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https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2368206/GolangPython-O(N)-timeor-O(N)-space-Quick-Select
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
array = [matrix[i][j] for i in range(len(matrix)) for j in range(len(matrix[0]))]
return quickselect(array,k)
def quickselect(array,k):
middle = len(array)//2
middle_item = array[middle]
left = []
right = []
for i,item in enumerate(array):
if i == middle:
continue
if item < middle_item:
left.append(item)
else:
right.append(item)
if len(left)+1 == k:
return middle_item
elif len(left) >= k:
return quickselect(left,k)
else:
return quickselect(right,k-len(left)-1)
|
kth-smallest-element-in-a-sorted-matrix
|
Golang/Python O(N) time| O(N) space Quick Select
|
vtalantsev
| 0 | 44 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,500 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2367833/378.-Stupid-215ms-Python3-in-place-sort
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
row0 = matrix[0]
for i in range(1, len(matrix)):
row0, matrix[i] = row0+matrix[i], []
row0.sort()
return row0[k-1]
|
kth-smallest-element-in-a-sorted-matrix
|
378. Stupid 215ms Python3 in-place sort
|
leetavenger
| 0 | 3 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,501 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2367832/Python-Solution-or-One-Liner-or-99-Faster-or-Sorting-Based
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
return sorted([cell for row in matrix for cell in row])[k-1]
|
kth-smallest-element-in-a-sorted-matrix
|
Python Solution | One-Liner | 99% Faster | Sorting Based
|
Gautam_ProMax
| 0 | 14 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,502 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2367524/python-solution-99-faster
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
sus = [i for k in matrix for i in k]
return sorted(sus)[k-1]
|
kth-smallest-element-in-a-sorted-matrix
|
python solution 99% faster
|
Noicelikeice
| 0 | 14 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,503 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2367493/Python-easy-solution-faster-than-95
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
one_dim_mat = []
for i in range(len(matrix)):
one_dim_mat += matrix[i]
return sorted(one_dim_mat)[k-1]
|
kth-smallest-element-in-a-sorted-matrix
|
Python easy solution faster than 95%
|
alishak1999
| 0 | 26 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,504 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2367396/Python-Simple-Python-Solution
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
result = []
for row in matrix:
result = result + row
result = sorted(result)
return result[k - 1]
|
kth-smallest-element-in-a-sorted-matrix
|
[ Python ] ✅✅ Simple Python Solution 🥳✌👍
|
ASHOK_KUMAR_MEGHVANSHI
| 0 | 115 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,505 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2367117/2-lines-of-code-using-list-flatten-technique-and-sorting
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
mat = [j for i in matrix for j in i]
mat.sort()
return mat[k - 1]
|
kth-smallest-element-in-a-sorted-matrix
|
2 lines of code using list flatten technique and sorting
|
ankurbhambri
| 0 | 31 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,506 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2367021/Python-1-Line-List-Comprehension-(beats-98.95)
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
return sorted([val for sublist in matrix for val in sublist])[k - 1]
|
kth-smallest-element-in-a-sorted-matrix
|
Python 1 Line List Comprehension (beats 98.95%)
|
Jonathanace
| 0 | 29 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,507 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2318549/Short-solution-python
|
class Solution(object):
def kthSmallest(self, matrix, k):
"""
:type matrix: List[List[int]]
:type k: int
:rtype: int
"""
sol=matrix[0]
for i in matrix[1:]:
sol+=i
sol.sort()
return sol[k-1]
|
kth-smallest-element-in-a-sorted-matrix
|
Short solution python
|
henryhe707
| 0 | 71 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,508 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2193377/easy-python-solution-in-5-lines
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
ans = []
for i in matrix:
ans += i
ans = sorted(ans)
return ans[k-1]
|
kth-smallest-element-in-a-sorted-matrix
|
easy python solution in 5 lines
|
rohansardar
| 0 | 89 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,509 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2073052/Simple-Short-Python-Solution-Lists
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
N = len(matrix)
for i in range(1, N):
for j in range(N):
matrix[0].append(matrix[i][j])
matrix[0].sort()
return matrix[0][k-1]
|
kth-smallest-element-in-a-sorted-matrix
|
Simple Short Python Solution - Lists
|
kn_vardhan
| 0 | 79 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,510 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2073052/Simple-Short-Python-Solution-Lists
|
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
values = []
for row in matrix:
values += row
values.sort()
return values[k - 1]
|
kth-smallest-element-in-a-sorted-matrix
|
Simple Short Python Solution - Lists
|
kn_vardhan
| 0 | 79 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,511 |
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/1971474/Python-oror-Easy-Solution-Using-Heap-oror-O(k*logk)
|
class Solution:
import heapq
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
heap = []
for arr in matrix:
heapq.heappush(heap,(arr[0],0,arr))
while k > 0 and heap:
val,idx,arr = heapq.heappop(heap)
ans = val
if idx+1<len(arr):
heapq.heappush(heap,(arr[idx+1],idx+1,arr))
k-=1
return ans
|
kth-smallest-element-in-a-sorted-matrix
|
Python || Easy Solution Using Heap || O(k*logk)
|
gamitejpratapsingh998
| 0 | 210 |
kth smallest element in a sorted matrix
| 378 | 0.616 |
Medium
| 6,512 |
https://leetcode.com/problems/linked-list-random-node/discuss/811617/Python3-reservoir-sampling
|
class Solution:
def __init__(self, head: ListNode):
"""
@param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node.
"""
self.head = head # store head of linked list
def getRandom(self) -> int:
"""
Returns a random node's value.
"""
cnt = 0
node = self.head
while node:
cnt += 1
if randint(1, cnt) == cnt: ans = node.val # reservoir sampling
node = node.next
return ans
|
linked-list-random-node
|
[Python3] reservoir sampling
|
ye15
| 4 | 395 |
linked list random node
| 382 | 0.596 |
Medium
| 6,513 |
https://leetcode.com/problems/linked-list-random-node/discuss/811617/Python3-reservoir-sampling
|
class Solution:
def __init__(self, head: ListNode):
self.head = head # store head
def getRandom(self) -> int:
node = self.head
n = 0
while node:
if randint(0, n) == 0: ans = node.val
n += 1
node = node.next
return ans
|
linked-list-random-node
|
[Python3] reservoir sampling
|
ye15
| 4 | 395 |
linked list random node
| 382 | 0.596 |
Medium
| 6,514 |
https://leetcode.com/problems/linked-list-random-node/discuss/1672294/Python3-solution
|
class Solution:
def __init__(self, head: Optional[ListNode]):
# store nodes' value in an array
self.array = []
curr = head
while curr:
self.array.append(curr.val)
curr = curr.next
def getRandom(self) -> int:
num = int(random.random() * len(self.array))
return self.array[num]
|
linked-list-random-node
|
Python3 solution
|
Janetcxy
| 1 | 28 |
linked list random node
| 382 | 0.596 |
Medium
| 6,515 |
https://leetcode.com/problems/linked-list-random-node/discuss/1673785/Python-Intuition
|
class Solution:
def __init__(self, head: Optional[ListNode]):
self.nums = []
while head:
self.nums.append(head.val)
head = head.next
def getRandom(self) -> int:
return random.choice(self.nums)
|
linked-list-random-node
|
Python Intuition
|
princess_asante
| 0 | 26 |
linked list random node
| 382 | 0.596 |
Medium
| 6,516 |
https://leetcode.com/problems/linked-list-random-node/discuss/1338778/Python-code-Beats-96-in-time
|
class Solution:
def __init__(self, head: ListNode):
"""
@param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node.
"""
self.mylist=[]
self.insert(head)
def insert(self, head):
while head:
self.mylist.append(head.val)
head=head.next
def getRandom(self) -> int:
"""
Returns a random node's value.
"""
return random.choice(self.mylist)
|
linked-list-random-node
|
Python code, Beats 96% in time
|
prajwal_vs
| 0 | 89 |
linked list random node
| 382 | 0.596 |
Medium
| 6,517 |
https://leetcode.com/problems/linked-list-random-node/discuss/976242/4Sum-II-Python3-one-liner
|
class Solution:
def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
return sum(xcount * cdcounts[-x]
for cdcounts in [Counter(x+y for x in C for y in D)]
for x, xcount in Counter(x+y for x in A for y in B).items()
if -x in cdcounts)
|
linked-list-random-node
|
4Sum II Python3 one-liner
|
leetavenger
| 0 | 46 |
linked list random node
| 382 | 0.596 |
Medium
| 6,518 |
https://leetcode.com/problems/linked-list-random-node/discuss/957190/Python3-beats-98.-%22random.choice(-)%22.-Very-Easy
|
class Solution:
def __init__(self, head: ListNode):
"""
@param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node.
"""
self.res = []
while head:
self.res.append(head.val)
head = head.next
def getRandom(self) -> int:
"""
Returns a random node's value.
"""
return random.choice(self.res)
|
linked-list-random-node
|
[Python3] beats 98%. "random.choice( )". Very-Easy
|
tilak_
| 0 | 62 |
linked list random node
| 382 | 0.596 |
Medium
| 6,519 |
https://leetcode.com/problems/ransom-note/discuss/1346131/Easiest-python-solution-faster-than-95
|
class Solution:
def canConstruct(self, ransomNote, magazine):
for i in set(ransomNote):
if magazine.count(i) < ransomNote.count(i):
return False
return True
|
ransom-note
|
Easiest python solution, faster than 95%
|
mqueue
| 35 | 3,500 |
ransom note
| 383 | 0.576 |
Easy
| 6,520 |
https://leetcode.com/problems/ransom-note/discuss/2500721/Very-Easy-oror-100-oror-Fully-Explained-oror-C%2B%2B-Java-Python-Python3
|
class Solution(object):
def canConstruct(self, ransomNote, magazine):
st1, st2 = Counter(ransomNote), Counter(magazine)
if st1 & st2 == st1:
return True
return False
|
ransom-note
|
Very Easy || 100% || Fully Explained || C++, Java, Python, Python3
|
PratikSen07
| 29 | 1,500 |
ransom note
| 383 | 0.576 |
Easy
| 6,521 |
https://leetcode.com/problems/ransom-note/discuss/2550434/Easy-and-Faster-than-98-python3-solutions
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
if len(magazine) < len(ransomNote):
return False
else:
for ransomNote_char in set(ransomNote):
if ransomNote.count(ransomNote_char) > magazine.count(ransomNote_char):
return False
return True
|
ransom-note
|
Easy and Faster than 98% python3 solutions
|
LucasHarim
| 9 | 302 |
ransom note
| 383 | 0.576 |
Easy
| 6,522 |
https://leetcode.com/problems/ransom-note/discuss/2359975/Python-One-Line-Solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
return Counter(ransomNote)-Counter(magazine)=={}
|
ransom-note
|
Python One Line Solution
|
dhruva3223
| 6 | 428 |
ransom note
| 383 | 0.576 |
Easy
| 6,523 |
https://leetcode.com/problems/ransom-note/discuss/1718553/Python-simple-solution-6-lines
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
dict1=collections.Counter(ransomNote)
dict2=collections.Counter(magazine)
for key in dict1:
if key not in dict2 or dict2[key]<dict1[key]:
return False
return True
|
ransom-note
|
Python simple solution 6 lines
|
amannarayansingh10
| 5 | 459 |
ransom note
| 383 | 0.576 |
Easy
| 6,524 |
https://leetcode.com/problems/ransom-note/discuss/2671667/Python-solution.-Clean-code-with-full-comments-93.55-space
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
dict_1 = from_str_to_dict(ransomNote)
dict_2 = from_str_to_dict(magazine)
return check_compatibility(dict_1, dict_2)
# Define helper method that checks if to dictionaries have keys in common, and
# if the ransomNote needs more letters then what the magazine can provide.
def check_compatibility(dict_1, dict_2):
# Check for common keys.
for key in list(dict_1.keys()):
if not key in dict_2:
return False
# Check for valid quantity.
if dict_1[key] > dict_2[key]:
return False
return True
# Convert a string into a dictionary.
def from_str_to_dict(string: str):
dic = {}
for i in string:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
return dic
# Runtime: 134 ms, faster than 24.02% of Python3 online submissions for Ransom Note.
# Memory Usage: 14.1 MB, less than 93.55% of Python3 online submissions for Ransom Note.
# If you like my work and found it helpful, then I'll appreciate a like. Thanks!
|
ransom-note
|
Python solution. Clean code with full comments 93.55% space
|
375d
| 4 | 191 |
ransom note
| 383 | 0.576 |
Easy
| 6,525 |
https://leetcode.com/problems/ransom-note/discuss/2480010/Python3-95.20-O(N)-Solution-or-Beginner-Friendly
|
class Solution:
def canConstruct(self, a: str, b: str) -> bool:
letter = [0 for _ in range(26)]
for c in b:
letter[ord(c) - 97] += 1
for c in a:
letter[ord(c) - 97] -= 1
return not any((i < 0 for i in letter))
|
ransom-note
|
Python3 95.20% O(N) Solution | Beginner Friendly
|
doneowth
| 3 | 88 |
ransom note
| 383 | 0.576 |
Easy
| 6,526 |
https://leetcode.com/problems/ransom-note/discuss/2270590/Simple-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
if len(magazine) < len(ransomNote):
return False
for letter in ransomNote:
if(letter not in magazine):
return False
magazine = magazine.replace(letter,'',1)
return True
|
ransom-note
|
Simple solution
|
jivanbasurtom
| 3 | 126 |
ransom note
| 383 | 0.576 |
Easy
| 6,527 |
https://leetcode.com/problems/ransom-note/discuss/1977169/Python3-Solution
|
class Solution(object):
def canConstruct(self, ransomNote, magazine):
for i in set(ransomNote):
if ransomNote.count(i) > magazine.count(i):
return False
return True
|
ransom-note
|
Python3 Solution
|
nomanaasif9
| 3 | 211 |
ransom note
| 383 | 0.576 |
Easy
| 6,528 |
https://leetcode.com/problems/ransom-note/discuss/610555/Python3-Simple-Solution-Beats-99.5
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
for l in ransomNote:
if l not in magazine:
return False
magazine = magazine.replace(l, '', 1)
return True
|
ransom-note
|
[Python3] Simple Solution Beats 99.5%
|
naraB
| 3 | 419 |
ransom note
| 383 | 0.576 |
Easy
| 6,529 |
https://leetcode.com/problems/ransom-note/discuss/2594211/Python-simple-solution-or-no-hashtable-or-beats-98.63-in-runtime
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
out = False
for elem in ransomNote:
if elem in magazine:
out = True
magazine = magazine.replace(elem,'',1)
else:
out = False
return out
return out
|
ransom-note
|
Python simple solution | no hashtable | beats 98.63% in runtime
|
shwetachandole
| 2 | 147 |
ransom note
| 383 | 0.576 |
Easy
| 6,530 |
https://leetcode.com/problems/ransom-note/discuss/2585361/Python-99.5-Faster.-Simple-Solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
#iterate the unqiue letter in ransom note
for i in set(ransomNote):
#check if the letter count in magazine is less than ransom note
if magazine.count(i) < ransomNote.count(i):
return(False)
return(True)
|
ransom-note
|
Python 99.5% Faster. Simple Solution
|
ovidaure
| 2 | 213 |
ransom note
| 383 | 0.576 |
Easy
| 6,531 |
https://leetcode.com/problems/ransom-note/discuss/1825705/Python-3-or-99.60-Faster-or-Easy-to-understand-or-Six-lines-of-code
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
c = 0
for i in ransomNote:
if i in magazine:
magazine = magazine.replace(i,"",1)
c += 1
return c == len(ransomNote)
|
ransom-note
|
✔Python 3 | 99.60% Faster | Easy to understand | Six lines of code
|
Coding_Tan3
| 2 | 182 |
ransom note
| 383 | 0.576 |
Easy
| 6,532 |
https://leetcode.com/problems/ransom-note/discuss/1433367/python-3-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
r=sorted(list(ransomNote))
m=sorted(list(magazine))
for char in m:
if r and char==r[0]:
r.pop(0)
if r:
return False
else:
return True
```
|
ransom-note
|
python 3 solution
|
abhi_n_001
| 2 | 207 |
ransom note
| 383 | 0.576 |
Easy
| 6,533 |
https://leetcode.com/problems/ransom-note/discuss/1266898/Easy-Python-Solution(94.14)
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
x=Counter(ransomNote)
y=Counter(magazine)
for i,v in x.items():
if(x[i]<=y[i]):
continue
else:
return False
return True
|
ransom-note
|
Easy Python Solution(94.14%)
|
Sneh17029
| 2 | 758 |
ransom note
| 383 | 0.576 |
Easy
| 6,534 |
https://leetcode.com/problems/ransom-note/discuss/2480474/Python-1-line-%2B-Set-%2B-Brute-Force-or-Detailed-explanation-or-Easy-understand-or-Detailed
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
for i in set(ransomNote):
if ransomNote.count(i) > magazine.count(i):
return False
return True
|
ransom-note
|
✅ Python 1 line + Set + Brute Force | Detailed explanation | Easy understand | Detailed
|
sezanhaque
| 1 | 20 |
ransom note
| 383 | 0.576 |
Easy
| 6,535 |
https://leetcode.com/problems/ransom-note/discuss/2480474/Python-1-line-%2B-Set-%2B-Brute-Force-or-Detailed-explanation-or-Easy-understand-or-Detailed
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
magazine_dict, ransomNote_dict = {}, {}
for letter in magazine:
if letter in magazine_dict:
magazine_dict[letter] += 1
else:
magazine_dict[letter] = 1
for letter in ransomNote:
if letter in ransomNote_dict:
ransomNote_dict[letter] += 1
else:
ransomNote_dict[letter] = 1
for letter in ransomNote_dict:
if letter not in magazine_dict:
return False
if ransomNote_dict[letter] > magazine_dict[letter]:
return False
return True
|
ransom-note
|
✅ Python 1 line + Set + Brute Force | Detailed explanation | Easy understand | Detailed
|
sezanhaque
| 1 | 20 |
ransom note
| 383 | 0.576 |
Easy
| 6,536 |
https://leetcode.com/problems/ransom-note/discuss/2418464/Low-time-complexity-approach
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
res = True
# use set() to get unique characters in string
for uniqueChar in set(ransomNote):
if ransomNote.count(uniqueChar) <= magazine.count(uniqueChar):
pass
else:
res = False
return res
|
ransom-note
|
Low time complexity approach
|
Wok2882
| 1 | 79 |
ransom note
| 383 | 0.576 |
Easy
| 6,537 |
https://leetcode.com/problems/ransom-note/discuss/2366128/Python-Short-Faster-Solution-using-str.replace()-oror-Documented
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
for c in ransomNote:
# if letter not in magazine, return True
if c not in magazine:
return False
else:
# remove first occurrence of letter
magazine = magazine.replace(c,'',1)
return True
|
ransom-note
|
[Python] Short Faster Solution using str.replace() || Documented
|
Buntynara
| 1 | 66 |
ransom note
| 383 | 0.576 |
Easy
| 6,538 |
https://leetcode.com/problems/ransom-note/discuss/2327790/Python-Simple-Solution-oror-2-Dicts-or-HashMaps-oror-Explained
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
dictRansom = {}
dictMagazine = {}
# construct dictionary with initial value of 0 for all letters
for c in 'abcdefghijklmnopqrstuvwxyz': dictRansom[c] = 0; dictMagazine[c] = 0
# increase count for each letter
for v in ransomNote: dictRansom[v] += 1
for v in magazine: dictMagazine[v] += 1
# if any letter-count in ransome is greater than on in magazine, return False
for k, v in dictRansom.items():
if v > dictMagazine[k]:
return False
# all counts in limit
return True
|
ransom-note
|
[Python] Simple Solution || 2 Dicts or HashMaps || Explained
|
Buntynara
| 1 | 89 |
ransom note
| 383 | 0.576 |
Easy
| 6,539 |
https://leetcode.com/problems/ransom-note/discuss/2282408/Python3-or-O(n)-with-list
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
letters = [0]*26
for letter in list(ransomNote):
letters[ord(letter)-ord('a')] += 1
for letter in list(magazine):
if letters[ord(letter)-ord('a')] > 0:
letters[ord(letter)-ord('a')] -= 1
check = [0]*26
return check == letters
|
ransom-note
|
Python3 | O(n) with list
|
muradbay
| 1 | 90 |
ransom note
| 383 | 0.576 |
Easy
| 6,540 |
https://leetcode.com/problems/ransom-note/discuss/2254814/Python-Easy-and-Short-Solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
for i in set(ransomNote):
if i not in magazine or ransomNote.count(i) > magazine.count(i):
return False
return True
|
ransom-note
|
Python Easy & Short Solution
|
Skiper228
| 1 | 87 |
ransom note
| 383 | 0.576 |
Easy
| 6,541 |
https://leetcode.com/problems/ransom-note/discuss/2187284/Python-FAST-and-Simple-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
for i in ransomNote:
if i in magazine:
magazine = magazine.replace(i, '', 1)
else:
return False
return True
|
ransom-note
|
Python FAST and Simple solution
|
ingli
| 1 | 78 |
ransom note
| 383 | 0.576 |
Easy
| 6,542 |
https://leetcode.com/problems/ransom-note/discuss/2019703/Python3-Solution-with-using-counting
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
c = collections.Counter(magazine)
for char in ransomNote:
if c[char] == 0:
return False
c[char] -= 1
return True
|
ransom-note
|
[Python3] Solution with using counting
|
maosipov11
| 1 | 81 |
ransom note
| 383 | 0.576 |
Easy
| 6,543 |
https://leetcode.com/problems/ransom-note/discuss/1825041/2-Lines-Python-Solution-oror-75-Faster-oror-Memory-less-than-75
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
if Counter(ransomNote) <= Counter(magazine): return True
return False
|
ransom-note
|
2-Lines Python Solution || 75% Faster || Memory less than 75%
|
Taha-C
| 1 | 55 |
ransom note
| 383 | 0.576 |
Easy
| 6,544 |
https://leetcode.com/problems/ransom-note/discuss/1793445/Python-Simple-Python-Solution-With-Two-Approach
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
t=list(dict.fromkeys(ransomNote))
s=0
for i in t:
if ransomNote.count(i)<=magazine.count(i):
s=s+1
if s==len(t):
return True
else:
return False
|
ransom-note
|
[ Python ] ✅✅ Simple Python Solution With Two Approach 🔥✌🥳👍
|
ASHOK_KUMAR_MEGHVANSHI
| 1 | 205 |
ransom note
| 383 | 0.576 |
Easy
| 6,545 |
https://leetcode.com/problems/ransom-note/discuss/1793445/Python-Simple-Python-Solution-With-Two-Approach
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
frequecy_r = {}
for i in range(len(ransomNote)):
if ransomNote[i] not in frequecy_r:
frequecy_r[ransomNote[i]] = 1
else:
frequecy_r[ransomNote[i]] = frequecy_r[ransomNote[i]] + 1
frequecy_m = {}
for i in range(len(magazine)):
if magazine[i] not in frequecy_m:
frequecy_m[magazine[i]] = 1
else:
frequecy_m[magazine[i]] = frequecy_m[magazine[i]] + 1
for i in ransomNote:
if i not in frequecy_m:
return False
else:
if frequecy_m[i] < frequecy_r[i]:
return False
return True
|
ransom-note
|
[ Python ] ✅✅ Simple Python Solution With Two Approach 🔥✌🥳👍
|
ASHOK_KUMAR_MEGHVANSHI
| 1 | 205 |
ransom note
| 383 | 0.576 |
Easy
| 6,546 |
https://leetcode.com/problems/ransom-note/discuss/1779868/Python-Easy-Step-by-Step-Solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
char_dict = {}
for char in magazine:
if char in char_dict:
char_dict[char] += 1
else:
char_dict[char] = 1
for char in ransomNote:
if not char in char_dict or char_dict[char] == 0:
return False
else:
char_dict[char] -= 1
return True
|
ransom-note
|
Python Easy Step by Step Solution
|
EnergyBoy
| 1 | 36 |
ransom note
| 383 | 0.576 |
Easy
| 6,547 |
https://leetcode.com/problems/ransom-note/discuss/1640283/Python3-super-easy-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
a = collections.Counter(ransomNote)
b = collections.Counter(magazine)
x = a - b
return len(x) == 0
|
ransom-note
|
Python3 super easy solution
|
freakleesin
| 1 | 91 |
ransom note
| 383 | 0.576 |
Easy
| 6,548 |
https://leetcode.com/problems/ransom-note/discuss/1532796/WEEB-DOES-PYTHON
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
Note = Counter(ransomNote)
Maga = Counter(magazine)
for char in Note:
if char not in Maga or Maga[char] < Note[char]: return False
return True
|
ransom-note
|
WEEB DOES PYTHON
|
Skywalker5423
| 1 | 145 |
ransom note
| 383 | 0.576 |
Easy
| 6,549 |
https://leetcode.com/problems/ransom-note/discuss/353415/Solution-in-Python-3-(beats-~97)-(one-line)
|
class Solution:
def canConstruct(self, r: str, m: str) -> bool:
return not any(r.count(i) > m.count(i) for i in set(r))
- Junaid Mansuri
(LeetCode ID)@hotmail.com
|
ransom-note
|
Solution in Python 3 (beats ~97%) (one line)
|
junaidmansuri
| 1 | 501 |
ransom note
| 383 | 0.576 |
Easy
| 6,550 |
https://leetcode.com/problems/ransom-note/discuss/2847994/Py-Hash-O(N)
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
dic_magazine = Counter(magazine)
for char in ransomNote:
dic_magazine[char] -= 1
if dic_magazine[char] < 0:
return False
return True
|
ransom-note
|
Py Hash O(N)
|
haly-leshchuk
| 0 | 2 |
ransom note
| 383 | 0.576 |
Easy
| 6,551 |
https://leetcode.com/problems/ransom-note/discuss/2839499/Python-using-Counter
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
r = Counter(ransomNote)
m = Counter(magazine)
for k in r.keys():
if r[k] > m.get(k,0):
return False
return True
|
ransom-note
|
Python using Counter
|
ajayedupuganti18
| 0 | 2 |
ransom note
| 383 | 0.576 |
Easy
| 6,552 |
https://leetcode.com/problems/ransom-note/discuss/2827103/python-easy-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
count = 0
for i in range(len(ransomNote)):
if ransomNote[i] in magazine:
magazine = magazine.replace(ransomNote[i], "", 1)
count += 1
return count == len(ransomNote)
|
ransom-note
|
python easy solution
|
skywonder
| 0 | 6 |
ransom note
| 383 | 0.576 |
Easy
| 6,553 |
https://leetcode.com/problems/ransom-note/discuss/2818292/Using-dicts-and-list-of-flags
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
ransom_dict = {}
magazine_dict = {}
flags = []
for i in ransomNote:
if i in ransom_dict:
ransom_dict[i] += 1
continue
ransom_dict[i] = 1
for i in magazine:
if i in magazine_dict:
magazine_dict[i] += 1
continue
magazine_dict[i] = 1
for symbol in ransom_dict.keys():
if symbol not in magazine_dict.keys():
return False
elif magazine_dict[symbol] >= ransom_dict[symbol]:
flags.append(1)
else:
flags.append(0)
for i in flags:
if i != 1:
return False
return True
|
ransom-note
|
Using dicts and list of flags
|
pkozhem
| 0 | 4 |
ransom note
| 383 | 0.576 |
Easy
| 6,554 |
https://leetcode.com/problems/ransom-note/discuss/2814520/simple-python-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
for x in list(ransomNote):
if x not in list(magazine):
return False
for x in list(ransomNote):
if list(ransomNote).count(x)>list(magazine).count(x):
return False
return True
|
ransom-note
|
simple python solution
|
sahityasetu1996
| 0 | 3 |
ransom note
| 383 | 0.576 |
Easy
| 6,555 |
https://leetcode.com/problems/ransom-note/discuss/2811939/Dictionary-count
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
letters = Counter(magazine)
for letter in ransomNote:
if letters[letter] == 0:
return False
letters[letter] -= 1
return True
|
ransom-note
|
Dictionary count
|
bourgeoibee
| 0 | 3 |
ransom note
| 383 | 0.576 |
Easy
| 6,556 |
https://leetcode.com/problems/ransom-note/discuss/2808000/Python-solutions-for-Ransom-Note
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
magazine_lettera = {i:magazine.count(i) for i in set(magazine)}
for letter in ransomNote:
if letter not in magazine_lettera.keys():
return False
else:
magazine_lettera[letter] -= 1
if magazine_lettera[letter] <= 0:
magazine_lettera.pop(letter)
return True
|
ransom-note
|
Python solutions for Ransom Note
|
Bassel_Alf
| 0 | 3 |
ransom note
| 383 | 0.576 |
Easy
| 6,557 |
https://leetcode.com/problems/ransom-note/discuss/2808000/Python-solutions-for-Ransom-Note
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
magazine_lettera = dict()
for letter in magazine:
magazine_lettera[letter] = magazine_lettera.get(letter, 0) + 1
for letter in ransomNote:
if letter not in magazine_lettera.keys():
return False
else:
magazine_lettera[letter] -= 1
if magazine_lettera[letter] <= 0:
magazine_lettera.pop(letter)
return True
|
ransom-note
|
Python solutions for Ransom Note
|
Bassel_Alf
| 0 | 3 |
ransom note
| 383 | 0.576 |
Easy
| 6,558 |
https://leetcode.com/problems/ransom-note/discuss/2804911/Python-(runtime-98.31-memory-93.97)-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
originalMagazineLength = len(magazine)
for c in ransomNote:
magazine = magazine.replace(c, '', 1)
return len(magazine) == originalMagazineLength - len(ransomNote)
|
ransom-note
|
Python (runtime 98.31%, memory 93.97%) solution
|
huseyinkeles
| 0 | 3 |
ransom note
| 383 | 0.576 |
Easy
| 6,559 |
https://leetcode.com/problems/ransom-note/discuss/2799026/PYTHON-Easy-Solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
a=Counter(ransomNote)
#count the occurance of char of ransomnote
b=Counter(magazine)
#count the occurance of char of magazine
if a & b ==a:
return True
return False
|
ransom-note
|
[PYTHON] Easy Solution
|
user9516zM
| 0 | 1 |
ransom note
| 383 | 0.576 |
Easy
| 6,560 |
https://leetcode.com/problems/ransom-note/discuss/2797474/Python-solution-using-Counter-(short-and-simple)
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
counter_ransomNote = Counter(ransomNote)
counter_magazine = Counter(magazine)
for char in counter_ransomNote.keys():
if counter_ransomNote[char] > counter_magazine[char]:
return False
return True
|
ransom-note
|
Python solution using Counter (short and simple)
|
hungqpham
| 0 | 3 |
ransom note
| 383 | 0.576 |
Easy
| 6,561 |
https://leetcode.com/problems/ransom-note/discuss/2779474/Very-easy-solution-without-any-additional-method-but-beats-89.77
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
for char in ransomNote:
char_found = False
for i in range(len(magazine)):
if char == magazine[i]:
magazine = magazine[:i] + magazine[i+1:]
char_found = True
break
if char_found == False:
return False
return True
|
ransom-note
|
Very easy solution without any additional method but beats 89.77%
|
sdsahil12
| 0 | 4 |
ransom note
| 383 | 0.576 |
Easy
| 6,562 |
https://leetcode.com/problems/ransom-note/discuss/2777466/Ransome-note-or-Easy
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
ransomNote = list(ransomNote)
tracker = list(set(ransomNote))
magazine = list(magazine)
while tracker:
if tracker and ransomNote.count(tracker[0]) <= magazine.count(tracker[0]):
tracker.remove(tracker[0])
else:
return False
return True
|
ransom-note
|
Ransome note | Easy
|
Gangajyoti
| 0 | 1 |
ransom note
| 383 | 0.576 |
Easy
| 6,563 |
https://leetcode.com/problems/ransom-note/discuss/2774991/Python-1-pointer
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
p_m = 0
note, mag = sorted(ransomNote), sorted(magazine)
while p_m < len(magazine) and note[0] != mag[p_m]:
p_m += 1
while len(note) and p_m < len(magazine):
if note[0] == mag[p_m]:
del note[0]
p_m += 1
return not len(note)
|
ransom-note
|
Python, 1 pointer
|
Gagampy
| 0 | 5 |
ransom note
| 383 | 0.576 |
Easy
| 6,564 |
https://leetcode.com/problems/ransom-note/discuss/2768180/383.-Ransom-Note-or-Python
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
count = {}
for c in magazine:
count[c] = 1 + count.get(c,0)
for ch in ransomNote:
if ch not in count or count[ch]==0:
return False
else:
count[ch]-=1
return True
|
ransom-note
|
383. Ransom Note | Python
|
rishabh_055
| 0 | 8 |
ransom note
| 383 | 0.576 |
Easy
| 6,565 |
https://leetcode.com/problems/ransom-note/discuss/2764531/Python3-oror-simple-solution-oror-easy-understanding
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
new = list(magazine)
for r in ransomNote:
if r not in new:
return False
i = new.index(r)
new.remove(new[i])
return True
|
ransom-note
|
Python3 || simple solution || easy understanding
|
amangupta3073
| 0 | 2 |
ransom note
| 383 | 0.576 |
Easy
| 6,566 |
https://leetcode.com/problems/ransom-note/discuss/2750639/Easiest-solution-with-python-two-dictionaries-one-comparison
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
have = dict()
for letter in magazine:
have[letter] = have.get(letter,0) + 1
needed = dict()
for letter in ransomNote:
needed[letter] = needed.get(letter,0) + 1
total = 0
for key, value in needed.items():
total += max(value - have.get(key, 0),0)
if total >0:
return False
return True
|
ransom-note
|
Easiest solution with python - two dictionaries one comparison
|
DavidCastillo
| 0 | 3 |
ransom note
| 383 | 0.576 |
Easy
| 6,567 |
https://leetcode.com/problems/ransom-note/discuss/2748373/Easy-python3-Counter-solution
|
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
note_freq = collections.Counter(ransomNote)
mag_freq = collections.Counter(magazine)
for letter in note_freq:
if note_freq[letter] > mag_freq[letter]:
return False
return True
|
ransom-note
|
Easy python3 Counter solution
|
Delacrua
| 0 | 4 |
ransom note
| 383 | 0.576 |
Easy
| 6,568 |
https://leetcode.com/problems/shuffle-an-array/discuss/1673643/Python-or-Best-Optimal-Approach
|
class Solution:
def __init__(self, nums: List[int]):
self.arr = nums[:] # Deep Copy, Can also use Shallow Copy concept!
# self.arr = nums # Shallow Copy would be something like this!
def reset(self) -> List[int]:
return self.arr
def shuffle(self) -> List[int]:
ans = self.arr[:]
for i in range(len(ans)):
swp_num = random.randrange(i, len(ans)) # Fisher-Yates Algorithm
ans[i], ans[swp_num] = ans[swp_num], ans[i]
return ans
|
shuffle-an-array
|
{ Python } | Best Optimal Approach ✔
|
leet_satyam
| 6 | 725 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,569 |
https://leetcode.com/problems/shuffle-an-array/discuss/1354489/Partition-Array-into-Disjoint-Intervals-Python-Easy-solution
|
class Solution:
def partitionDisjoint(self, nums: List[int]) -> int:
a = list(accumulate(nums, max))
b = list(accumulate(nums[::-1], min))[::-1]
for i in range(1, len(nums)):
if a[i-1] <= b[i]:
return i
|
shuffle-an-array
|
Partition Array into Disjoint Intervals , Python , Easy solution
|
user8744WJ
| 1 | 120 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,570 |
https://leetcode.com/problems/shuffle-an-array/discuss/1354450/Partition-Array-into-Disjoint-Intervals%3A-Simple-Python-Solution
|
class Solution:
def partitionDisjoint(self, nums: List[int]) -> int:
n = len(nums)
left_length = 1
left_max = curr_max = nums[0]
for i in range(1, n-1):
if nums[i] < left_max:
left_length = i+1
left_max = curr_max
else:
curr_max = max(curr_max, nums[i])
return left_length
|
shuffle-an-array
|
Partition Array into Disjoint Intervals: Simple Python Solution
|
user6820GC
| 1 | 81 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,571 |
https://leetcode.com/problems/shuffle-an-array/discuss/1350083/Shuffle-An-Array-Python-Easy-Solution
|
class Solution:
def __init__(self, nums: List[int]):
self.a = nums
def reset(self) -> List[int]:
"""
Resets the array to its original configuration and return it.
"""
return list(self.a)
def shuffle(self) -> List[int]:
"""
Returns a random shuffling of the array.
"""
x = list(self.a)
ans= []
while x:
i = random.randint(0,len(x)-1)
x[i],x[-1] = x[-1],x[i]
ans.append(x.pop())
return ans
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()
|
shuffle-an-array
|
Shuffle An Array , Python , Easy Solution
|
user8744WJ
| 1 | 144 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,572 |
https://leetcode.com/problems/shuffle-an-array/discuss/1103904/Python-A-simple-O(N)-solution-that-is-NOT-Fisher-Yates-Algorithm
|
class Solution:
def __init__(self, nums: List[int]):
self.original = nums[:]
self.nums = nums
def reset(self) -> List[int]:
return self.original
def shuffle(self) -> List[int]:
tmp = list(self.nums) # make a copy
self.nums = []
while tmp:
index = random.randint(0, len(tmp) - 1) # get random index
tmp[-1], tmp[index] = tmp[index], tmp[-1] # switch places
self.nums.append(tmp.pop()) # pop the last index
return self.nums
|
shuffle-an-array
|
[Python] A simple O(N) solution that is NOT Fisher-Yates Algorithm
|
JummyEgg
| 1 | 390 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,573 |
https://leetcode.com/problems/shuffle-an-array/discuss/2726092/Python-oror-Fisher-Yates-algorithm
|
class Solution:
def __init__(self, nums: List[int]):
self.nums = nums
def reset(self) -> List[int]:
return self.nums
def shuffle(self) -> List[int]:
# fisher yates shuffle
copy = self.nums[:]
m = len(self.nums)
# while there remain elements to shuffle
while(m):
i = random.randint(0,m-1) # pick a random element from the remaining elements
m-=1
copy[i],copy[m]=copy[m],copy[i] # swap current and random element
return copy
|
shuffle-an-array
|
Python || Fisher-Yates algorithm
|
Graviel77
| 0 | 12 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,574 |
https://leetcode.com/problems/shuffle-an-array/discuss/2710888/EXTREMELY-SIMPLE-and-EFFICIENT-python-solution
|
class Solution:
def __init__(self, nums: List[int]):
# "Contruct" the object:
# Init the class variable(s)
self.nums = nums
def reset(self) -> List[int]:
# Return the inital variable(s)
# This assumes the variable(s) didn't change!
return self.nums
def shuffle(self) -> List[int]:
# Pick a random index from the list & add
# it to a new list. Remove that item, then loop.
# DON'T MODIFY THE ORIGINAL LIST!
nums = []
temp = self.nums.copy()
while True:
length = len(temp) - 1
if length == -1:
return nums
index = random.randint(0, length)
nums.append(temp[index])
temp.pop(index)
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()
|
shuffle-an-array
|
EXTREMELY SIMPLE & EFFICIENT python solution
|
rosey003
| 0 | 7 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,575 |
https://leetcode.com/problems/shuffle-an-array/discuss/2667750/FisherYates-shuffle
|
class Solution:
def __init__(self, nums: List[int]):
self.orig = list(nums)
self.perm = nums
def reset(self) -> List[int]:
self.perm = list(self.orig)
return self.perm
def shuffle(self) -> List[int]:
for i in range(0, len(self.perm)):
j = randint(i, len(self.perm)-1)
self.perm[i], self.perm[j] = self.perm[j], self.perm[i]
return self.perm
|
shuffle-an-array
|
Fisher–Yates shuffle
|
tomfran
| 0 | 5 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,576 |
https://leetcode.com/problems/shuffle-an-array/discuss/2652958/Python-and-Golang
|
class Solution:
def __init__(self, nums: List[int]):
self.nums = nums
def reset(self) -> List[int]:
return self.nums
def shuffle(self) -> List[int]:
import random
return random.sample(self.nums, k=len(self.nums))
|
shuffle-an-array
|
Python and Golang答え
|
namashin
| 0 | 15 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,577 |
https://leetcode.com/problems/shuffle-an-array/discuss/1819524/Python3-Solution-with-using-FisherYates-shuffle
|
class Solution:
def __init__(self, nums: List[int]):
self.arr = nums
self.nums = nums[:]
def reset(self) -> List[int]:
self.arr = self.nums
self.nums = self.nums[:]
return self.arr
def shuffle(self) -> List[int]:
for i in range(len(self.arr)):
rand_idx = random.randrange(i, len(self.arr))
self.arr[i], self.arr[rand_idx] = self.arr[rand_idx], self.arr[i]
return self.arr
|
shuffle-an-array
|
[Python3] Solution with using Fisher–Yates shuffle
|
maosipov11
| 0 | 89 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,578 |
https://leetcode.com/problems/shuffle-an-array/discuss/1813908/Python-solution-backtracking-and-shuffle
|
class Solution:
def __init__(self, nums: List[int]):
self.nums=nums
self.lists=self.getList()
def reset(self) -> List[int]:
a=sorted(self.nums)
return a
def getList(self):
nums=self.nums
res=[]
path=[]
used=[0]*len(nums)
def backtracking(nums):
if len(path)==len(nums):
res.append(path[:])
return
for i in range(len(nums)):
if used[i]==0:
if i>0 and nums[i]==nums[i-1] and used[i-1]==0:
continue
used[i]=1
path.append(nums[i])
backtracking(nums)
used[i]=0
path.pop()
return res
backtracking(nums)
return res
def shuffle(self) -> List[int]:
n=random.randint(0,len(self.nums)-1)
return self.lists[n]
|
shuffle-an-array
|
Python solution backtracking and shuffle
|
FangyuanXie
| 0 | 58 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,579 |
https://leetcode.com/problems/shuffle-an-array/discuss/1813908/Python-solution-backtracking-and-shuffle
|
class Solution:
def __init__(self, nums: List[int]):
self.nums=nums
self.orgin=nums[:]
def reset(self) -> List[int]:
self.nums=self.orgin[:]
return self.nums
def shuffle(self) -> List[int]:
n=len(self.nums)
for l in range(n):
r=random.randint(l,n-1)
self.nums[l],self.nums[r]=self.nums[r],self.nums[l]
return self.nums
|
shuffle-an-array
|
Python solution backtracking and shuffle
|
FangyuanXie
| 0 | 58 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,580 |
https://leetcode.com/problems/shuffle-an-array/discuss/1374919/Trapping-Rainwater-Python3-O(n)
|
class Solution:
def trap(self, height: List[int]) -> int:
if not height: return 0
def addwater(range_):
mx, out = 0, 0
for i in range_:
if height[i] > mx: mx = height[i]
else: out += mx - height[i]
return out
#Get midpoint
mid = height.index(max(height))
#Add up all water values at each height left of mid
left = addwater(range(mid))
#Add up all water value at each height right of mid
right = addwater(range(len(height)-1, mid, -1))
return left + right
|
shuffle-an-array
|
Trapping Rainwater [Python3] O(n)
|
vscala
| 0 | 75 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,581 |
https://leetcode.com/problems/shuffle-an-array/discuss/1374919/Trapping-Rainwater-Python3-O(n)
|
class Solution:
def trap(self, height: List[int]) -> int:
if len(height) < 3: return 0
l, r = 0, len(height)-1
lmax, rmax = height[l], height[r]
out = 0
while l+1 != r:
if lmax < rmax:
l += 1
if height[l] > lmax: lmax = height[l]
else: out += lmax - height[l]
else:
r -= 1
if height[r] > rmax: rmax = height[r]
else: out += rmax - height[r]
return out
|
shuffle-an-array
|
Trapping Rainwater [Python3] O(n)
|
vscala
| 0 | 75 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,582 |
https://leetcode.com/problems/shuffle-an-array/discuss/1350445/Shuffle-Array-or-Python-Random-Swapping-Approach
|
class Solution:
def __init__(self, nums: List[int]):
self.original = nums
def reset(self) -> List[int]:
return self.original
def shuffle(self) -> List[int]:
result, n = self.original.copy(), len(self.original)
for i in range(n):
ind = random.randint(0, n - 1)
result[i], result[ind] = result[ind], result[i]
return result
|
shuffle-an-array
|
Shuffle Array | Python Random Swapping Approach
|
yiseboge
| 0 | 85 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,583 |
https://leetcode.com/problems/shuffle-an-array/discuss/811552/Python3-Knuth-shuffle
|
class Solution:
def __init__(self, nums: List[int]):
self.nums = nums
self.orig = nums.copy() # original array
def reset(self) -> List[int]:
"""
Resets the array to its original configuration and return it.
"""
return self.orig
def shuffle(self) -> List[int]:
"""
Returns a random shuffling of the array.
"""
for i in range(1, len(self.nums)):
ii = randint(0, i)
self.nums[ii], self.nums[i] = self.nums[i], self.nums[ii]
return self.nums
|
shuffle-an-array
|
[Python3] Knuth shuffle
|
ye15
| 0 | 141 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,584 |
https://leetcode.com/problems/shuffle-an-array/discuss/603080/Python-sol-by-native-tools.-85%2B-w-Comment
|
class Solution:
def __init__(self, nums: List[int]):
self.size = len(nums)
self.origin = nums
self.array = [*nums]
def reset(self) -> List[int]:
"""
Resets the array to its original configuration and return it.
"""
return self.origin
def shuffle(self) -> List[int]:
"""
Returns a random shuffling of the array.
"""
size, arr = self.size, self.array
for i in range(size):
j = randint(i, size-1)
arr[i], arr[j] = arr[j], arr[i]
return arr
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()
|
shuffle-an-array
|
Python sol by native tools. 85%+ [w/ Comment ]
|
brianchiang_tw
| 0 | 285 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,585 |
https://leetcode.com/problems/shuffle-an-array/discuss/514309/Python3-use-random-shuffle-function
|
class Solution:
def __init__(self, nums: List[int]):
self.nums=nums
def reset(self) -> List[int]:
"""
Resets the array to its original configuration and return it.
"""
return self.nums
def shuffle(self) -> List[int]:
"""
Returns a random shuffling of the array.
"""
import random
res = list(self.nums)
random.shuffle(res)
return res
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()
|
shuffle-an-array
|
Python3 use random shuffle function
|
jb07
| 0 | 129 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,586 |
https://leetcode.com/problems/shuffle-an-array/discuss/485921/Python-3-(six-lines)-(beats-~98)
|
class Solution:
def __init__(self, n): self.array = n
def reset(self): return self.array
def shuffle(self):
s = self.array[:]
random.shuffle(s)
return s
- Junaid Mansuri
- Chicago, IL
|
shuffle-an-array
|
Python 3 (six lines) (beats ~98%)
|
junaidmansuri
| 0 | 420 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,587 |
https://leetcode.com/problems/shuffle-an-array/discuss/422655/EXACTLY-equally-likely-permutations
|
class Solution:
def __init__(self, nums: List[int]):
self.original = nums
def reset(self) -> List[int]:
return self.original
def shuffle(self) -> List[int]:
try:
return next(self.perm)
except:
self.perm = itertools.permutations(self.original)
return next(self.perm)
|
shuffle-an-array
|
EXACTLY equally likely permutations
|
SkookumChoocher
| 0 | 95 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,588 |
https://leetcode.com/problems/shuffle-an-array/discuss/1365722/3Sum-Closest-Easy-Python-2-pointer-Binary-Search
|
class Solution:
def threeSumClosest(self, nums: List[int], target: int) -> int:
a = nums[0]+nums[1]+nums[len(nums)-1]
nums.sort()
for i in range(len(nums)-2):
l = i+1
h = len(nums)-1
while l<h:
b = nums[i]+nums[l]+nums[h]
if b > target:
h -= 1
else:
l += 1
if abs(target - b)<abs(target - a):
a = b
return a
|
shuffle-an-array
|
3Sum Closest , Easy , Python , 2 pointer , Binary Search
|
user8744WJ
| -1 | 181 |
shuffle an array
| 384 | 0.577 |
Medium
| 6,589 |
https://leetcode.com/problems/mini-parser/discuss/875743/Python3-a-concise-recursive-solution
|
class Solution:
def deserialize(self, s: str) -> NestedInteger:
if not s: return NestedInteger()
if not s.startswith("["): return NestedInteger(int(s)) # integer
ans = NestedInteger()
s = s[1:-1] # strip outer "[" and "]"
if s:
ii = op = 0
for i in range(len(s)):
if s[i] == "[": op += 1
if s[i] == "]": op -= 1
if s[i] == "," and op == 0:
ans.add(self.deserialize(s[ii:i]))
ii = i+1
ans.add(self.deserialize(s[ii:i+1]))
return ans
|
mini-parser
|
[Python3] a concise recursive solution
|
ye15
| 2 | 178 |
mini parser
| 385 | 0.366 |
Medium
| 6,590 |
https://leetcode.com/problems/mini-parser/discuss/1638534/Elegant-Python-One-Pass-Solution-(using-stack)
|
class Solution:
def deserialize(self, s: str) -> NestedInteger:
stack = []
integerStr = ''
for c in s:
if c == '[':
stack.append(NestedInteger())
elif c == ']':
if len(integerStr)>0:
stack[-1].add(NestedInteger(int(integerStr)))
integerStr = ''
poppedList = stack.pop()
if len(stack)==0:
return poppedList
stack[-1].add(poppedList)
elif c == ',':
if len(integerStr)>0:
stack[-1].add(NestedInteger(int(integerStr)))
integerStr = ''
else:
integerStr += c
return NestedInteger(int(s))
|
mini-parser
|
Elegant Python One Pass Solution (using stack)
|
RG97
| 1 | 152 |
mini parser
| 385 | 0.366 |
Medium
| 6,591 |
https://leetcode.com/problems/lexicographical-numbers/discuss/2053392/Python-oneliner
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
return sorted([x for x in range(1,n+1)],key=lambda x: str(x))
|
lexicographical-numbers
|
Python oneliner
|
StikS32
| 1 | 135 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,592 |
https://leetcode.com/problems/lexicographical-numbers/discuss/1879853/Single-line-python-3-solution-98-faster
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
return sorted(list(map(str,list(range(1,n+1)))))
|
lexicographical-numbers
|
Single line python 3 solution 98% faster
|
amannarayansingh10
| 1 | 112 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,593 |
https://leetcode.com/problems/lexicographical-numbers/discuss/1787311/O(n)-time-O(1)-space-stack-solution-in-Python
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
stack = []
for i in range(min(n, 9), 0, -1):
stack.append(i)
res = []
while stack:
last = stack.pop()
if last > n:
continue
else:
res.append(last)
for i in range(9, -1, -1):
stack.append(last * 10 + i)
return res
|
lexicographical-numbers
|
O(n) time, O(1) space stack solution in Python
|
kryuki
| 1 | 124 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,594 |
https://leetcode.com/problems/lexicographical-numbers/discuss/817573/Python3-two-approaches-(sorting-and-dfs)
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
return sorted(range(1, n+1), key=str)
|
lexicographical-numbers
|
[Python3] two approaches (sorting & dfs)
|
ye15
| 1 | 74 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,595 |
https://leetcode.com/problems/lexicographical-numbers/discuss/817573/Python3-two-approaches-(sorting-and-dfs)
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
def dfs(x):
"""Pre-order traverse the tree."""
if x <= n:
ans.append(x)
for xx in range(10): dfs(10*x + xx)
ans = []
for x in range(1, 10): dfs(x)
return ans
|
lexicographical-numbers
|
[Python3] two approaches (sorting & dfs)
|
ye15
| 1 | 74 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,596 |
https://leetcode.com/problems/lexicographical-numbers/discuss/810448/Python-3-or-One-liner-cheat-or-Sort-String
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
return [int(i) for i in sorted([str(i) for i in range(1, n+1)])]
|
lexicographical-numbers
|
Python 3 | One-liner cheat | Sort String
|
idontknoooo
| 1 | 290 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,597 |
https://leetcode.com/problems/lexicographical-numbers/discuss/2848655/Brute-force-solution-from-empty-list
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
out = []
i = 1
nlen = len(str(n))
while len(out) < n:
out.append(i)
if i*10 <= n:
i = i*10
elif i < n and (i+1)//10 < (i//10)+1:
i += 1
else:
while str(i//10)[-1] == '9':
i = i // 10
i = i // 10 + 1
return out
|
lexicographical-numbers
|
Brute force solution from empty list
|
yoyo10
| 0 | 1 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,598 |
https://leetcode.com/problems/lexicographical-numbers/discuss/2829526/No-Trie-or-Recursion%3A-Beats-97.4-Time%3A-O(n)-Space%3A-O(1)O(n)
|
class Solution:
def lexicalOrder(self, n: int) -> List[int]:
r = []
a = 1
while a <= n and a < 10:
r.append(a)
b = a * 10
while b <= n and b < (a + 1) * 10:
r.append(b)
c = b * 10
while c <= n and c < (b + 1) * 10:
r.append(c)
d = c * 10
while d <= n and d < (c + 1) * 10:
r.append(d)
e = d * 10
while e <= n and e < (d + 1) * 10:
r.append(e)
e += 1
d += 1
c += 1
b += 1
a += 1
return r
|
lexicographical-numbers
|
No Trie or Recursion: Beats 97.4%, Time: O(n), Space: O(1)/O(n)
|
Squidster777
| 0 | 4 |
lexicographical numbers
| 386 | 0.608 |
Medium
| 6,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.