker can be at throughout their journey. There are ```len(gain) + 1``` different points in the roadtrip (since the start is considered a point as well). Let us create an array ```heights``` of size ```len(gain) + 1``` which represents the altitude of the biker at the $i$\'th point. Initially, the array will be filled with 0\'s. \n\nNow, let use traverse through ```gain```. The altitude at ```heights[i]``` is the change in altitude from the altitude of the previous point in the journey. In other words, ```heights[i+1] = heights[i] + gain[i]```. After traversing through the entire input array, we can return the maximum altitude we come across as our answer. \n\nFor a slight optimization, we could define our result ```res``` to be 0 at the start of the loop and at each iteration, take the maximum between ```res``` and ```heights[i+1]``` and return res in the end. \n\nNote that there is no need to maintain the full array of the heights and we can initialize height to be 0 and keep adding the change in altitudes and then comparing them to res which would increase the memory effiency but this approach is more explicit.\n\n# Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n # Initialze the result and the heights array\n res = 0\n heights = [0] * (len(gain) + 1)\n\n # Iterate through the input array and calculate the altitude\n # for the i\'th point in the journey\n for i in range(len(gain)):\n heights[i+1] = heights[i] + gain[i]\n # Update res to be the maximum altitude we have seen thus far\n res = max(res, heights[i+1])\n \n return res\n```
| 3 | 0 |
['Python3']
| 0 |
find-the-highest-altitude
|
C++||5 LINES|| EASIEST CODE || SIMPLE LOGIC||O(N)
|
c5-lines-easiest-code-simple-logicon-by-iqdma
|
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
|
_manishh12
|
NORMAL
|
2023-06-19T05:03:33.885810+00:00
|
2023-06-19T05:03:33.885833+00:00
| 13 | 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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n// INITIALIZE VARI TO STORE SUM AND ANS\n int ans = 0, sm = 0;\n//TRAVERSE GIVEN ARRAY\n for(auto &x: gain){\n// ADD CURENT ELEMENT TO SUM\n sm += x;\n//STORING MAX AMONG SUM IN ANS\n ans = max(ans,sm);\n }\n// RETURN ANS\n return ans;\n }\n};\n```
| 3 | 0 |
['Math', 'Prefix Sum', 'C++']
| 0 |
find-the-highest-altitude
|
Very Easy Image Explanation Prefix sum type approach||C++,Java
|
very-easy-image-explanation-prefix-sum-t-wygp
|
Intuition and Approach\nIn the given question we have the difference between two heights which is i->i+1 and hence they are depending on their original height
|
Rarma
|
NORMAL
|
2023-06-19T04:02:55.845859+00:00
|
2023-06-19T04:17:42.394143+00:00
| 1,076 | false |
# Intuition and Approach\nIn the given question we have the difference between two heights which is `i->i+1` and hence they are depending on their original height and not on the net height which is given,\n****solving first test case given in question***\n\nSo as to find the original height values we will do the following and understand what is going on behind the question,\n*(g represent gain)*\n\nConsidering above you can get to know that we can easily compute the altitude values one by one,\n```\ni.e. altitude[0]=0 ;given in question\n g[1]-0=-5 -> altitude[1]=g[1]=-5+0=-5\n g[2]-(-5)=1 -> altitude[2]=g[2]=1-5=-4\n g[3]-(-4)=5 -> altitude[3]=g[3]=5-4=1\n g[4]-(1)=0 -> altitude[4]=g[4]=0-1=-1\n```\n\nSo, we have to find maximum of all altitudes which we can do by using inbuilt function *max_element\n\n**(Get the Question? Upvote:comment)**\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```C++ []\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int ans=INT_MIN;\n vector<int> altitude;\n\n //pushing the starting altiude value\n altitude.push_back(0);\n\n for(int i=1;i<=gain.size();i++){\n altitude.push_back(gain[i-1]+altitude[i-1]);\n }\n\n //check the altitude vector by uncommenting below\n // for(auto val:altitude){\n // cout<<val <<" ";\n // }\n\n ans=*max_element(altitude.begin(),altitude.end());\n\n return ans;\n }\n};\n```\n```Java []\nclass Solution {\n public int largestAltitude(int[] gain) {\n int ans = Integer.MIN_VALUE;\n List<Integer> altitude = new ArrayList<>();\n\n // Pushing the starting altitude value\n altitude.add(0);\n\n for (int i = 1; i <= gain.length; i++) {\n altitude.add(gain[i - 1] + altitude.get(i - 1));\n }\n\n ans = Collections.max(altitude);\n\n return ans;\n }\n}\n\n```\n```python []\nfrom typing import List\n\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n ans = float(\'-inf\')\n altitude = [0]\n\n # Pushing the starting altitude value\n altitude.append(0)\n\n for i in range(1, len(gain) + 1):\n altitude.append(gain[i - 1] + altitude[i - 1]) \n\n ans = max(altitude)\n\n return ans\n\n```\n\n
| 3 | 0 |
['Array', 'Math', 'Prefix Sum', 'C++']
| 2 |
find-the-highest-altitude
|
JAVA Solution || 0 MS
|
java-solution-0-ms-by-sanjay1305-j3hh
|
\nclass Solution {\n public int largestAltitude(int[] gain) {\n int max=0;\n int sum=0;\n for(int i=0;i<gain.length;i++){\n sum+=gain[i];\n
|
sanjay1305
|
NORMAL
|
2023-06-19T03:33:28.079559+00:00
|
2023-06-19T03:33:28.079583+00:00
| 283 | false |
```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int max=0;\n int sum=0;\n for(int i=0;i<gain.length;i++){\n sum+=gain[i];\n max=Math.max(max,sum);\n }\n return max;\n }\n}\n```
| 3 | 0 |
['Java']
| 1 |
find-the-highest-altitude
|
[ C++ ] [ Easy Solution ] [ Maximum Prefix Sum Value ]
|
c-easy-solution-maximum-prefix-sum-value-lpuy
|
Code\n\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int sum = 0;\n int ans = 0;\n for(int i = 0; i < gain.siz
|
Sosuke23
|
NORMAL
|
2023-06-19T03:14:29.943033+00:00
|
2023-06-19T03:14:29.943070+00:00
| 253 | false |
# Code\n```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int sum = 0;\n int ans = 0;\n for(int i = 0; i < gain.size(); i++) {\n sum += gain[i];\n ans = max(ans, sum);\n }\n return ans;\n }\n};\n```
| 3 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
Python3 Solution
|
python3-solution-by-motaharozzaman1996-bjdh
|
\n\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n curr=0\n highest=0\n for x in gain:\n curr+=x\n
|
Motaharozzaman1996
|
NORMAL
|
2023-06-19T00:22:12.869426+00:00
|
2023-06-19T00:22:12.869444+00:00
| 1,923 | false |
\n```\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n curr=0\n highest=0\n for x in gain:\n curr+=x\n highest=max(curr,highest)\n\n return highest \n```
| 3 | 0 |
['Python', 'Python3']
| 0 |
find-the-highest-altitude
|
c++ solution
|
c-solution-by-suhani_29-fe2x
|
\n\n# Approach\n Describe your approach to solving the problem. \nInitialize a vector of length n+1 with all values 0. This vector denotes all the altitudes. \n
|
suhani_29
|
NORMAL
|
2023-04-29T18:20:53.348184+00:00
|
2023-04-29T18:20:53.348231+00:00
| 105 | false |
\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInitialize a vector of length n+1 with all values 0. This vector denotes all the altitudes. \nCalculate the net gain in altitudes at all points and return the max value from it.\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```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int n = gain.size();\n vector<int> altitude(n+1, 0);\n for(int i = 0; i<gain.size(); i++){\n altitude[i+1] = altitude[i] + gain[i];\n }\n return *max_element(altitude.begin(), altitude.end());\n }\n};\n```
| 3 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
0 ms RunTime, Naive Approach
|
0-ms-runtime-naive-approach-by-sahilaror-q0g9
|
Do hit Upvote if you like the solution.\n\n# Code\n\nclass Solution {\n public int largestAltitude(int[] gain) {\n int len = gain.length;\n int
|
sahilaroraesl
|
NORMAL
|
2023-03-25T08:02:08.178450+00:00
|
2023-03-25T08:02:08.178488+00:00
| 730 | false |
Do hit Upvote if you like the solution.\n\n# Code\n```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int len = gain.length;\n int[] res = new int[len+1]; \n int sum = 0;\n int a = Integer.MIN_VALUE;\n res[0] = 0;\n for (int i = 1; i < len+1; i++) {\n sum += gain[i-1];\n res[i] = sum;\n }\n for (int i = 0; i < res.length; i++) {\n if(a < res[i]) {\n a = res[i];\n }\n }\n return a;\n }\n}\n```
| 3 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
Java easy soln ( 0 ms runtime)
|
java-easy-soln-0-ms-runtime-by-akg56-lxmw
|
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
|
akg56
|
NORMAL
|
2022-12-14T02:00:35.555755+00:00
|
2022-12-14T02:00:35.555793+00:00
| 798 | 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<!-- 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```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int max=0;\n int current=0;\n for(int i=0;i<gain.length;i++){\n current+=gain[i];\n max=Math.max(current,max);\n }\n return max;\n \n }\n}\n```
| 3 | 0 |
['Java']
| 1 |
find-the-highest-altitude
|
Python3 ✅☑✅ || Faster than 86% || Memory beats 96.48%
|
python3-faster-than-86-memory-beats-9648-l3ok
|
Code\n\nclass Solution:\n def loadAltitude(self, gain):\n altitude = [0]\n for i in range(len(gain)+1):\n altitude.append(sum(gain[:
|
Little_Tiub
|
NORMAL
|
2022-12-04T00:34:35.775051+00:00
|
2022-12-04T00:34:35.775085+00:00
| 715 | false |
# Code\n```\nclass Solution:\n def loadAltitude(self, gain):\n altitude = [0]\n for i in range(len(gain)+1):\n altitude.append(sum(gain[:i]))\n return altitude\n def largestAltitude(self, gain: List[int]) -> int:\n altitude = self.loadAltitude(gain)\n return max(altitude)\n```\n\n\n\n
| 3 | 0 |
['Python3']
| 1 |
find-the-highest-altitude
|
Java solution mera bhi dekho
|
java-solution-mera-bhi-dekho-by-kcoder_0-hybn
|
\nclass Solution {\n public int largestAltitude(int[] gain) {\n \n int []result = new int[gain.length+1];\n result[0]=0;\n for(in
|
Kcoder_01
|
NORMAL
|
2022-10-20T16:07:59.998120+00:00
|
2022-10-20T16:07:59.998153+00:00
| 476 | false |
```\nclass Solution {\n public int largestAltitude(int[] gain) {\n \n int []result = new int[gain.length+1];\n result[0]=0;\n for(int i=1;i<=gain.length;i++){\n result[i]=result[i-1]+gain[i-1];\n }\n \n int max = result[0];\n for(int i=0;i<result.length;i++){\n if(max<result[i]){\n max = result[i];\n }\n }\n return max; \n \n }\n}\n```
| 3 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
Java || Easy || 0 ms Solution || Beats 100%
|
java-easy-0-ms-solution-beats-100-by-sak-6w1n
|
`\n \xA0 class Solution {\n public int largestAltitude(int[] gain) {\n int alt[]=new int[gain.length+1];\n alt[0]= 0;\n int max=alt[0];
|
sakshi_sinha_
|
NORMAL
|
2022-10-06T06:21:58.275758+00:00
|
2022-10-06T06:21:58.275798+00:00
| 536 | false |
````\n \xA0 class Solution {\n public int largestAltitude(int[] gain) {\n int alt[]=new int[gain.length+1];\n alt[0]= 0;\n int max=alt[0];\n for(int i=1;i<alt.length;i++){\n alt[i]=alt[i-1]+gain[i-1]; \n if(alt[i]>max)\n max=alt[i];\n } \n return(max);\n }\n}```
| 3 | 0 |
[]
| 1 |
find-the-highest-altitude
|
100% faster solution | Java
|
100-faster-solution-java-by-riyagarwal23-2l4j
|
Runtime: 0 ms, faster than 100.00% of Java online submissions...\n\nMaintain array for altitude.\nFind max altitiude in the same iteration as updating altitiude
|
riyagarwal23
|
NORMAL
|
2022-09-20T13:44:31.581421+00:00
|
2022-09-20T13:44:31.581464+00:00
| 215 | false |
**Runtime: 0 ms, faster than 100.00% of Java online submissions...**\n\n*Maintain array for altitude.\nFind max altitiude in the same iteration as updating altitiude array elements.*\n\n```class Solution {\n public int largestAltitude(int[] gain) {\n \n int n = gain.length+1;\n \n int[] alt = new int[n];\n \n alt[0] = 0;\n int ans = 0;\n int k = 1; //index for altitude array\n int maxx = 0;\n \n for(int i = 0; i<n-1; i++){\n \n alt[k] = alt[k-1] + gain[i];\n maxx = Math.max(maxx, alt[k]);\n k++;\n }\n \n return maxx;\n }\n}
| 3 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
Java Easy solution with (n) 0ms
|
java-easy-solution-with-n-0ms-by-deleted-7l7p
|
\nclass Solution {\n public int largestAltitude(int[] gain) {\n int highestPoint = 0;\n\n int point = 0;\n\n for (int i = 0; i < gain.le
|
deleted_user
|
NORMAL
|
2022-08-30T13:32:31.155674+00:00
|
2022-08-30T13:34:12.012782+00:00
| 392 | false |
```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int highestPoint = 0;\n\n int point = 0;\n\n for (int i = 0; i < gain.length; i++) {\n point += gain[i];\n\n highestPoint = Math.max(highestPoint, point);\n }\n return highestPoint;\n }\n}\n```
| 3 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
Runtime: 0 ms, faster than 100.00% of Java online submissions
|
runtime-0-ms-faster-than-10000-of-java-o-j0al
|
\n/** 1732. Find the Highest Altitude **/\n\nclass Solution {\n public int largestAltitude(int[] gain) {\n int n = gain.length;\n \n int max =
|
swangi_kumari
|
NORMAL
|
2022-08-28T07:17:58.617316+00:00
|
2022-08-28T07:17:58.617353+00:00
| 237 | false |
```\n/** 1732. Find the Highest Altitude **/\n\nclass Solution {\n public int largestAltitude(int[] gain) {\n int n = gain.length;\n \n int max = Integer.MIN_VALUE;\n int ans = 0;\n int[] arr = new int[n+1];\n arr[0]=0;\n for(int i = 1; i<n+1; i++){\n arr[i]=gain[i-1];\n }\n \n for(int i = 0; i< n+1; i++){\n \n ans += arr[i];\n if(ans>max){\n max = ans;\n }\n }\n return max;\n }\n}\n\n```\n
| 3 | 0 |
['Array', 'Prefix Sum', 'Java']
| 0 |
find-the-highest-altitude
|
C# Easy solution without using built-in functions(Easy-Understanding)
|
c-easy-solution-without-using-built-in-f-cqmr
|
\npublic class Solution {\n public int LargestAltitude(int[] gain) {\n int highestAltitude = 0, temp = 0;\n\n for (int i = 0; i < gain.Lengt
|
SaleemKassab
|
NORMAL
|
2022-06-11T10:21:56.514887+00:00
|
2022-06-11T10:21:56.514937+00:00
| 375 | false |
```\npublic class Solution {\n public int LargestAltitude(int[] gain) {\n int highestAltitude = 0, temp = 0;\n\n for (int i = 0; i < gain.Length; i++)\n {\n temp += gain[i];\n if (temp > highestAltitude)\n highestAltitude = temp;\n }\n return highestAltitude; \n }\n}\n```
| 3 | 0 |
['C', 'C#']
| 0 |
find-the-highest-altitude
|
Easy C++ Solution
|
easy-c-solution-by-priyesh_raj_singh-kxex
|
\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int sum = 0 ;\n vector<int> v;\n for(int i = 0 ; i<gain.size()
|
priyesh_raj_singh
|
NORMAL
|
2022-01-25T20:47:34.443941+00:00
|
2022-01-25T20:47:34.443984+00:00
| 122 | false |
```\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int sum = 0 ;\n vector<int> v;\n for(int i = 0 ; i<gain.size() ; i++){\n sum+=gain[i];\n v.push_back(sum);\n }\n int x = *max_element(v.begin() , v.end());\n if(x>0)\n return x;\n return 0;\n }\n};\n```
| 3 | 1 |
['C']
| 0 |
find-the-highest-altitude
|
100% faster solution
|
100-faster-solution-by-aayush0307-3e89
|
\nclass Solution {\n public int largestAltitude(int[] gain) {\n int sum=0; \n int max=0;\n for(int i=0;i<gain.length;i++){\n s
|
aayush0307
|
NORMAL
|
2022-01-18T20:30:55.655442+00:00
|
2022-01-18T20:30:55.655476+00:00
| 258 | false |
```\nclass Solution {\n public int largestAltitude(int[] gain) {\n int sum=0; \n int max=0;\n for(int i=0;i<gain.length;i++){\n sum=sum + gain[i]; \n if(sum>max)\n max=sum;\n }\n return max;\n }\n}\n\n\n```
| 3 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
Swift | 1732. Find the Highest Altitude | One-liner
|
swift-1732-find-the-highest-altitude-one-p5be
|
Using a tuple to keep track of the (height, maxSoFar) and reducing over the gains:\nswift \nfunc largestAltitude(_ gain: [Int]) -> Int {\n gain.reduce((0,0))
|
iosdevzone
|
NORMAL
|
2021-10-21T16:24:32.504744+00:00
|
2021-10-21T16:24:32.504770+00:00
| 167 | false |
Using a tuple to keep track of the (height, maxSoFar) and reducing over the gains:\n```swift \nfunc largestAltitude(_ gain: [Int]) -> Int {\n gain.reduce((0,0)) { ($0.0+$1, max($0.1, $0.0+$1)) }.1\n}\n```
| 3 | 0 |
['Swift']
| 1 |
find-the-highest-altitude
|
Golang-solution
|
golang-solution-by-engineerkoder-p3xl
|
\nfunc largestAltitude(gain []int) int {\n\tvar max, alt int\n\tfor _, g := range gain {\n\t\talt += g\n\t\tif alt > max {\n\t\t\tmax = alt\n\t\t}\n\t}\n\tretur
|
engineerkoder
|
NORMAL
|
2021-02-22T19:31:50.446891+00:00
|
2021-02-22T19:31:50.446934+00:00
| 248 | false |
```\nfunc largestAltitude(gain []int) int {\n\tvar max, alt int\n\tfor _, g := range gain {\n\t\talt += g\n\t\tif alt > max {\n\t\t\tmax = alt\n\t\t}\n\t}\n\treturn max\n}\n```
| 3 | 0 |
['Go']
| 0 |
find-the-highest-altitude
|
JavaScript one line solution functional programming
|
javascript-one-line-solution-functional-n4wnv
|
\nvar largestAltitude = (gain, prev = 0) => gain.reduce((max, curr) => Math.max(max, prev += curr), 0);\n
|
ChaoWan_2021
|
NORMAL
|
2021-02-22T15:17:08.795168+00:00
|
2021-02-22T15:17:08.795216+00:00
| 429 | false |
```\nvar largestAltitude = (gain, prev = 0) => gain.reduce((max, curr) => Math.max(max, prev += curr), 0);\n```
| 3 | 1 |
['JavaScript']
| 0 |
find-the-highest-altitude
|
Simple trick!
|
simple-trick-by-1zkmhmzztg-ndip
|
IntuitionEach value in the gain array represents the net gain in altitude between two points. If we start at altitude 0, we can calculate the altitude after eac
|
1ZkMHmzzTG
|
NORMAL
|
2025-04-04T05:32:17.457334+00:00
|
2025-04-04T05:32:17.457334+00:00
| 175 | false |
# Intuition
Each value in the gain array represents the net gain in altitude between two points. If we start at altitude 0, we can calculate the altitude after each gain by summing it with the previous altitude.
We need to find the maximum altitude reached during this journey.
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
1.Initialize a list n with the starting altitude 0.
2.Loop through each gain[i], and compute the current altitude by adding it to the last value in n.
3.Append this new altitude to n.
4.Finally, return the maximum value in n.
<!-- 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 largestAltitude(self, gain: List[int]) -> int:
n=[0]
for i in range(len(gain)):
n.append(n[-1]+gain[i])
return max(n)
```
| 2 | 0 |
['Python3']
| 1 |
find-the-highest-altitude
|
🚴 **BEATS 100%** Prefix sum solution with explanation
|
beats-100-prefix-sum-solution-with-expla-cz02
|
🚴 Find the Highest Altitude During a Bike Trip🧠 IntuitionThe biker starts at altitude 0. As they travel through different segments, their altitude increases or
|
bob_dylan
|
NORMAL
|
2025-04-02T08:37:15.633406+00:00
|
2025-04-02T08:38:49.404902+00:00
| 198 | false |
# 🚴 **Find the Highest Altitude During a Bike Trip**
## 🧠 **Intuition**
The biker starts at altitude `0`. As they travel through different segments, their altitude increases or decreases based on the values in the `gain` array. Our goal is to find the highest altitude reached during the trip.
The altitude at each step can be computed as:
\[
\text{altitude}[i] = \text{altitude}[i-1] + \text{gain}[i-1]
\]
Instead of storing all altitude values, we only need to track the **current altitude** and **maximum altitude reached**.
---
## 🔍 **Approach**
1. **Initialize**:
- `maxGain = 0` (tracks highest altitude reached)
- `totalGain = 0` (tracks current altitude)
2. **Iterate through `gain[]`**:
- Add `gain[i]` to `totalGain` to update current altitude.
- Update `maxGain` if `totalGain` is higher than the previous maximum.
3. **Return `maxGain`**.
This approach ensures we compute the answer in a **single pass** without extra storage.
---
## ⏳ **Complexity Analysis**
- **Time Complexity:** \( O(n) \) (We iterate through the `gain` array once)
- **Space Complexity:** \( O(1) \) (Only a few integer variables are used)
---
## 📜 **C++ Solution**
```cpp []
class Solution {
public:
int largestAltitude(vector<int>& gain) {
int maxGain = 0;
int totalGain = 0;
for(int i = 0; i < gain.size(); i++) {
totalGain += gain[i];
maxGain = max(maxGain, totalGain);
}
return maxGain;
}
};
```
```java []
class Solution {
public int largestAltitude(int[] gain) {
int maxGain = 0;
int totalGain = 0;
for (int i = 0; i < gain.length; i++) {
totalGain += gain[i];
maxGain = Math.max(maxGain, totalGain);
}
return maxGain;
}
}
```
```python []
class Solution:
def largestAltitude(self, gain):
maxGain = 0
totalGain = 0
for g in gain:
totalGain += g
maxGain = max(maxGain, totalGain)
return maxGain
```
```javascript []
var largestAltitude = function(gain) {
let maxGain = 0, totalGain = 0;
for (let g of gain) {
totalGain += g;
maxGain = Math.max(maxGain, totalGain);
}
return maxGain;
};
```
## 📊 Example Walkthrough
### Example 1
Input:
```plaintext
gain = [-5, 1, 5, 0, -7]
```
Altitude Calculation
```
Step Gain Altitude
Start — 0
+(-5) -5 0 → -5
+1 -4 -5 → -4
+5 1 -4 → 1 ✅ (New max)
+0 1 1 → 1
-7 -6 1 → -6
```
```
Output: 1
```
### Example 2
Input:
```plaintext
gain = [-4, -3, -2, -1, 4, 3, 2]
```
Altitude Calculation
```
Step Gain Altitude
Start — 0
-4 -4 0 → -4
-3 -7 -4 → -7
-2 -9 -7 → -9
-1 -10 -9 → -10
+4 -6 -10 → -6
+3 -3 -6 → -3
+2 -1 -3 → -1
```
```
Output: 0
```
# ✅ Key Takeaways
1. Efficient computation with just one loop.
2. No extra space used (constant space complexity).
3. Iterate once and track the highest altitude in real-time.
🚀 Optimized for large inputs!
| 2 | 0 |
['Array', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3', 'JavaScript']
| 0 |
find-the-highest-altitude
|
0 ms solution beats 100%.
|
0-ms-solution-beats-100-by-anshgupta1234-fetm
|
IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(1)
Code
|
anshgupta1234
|
NORMAL
|
2025-02-02T09:35:51.432834+00:00
|
2025-02-02T09:35:51.432834+00:00
| 124 | 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(1)
# Code
```java []
class Solution {
public int largestAltitude(int[] gain) {
int maxAltitude = 0;
int currentAltitude = 0;
for (int i = 0; i < gain.length; i++) {
currentAltitude = currentAltitude + gain[i];
if (currentAltitude > maxAltitude) {
maxAltitude = currentAltitude;
}
}
return maxAltitude;
}
}
```
| 2 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
Easy Solution Using Java
|
easy-solution-using-java-by-prateeksutha-kupt
|
IntuitionApproachComplexity
Time complexity : O(n)
Space complexity : O(1)
Code
|
prateeksuthar2hzar
|
NORMAL
|
2025-02-01T07:32:26.707891+00:00
|
2025-02-01T07:32:26.707891+00:00
| 99 | 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(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int largestAltitude(int[] gain) {
int value = 0;
int max = 0;
for(int i = 0; i < gain.length; i++){
value = value + (gain[i]);
max = Math.max(value, max);
}
return max;
}
}
```
| 2 | 0 |
['Java']
| 0 |
find-the-highest-altitude
|
JavaScript - One-liner
|
javascript-one-liner-by-ov8cfej3su-o2pk
| null |
Ov8CfeJ3Su
|
NORMAL
|
2025-01-19T00:46:47.795963+00:00
|
2025-01-19T00:47:03.475068+00:00
| 243 | false |
```javascript []
var largestAltitude = function(gain) {
return gain.reduce((a, g) => [c = a[0] + g, c > a[1] ? c : a[1]], [0, 0])[1];
};
```
| 2 | 0 |
['JavaScript']
| 2 |
find-the-highest-altitude
|
Solution in C
|
solution-in-c-by-akcyyix0sj-ihfq
|
Code
|
vickyy234
|
NORMAL
|
2025-01-05T09:26:14.572673+00:00
|
2025-01-05T09:26:14.572673+00:00
| 101 | false |
# Code
```c []
int largestAltitude(int* gain, int gainSize) {
int ans = 0, temp = 0;
for (int i = 0; i < gainSize; i++) {
temp += gain[i];
if (temp >= ans)
ans = temp;
}
return ans;
}
```
| 2 | 0 |
['C']
| 0 |
find-the-highest-altitude
|
Very Simple C solution
|
very-simple-c-solution-by-aswath_1192-oxnj
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
Aswath_1192
|
NORMAL
|
2025-01-05T06:14:26.003418+00:00
|
2025-01-05T06:14:26.003418+00:00
| 57 | 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
```c []
int largestAltitude(int* gain, int size) {
int high=0;
int cur=0;
for(int i=0;i<size;i++){
cur+=gain[i];
if(cur>high)
high=cur;
}
return high;
}
```
| 2 | 0 |
['C']
| 0 |
find-the-highest-altitude
|
Highest Altitude using C++
|
highest-altitude-using-c-by-parodkarvais-gw53
|
Notes:
Altitude: It is generally the height achieved by an object from the reference level.
Gain: It is the altitude achieved during the journey, it can be posi
|
parodkarvaishnavi
|
NORMAL
|
2025-01-04T14:18:19.330415+00:00
|
2025-01-04T14:18:19.330415+00:00
| 180 | false |
# Notes:
1. Altitude: It is generally the height achieved by an object from the reference level.
2. Gain: It is the altitude achieved during the journey, it can be positive or negative.
3. Negative gain: when the biker moves downhill.
4. positive gain: when the biker moves uphill.
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
In the problem we are given an integer array gain of length n where gain[i] is the net gain in altitude between points i and i+1. we are expected to return the highest altitude of a point.
- we initially start from point 0(sea level).
- We are given an array gain which consists of altitudes achieved by biker, it can be positive or negative.
- We need to return the highest altitude of the biker in the complete journey.
- To find the max altitude we can add the altitude at the previous step plus the gain at the current step. Starting from 0 and adding the gain at each step.
- After each addition, we will update the max altitude.
# Pattern Used:
is Prefix Sum pattern in DSA.
# Approach
<!-- Describe your approach to solving the problem. -->
- 1st we initialise currentAltitude = 0 and maxAltitude = 0
- Then we iterate over the gains in altitude, add the gain[i] to currentAltitude
- next we update the maxAltitude.
- finally we return the maxAltitude which is been asked to find in the problem.
# Complexity
- Time complexity: It is O(n) where n is the size of the gain array, as we iterate over the elements in gain only once.
- Space complexity: It is O(1).
- Here gain array is provided as input and doesn't contribute to additional space complexity.
- currentAltitude, maxAltitude These take a fixed amount of space, regardless of the size of the input.
- Therefore overall space complexity is O(1).
# Code
```cpp []
class Solution {
public:
int largestAltitude(vector<int>& gain) {
int currentAltitude = 0;
int maxAltitude = 0;
for (int i = 0; i < gain.size(); i++) {
currentAltitude= currentAltitude + gain[i];
maxAltitude = max(maxAltitude, currentAltitude);
}
return maxAltitude;
}
};
```
| 2 | 0 |
['C++']
| 0 |
find-the-highest-altitude
|
Beats 100% using C++
|
beats-100-using-c-by-hriii11-s1ev
|
Intuition\nCumulative Calculation: The algorithm effectively keeps a running total of the altitude changes. By continuously adding each gain to the current alti
|
Hriii11
|
NORMAL
|
2024-10-09T14:20:24.165293+00:00
|
2024-10-09T14:20:24.165334+00:00
| 6 | false |
# Intuition\nCumulative Calculation: The algorithm effectively keeps a running total of the altitude changes. By continuously adding each gain to the current altitude, you can easily track how the altitude evolves over time.\nDynamic Tracking: By updating max whenever curr exceeds it, you ensure that you capture the highest point reached at any stage in the journey.\nLinear Time Complexity: The solution runs in O(n) time, where n is the length of the gain array. This is efficient since it only requires a single pass through the array\n\n# Approach\nExample Walkthrough\nConsider the gain array: [-5, 1, 5, 0, -7].\n\nStarting Altitude: 0\nChanges:\nAfter -5: curr = 0 - 5 = -5 \u2192 max = 0\nAfter 1: curr = -5 + 1 = -4 \u2192 max = 0\nAfter 5: curr = -4 + 5 = 1 \u2192 max = 1\nAfter 0: curr = 1 + 0 = 1 \u2192 max = 1\nAfter -7: curr = 1 - 7 = -6 \u2192 max = 1\nThe maximum altitude reached is 1.\n\n# Complexity\n- Time complexity: $$O(n)$$ \n\n- Space complexity: $$O(1)$$\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int largestAltitude(vector<int>& gain) {\n int max=0;\n int curr=0;\n for(int i =0 ; i<gain.size(); i++){\n curr+=gain[i];\n max=std::max(curr,max);\n }\n return max;\n }\n};\n```
| 2 | 0 |
['C++']
| 1 |
find-the-highest-altitude
|
Find the Highest Altitude
|
find-the-highest-altitude-by-tejdekiwadi-gqyh
|
Intuition\nThe problem is about finding the largest altitude reached after a series of altitude gains. Initially, the altitude is 0, and the gains are cumulativ
|
tejdekiwadiya
|
NORMAL
|
2024-09-11T15:01:02.022286+00:00
|
2024-09-11T15:01:02.022304+00:00
| 136 | false |
# Intuition\nThe problem is about finding the largest altitude reached after a series of altitude gains. Initially, the altitude is 0, and the gains are cumulative changes to the altitude. To find the highest altitude, we need to keep track of the cumulative altitude after each gain and record the maximum value.\n\n# Approach\n1. **Initialize Variables**: \n - Start with `minAltitude = 0` since the initial altitude is 0, and we are interested in the highest point reached.\n - A variable `sum` is used to keep track of the cumulative altitude, starting from 0.\n\n2. **Iterate Through the Gain Array**:\n - For each value in the `gain` array, add it to `sum` (which represents the current altitude).\n - After updating the altitude, compare it to `minAltitude`. If the current altitude is greater than `minAltitude`, update `minAltitude`.\n\n3. **Return the Result**:\n - At the end of the loop, return the highest altitude recorded (`minAltitude`).\n\n# Complexity\n- **Time Complexity**: \n - The algorithm iterates through the `gain` array exactly once. Therefore, the time complexity is **O(n)**, where **n** is the length of the `gain` array.\n\n- **Space Complexity**: \n - The algorithm uses a constant amount of extra space (`sum` and `minAltitude` variables). Thus, the space complexity is **O(1)**.\n\n# Example\nFor the input `gain = [-5, 1, 5, 0, -7]`:\n\n- Initially, `minAltitude = 0` and `sum = 0`.\n- Iterating through `gain`:\n - After adding `-5`: `sum = -5`, `minAltitude = max(0, -5) = 0`.\n - After adding `1`: `sum = -4`, `minAltitude = max(0, -4) = 0`.\n - After adding `5`: `sum = 1`, `minAltitude = max(0, 1) = 1`.\n - After adding `0`: `sum = 1`, `minAltitude = max(1, 1) = 1`.\n - After adding `-7`: `sum = -6`, `minAltitude = max(1, -6) = 1`.\n\nThus, the largest altitude reached is 1.\n\n# Code\n```java []\nclass Solution {\n public int largestAltitude(int[] gain) {\n int minAltitude = 0;\n int sum = 0;\n for (int i = 0; i < gain.length; i++) {\n sum = sum + gain[i];\n minAltitude = Math.max(sum, minAltitude);\n }\n return minAltitude;\n }\n}\n```
| 2 | 0 |
['Array', 'Prefix Sum', 'Java']
| 0 |
find-the-highest-altitude
|
largestAltitude solution with easy approach
|
largestaltitude-solution-with-easy-appro-hx28
|
Intuition\nTo find the highest altitude, we need to calculate the cumulative sum of the altitude gains at each step. Starting from the initial altitude of 0, we
|
Manas_Mittal_
|
NORMAL
|
2024-08-25T18:01:10.654327+00:00
|
2024-08-25T18:01:10.654364+00:00
| 1,049 | false |
# Intuition\nTo find the highest altitude, we need to calculate the cumulative sum of the altitude gains at each step. Starting from the initial altitude of 0, we keep adding the gain at each step to the current altitude. The highest altitude reached during this process is the answer we are looking for.\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nInitialize two variables: maxAltitude and currentAltitude to keep track of the maximum altitude encountered so far and the current altitude, respectively. Set both variables to 0.\nIterate over the gain vector, starting from the first element.\nAt each step, update the currentAltitude by adding the gain of the current step.\nCompare the currentAltitude with the maxAltitude. If it is greater, update maxAltitude to the new value.\nRepeat steps 3-4 for all elements in the gain vector.\nOnce the iteration is complete, maxAltitude will hold the highest altitude reached.\nReturn the value of maxAltitude as the result.\n\nThis approach ensures that we keep track of the maximum altitude encountered during the traversal and avoids unnecessary modifications to the input vector. By iterating over the gain vector only once, the solution is efficient and provides the correct result.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n**For python code**\n- Time complexity: O(n)\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**For java code**\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. $$O(n)$$ -->\nWe can do both either in same way but i used different approaches.\n\n# Code\n```python []\nclass Solution:\n def largestAltitude(self, gain: List[int]) -> int:\n max_alt = 0\n curr = 0\n for g in gain:\n curr += g\n max_alt = max(max_alt,curr)\n return max_alt\n \n \n```\n```java []\nclass Solution {\n public int largestAltitude(int[] gain) {\n int[] brr = new int[gain.length + 1];\n brr[0] = 0;\n brr[1]= brr[0] + gain[0];\n for(int i =2 ; i< brr.length ; i++){\n brr[i] = brr[i-1] + gain[i-1];\n }\n int max = Integer.MIN_VALUE;\n for(int i =0 ; i< brr.length;i++){\n max = Math.max(brr[i],max);\n }\n return max;\n \n }\n}\n```
| 2 | 0 |
['Array', 'Prefix Sum', 'Java', 'Python3']
| 0 |
find-the-highest-altitude
|
100 % beat and simple solution
|
100-beat-and-simple-solution-by-nirajbit-zomv
|
Intuition\n Describe your first thoughts on how to solve this problem. \n1.Inplace prefix sum calculation\n2.Use Math.max function\n# Approach\n Describe your a
|
nirajBits
|
NORMAL
|
2024-08-24T02:02:33.212783+00:00
|
2024-08-24T02:08:04.229696+00:00
| 5 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1.Inplace prefix sum calculation\n2.Use Math.max function\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. for i=1\narr[1]=arr[1]+arr[0]\n2. for i=2\narr[2]=arr[2]+arr[1] //here arr[1] is itself changed in previous formula if u r confused then think about substitution\n1. for i=3\narr[3]=arr[2]+arr[1]\n4. illy for upto i=n-1(length of array)\n5. After that iterate over every elements of prefix sum array and using math.max function compare and store the max element in max variable\n6. At last return max variable\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n), here n is length of given array\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1) ,No extra space used (Inplace operation)\n\n\n# Code\n```java []\nclass Solution {\n public int largestAltitude(int[] gain) {\n int n=gain.length,max=0;\n for(int i=1;i<n;i++){\n gain[i]+=gain[i-1];\n }\n for(int i=0;i<n;i++){\n max=Math.max(max,gain[i]);\n }\n return max;\n }\n}\n```\n\n
| 2 | 0 |
['Java']
| 0 |
break-a-palindrome
|
[Java/C++/Python] Easy and Concise
|
javacpython-easy-and-concise-by-lee215-3an4
|
Check half of the string,\nreplace a non \'a\' character to \'a\'.\n\nIf only one character, return empty string.\nOtherwise repalce the last character to \'b\'
|
lee215
|
NORMAL
|
2020-01-25T16:22:10.425156+00:00
|
2020-01-25T16:58:32.114652+00:00
| 35,407 | false |
Check half of the string,\nreplace a non `\'a\'` character to `\'a\'`.\n\nIf only one character, return empty string.\nOtherwise repalce the last character to `\'b\'`\n<br>\n\n## **Complexity**\nTime `O(N)`\nSpace `O(N)`\n<br>\n\n**Java**\nFrom @kniffina\n```java\n public String breakPalindrome(String palindrome) {\n char[] s = palindrome.toCharArray();\n int n = s.length;\n\n for (int i = 0; i < n / 2; i++) {\n if (s[i] != \'a\') {\n s[i] = \'a\';\n return String.valueOf(s);\n }\n }\n s[n - 1] = \'b\'; //if all \'a\'\n return n < 2 ? "" : String.valueOf(s);\n }\n```\n**C++**\n```cpp\n string breakPalindrome(string S) {\n int n = S.size();\n for (int i = 0; i < n / 2; ++i) {\n if (S[i] != \'a\') {\n S[i] = \'a\';\n return S;\n }\n }\n S[n - 1] = \'b\';\n return n < 2 ? "" : S;\n }\n```\n**Python**\n```py\n def breakPalindrome(self, S):\n for i in xrange(len(S) / 2):\n if S[i] != \'a\':\n return S[:i] + \'a\' + S[i + 1:]\n return S[:-1] + \'b\' if S[:-1] else \'\'\n```\n\n
| 362 | 7 |
[]
| 36 |
break-a-palindrome
|
🔥 [LeetCode The Hard Way] 🔥 Explained Line By Line
|
leetcode-the-hard-way-explained-line-by-y0rv4
|
Please check out LeetCode The Hard Way for more solution explanations and tutorials. \nI\'ll explain my solution line by line daily and you can find the full li
|
__wkw__
|
NORMAL
|
2022-10-10T02:08:00.103427+00:00
|
2022-10-10T02:14:28.286285+00:00
| 5,089 | false |
Please check out [LeetCode The Hard Way](https://wingkwong.github.io/leetcode-the-hard-way/) for more solution explanations and tutorials. \nI\'ll explain my solution line by line daily and you can find the full list in my [Discord](https://discord.gg/Nqm4jJcyBf).\nIf you like it, please give a star, watch my [Github Repository](https://github.com/wingkwong/leetcode-the-hard-way) and upvote this post.\n\n---\n\n<iframe src="https://leetcode.com/playground/kbDRbvuH/shared" frameBorder="0" width="100%" height="500"></iframe>
| 65 | 0 |
['Greedy', 'C', 'Python', 'Java', 'Python3']
| 4 |
break-a-palindrome
|
[Python] short greedy O(n) solution, explained
|
python-short-greedy-on-solution-explaine-de1o
|
We need to replace one symbol and to get the smallest string as possible. First, we try to change elements on the smaller positions: imagine we have string xyzy
|
dbabichev
|
NORMAL
|
2021-09-23T07:32:54.943917+00:00
|
2021-09-23T07:32:54.943982+00:00
| 4,785 | false |
We need to replace one symbol and to get the smallest string as possible. First, we try to change elements on the smaller positions: imagine we have string `xyzyx` - where `x, y, z` can be any symbols. If `x` is not equal to `"a"`, then if we replace this symbol with `"a"` we will break palindrome and it will be as small as possible. However if `x = "a"`, it is not optimal to replace it to say `"b"`, we can only make our string bigger. So, we move to the next element `y` and so on. Notice that we can not change `z` in this case, because we will not break palindrome property.\n\nWhat happens, if we reached the last element and were not able to apply strategy above? Then we have string like this `aaaaaa..aaaaaa` or `aaaaaa...z...aaaaaa`, where `z` can be any symbol. In this case to get the smalles string as possible we need to replace last symbol to `"b"`. Also there is case when `n = 1`, and we can not break palindrome property so we return `""`.\n\n#### Complexity\nIt is `O(n)` for time and for space.\n\n#### Code\n```python\nclass Solution:\n def breakPalindrome(self, pal):\n n = len(pal)\n for i in range(n//2):\n if pal[i] != "a": return pal[:i] + "a" + pal[i+1:]\n return pal[:-1] + "b" if n > 1 else ""\n```\n\nIf you have any questoins, feel free to ask. If you like the solution and explanation, please **upvote!**
| 49 | 0 |
['Greedy']
| 4 |
break-a-palindrome
|
[Java] 0 ms Intuitive Code w/ Explanation
|
java-0-ms-intuitive-code-w-explanation-b-gdyf
|
1328. Break a Palindrome\n\nclass Solution {\n public String breakPalindrome(String palindrome) {\n if(palindrome.length()<=1)\n return "";\
|
allenjue
|
NORMAL
|
2020-11-18T06:07:12.386732+00:00
|
2020-11-24T04:01:51.060945+00:00
| 3,704 | false |
**1328. Break a Palindrome**\n```\nclass Solution {\n public String breakPalindrome(String palindrome) {\n if(palindrome.length()<=1)\n return "";\n char[] arr = palindrome.toCharArray();\n \n for(int i=0; i<arr.length/2;i++){ \n if(arr[i] != \'a\'){ // if not a then change it to be lexographically smallest\n arr[i] = \'a\';\n return String.valueOf(arr);\n }\n }\n // if we reach here, there are ONLY \'a\' in palindrome string, so we should change the last character to a b\n arr[arr.length-1] = \'b\';\n return String.valueOf(arr);\n }\n}\n```\n**PROBELM OVERVIEW**\nWe are given a palindrome and want to make it the lexicographically *smallest* non-palindrome\n\n**SOLUTION ASSESSMENT**\nThere are one of 2 states:\n* If the characer is not an \'a,\' change it to be an \'a\' to break the palindrome **and** make it lexiographically smaller\n* if all the characters are \'a\' change the last letter to be a \'b\' to break the palindrome, and since it\'s the last letter, increasing it to a \'b\' will **minimize** the increase in lexicographical order.\n\t\n**PROCESS**\nWe do this by converting the string into a char[] array. Note, we only need to iterate until arr.length/2 because it is guaranteed that anything afterwards is a repetition of the characters we have already encountered. Also, arr.length/2 avoids the error of changing the center character in an odd-lengthed string, which has no effect on it.\n\nEX: "aabaa" -> "aaaaa" is still a palindrome **|** therefore, arr.length/2, which is 2, will not reach the middle character. \nEX: "aaaaa" -> "aaaab" is the smallest. We need to change the **LAST non-a** because anything else will be larger ( "aaaba" > "aaaab" in the dictionary) \n\nHopefully this helps. :)
| 41 | 0 |
['Java']
| 3 |
break-a-palindrome
|
Break Palindrome [C++]
|
break-palindrome-c-by-ashokaks-29j0
|
Hint 1 : Length of string == 1, return ""\nH2: Palindrome string is lexicographical, so return first non \'a\' by \'a\' to break sequence.\n\n100% accpeted solu
|
ashokaks
|
NORMAL
|
2020-01-27T13:43:32.274821+00:00
|
2020-01-27T13:43:32.274853+00:00
| 4,140 | false |
Hint 1 : Length of string == 1, return ""\nH2: Palindrome string is lexicographical, so return first non \'a\' by \'a\' to break sequence.\n\n100% accpeted solution.\n\n```\nstring breakPalindrome(string p) {\n \n size_t sz = p.size();\n if (sz <= 1) return "";\n \n for (size_t i=0; i < sz/2; i++) {\n if (p[i] > \'a\') {\n p[i] = \'a\';\n return p;\n }\n }\n \n p[sz-1] = \'b\';\n return p;\n }\n```
| 29 | 0 |
[]
| 6 |
break-a-palindrome
|
Python Solution. Readable. simple code.
|
python-solution-readable-simple-code-by-oz9dy
|
\nclass Solution(object):\n def breakPalindrome(self, palindrome):\n """\n :type palindrome: str\n :rtype: str\n """ \n p1
|
mayk3r
|
NORMAL
|
2020-10-13T11:50:32.288001+00:00
|
2020-10-15T13:51:51.844205+00:00
| 3,342 | false |
```\nclass Solution(object):\n def breakPalindrome(self, palindrome):\n """\n :type palindrome: str\n :rtype: str\n """ \n p1 = 0\n p2 = len(palindrome)-1\n new_pal = list(palindrome)\n while p1 < p2:\n if new_pal[p1] != "a":\n new_pal[p1] = "a"\n return "".join(new_pal)\n p1+=1\n p2-=1\n if len(new_pal) == 1: return ""\n new_pal[-1] = "b"\n return "".join(new_pal)\n```\nAlright.\n**Explanation**\nFirst of all you should know how to find a palindrom.\n**Given** : TENET\n**Using**: Two pointers. One far right, the other far left.\n```\nT E N E T\n^ ^\nT == T\n\nT E N E T\n ^ ^\nE == E\n\nT E N E T\n ^^\nN == N\nYay! valid palindrome\n```\nYou can break a pallindrome at any of the point T=T; E=E; N=N.\n**Rules** : \n1. Only choose / change one element .\n2. Only change element at the first pointer. \n3. If the above (rule 2) is not possible, change the element last element.\n\nNow, given the following condition we can\'t just change an element to any letter :\n\n**Condition**\n```...string becomes the lexicographically smallest possible```\nLexicographical order : a < b < c < d < e ..... < z . \na is smallest, followed by b then c... z is largest. As a consequence :\n*Reason for rule 2* : \nT E N E T : (changing element at first pointer vesus element at second pointer)\n(A) E N E T < T E N E (A) -: (A) E N E T lexicographically smallest\nT (A) N E T < T E N (A) T -: T (A) N E T lexicographically smallest\n\n*Reason for rule 3 :*\nGiven : A A A A C , lexicographically smallest : A A A A B \nGiven : A A A A A , lexicographically smallest RESULT - A A A A B \n\nGenerally, think about how words are arranged in the dictionary.\n\nIntuitively to get the lexicographically smallest string possible you have two options, a or b.\n\n**Intuition**\nYou can only choose one letter a or b. And please,**Remember rule 1** . \nIf you find a non-\'a\' element change it to \'a\'. **Remember rule 2**\nIf every element is an "a", then change the last element to "b". **Remember rule 3**\n\n**Algorithm**\n* Iterate through the palindrome string,\n* If the element at the first pointer is "a" you shouldnt change that. Continue.\n* When the element at the first pointer is not "a", change it to "a" return the modified string\n* If no change has occured and the iteration ends,\nCheck: \n* If the string contains only one letter, no possible change can be made return empty string\n* Else change the last letter to b and return the modified string.\n
| 26 | 1 |
['Two Pointers', 'Python']
| 3 |
break-a-palindrome
|
Python with Regex, two candidates
|
python-with-regex-two-candidates-by-stef-w7ew
|
Check two candidates:\n1) Replace the first non-\'a\' with \'a\'.\n2) Replace the last letter with \'b\'.\n\ndef breakPalindrome(self, p):\n for s in re.sub(
|
stefanpochmann
|
NORMAL
|
2020-01-25T19:41:38.517525+00:00
|
2020-01-25T19:41:38.517578+00:00
| 998 | false |
Check two candidates:\n1) Replace the first non-`\'a\'` with `\'a\'`.\n2) Replace the last letter with `\'b\'`.\n```\ndef breakPalindrome(self, p):\n for s in re.sub(\'[^a]\', \'a\', p, 1), p[:-1] + \'b\':\n if s < s[::-1]:\n return s\n return \'\' \n```
| 20 | 15 |
[]
| 2 |
break-a-palindrome
|
✅ Break a Palindrome || W/ Explanation || C++ | Python | JAVA
|
break-a-palindrome-w-explanation-c-pytho-erd6
|
IDEA\n\nThere are one of 2 states:\n\n If the characer is not an a change it to be an a to break the palindrome and then make it lexiographically smaller.\n If
|
Maango16
|
NORMAL
|
2021-09-23T12:08:10.785529+00:00
|
2021-09-23T12:08:10.785574+00:00
| 1,738 | false |
**IDEA**\n\nThere are one of 2 states:\n\n* If the characer is not an `a` change it to be an `a` to break the palindrome and then make it lexiographically smaller.\n* If all the characters are `a` change the last letter to be a `b` to break the palindrome.\n\t* Since it\'s the last letter, increasing it to a `b` will minimize the increase in lexicographical order.\n\nIf the palindrome string has only one character, return an `empty string`.\n\n**TIME COMPLEXITY - O(N)**\n**SPACE COMPLEXITY - O(N)**\n\n**SOLUTION**\n`IN C++`\n```\nclass Solution {\npublic:\n string breakPalindrome(string pal) {\n int n = pal.length();\n for (int i = 0; i < n / 2; i++) {\n if (pal[i] != \'a\') \n {\n pal[i] = \'a\';\n return pal ;\n }\n }\n pal[n - 1] = \'b\' ;\n return n < 2 ? "" : pal ;\n }\n};\n```\n`IN PYTHON`\n```\nclass Solution:\n def breakPalindrome(self, pal):\n n = len(pal)\n for i in range(n//2):\n if pal[i] != "a": return pal[:i] + "a" + pal[i+1:]\n return pal[:-1] + "b" if n > 1 else ""\n```\n`IN JAVA`\n```\nclass Solution {\n public String breakPalindrome(String palindrome) {\n char[] s = palindrome.toCharArray();\n int n = s.length;\n\n for (int i = 0; i < n / 2; i++) {\n if (s[i] != \'a\') \n {\n s[i] = \'a\';\n return String.valueOf(s);\n }\n }\n s[n - 1] = \'b\';\n return n < 2 ? "" : String.valueOf(s);\n }\n}\n```
| 19 | 0 |
[]
| 2 |
break-a-palindrome
|
BREAK A PALINDROME || EASY TO UNDERSTAND || WELL COMMENTED || SIMPLE SOLUTION
|
break-a-palindrome-easy-to-understand-we-uxpw
|
\n/*\nfor making it lexographically smaller non-palindromic after replacing a character , we will simply look for a character which is not \'a\' from i=0 to i<n
|
_chintu_bhai
|
NORMAL
|
2022-10-10T06:52:21.590276+00:00
|
2022-10-10T06:52:21.590318+00:00
| 1,943 | false |
```\n/*\nfor making it lexographically smaller non-palindromic after replacing a character , we will simply look for a character which is not \'a\' from i=0 to i<n/2 .\n1.If such character exists , we will simply put palindrome[i]=\'a\' and exists out of the loop while making flag=1 (there is such character exists)\n2. If such character doesn\'t exist then flag will remain equal to 0 .In this case , we will simply put palinrome[n-1]=\'a\';\n*/\nclass Solution {\npublic:\n string breakPalindrome(string palindrome) {\n int n=palindrome.size();//calculating size\n//there is no sense of performing if size<2 as there is no way to make it non-palindrome\n if(n<2)\n {\n return "";\n }\n int flag=0;\n \n for(int i=0;i<n/2;i++)\n {\n if(palindrome[i]!=\'a\')\n {\n flag=1;\n palindrome[i]=\'a\';\n break;\n }\n }\n if(flag==0)\n {\n palindrome[n-1]=\'b\';\n }\n return palindrome;\n }\n};\n```\n**IF YOU FOUND THIS HELPFUL , PLEASE DO UPVOTE IT**
| 12 | 0 |
['String', 'C']
| 0 |
break-a-palindrome
|
JAVA || Easy Solution||100% Faster Code || Beginner Friendly
|
java-easy-solution100-faster-code-beginn-o3lp
|
\tPLEASE UPVOTE IF YOU LIKE.\n\nclass Solution {\n public String breakPalindrome(String palindrome) {\n if(palindrome.length() == 0 || palindrome.leng
|
shivrastogi
|
NORMAL
|
2022-10-10T03:38:55.784074+00:00
|
2022-10-10T03:41:30.376533+00:00
| 814 | false |
\tPLEASE UPVOTE IF YOU LIKE.\n```\nclass Solution {\n public String breakPalindrome(String palindrome) {\n if(palindrome.length() == 0 || palindrome.length() == 1){\n return "";\n }\n char[] ch = palindrome.toCharArray();\n for(int i = 0;i<ch.length/2;i++){\n if(ch[i]-\'a\' != 0){\n ch[i] = \'a\';\n return new String(ch);\n }\n \n }\n ch[ch.length-1] = \'b\';\n return new String(ch);\n }\n}\n```
| 11 | 0 |
['Java']
| 1 |
break-a-palindrome
|
Very simple python 🐍 solution O(N) time and space
|
very-simple-python-solution-on-time-and-asxa6
|
\nclass Solution:\n def breakPalindrome(self, palindrome: str) -> str:\n \n \n for i in range(len(palindrome)//2):\n if
|
injysarhan
|
NORMAL
|
2020-10-24T06:34:26.245404+00:00
|
2020-10-24T06:51:16.817542+00:00
| 2,155 | false |
```\nclass Solution:\n def breakPalindrome(self, palindrome: str) -> str:\n \n \n for i in range(len(palindrome)//2):\n if palindrome[i]!=\'a\':\n return palindrome[:i]+\'a\'+palindrome[i+1:]\n \n return palindrome[:-1]+\'b\' if len(palindrome)>1 else ""\n \n \n```
| 11 | 0 |
['Python', 'Python3']
| 1 |
break-a-palindrome
|
Java [0 ms]
|
java-0-ms-by-noah_21-oyij
|
\n\nclass Solution {\n public String breakPalindrome(String palindrome) {\n \n char[] ch = palindrome.toCharArray(); \n int n = ch.lengt
|
noah_21
|
NORMAL
|
2020-01-25T16:09:15.103515+00:00
|
2020-01-25T16:09:15.103571+00:00
| 1,425 | false |
\n```\nclass Solution {\n public String breakPalindrome(String palindrome) {\n \n char[] ch = palindrome.toCharArray(); \n int n = ch.length/2;\n \n for (int i = 0; i < n; i++) {\n \n if (ch[i] != \'a\') {\n ch[i] = \'a\';\n return String.valueOf(ch);\n }\n }\n \n if (palindrome.length() > 1) {\n ch[ch.length - 1] = \'b\';\n return String.valueOf(ch);\n }\n \n return "";\n \n }\n}\n```
| 10 | 0 |
[]
| 1 |
break-a-palindrome
|
Python 3 | Greedy one pass | Explanations
|
python-3-greedy-one-pass-explanations-by-vnzk
|
Explanation\n- If length of palindrome == 1, return \'\'\n- For even length string, if we found a char that is not a, replace it with a and return\n- For odd le
|
idontknoooo
|
NORMAL
|
2020-09-14T21:22:00.095777+00:00
|
2020-09-14T21:22:00.095839+00:00
| 1,841 | false |
### Explanation\n- If `length of palindrome == 1`, return `\'\'`\n- For even length string, if we found a char that is not `a`, replace it with `a` and return\n- For odd length string, if we find a char that is not `a` and it\'s not the middle of string, replace it with `a` and return\n- If all `a` in the string, replace the last char to `b`\n### Implementation\n```\nclass Solution:\n def breakPalindrome(self, palindrome: str) -> str:\n n = len(palindrome)\n if n == 1: return \'\'\n for i, c in enumerate(palindrome):\n if c != \'a\' and ((i != n // 2 and n % 2) or not n % 2): return palindrome[:i] + \'a\' + palindrome[i+1:] \n else: return palindrome[:-1] + \'b\'\n```
| 9 | 1 |
['Greedy', 'Python', 'Python3']
| 3 |
break-a-palindrome
|
Straightforward Python
|
straightforward-python-by-sushanthsamala-scow
|
\n\nclass Solution:\n def breakPalindrome(self, palindrome: str) -> str:\n n = len(palindrome)\n if n == 1: return \'\'\n for i, char in
|
sushanthsamala
|
NORMAL
|
2020-01-25T23:27:49.680733+00:00
|
2020-10-14T00:18:58.202545+00:00
| 1,373 | false |
\n```\nclass Solution:\n def breakPalindrome(self, palindrome: str) -> str:\n n = len(palindrome)\n if n == 1: return \'\'\n for i, char in enumerate(palindrome):\n if char != \'a\' and i != n//2:\n return palindrome[:i] + \'a\' + palindrome[i+1:]\n elif char == \'a\' and i == n - 1:\n return palindrome[:-1] + \'b\'\n```
| 8 | 0 |
[]
| 0 |
break-a-palindrome
|
python 3 || 8 lines, one-pass || T/S: 99% / 99%
|
python-3-8-lines-one-pass-ts-99-99-by-sp-yhcy
|
\nclass Solution:\n def breakPalindrome(self, p: str) -> str:\n \n n = len(p)\n if n < 2: return \'\'\n \n for i in range(
|
Spaulding_
|
NORMAL
|
2022-10-10T00:19:11.843288+00:00
|
2024-06-25T06:32:15.156478+00:00
| 468 | false |
```\nclass Solution:\n def breakPalindrome(self, p: str) -> str:\n \n n = len(p)\n if n < 2: return \'\'\n \n for i in range(n//2):\n \n if p[i] != \'a\':\n p = p[:i] + \'a\' + p[i+1:]\n break\n \n else: p = p[:-1] + \'b\'\n \n return p\n```\n[https://leetcode.com/problems/break-a-palindrome/submissions/559838984/](https://leetcode.com/problems/break-a-palindrome/submissions/559838984/)\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(1), in which *N* ~ `len(p)`.
| 7 | 0 |
['Python', 'Python3']
| 0 |
break-a-palindrome
|
Easy Java Solution- 100% fast with explanation
|
easy-java-solution-100-fast-with-explana-qcgv
|
There are three test cases which helped me for arriving at solution\n1. abccba - Here we are replacing it to aaccba. This should be lexographically small, hence
|
apoorvajoshi66
|
NORMAL
|
2021-05-11T07:56:20.035142+00:00
|
2021-05-11T07:56:20.035177+00:00
| 950 | false |
There are three test cases which helped me for arriving at solution\n1. abccba - Here we are replacing it to aaccba. This should be lexographically small, hence we have to search for first non occurence of a and replace it with \'a\'. \n2. aaabaaa - Here it will be replaced as aaabaab which will be the lexographically small string. Here we replace end of the string with next smaller alphabet after \'a\' which will be \'b\'.\n3. a/b/c - These cannot be made as non-pallindrome, hence return empty string.\n```\nclass Solution {\n public String breakPalindrome(String palindrome) {\n if(palindrome.length() <= 1) return new String();\n int n = palindrome.length();\n StringBuilder sb = new StringBuilder(palindrome);\n for(int i=0; i<n/2; i++){\n if(palindrome.charAt(i)!=\'a\'){\n sb.setCharAt(i , \'a\');\n return sb.toString();\n }\n }\n sb.setCharAt(n-1 , \'b\');\n return sb.toString();\n }\n}\n```
| 7 | 0 |
['Java']
| 2 |
break-a-palindrome
|
Python brute force
|
python-brute-force-by-stefanpochmann-2073
|
Just using how fast Python is with string operations. Got accepted in 28 ms.\n\nTry replacing each character (except a middle one) with \'a\' if it\'s not \'a\'
|
stefanpochmann
|
NORMAL
|
2020-01-25T18:38:54.567609+00:00
|
2020-01-25T18:38:54.567645+00:00
| 992 | false |
Just using how fast Python is with string operations. Got accepted in 28 ms.\n\nTry replacing each character (except a middle one) with `\'a\'` if it\'s not `\'a`\' already, and with `\'b\'` otherwise. Take the minimum such modified string, default to empty string.\n```\ndef breakPalindrome(self, p):\n return min((p[:i] + \'ba\'[c > \'a\'] + p[i+1:]\n for i, c in enumerate(p)\n if 2*i != len(p)-1),\n default=\'\')\n```
| 7 | 5 |
[]
| 1 |
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