post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/plates-between-candles/discuss/1605103/Python3-Easy-and-Simple-to-understand-Solution-Good-example-(O(N%2BQ)) | class Solution:
def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]:
# accumulated sum of '*'
accumulated = []
accumulated.append(int(s[0] == '|'))
for char in s[1:]: # ex. "* * | * * | * * * | "
... | plates-between-candles | [Python3] Easy & Simple-to-understand Solution, Good example (O(N+Q)) | Otto_Kwon | 1 | 164 | plates between candles | 2,055 | 0.444 | Medium | 28,500 |
https://leetcode.com/problems/plates-between-candles/discuss/2699048/Python-O(N-%2B-MlogN)-O(N) | class Solution:
def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]:
mapping, plates = OrderedDict(), 0
candles = []
for i in range(len(s)):
if s[i] == "*": plates += 1
else:
mapping[i] = plates
candl... | plates-between-candles | Python - O(N + MlogN), O(N) | Teecha13 | 0 | 23 | plates between candles | 2,055 | 0.444 | Medium | 28,501 |
https://leetcode.com/problems/plates-between-candles/discuss/1554464/Python-3-or-Prefix-Sum-Math-O(N)-or-Explanation | class Solution:
def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]:
cur, left, n = 0, -1, len(s)
right = n
pre_sum = [0] * n # pre_sum[i]: number of candles before `i` (including `i`)
left_idx = [-1] * n ... | plates-between-candles | Python 3 | Prefix Sum, Math, O(N) | Explanation | idontknoooo | 0 | 192 | plates between candles | 2,055 | 0.444 | Medium | 28,502 |
https://leetcode.com/problems/plates-between-candles/discuss/1554464/Python-3-or-Prefix-Sum-Math-O(N)-or-Explanation | class Solution:
def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]:
n = len(s)
prev = -1
left_idx = [-1] * n
for i, c in enumerate(s):
if c == '|':
prev = i
left_idx[i] = prev
prev = n
rig... | plates-between-candles | Python 3 | Prefix Sum, Math, O(N) | Explanation | idontknoooo | 0 | 192 | plates between candles | 2,055 | 0.444 | Medium | 28,503 |
https://leetcode.com/problems/plates-between-candles/discuss/1554464/Python-3-or-Prefix-Sum-Math-O(N)-or-Explanation | class Solution:
def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]:
cur, left, n = 0, -1, len(s)
right = n
pre_sum = [0] * n
left_idx = [-1] * n
right_idx = [-1] * n
for i, c in enumerate(s):
if c == '|': ... | plates-between-candles | Python 3 | Prefix Sum, Math, O(N) | Explanation | idontknoooo | 0 | 192 | plates between candles | 2,055 | 0.444 | Medium | 28,504 |
https://leetcode.com/problems/plates-between-candles/discuss/1549047/Python3-O(N%2BQ) | class Solution:
def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]:
pref=[0]
#prefix sum array for calculating *
for x in s:
if x=='*':
pref.append(pref[-1]+1)
else:
pref.append(pref[-1])
ans=[]
#array for... | plates-between-candles | Python3 O(N+Q) | abhinav_201 | 0 | 80 | plates between candles | 2,055 | 0.444 | Medium | 28,505 |
https://leetcode.com/problems/number-of-valid-move-combinations-on-chessboard/discuss/2331905/Python3-or-DFS-with-backtracking-or-Clean-code-with-comments | class Solution:
BOARD_SIZE = 8
def diag(self, r, c):
# Return all diagonal indices except (r, c)
# Diagonal indices has the same r - c
inv = r - c
result = []
for ri in range(self.BOARD_SIZE):
ci = ri - inv
if 0 <= ci < self.BOARD_SIZE and ri ... | number-of-valid-move-combinations-on-chessboard | Python3 | DFS with backtracking | Clean code with comments | snorkin | 1 | 58 | number of valid move combinations on chessboard | 2,056 | 0.59 | Hard | 28,506 |
https://leetcode.com/problems/number-of-valid-move-combinations-on-chessboard/discuss/1557870/Python3-traversal | class Solution:
def countCombinations(self, pieces: List[str], positions: List[List[int]]) -> int:
n = len(pieces)
mp = {"bishop": ((-1, -1), (-1, 1), (1, -1), (1, 1)),
"queen" : ((-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)),
"rook" : ((-1, 0), ... | number-of-valid-move-combinations-on-chessboard | [Python3] traversal | ye15 | 0 | 335 | number of valid move combinations on chessboard | 2,056 | 0.59 | Hard | 28,507 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1549993/Python3-1-line | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
return next((i for i, x in enumerate(nums) if i%10 == x), -1) | smallest-index-with-equal-value | [Python3] 1-line | ye15 | 10 | 646 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,508 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1558084/Python-1-line-99-speed-95-memory-usage | class Solution(object):
def smallestEqual(self, nums, i=0):
return -1 if i == len(nums) else ( i if i%10 == nums[i] else self.smallestEqual(nums, i+1) ) | smallest-index-with-equal-value | Python 1 line - 99% speed 95% memory usage | SmittyWerbenjagermanjensen | 5 | 340 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,509 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2640991/Python-O(N) | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
n=len(nums)
for i in range(n):
if i%10==nums[i]:
return i
return -1 | smallest-index-with-equal-value | Python O(N) | Sneh713 | 1 | 50 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,510 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2840410/Just-1-line-code-in-Python3 | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
return min([k[0] for k in list(enumerate (nums)) if k[0]%10 == k[1]], default=-1) | smallest-index-with-equal-value | Just 1 line code in Python3 | DNST | 0 | 1 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,511 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2825297/Fast-Simple-Python-Solution-Beats-99.9 | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(0, len(nums)):
if i % 10 == nums[i]:
return i
return -1 | smallest-index-with-equal-value | Fast, Simple Python Solution - Beats 99.9% | PranavBhatt | 0 | 2 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,512 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2805456/Python3-list-comprehension | class Solution:
def smallestEqual(self, nums):
idxs = [idx for idx, ele in enumerate(nums) if idx % 10 == ele]
return -1 if not idxs else min(idxs) | smallest-index-with-equal-value | Python3-list comprehension | alamwasim29 | 0 | 4 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,513 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2780464/Easy-Python-Solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
count = 0
for i in range(len(nums)):
if i % 10 == nums[i]:
return i
return -1 | smallest-index-with-equal-value | Easy Python Solution | danishs | 0 | 3 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,514 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2672248/v | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
n=len(nums)
for i in range(n):
if i%10==nums[i]:
return i
return -1 | smallest-index-with-equal-value | v | adityabakshi45 | 0 | 2 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,515 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2670790/Easy-and-optimal-faster | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(0,len(nums)):
if(nums[i]==i %10):
return i
return -1 | smallest-index-with-equal-value | Easy and optimal faster | Raghunath_Reddy | 0 | 3 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,516 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2560519/Easy-python-solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
result = []
for i in range(len(nums)):
if i % 10 == nums[i]:
result.append(i)
if len(result) == 0:
return -1
return min(result) | smallest-index-with-equal-value | Easy python solution | samanehghafouri | 0 | 7 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,517 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2536018/Simple-python-solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(len(nums)):
if i % 10 == nums[i]:
return i
else:
return -1 | smallest-index-with-equal-value | Simple python solution | aruj900 | 0 | 20 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,518 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2520388/Python-solution-or-99-Faster-or-O(n)-complexity | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(0,len(nums)):
if i%10==nums[i]:
return i
break
return -1 | smallest-index-with-equal-value | Python solution | 99% Faster | O(n) complexity | KavitaPatel966 | 0 | 15 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,519 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2180507/Python-Solution-Easy-To-Understand | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(len(nums)):
if i%10==nums[i]:
return i
return -1 | smallest-index-with-equal-value | Python Solution- Easy To Understand✅ | T1n1_B0x1 | 0 | 27 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,520 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2091949/Python-Simple-Solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for idx, val in enumerate(nums):
if idx % 10 == val:
return idx
return -1 | smallest-index-with-equal-value | Python Simple Solution | Nk0311 | 0 | 54 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,521 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1972392/Simple-solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
nums = [i for i, n in enumerate(nums) if i % 10 == n]
return min(nums) if nums else -1 | smallest-index-with-equal-value | Simple solution | andrewnerdimo | 0 | 30 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,522 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1934730/easy-python-code | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(len(nums)):
if i%10 == nums[i]:
return i
return -1 | smallest-index-with-equal-value | easy python code | dakash682 | 0 | 18 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,523 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1886752/Python-dollarolution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i,j in enumerate(nums):
if i%10 == j:
return i
return -1 | smallest-index-with-equal-value | Python $olution | AakRay | 0 | 28 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,524 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1858061/Python-easy-straightforward-solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(len(nums)):
if i % 10 == nums[i]:
return i
return -1 | smallest-index-with-equal-value | Python easy straightforward solution | alishak1999 | 0 | 26 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,525 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1830480/1-line-solution-in-Python-3 | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
return min((i for i, n in enumerate(nums) if i % 10 == n), default=-1) | smallest-index-with-equal-value | 1-line solution in Python 3 | mousun224 | 0 | 35 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,526 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1810607/Python-O(n)-Solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
# Use the enumerate() function
# to look at both index and its associated value at once
for index, num in enumerate(nums):
# We can just return as soon as we see a match
if index % 10 == num:
return index
# Othe... | smallest-index-with-equal-value | Python O(n) Solution | White_Frost1984 | 0 | 20 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,527 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1721097/Python3-accepted-solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(len(nums)):
if(i%10 == nums[i]):
return i
return -1 | smallest-index-with-equal-value | Python3 accepted solution | sreeleetcode19 | 0 | 46 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,528 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1589587/Python3-Brute-Force-Solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(len(nums)):
if i % 10 == nums[i]:
return i
else:
return -1 | smallest-index-with-equal-value | [Python3] Brute Force Solution | terrencetang | 0 | 53 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,529 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1549966/Python-solution | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i in range(len(nums)):
if i %10 == nums[i]:
return i
return -1 | smallest-index-with-equal-value | Python solution | abkc1221 | 0 | 54 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,530 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2003813/Python-Solution-%2B-One-Liner! | class Solution:
def smallestEqual(self, nums):
for i,n in enumerate(nums):
if i % 10 == n: return i
return -1 | smallest-index-with-equal-value | Python - Solution + One Liner! | domthedeveloper | -1 | 38 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,531 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/2003813/Python-Solution-%2B-One-Liner! | class Solution:
def smallestEqual(self, nums):
return min((i for i,n in enumerate(nums) if i%10 == n), default=-1) | smallest-index-with-equal-value | Python - Solution + One Liner! | domthedeveloper | -1 | 38 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,532 |
https://leetcode.com/problems/smallest-index-with-equal-value/discuss/1550918/Python | class Solution:
def smallestEqual(self, nums: List[int]) -> int:
for i, n in enumerate(nums):
if n == i % 10:
return i
return -1 | smallest-index-with-equal-value | Python | blue_sky5 | -1 | 42 | smallest index with equal value | 2,057 | 0.712 | Easy | 28,533 |
https://leetcode.com/problems/find-the-minimum-and-maximum-number-of-nodes-between-critical-points/discuss/1551353/Python-O(1)-memory-O(n)-time-beats-100.00 | class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
min_res = math.inf
min_point = max_point = last_point = None
prev_val = head.val
head = head.next
i = 1
while head.next:
if ((head.next.val < head.val and prev_val... | find-the-minimum-and-maximum-number-of-nodes-between-critical-points | Python O(1) memory, O(n) time beats 100.00% | dereky4 | 1 | 125 | find the minimum and maximum number of nodes between critical points | 2,058 | 0.57 | Medium | 28,534 |
https://leetcode.com/problems/find-the-minimum-and-maximum-number-of-nodes-between-critical-points/discuss/2246179/not-easy-to-understand-but-easy-solution-(90-faster) | class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
arr,dist=[],[]
diff=10 ** 5
index=0
while head:
arr.append(head.val)
head=head.next
for i in range(1,len(arr)-1):
if (arr[i-1] > arr[i] and arr[i+... | find-the-minimum-and-maximum-number-of-nodes-between-critical-points | not easy to understand, but easy solution (90% faster) | alenov | 0 | 15 | find the minimum and maximum number of nodes between critical points | 2,058 | 0.57 | Medium | 28,535 |
https://leetcode.com/problems/find-the-minimum-and-maximum-number-of-nodes-between-critical-points/discuss/1552796/One-pass-100-speed | class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
i = prev_val = first_critical_idx = prev_critical_idx = critical_idx = 0
min_diff = inf
found_two_critical = False
while head:
if i == 0:
prev_val = head.val
... | find-the-minimum-and-maximum-number-of-nodes-between-critical-points | One pass, 100% speed | EvgenySH | 0 | 45 | find the minimum and maximum number of nodes between critical points | 2,058 | 0.57 | Medium | 28,536 |
https://leetcode.com/problems/find-the-minimum-and-maximum-number-of-nodes-between-critical-points/discuss/1551192/Python3-one-pass-O(1)-space | class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
prev = head.val
node = head.next
dmin = inf
first = i = last = 0
while node and node.next:
i += 1
if prev < node.val > node.next.val or prev > node.val < n... | find-the-minimum-and-maximum-number-of-nodes-between-critical-points | [Python3] one-pass O(1) space | ye15 | 0 | 43 | find the minimum and maximum number of nodes between critical points | 2,058 | 0.57 | Medium | 28,537 |
https://leetcode.com/problems/find-the-minimum-and-maximum-number-of-nodes-between-critical-points/discuss/1550791/Python-3-(Self-Explanatory) | class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
if not head:
return [-1, -1]
prev, curr = head, head.next
idx = 1
res = []
while curr and curr.next:
if self.is_critical(prev.val, curr.val, curr.next... | find-the-minimum-and-maximum-number-of-nodes-between-critical-points | Python 3 (Self-Explanatory) | 1nnOcent | 0 | 21 | find the minimum and maximum number of nodes between critical points | 2,058 | 0.57 | Medium | 28,538 |
https://leetcode.com/problems/find-the-minimum-and-maximum-number-of-nodes-between-critical-points/discuss/1561343/PythonSolution-beats-97-Find-Minimum-and-Max-No.-of-Nodes | class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
'''
5->3->1->2->5->1->2
C C. C
c = [2,4,5]
'''
if head.next.next is None:
return [-1,-1]
def calc(head):
prev = head
... | find-the-minimum-and-maximum-number-of-nodes-between-critical-points | [Python]Solution beats 97% - Find Minimum and Max No. of Nodes | SaSha59 | -1 | 50 | find the minimum and maximum number of nodes between critical points | 2,058 | 0.57 | Medium | 28,539 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1550007/Python3-bfs | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
ans = 0
seen = {start}
queue = deque([start])
while queue:
for _ in range(len(queue)):
val = queue.popleft()
if val == goal: return ans
... | minimum-operations-to-convert-number | [Python3] bfs | ye15 | 17 | 1,200 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,540 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1550007/Python3-bfs | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
ans = 0
queue = deque([start])
visited = [False]*1001
while queue:
for _ in range(len(queue)):
val = queue.popleft()
if val == goal: return ans
... | minimum-operations-to-convert-number | [Python3] bfs | ye15 | 17 | 1,200 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,541 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1551110/BFS-Approach-oror-Well-coded-oror-Easy-to-understand | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
if start==goal:
return 0
q = [(start,0)]
seen = {start}
while q:
n,s = q.pop(0)
for num in nums:
for cand in [n+num,n-num,n^num]:
if cand==goal:
... | minimum-operations-to-convert-number | 📌📌 BFS Approach || Well-coded || Easy-to-understand 🐍 | abhi9Rai | 5 | 198 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,542 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1549955/py3-Simple-BFS | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
seen = set()
que = deque([(start, 0)])
while que:
item, cnt = que.popleft()
if item == goal:
return cnt
if item in seen:
co... | minimum-operations-to-convert-number | [py3] Simple BFS | mtx2d | 3 | 196 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,543 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/2211975/python-3-or-bfs-or-O(n)O(1) | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
q = collections.deque([(start, 0)])
visited = {start}
while q:
x, count = q.popleft()
count += 1
for num in nums:
for newX in (x + num, x -... | minimum-operations-to-convert-number | python 3 | bfs | O(n)/O(1) | dereky4 | 2 | 83 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,544 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1707143/Using-sets-for-new-numbers-84-speed | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
seen, level, ops = {start}, {start}, 0
while level:
new_level = set()
ops += 1
for x in level:
for n in nums:
for new_x in [x + n, x - n... | minimum-operations-to-convert-number | Using sets for new numbers, 84% speed | EvgenySH | 0 | 63 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,545 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1551203/Python3-BFS-100 | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
n = len(nums)
que = [start]
steps = 0
visited = set()
while que:
# print(que)
nxt = []
for x in que:
if x == goal:
... | minimum-operations-to-convert-number | [Python3] BFS 100% | oliver33697 | 0 | 43 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,546 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1551106/Python-3-Two-head-BFS | class Solution:
def minimumOperations(self, nums: List[int], start: int, target: int) -> int:
n = len(nums)
vis = defaultdict(lambda: float('inf'), {start: 0})
vis2 = defaultdict(lambda: float('inf'), {target: 0})
q = deque([(0, start, 0), (0, target, 1)])
nums = set(nums)
... | minimum-operations-to-convert-number | [Python 3] Two head BFS | chestnut890123 | 0 | 29 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,547 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1550084/Python-Efficient-%22BFS%22-Using-Set | class Solution:
def f(self, nums = [2,4,12], o = 2):
r = set()
for n in nums:
r.update([o+n, o-n, o^n])
return r
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
seen = set()
todo = set([start])
d = 1
while True:
r = set()
for n in todo:
r.update(self.f(nu... | minimum-operations-to-convert-number | [Python] Efficient "BFS" Using Set | datafireball | 0 | 36 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,548 |
https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/1550059/Python-Simple-BFS-solution | class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
nums.sort()
q = collections.deque([(start, 0)])
visited = set()
while q:
for _ in range(len(q)):
x, step = q.popleft()
if x == goal:
... | minimum-operations-to-convert-number | [Python] Simple BFS solution | nightybear | 0 | 24 | minimum operations to convert number | 2,059 | 0.473 | Medium | 28,549 |
https://leetcode.com/problems/check-if-an-original-string-exists-given-two-encoded-strings/discuss/1550012/Python3-dp | class Solution:
def possiblyEquals(self, s1: str, s2: str) -> bool:
def gg(s):
"""Return possible length"""
ans = [int(s)]
if len(s) == 2:
if s[1] != '0': ans.append(int(s[0]) + int(s[1]))
return ans
elif len(s) == 3:... | check-if-an-original-string-exists-given-two-encoded-strings | [Python3] dp | ye15 | 82 | 6,200 | check if an original string exists given two encoded strings | 2,060 | 0.407 | Hard | 28,550 |
https://leetcode.com/problems/check-if-an-original-string-exists-given-two-encoded-strings/discuss/1550012/Python3-dp | class Solution:
def possiblyEquals(self, s1: str, s2: str) -> bool:
def gg(s):
"""Return possible length."""
ans = {int(s)}
for i in range(1, len(s)):
ans |= {x+y for x in gg(s[:i]) for y in gg(s[i:])}
return ans
@ca... | check-if-an-original-string-exists-given-two-encoded-strings | [Python3] dp | ye15 | 82 | 6,200 | check if an original string exists given two encoded strings | 2,060 | 0.407 | Hard | 28,551 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1563707/Python3-sliding-window-O(N) | class Solution:
def countVowelSubstrings(self, word: str) -> int:
ans = 0
freq = defaultdict(int)
for i, x in enumerate(word):
if x in "aeiou":
if not i or word[i-1] not in "aeiou":
jj = j = i # set anchor
freq.clear()
... | count-vowel-substrings-of-a-string | [Python3] sliding window O(N) | ye15 | 16 | 1,900 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,552 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1594931/Python.-Time%3A-one-pass-O(N)-space%3A-O(1).-Not-a-sliding-window. | class Solution:
def countVowelSubstrings(self, word: str) -> int:
vowels = ('a', 'e', 'i', 'o', 'u')
result = 0
start = 0
vowel_idx = {}
for idx, c in enumerate(word):
if c in vowels:
if not vowel_idx:
start = idx ... | count-vowel-substrings-of-a-string | Python. Time: one pass O(N), space: O(1). Not a sliding window. | blue_sky5 | 12 | 654 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,553 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1563978/Python3-One-line-Solution | class Solution:
def countVowelSubstrings(self, word: str) -> int:
return sum(set(word[i:j+1]) == set('aeiou') for i in range(len(word)) for j in range(i+1, len(word))) | count-vowel-substrings-of-a-string | [Python3] One-line Solution | leefycode | 8 | 871 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,554 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1574341/python-straightforward-solution | class Solution:
def countVowelSubstrings(self, word: str) -> int:
count = 0
current = set()
for i in range(len(word)):
if word[i] in 'aeiou':
current.add(word[i])
for j in range(i+1, len(word)):
if word[j] in... | count-vowel-substrings-of-a-string | python straightforward solution | mqueue | 6 | 718 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,555 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1564159/Python3-Sliding-Window-O(N)-with-explanations | class Solution:
def countVowelSubstrings(self, word: str) -> int:
result = 0
vowels = 'aeiou'
# dictionary to record counts of each char
mp = defaultdict(lambda: 0)
for i, char in enumerate(word):
# if the current letter is a vowel
if char in vowels:
... | count-vowel-substrings-of-a-string | [Python3] Sliding Window O(N) with explanations | idzhanghao | 4 | 360 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,556 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/2670441/Python | class Solution:
def countVowelSubstrings(self, word: str) -> int:
fin=0
for i in range(len(word)):
res = set()
for j in range(i,len(word)):
if word[j] in 'aeiou':
res.add(word[j])
if len(res)>=5:
... | count-vowel-substrings-of-a-string | Python | naveenraiit | 1 | 97 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,557 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1626076/Easy-python3-solution-using-counter | class Solution:
def countVowelSubstrings(self, word: str) -> int:
n=len(word)
vowels="aeiou"
from collections import Counter
res=0
for i in range(n):
for j in range(i+4,n):
counter=Counter(word[i:j+1])
if all(count... | count-vowel-substrings-of-a-string | Easy python3 solution using counter | Karna61814 | 1 | 211 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,558 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/2843354/Python3-using-Set | class Solution:
def countVowelSubstrings(self, word: str) -> int:
if len(word) < 5: return 0
result = 0
for i in range (len(word)-4):
r = i + 5
while r <= len(word):
if "".join(sorted(set(word[i:r]))) == "aeiou":
result += 1
... | count-vowel-substrings-of-a-string | Python3 using Set | DNST | 0 | 2 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,559 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/2775965/Simple-approach-using-counter-%3A-) | class Solution:
def countVowelSubstrings(self, word: str) -> int:
# creating a list of vowels
v = list('aeiou')
# counter for vowels
counter = { i: 0 for i in v }
count = 0
# Outer loop runs 0 to n
for i in range(len(word)):
# checking either ... | count-vowel-substrings-of-a-string | Simple approach using counter : ) | aadarshvelu | 0 | 4 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,560 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1798228/1-Line-Python-Solution-oror-30-Faster-oror-Memory-less-than-99 | class Solution:
def countVowelSubstrings(self, word: str) -> int:
return sum(set(word[i:j]) == set('aeiou') for i in range(len(word)-4) for j in range(i+5,len(word)+1)) | count-vowel-substrings-of-a-string | 1-Line Python Solution || 30% Faster || Memory less than 99% | Taha-C | 0 | 267 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,561 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1563878/Sliding-window-with-subset-check | class Solution:
vowels = frozenset(["a", "e", "i", "o", "u"])
def countVowelSubstrings(self, word: str) -> int:
n = len(word)
counter = 0
for i in range(n - 4):
end = i + 4
check = word[i:end+1]
current = set(check)
only_vowels = current.issub... | count-vowel-substrings-of-a-string | Sliding window with subset check | 0hh98h55f | 0 | 86 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,562 |
https://leetcode.com/problems/count-vowel-substrings-of-a-string/discuss/1563777/Just-another-beginner's-Python3-solution | class Solution:
def countVowels(self, word: str) -> int:
vowels = ["a", "e", "i", "o", "u"]
res = []
vowels_cop = set(vowels.copy())
for j in range(len(word)):
if word[j] not in vowels:
continue
string = ""
for i in range(j, len(wo... | count-vowel-substrings-of-a-string | Just another beginner's Python3 solution | WoodlandXander | 0 | 229 | count vowel substrings of a string | 2,062 | 0.659 | Easy | 28,563 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1564075/Detailed-explanation-of-why-(len-pos)-*-(pos-%2B-1)-works | class Solution:
def countVowels(self, word: str) -> int:
count = 0
sz = len(word)
for pos in range(sz):
if word[pos] in 'aeiou':
count += (sz - pos) * (pos + 1)
return count | vowels-of-all-substrings | Detailed explanation of why (len - pos) * (pos + 1) works | bitmasker | 77 | 2,000 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,564 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1563701/(i-%2B-1)-*-(sz-i) | class Solution:
def countVowels(self, word: str) -> int:
return sum((i + 1) * (len(word) - i) for i, ch in enumerate(word) if ch in 'aeiou') | vowels-of-all-substrings | (i + 1) * (sz - i) | votrubac | 34 | 2,800 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,565 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1564031/Fastest-Python-Solution-or-108ms | class Solution:
def countVowels(self, word: str) -> int:
c, l = 0, len(word)
d = {'a':1, 'e':1,'i':1,'o':1,'u':1}
for i in range(l):
if word[i] in d:
c += (l-i)*(i+1)
return c | vowels-of-all-substrings | Fastest Python Solution | 108ms | the_sky_high | 5 | 270 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,566 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1794416/Python-oror-Simple-Math-and-Combinatorics | class Solution:
def countVowels(self, word: str) -> int:
vows = set("aeiou")
l = len(word)
s = 0
for i in range(l):
if word[i] in vows:
s += (i + 1) * (l - i)
return s | vowels-of-all-substrings | Python || Simple Math and Combinatorics | cherrysri1997 | 1 | 80 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,567 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1564858/Python-3-two-prefix-sums | class Solution:
def countVowels(self, word: str) -> int:
n = len(word)
# mark vowel
# 'aba' vowels = [1, 0, 1]
vowels = list(map(lambda x: int(x in 'aeiou'), word))
# add vowel count in each substring
# acc = [0, 1, 1, 2]
acc = list(accumulat... | vowels-of-all-substrings | [Python 3] two prefix sums | chestnut890123 | 1 | 68 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,568 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/2773836/Python3-or-O(n)-Solution | class Solution:
def countVowels(self, word: str) -> int:
counter = 0
ans = 0;n = len(word)
for i in range(len(word)):
if(word[i] in ["a","e","i","o","u"]):
counter+=1
ans+=(i+1)*counter-(n-i-1)*counter
return ans | vowels-of-all-substrings | Python3 | O(n) Solution | ty2134029 | 0 | 5 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,569 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/2257479/python-3-or-simple-O(n)-time-O(1)-space-solution | class Solution:
def countVowels(self, word: str) -> int:
count = vowelIndexSum = 0
vowels = {'a', 'e', 'i', 'o', 'u'}
for i, c in enumerate(word, start=1):
if c in vowels:
vowelIndexSum += i
count += vowelIndexSum
return count | vowels-of-all-substrings | python 3 | simple O(n) time, O(1) space solution | dereky4 | 0 | 143 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,570 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1565085/Simple-formula-100-speed | class Solution:
vowels = set("aeiou")
def countVowels(self, word: str) -> int:
len_word = len(word)
return sum((i + 1) * (len_word - i) for i, c in enumerate(word)
if c in Solution.vowels) | vowels-of-all-substrings | Simple formula, 100% speed | EvgenySH | 0 | 60 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,571 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1564177/Python3-one-line-solution | class Solution:
def countVowels(self, word: str) -> int:
return sum((i+1)*(len(word)-i) for i, val in enumerate(word) if val in 'aeiou') | vowels-of-all-substrings | [Python3] one line solution | idzhanghao | 0 | 34 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,572 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1564123/Python-Solution-Easy | class Solution:
def countVowels(self, word: str) -> int:
count = 0
vowel = ['a','e','i','o','u']
n = len(word)
for i,x in enumerate(word):
if x in vowel:
count += (n-i)*(i+1)
return count | vowels-of-all-substrings | [Python] Solution -Easy | SaSha59 | 0 | 50 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,573 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1563876/Clean-and-Concise-oror-Well-Explained-oror-Easy-Approach | class Solution:
def countVowels(self, word: str) -> int:
vowel = "aeiou"
n, res = len(word), 0
for i,c in enumerate(word):
if c in vowel:
res+=(n-i)*(i+1)
return res | vowels-of-all-substrings | 📌📌 Clean & Concise || Well-Explained || Easy Approach 🐍 | abhi9Rai | 0 | 40 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,574 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1563837/Python3-greedy-1-line-O(N) | class Solution:
def countVowels(self, word: str) -> int:
return sum((i+1)*(len(word)-i) for i, x in enumerate(word) if x in "aeiou") | vowels-of-all-substrings | [Python3] greedy 1-line O(N) | ye15 | 0 | 61 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,575 |
https://leetcode.com/problems/vowels-of-all-substrings/discuss/1626199/Easy-python3-solution-using-prefix-sum | class Solution:
def countVowels(self, word: str) -> int:
vowels="aeiou"
n=len(word)
prefix=[0]
for s in word:
prefix.append(prefix[-1]+ (s in vowels))
prefix.pop(0)
total=sum(prefix)
diff=0
res=total
for i in range(1,n):
... | vowels-of-all-substrings | Easy python3 solution using prefix sum | Karna61814 | -1 | 99 | vowels of all substrings | 2,063 | 0.551 | Medium | 28,576 |
https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/1563731/Python3-binary-search | class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
lo, hi = 1, max(quantities)
while lo < hi:
mid = lo + hi >> 1
if sum(ceil(qty/mid) for qty in quantities) <= n: hi = mid
else: lo = mid + 1
return lo | minimized-maximum-of-products-distributed-to-any-store | [Python3] binary search | ye15 | 9 | 296 | minimized maximum of products distributed to any store | 2,064 | 0.5 | Medium | 28,577 |
https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/1563991/Python-Binary-Search-with-explain | class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
l, r = 1, max(quantities)
while l <= r:
mid = (l+r)//2
# count the number of stores needed if the max distribution is mid
# one store can has max one product only and we need to dis... | minimized-maximum-of-products-distributed-to-any-store | [Python] Binary Search with explain | nightybear | 2 | 133 | minimized maximum of products distributed to any store | 2,064 | 0.5 | Medium | 28,578 |
https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/1563800/Heap-oror-Well-Explained-oror-Visualizable | class Solution:
def minimizedMaximum(self, n, A):
heap = []
for q in A:
heapq.heappush(heap,(-q,1))
n-=1
while n>0:
quantity,num_shops = heapq.heappop(heap)
total = (-1)*quantity*num_shops
num_shops+=1
n-=1
heapq.heappush(heap,(-1*(total/... | minimized-maximum-of-products-distributed-to-any-store | 📌📌 Heap || Well-Explained || Visualizable 🐍 | abhi9Rai | 2 | 168 | minimized maximum of products distributed to any store | 2,064 | 0.5 | Medium | 28,579 |
https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/2808229/Python-Readable-solution-using-the-bisect-library | class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
search_range = range(1, max(quantities))
return bisect.bisect_left(search_range, 1,
key=lambda x: self.canDistribute(n, quantities, x)) + 1
def canDistribute(self, n: int, quantities: List[int], k: in... | minimized-maximum-of-products-distributed-to-any-store | [Python] Readable solution using the bisect library | lefteris12 | 0 | 3 | minimized maximum of products distributed to any store | 2,064 | 0.5 | Medium | 28,580 |
https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/1564122/Python-binary-search | class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
l, h = math.ceil(sum(quantities)/n), max(quantities)
getCount = lambda x: sum(math.ceil(quantities[i]/x) for i in range(len(quantities)))
while l < h:
mid = l + (h - l)//2
# if getting a l... | minimized-maximum-of-products-distributed-to-any-store | Python binary search | abkc1221 | 0 | 47 | minimized maximum of products distributed to any store | 2,064 | 0.5 | Medium | 28,581 |
https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/1580955/Binary-search-96-speed | class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
left, right = 1, max(quantities)
while left < right:
middle = (left + right) // 2
if sum(ceil(q / middle) for q in quantities) > n:
left = middle + 1
else:
... | minimized-maximum-of-products-distributed-to-any-store | Binary search, 96% speed | EvgenySH | -1 | 101 | minimized maximum of products distributed to any store | 2,064 | 0.5 | Medium | 28,582 |
https://leetcode.com/problems/maximum-path-quality-of-a-graph/discuss/1564102/Python-DFS-and-BFS-solution | class Solution:
def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
ans = 0
graph = collections.defaultdict(dict)
for u, v, t in edges:
graph[u][v] = t
graph[v][u] = t
def dfs(curr, visited, score, cost):
... | maximum-path-quality-of-a-graph | [Python] DFS and BFS solution | nightybear | 2 | 424 | maximum path quality of a graph | 2,065 | 0.576 | Hard | 28,583 |
https://leetcode.com/problems/maximum-path-quality-of-a-graph/discuss/1564102/Python-DFS-and-BFS-solution | class Solution:
def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
ans = 0
graph = collections.defaultdict(dict)
for u,v,t in edges:
graph[u][v] = t
graph[v][u] = t
# node, cost, visited, score
q = co... | maximum-path-quality-of-a-graph | [Python] DFS and BFS solution | nightybear | 2 | 424 | maximum path quality of a graph | 2,065 | 0.576 | Hard | 28,584 |
https://leetcode.com/problems/maximum-path-quality-of-a-graph/discuss/1563743/Python3-iterative-dfs | class Solution:
def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
graph = [[] for _ in values]
for u, v, t in edges:
graph[u].append((v, t))
graph[v].append((u, t))
ans = 0
stack = [(0, values[0], 0, 1)]
... | maximum-path-quality-of-a-graph | [Python3] iterative dfs | ye15 | 2 | 210 | maximum path quality of a graph | 2,065 | 0.576 | Hard | 28,585 |
https://leetcode.com/problems/maximum-path-quality-of-a-graph/discuss/2812642/Python-BFS-%2B-Pruning-with-Dijktra | class Solution:
def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
graph = defaultdict(list)
for outgoing, incoming, distance in edges:
graph[outgoing].append((incoming, distance))
graph[incoming].append((outgoing, distance))
... | maximum-path-quality-of-a-graph | Python BFS + Pruning with Dijktra | PIG208 | 0 | 5 | maximum path quality of a graph | 2,065 | 0.576 | Hard | 28,586 |
https://leetcode.com/problems/maximum-path-quality-of-a-graph/discuss/2206944/Python-DFS-Recursive | class Solution:
def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
d = defaultdict(list)
n = len(values)
for edge in edges:
d[edge[0]].append([edge[1], edge[2]])
d[edge[1]].append([edge[0], edge[2]])
... | maximum-path-quality-of-a-graph | [Python] DFS Recursive | happysde | 0 | 76 | maximum path quality of a graph | 2,065 | 0.576 | Hard | 28,587 |
https://leetcode.com/problems/maximum-path-quality-of-a-graph/discuss/1818481/Python-DFS-%2B-Djikstra-or-faster-than-88.89-1095ms | class Solution:
def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
graph = defaultdict(list)
dist = defaultdict(int)
parent = defaultdict(int)
R2Bpath = defaultdict(list)
for u, v, w in edges:
graph[u].append( (v, w) )
... | maximum-path-quality-of-a-graph | [Python] DFS + Djikstra | faster than 88.89%, 1095ms | ark10806 | 0 | 252 | maximum path quality of a graph | 2,065 | 0.576 | Hard | 28,588 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/1576339/Python3-freq-table | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
freq = [0]*26
for x in word1: freq[ord(x)-97] += 1
for x in word2: freq[ord(x)-97] -= 1
return all(abs(x) <= 3 for x in freq) | check-whether-two-strings-are-almost-equivalent | [Python3] freq table | ye15 | 2 | 140 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,589 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2513742/Python-Concise-Array-Solution-No-CounterHash-Map-Beats-98 | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
# make an array to track occurences for every letter of the
# alphabet
alphabet = [0]*26
# go through both words and count occurences
# word 1 add and word 2 subtract
# ... | check-whether-two-strings-are-almost-equivalent | [Python] - Concise Array Solution - No Counter/Hash-Map - Beats 98% | Lucew | 1 | 314 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,590 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2840266/Python-1-line%3A-Optimal-and-Clean-with-explanation-2-ways%3A-O(n)-time-and-O(1)-space | class Solution:
# use two frequency counters.
# O(n) time : O(26) = O(1) space
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
freq1, freq2 = Counter(word1), Counter(word2)
for c,f in freq1.items():
if abs(f - freq2[c]) > 3: return False
fo... | check-whether-two-strings-are-almost-equivalent | Python 1 line: Optimal and Clean with explanation - 2 ways: O(n) time and O(1) space | topswe | 0 | 4 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,591 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2836607/Simple-python-solution-using-hashmap | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
dic1 = defaultdict(int)
dic2 = defaultdict(int)
for i in word1:
dic1[i] += 1
for j in word2:
dic2[j] += 1
for k,v in dic1.items():
if abs(v-dic2[k]) >... | check-whether-two-strings-are-almost-equivalent | Simple python solution using hashmap | aruj900 | 0 | 5 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,592 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2829247/Python-Solution-explained-oror-HASHMAP | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
d1={}
d2={}
#FREQUENCIES OF WORD1 CHARACTERS
for i in word1:
if i in d1:
d1[i]+=1
else:
d1[i]=1
#FREQUENCIES OF WORD2 CHARACTERS
... | check-whether-two-strings-are-almost-equivalent | Python Solution - explained || HASHMAP✔ | T1n1_B0x1 | 0 | 6 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,593 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2705949/97-faster-easy-using-counter | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
f=list(set(word1+word2))
word1=Counter(word1)
word2=Counter(word2)
for i in f:
k=word1[i]
l=word2[i]
if(abs(k-l)>3):
return False
return True | check-whether-two-strings-are-almost-equivalent | 97% faster easy using counter | Raghunath_Reddy | 0 | 37 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,594 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2674202/Frequency-%22Tug-of-War%22-Between-Two-Words-Without-a-Dictionary | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
counts = [0] * 26
for char1, char2 in zip(word1, word2):
counts[ord(char1) - 97] += 1
counts[ord(char2) - 97] -= 1
return all(abs(count) <= 3 for count in counts) | check-whether-two-strings-are-almost-equivalent | Frequency "Tug of War" Between Two Words Without a Dictionary | kcstar | 0 | 8 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,595 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2664075/Easy-Python-Solution-Using-Dictionary | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
dic1={}
for i in word1:
if i not in dic1:
dic1[i]=1
else:
dic1[i]+=1
dic2={}
for i in word2:
if i not in dic2:
dic2[i]=... | check-whether-two-strings-are-almost-equivalent | Easy Python Solution Using Dictionary | ankitr8055 | 0 | 11 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,596 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2646466/Python3-Just-Look-At-Each-Character-(Counter) | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
c1, c2 = Counter(word1), Counter(word2)
for ch in set(word1+word2):
if abs(c1[ch]-c2[ch]) > 3:
return False
return True | check-whether-two-strings-are-almost-equivalent | Python3 Just Look At Each Character (Counter) | godshiva | 0 | 11 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,597 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2616741/Python-Simple-Python-Solution-By-Merging-Both-Word-or-92-Faster | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
merge_string = word1 + word2
for index in range(len(merge_string)):
frequency_count_in_word1 = word1.count(merge_string[index])
frequency_count_in_word2 = word2.count(merge_string[index])
if abs(frequency_count_in_word1 ... | check-whether-two-strings-are-almost-equivalent | [ Python ] ✅✅ Simple Python Solution By Merging Both Word | 92% Faster 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 0 | 45 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,598 |
https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/discuss/2586448/Python-Using-two-lists-O(n) | class Solution:
def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
if len(word1) != len(word2):
return False
freq1 = [0] * 256
freq2 = [0] * 256
for char in word1:
freq1[ord(char)] += 1
for char in word2:
fre... | check-whether-two-strings-are-almost-equivalent | Python - Using two lists - O(n) | aj1904 | 0 | 39 | check whether two strings are almost equivalent | 2,068 | 0.646 | Easy | 28,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.