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/rearrange-spaces-between-words/discuss/1686209/Easy-to-understand-python-solution | class Solution:
def reorderSpaces(self, text: str) -> str:
countSpace = text.count(" ")
countWords = len(text.split())
## If only one word is present, then add all the spaces at the end
if(countWords == 1):
return ''.join(text.split()[0] + " "*countSpace)
## If numb... | rearrange-spaces-between-words | Easy to understand python solution | htalanki2211 | 0 | 55 | rearrange spaces between words | 1,592 | 0.437 | Easy | 23,400 |
https://leetcode.com/problems/rearrange-spaces-between-words/discuss/1442988/Python3-Faster-Than-96.81 | class Solution:
def reorderSpaces(self, text: str) -> str:
t, spaces = text.split(), text.count(" ")
not_spaces = len(t)
if spaces == 0:
return text
if not_spaces == 1:
return text.strip() + " " * spaces
ratio, left = divmod(spaces, ... | rearrange-spaces-between-words | Python3 Faster Than 96.81% | Hejita | 0 | 73 | rearrange spaces between words | 1,592 | 0.437 | Easy | 23,401 |
https://leetcode.com/problems/rearrange-spaces-between-words/discuss/1335450/python-3-easy-solution-for-beginners | class Solution:
def reorderSpaces(self, text: str) -> str:
count=0
newt=""
rem=0
words=text.split()
for i in text:
if i==" ":
count+=1
if len(words)<=1:
newt=words[0]+" "*(count)
else:
each_sp=count... | rearrange-spaces-between-words | python 3 easy solution for beginners | minato_namikaze | 0 | 46 | rearrange spaces between words | 1,592 | 0.437 | Easy | 23,402 |
https://leetcode.com/problems/rearrange-spaces-between-words/discuss/1246119/Python3-simple-solution-using-split()-and-count() | class Solution:
def reorderSpaces(self, text: str) -> str:
x = text.count(' ')
n = [i for i in text.split(' ') if len(i) >= 1]
if len(n) == 1:
return n[0] + ' '*x
spaces = x//(len(n)-1)
extra = x%(len(n)-1)
return (' '*spaces).join(n) + extra*' ' | rearrange-spaces-between-words | Python3 simple solution using split() and count() | EklavyaJoshi | 0 | 44 | rearrange spaces between words | 1,592 | 0.437 | Easy | 23,403 |
https://leetcode.com/problems/rearrange-spaces-between-words/discuss/1105505/Easy-to-Understand-no-headache | class Solution:
def reorderSpaces(self, text: str) -> str:
spaces=0
for i in text:
if i==" ":
spaces=spaces+1
words=text.split()
if len(words)==1:
return (" ".join(words)+" "*spaces)
equal=(spaces//(len(words)-1))*" "
ext... | rearrange-spaces-between-words | Easy to Understand no headache | ashish87 | 0 | 56 | rearrange spaces between words | 1,592 | 0.437 | Easy | 23,404 |
https://leetcode.com/problems/rearrange-spaces-between-words/discuss/944342/Intuitive-approach | class Solution:
def reorderSpaces(self, text: str) -> str:
words = text.strip().split()
if len(words) == 1:
return words[0] + (" " * text.count(" "))
else:
space_num = text.count(" ")
d, r = divmod(space_num, len(words)-1)
return (" " * d).join... | rearrange-spaces-between-words | Intuitive approach | puremonkey2001 | 0 | 38 | rearrange spaces between words | 1,592 | 0.437 | Easy | 23,405 |
https://leetcode.com/problems/rearrange-spaces-between-words/discuss/855387/Python-3-or-String-split-count-6-lines-or-Explanations | class Solution:
def reorderSpaces(self, text: str) -> str:
space = text.count(' ') # count how many space in total
text = [word for word in text.split(' ') if word] # split text to individual word in a list
n = len(text) # coun... | rearrange-spaces-between-words | Python 3 | String, split, count, 6 lines | Explanations | idontknoooo | 0 | 94 | rearrange spaces between words | 1,592 | 0.437 | Easy | 23,406 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/855405/Python-3-or-Backtracking-DFS-clean-or-Explanations | class Solution:
def maxUniqueSplit(self, s: str) -> int:
ans, n = 0, len(s)
def dfs(i, cnt, visited):
nonlocal ans, n
if i == n: ans = max(ans, cnt); return # stop condition
for j in range(i+1, n+1):
if s[i:j] in visited: continue # avoid... | split-a-string-into-the-max-number-of-unique-substrings | Python 3 | Backtracking, DFS, clean | Explanations | idontknoooo | 5 | 849 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,407 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/855071/Python3-backtracking | class Solution:
def maxUniqueSplit(self, s: str) -> int:
def fn(i):
"""Find max length via backtracking."""
nonlocal ans
if i == len(s): return (ans := max(ans, len(tabu)))
for ii in range(i+1, len(s)+1):
if s[i:ii] not in tabu:
... | split-a-string-into-the-max-number-of-unique-substrings | [Python3] backtracking | ye15 | 4 | 801 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,408 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/855071/Python3-backtracking | class Solution:
def maxUniqueSplit(self, s: str) -> int:
def fn(i, seen=set()):
"""Find max length via backtracking."""
ans = 0
if i < len(s): # boundary condition when i == len(s)
for ii in range(i+1, len(s)+1):
if s[i:ii] no... | split-a-string-into-the-max-number-of-unique-substrings | [Python3] backtracking | ye15 | 4 | 801 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,409 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/1749644/Easy-using-python3-and-backtracking | class Solution:
def maxUniqueSplit(self, s: str) -> int:
ans = 0
n = len(s)
def solve(index,current,vis):
nonlocal ans,n
if(index == n):
ans = max(ans,current)
return
for i in range(index,n):
# print(s[index:i])
if(s[index:i+1] not in vis):
solve(i+1,current+1,vis+(s[index:i+1],))
... | split-a-string-into-the-max-number-of-unique-substrings | Easy using python3 and backtracking | jagdishpawar8105 | 1 | 209 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,410 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/855790/Simple-python3-backtracking-solution-w-explanation | class Solution:
def maxUniqueSplit(self, s: str) -> int:
def maxU(s, soFar=set()):
if len(s) == 1 and s in soFar:
return 0
maxSplit = len(soFar)+1
for partition in range(1, len(s)):
a = s[:partition]
b = s[partition:]
... | split-a-string-into-the-max-number-of-unique-substrings | Simple python3 backtracking solution w/ explanation | popestr | 1 | 132 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,411 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/2786620/Python-backtracking-and-pruning | class Solution:
def maxUniqueSplit_fun(self, s: str, cur_list, sub_set, res) -> int:
if len(cur_list) > 0:
sub = s[cur_list[-1]:]
if len(sub) > 0 and sub not in sub_set:
res[0] = max(res[0], len(cur_list) + 1)
if len(cur_list) + len(s) - cur_list[-1] - 1 ... | split-a-string-into-the-max-number-of-unique-substrings | Python backtracking and pruning | laofu_laoluzi | 0 | 4 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,412 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/2473823/Python3-or-Intuitive-Recursion-%2B-Backtracking-Approach | class Solution:
#Time-Complexity: O(n^n * n*n), since rec. tree has worst case branching factor and depth of both n! #-> Also, every rec. call will loop at most n times and the splicing operation takes linear time, which
#will copy in worst case each and every n characters! -> O(n^(n+2)) -> O(n^n)
#Space-Complexity: O... | split-a-string-into-the-max-number-of-unique-substrings | Python3 | Intuitive Recursion + Backtracking Approach | JOON1234 | 0 | 45 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,413 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/1899045/Python-3-BackTracking | class Solution:
def maxUniqueSplit(self, s: str) -> int:
mx = [0]
def dfs(ansset, remain):
nonlocal mx
if remain == "":
if len(ansset) >= mx[0]:
mx[0] = len(ansset)
for i in range(len(remain)+1):
insert ... | split-a-string-into-the-max-number-of-unique-substrings | Python 3 BackTracking | DietCoke777 | 0 | 88 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,414 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/1847270/Python3-Easy-understanding-backtracking-solution-beats-90-speed-and-100-memory | class Solution:
def maxUniqueSplit(self, s: str) -> int:
def backtrack(string, tempSet):
nonlocal cnt
if string == "": cnt = max(cnt, len(tempSet))
for i in range(1, len(string)+1):
substring = string[:i]
if substring in tempSet: co... | split-a-string-into-the-max-number-of-unique-substrings | [Python3] Easy understanding backtracking solution, beats 90% speed and 100% memory | freetochoose | 0 | 171 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,415 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/1847270/Python3-Easy-understanding-backtracking-solution-beats-90-speed-and-100-memory | class Solution:
def maxUniqueSplit(self, s: str) -> int:
def backtrack(string, tempSet):
nonlocal cnt
if string == "":
cnt = max(cnt, len(tempSet))
for offsetIdx in range(len(string)):
for i in range(len(string)):
... | split-a-string-into-the-max-number-of-unique-substrings | [Python3] Easy understanding backtracking solution, beats 90% speed and 100% memory | freetochoose | 0 | 171 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,416 |
https://leetcode.com/problems/split-a-string-into-the-max-number-of-unique-substrings/discuss/1402642/Simple-Backtracking-Easy-to-understand-python-3 | class Solution:
def maxUniqueSplit(self, s: str) -> int:
strings = set()
self.max_len = 0
def split(s) :
if not s :
self.max_len = max(self.max_len, len(strings))
return
for i in range(1,len(s)+1) :
cs = s... | split-a-string-into-the-max-number-of-unique-substrings | Simple Backtracking Easy to understand python 3 | Manideep8 | -1 | 126 | split a string into the max number of unique substrings | 1,593 | 0.55 | Medium | 23,417 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/855131/Python3-top-down-dp | class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
@lru_cache(None)
def fn(i, j):
"""Return maximum & minimum products ending at (i, j)."""
if i == 0 and j == 0: return grid[0][0], grid[0][0]
... | maximum-non-negative-product-in-a-matrix | [Python3] top-down dp | ye15 | 36 | 1,800 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,418 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/855377/Python-3-or-DP-O(m*n)-time-In-place-or-Explanation | class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
grid[0][0] = grid[0][0], grid[0][0] # (small, large) for starting point
for j in range(1, n):
grid[0][j] = grid[0][j-1][0]*grid[0][j], grid[... | maximum-non-negative-product-in-a-matrix | Python 3 | DP O(m*n) time, In place | Explanation | idontknoooo | 6 | 327 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,419 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/2373966/DFS-oror-Dynamic-Programming-oror-Memoization-oror-Python3 | class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
def isBound(row,col):
return 0<=row<len(grid) and 0<=col<len(grid[0])
@lru_cache(None)
def dfs(row,col,product):
if not isBound(row, col):
return -1
if row... | maximum-non-negative-product-in-a-matrix | DFS || Dynamic Programming || Memoization || Python3 | Sefinehtesfa34 | 1 | 133 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,420 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/2816531/Python-(Simple-DP) | class Solution:
def maxProductPath(self, grid):
m, n = len(grid), len(grid[0])
max_ = [[0]*n for _ in range(m)]
min_ = [[0]*n for _ in range(m)]
max_[0][0] = min_[0][0] = grid[0][0]
for i in range(1,m):
max_[i][0] = grid[i][0]*max_[i-1][0]
min_[i][0... | maximum-non-negative-product-in-a-matrix | Python (Simple DP) | rnotappl | 0 | 2 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,421 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/2701373/Python-Simple-DP-Solution | class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
numRows, numCols = len(grid), len(grid[0])
maxProduct, minProduct = [[0] * numCols for _ in range(numRows)], [[0] * numCols for _ in range(numRows)]
maxProduct[0][0], minProduct[0][0] = grid[0][0], grid[0][0]
for... | maximum-non-negative-product-in-a-matrix | Python Simple DP Solution | hemantdhamija | 0 | 13 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,422 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/1849714/Python-easy-to-read-and-understand-or-DP | class Solution:
def dfs(self, grid, row, col, d):
if row == len(grid)-1 and col == len(grid[0])-1:
return (grid[row][col], grid[row][col])
if row == len(grid) or col == len(grid[0]):
return (float("-inf"), float("inf"))
if grid[row][col] == 0:
return (0, 0... | maximum-non-negative-product-in-a-matrix | Python easy to read and understand | DP | sanial2001 | 0 | 92 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,423 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/1809378/Runtime%3A-40-ms-faster-than-100.00-of-Python3-online-submissions | class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
m,n=len(grid),len(grid[0])
dp=[[[float('-inf'),float('inf')]]*(n+1) for i in range(m+1) ]
dp[m-1][n-1]=[grid[-1][-1],grid[-1][-1]]
for i in range(m-1,-1,-1):
for j in range(n-1,-1,-1):
... | maximum-non-negative-product-in-a-matrix | Runtime: 40 ms, faster than 100.00% of Python3 online submissions | laxminarayanak | 0 | 41 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,424 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/1289365/Intuitive-Python-Memoization-Easy-understand-Solution | class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
m=len(grid)
n=len(grid[0])
dp=[[-1]*n for i in range(m)]
def solve(i,j):
if i==m-1 and j==n-1:
return grid[m-1][n-1],grid[m-1][n-1]
if i>=m or j>=n:
... | maximum-non-negative-product-in-a-matrix | Intuitive Python Memoization - Easy understand Solution | coder1311 | 0 | 139 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,425 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/855140/Python3-Dynamic-Programming-with-explanation | class Solution:
mod=pow(10,9)+7
def maxProductPath(self, grid: List[List[int]]) -> int:
n,m=len(grid),len(grid[0])
maxPath = [[0 for i in range(m)] for j in range(n)]
minPath = [[0 for i in range(m)] for j in range(n)]
for i in range(n):
for j in range(m):
... | maximum-non-negative-product-in-a-matrix | [Python3] Dynamic Programming with explanation | pranav05 | 0 | 57 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,426 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/855094/Python-DP-Top-Down | class Solution:
def maxProductPath(self, mat: List[List[int]]) -> int:
mod = 10**9+7
m, n = len(mat), len(mat[0])
@lru_cache(None)
def dp(i, j):
if i >= m or j >= n: return (-math.inf, math.inf)
curr = mat[i][j]
if i == m-1 and j ... | maximum-non-negative-product-in-a-matrix | [Python] DP Top Down | carloscerlira | 0 | 98 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,427 |
https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/1493852/DP-oror-For-Beginners-oror-Well-Explained-oror-97-faster | class Solution:
def maxProductPath(self, grid: List[List[int]]) -> int:
MOD = 10**9+7
m,n = len(grid),len(grid[0])
dp = [[[0,0] for _ in range(n)] for _ in range(m)]
dp[0][0][0]=grid[0][0]
dp[0][0][1]=grid[0][0]
for i in range(1,m): #First Column
dp[i][0][0] = dp... | maximum-non-negative-product-in-a-matrix | 📌📌 DP || For Beginners || Well-Explained || 97% faster 🐍 | abhi9Rai | -1 | 240 | maximum non negative product in a matrix | 1,594 | 0.33 | Medium | 23,428 |
https://leetcode.com/problems/minimum-cost-to-connect-two-groups-of-points/discuss/858187/Python3-top-down-dp | class Solution:
def connectTwoGroups(self, cost: List[List[int]]) -> int:
m, n = len(cost), len(cost[0])
mn = [min(x) for x in zip(*cost)] # min cost of connecting points in 2nd group
@lru_cache(None)
def fn(i, mask):
"""Return min cost of connecting group1[i:] ... | minimum-cost-to-connect-two-groups-of-points | [Python3] top-down dp | ye15 | 1 | 169 | minimum cost to connect two groups of points | 1,595 | 0.463 | Hard | 23,429 |
https://leetcode.com/problems/minimum-cost-to-connect-two-groups-of-points/discuss/858187/Python3-top-down-dp | class Solution:
def connectTwoGroups(self, cost: List[List[int]]) -> int:
m, n = len(cost), len(cost[0])
mn = [min(x) for x in cost] # min cost of connecting points in 1st group
@lru_cache(None)
def fn(j, mask):
"""Return min cost of connecting group1[i:] and gr... | minimum-cost-to-connect-two-groups-of-points | [Python3] top-down dp | ye15 | 1 | 169 | minimum cost to connect two groups of points | 1,595 | 0.463 | Hard | 23,430 |
https://leetcode.com/problems/crawler-log-folder/discuss/866343/Python3-straightforward | class Solution:
def minOperations(self, logs: List[str]) -> int:
ans = 0
for log in logs:
if log == "./": continue
elif log == "../": ans = max(0, ans-1) # parent directory
else: ans += 1 # child directory
return ans | crawler-log-folder | [Python3] straightforward | ye15 | 11 | 627 | crawler log folder | 1,598 | 0.644 | Easy | 23,431 |
https://leetcode.com/problems/crawler-log-folder/discuss/2102858/PYTHON-or-Super-Easy-python-solution | class Solution:
def minOperations(self, logs: List[str]) -> int:
res = 0
for i in logs:
if i == '../' and res > 0:
res -= 1
elif i != './' and i != '../':
res += 1
return res | crawler-log-folder | PYTHON | Super Easy python solution | shreeruparel | 3 | 123 | crawler log folder | 1,598 | 0.644 | Easy | 23,432 |
https://leetcode.com/problems/crawler-log-folder/discuss/2507874/Intuitive-code-and-easily-understandable-in-Python | class Solution:
def minOperations(self, logs: List[str]) -> int:
res = 0
for i in logs:
if i == './':
continue
elif i == '../':
if res > 0:
res -= 1
else:
res += 1
return res | crawler-log-folder | Intuitive code and easily understandable in Python | ankurbhambri | 1 | 46 | crawler log folder | 1,598 | 0.644 | Easy | 23,433 |
https://leetcode.com/problems/crawler-log-folder/discuss/1291927/Easy-Python-Solution(93.94) | class Solution:
def minOperations(self, logs: List[str]) -> int:
stack=[]
for i in logs:
if(i=='../' and stack):
stack.pop()
elif(i=='./'):
continue
elif(i!='../' ):
stack.append(i)
return len(stack) | crawler-log-folder | Easy Python Solution(93.94%) | Sneh17029 | 1 | 140 | crawler log folder | 1,598 | 0.644 | Easy | 23,434 |
https://leetcode.com/problems/crawler-log-folder/discuss/2836646/Simple-Python-solution-beats-90 | class Solution:
def minOperations(self, logs: List[str]) -> int:
steps = 0
for i in logs:
if i == "../" and steps != 0:
steps -= 1
elif i == "./" or i == "../" and steps == 0:
continue
else:
steps += 1
return... | crawler-log-folder | Simple Python solution beats 90% | aruj900 | 0 | 1 | crawler log folder | 1,598 | 0.644 | Easy | 23,435 |
https://leetcode.com/problems/crawler-log-folder/discuss/2703615/Easy-or-Python-or-1-loop | class Solution:
def minOperations(self, logs: List[str]) -> int:
check=[]
for i in logs:
if i=="../" and len(check)!=0:
check.pop()
elif i=="./":
continue
else:
if i!="./" and i!="../":
check.appe... | crawler-log-folder | Easy | Python | 1 loop | pranshuvishnoi85 | 0 | 4 | crawler log folder | 1,598 | 0.644 | Easy | 23,436 |
https://leetcode.com/problems/crawler-log-folder/discuss/2673874/Python-solution | class Solution:
def minOperations(self, logs: List[str]) -> int:
listof= []
for l in logs:
if l == '../':
if not (len(listof) == 0):
listof.pop()
elif l == './':
pass
else:
listof.append(l)
... | crawler-log-folder | Python solution | Sheeza | 0 | 4 | crawler log folder | 1,598 | 0.644 | Easy | 23,437 |
https://leetcode.com/problems/crawler-log-folder/discuss/2650218/Easy-Python-solution-with-explanation | class Solution:
def minOperations(self, logs: List[str]) -> int:
# Initiate a blank list
shortest_dir_path = []
for i in range(len(logs)):
if logs[i] == "../":
# If str element = ../ , then remove last element of the shortestdirpath list only if it ain't empty. Do absolutely not... | crawler-log-folder | Easy Python solution with explanation | code_snow | 0 | 12 | crawler log folder | 1,598 | 0.644 | Easy | 23,438 |
https://leetcode.com/problems/crawler-log-folder/discuss/2632873/Python-solution-(easy) | class Solution:
def minOperations(self, logs: List[str]) -> int:
ans = 0
for i in logs:
if i == "./":
continue
elif i == "../":
if ans > 0:
ans -= 1
else:
ans += 1
return ans | crawler-log-folder | Python solution (easy) | rohansardar | 0 | 10 | crawler log folder | 1,598 | 0.644 | Easy | 23,439 |
https://leetcode.com/problems/crawler-log-folder/discuss/2614540/Simple-and-easy-way-or-Python-or-faster-than-97 | class Solution(object):
def minOperations(self, logs):
#only define two entries, one have to move folder back, other have to do nothing, means we don't need to count it. If no statment is in dictionary, it means we have created folder
#We must have to define "./" in dictionary, otherwise it will count and conside... | crawler-log-folder | Simple and easy way | Python | faster than 97% | msherazedu | 0 | 26 | crawler log folder | 1,598 | 0.644 | Easy | 23,440 |
https://leetcode.com/problems/crawler-log-folder/discuss/2484737/Python-Stack-99.54-faster-or-Simplest-solution-with-explanation-or-Beg-to-Adv-or-Stack | class Solution:
def minOperations(self, logs: List[str]) -> int:
stack = [] # taking empty stack
for elem in logs: # traversing through the list.
if elem!="../" and elem!="./": # As these are operations as we cant have them in stack as a directory.
stack.append... | crawler-log-folder | Python Stack 99.54% faster | Simplest solution with explanation | Beg to Adv | Stack | rlakshay14 | 0 | 39 | crawler log folder | 1,598 | 0.644 | Easy | 23,441 |
https://leetcode.com/problems/crawler-log-folder/discuss/2319446/Easy-python-stack | class Solution:
def minOperations(self, logs: List[str]) -> int:
stack=[]
for i in logs:
if stack:
if i=="../":
stack.pop()
elif i=="./":
pass
else:
stack.append(i)
... | crawler-log-folder | Easy python stack | sunakshi132 | 0 | 49 | crawler log folder | 1,598 | 0.644 | Easy | 23,442 |
https://leetcode.com/problems/crawler-log-folder/discuss/2032657/Python3-or-Easy-Solution-or-Steps-Counter-or-Time-O(N) | class Solution:
def minOperations(self, logs: List[str]) -> int:
steps = 0
for x in logs:
if x == "../":
if steps != 0:
steps -= 1
elif x == "./":
pass
else:
steps += 1
... | crawler-log-folder | Python3 | Easy Solution | Steps Counter | Time O(N) | PeterPierinakos | 0 | 19 | crawler log folder | 1,598 | 0.644 | Easy | 23,443 |
https://leetcode.com/problems/crawler-log-folder/discuss/2032633/Python-Stack-implementation | class Solution:
def minOperations(self, logs: List[str]) -> int:
stack = []
for i in logs:
if i == "./":
pass
elif i=="../":
if len(stack)>0:
stack.pop()
else:
stack.append(i[0:len(i)-1])
... | crawler-log-folder | Python Stack implementation | Vamsidhar01 | 0 | 27 | crawler log folder | 1,598 | 0.644 | Easy | 23,444 |
https://leetcode.com/problems/crawler-log-folder/discuss/1998240/Python-Solution-or-Over-96-Faster-or-Straightforward-Logic | class Solution:
def minOperations(self, logs: List[str]) -> int:
deep = 0
for op in logs:
if op == "./":
continue
elif op == "../":
if deep > 0:
deep -= 1
else:
... | crawler-log-folder | Python Solution | Over 96% Faster | Straightforward Logic | Gautam_ProMax | 0 | 19 | crawler log folder | 1,598 | 0.644 | Easy | 23,445 |
https://leetcode.com/problems/crawler-log-folder/discuss/1862411/Python-Simple-Solution | class Solution:
def minOperations(self, logs: List[str]) -> int:
result = 0
for op in logs:
# The if-else conditions are ordered this way to pass certain test cases
if op == "./":
continue
elif op == "../":
# Checks if there are multiple '../' ... | crawler-log-folder | Python Simple Solution | White_Frost1984 | 0 | 14 | crawler log folder | 1,598 | 0.644 | Easy | 23,446 |
https://leetcode.com/problems/crawler-log-folder/discuss/1759358/Python-dollarolution-(Faster-than-97) | class Solution:
def minOperations(self, logs: List[str]) -> int:
count = 0
for i in logs:
if i[0:2] == '..':
if count > 0:
count -= 1
elif i[0:2] != './':
count += 1
return count | crawler-log-folder | Python $olution (Faster than 97%) | AakRay | 0 | 35 | crawler log folder | 1,598 | 0.644 | Easy | 23,447 |
https://leetcode.com/problems/crawler-log-folder/discuss/1575498/44-ms-and-14.4-MB-easy-python-solution | class Solution:
def minOperations(self, logs: List[str]) -> int:
operations=0
for i in logs:
if i!="./" and i!="../":
operations+=1
if i=="../" and operations>0:
operations-=1
return operations | crawler-log-folder | 44 ms and 14.4 MB easy python solution | fluturandra2 | 0 | 20 | crawler log folder | 1,598 | 0.644 | Easy | 23,448 |
https://leetcode.com/problems/crawler-log-folder/discuss/1420963/O(1)-space-solution-in-Python | class Solution:
def minOperations(self, logs: List[str]) -> int:
i, n, cnt = 0, len(logs), 0
while i < n:
if logs[i] == "../":
if cnt > 0:
cnt -= 1
elif logs[i] != "./":
cnt += 1
i += 1
retur... | crawler-log-folder | O(1) space solution in Python | mousun224 | 0 | 52 | crawler log folder | 1,598 | 0.644 | Easy | 23,449 |
https://leetcode.com/problems/crawler-log-folder/discuss/1327951/PythonororSimple-logicoror-o(1)-space-o(n)-time-oror-No-length-of-array-No-stack | class Solution:
def minOperations(self, logs: List[str]) -> int:
step=0
# steps till main
for x in logs:
if x=="../":
if step:
step-=1
#one folder removed move one step back or stay at main
elif x!="./":
step+=1
#... | crawler-log-folder | Python||Simple logic|| o(1) space o(n) time || No length of array No stack | ana_2kacer | 0 | 26 | crawler log folder | 1,598 | 0.644 | Easy | 23,450 |
https://leetcode.com/problems/crawler-log-folder/discuss/1312588/Python-O(1)-Space-O(n)-time-Solution | class Solution:
def minOperations(self, logs: List[str]) -> int:
operations = 0
for log in logs:
string = log[0:2]
if string == "..":
operations -= 1 if operations > 0 else 0
elif string != "./":
operations += 1
return opera... | crawler-log-folder | Python O(1) Space, O(n) time Solution | ramit_kumar | 0 | 34 | crawler log folder | 1,598 | 0.644 | Easy | 23,451 |
https://leetcode.com/problems/crawler-log-folder/discuss/1267841/python-or-easy-or-beats-99.22-runtime-and-99.22-less-memory-or | class Solution:
def minOperations(self, logs: List[str]) -> int:
ans=[]
while logs:
a= logs.pop(0)
if a=='../':
if len(ans)>0:
ans.pop()
else:
continue
... | crawler-log-folder | python | easy | beats 99.22% runtime and 99.22% less memory | | chikushen99 | 0 | 50 | crawler log folder | 1,598 | 0.644 | Easy | 23,452 |
https://leetcode.com/problems/crawler-log-folder/discuss/1159046/Python-using-stack | class Solution:
def minOperations(self, logs: List[str]) -> int:
st = list()
for op in logs:
if not st and op == "../":
continue
elif op == "../":
st.pop()
elif op == "./":
continue
else:
... | crawler-log-folder | Python using stack | keewook2 | 0 | 28 | crawler log folder | 1,598 | 0.644 | Easy | 23,453 |
https://leetcode.com/problems/crawler-log-folder/discuss/1139489/Python-93-faster-(Stacks)-98-faster-(without-Stacks) | class Solution:
def minOperations(self, logs: List[str]) -> int:
Operations = ['./', '../']
Stack = []
for op in logs:
if op not in Operations:
Stack.append(op)
else:
if op == '../':
if... | crawler-log-folder | Python - 93% faster (Stacks) - 98% faster (without Stacks) | piyushagg19 | 0 | 40 | crawler log folder | 1,598 | 0.644 | Easy | 23,454 |
https://leetcode.com/problems/crawler-log-folder/discuss/1139489/Python-93-faster-(Stacks)-98-faster-(without-Stacks) | class Solution:
def minOperations(self, logs: List[str]) -> int:
steps = 0
for op in logs:
if op == './':
continue
elif op == '../':
if not steps == 0:
steps -= 1
... | crawler-log-folder | Python - 93% faster (Stacks) - 98% faster (without Stacks) | piyushagg19 | 0 | 40 | crawler log folder | 1,598 | 0.644 | Easy | 23,455 |
https://leetcode.com/problems/crawler-log-folder/discuss/1060102/Python3-simple-solution | class Solution:
def minOperations(self, logs: List[str]) -> int:
count = 0
for i in logs:
if i.startswith('..'):
if count > 0:
count -= 1
elif i.startswith('.'):
pass
else:
count += 1
retu... | crawler-log-folder | Python3 simple solution | EklavyaJoshi | 0 | 21 | crawler log folder | 1,598 | 0.644 | Easy | 23,456 |
https://leetcode.com/problems/crawler-log-folder/discuss/915202/Easy-Python-O(n)-Solution-with-Explanation | class Solution:
def minOperations(self, logs: List[str]) -> int:
res = 0
for i in logs:
if i == './': pass
elif i == '../': res = max(0, res-1)
else: res += 1
return res | crawler-log-folder | Easy Python O(n) Solution with Explanation | shadow_sm36 | 0 | 30 | crawler log folder | 1,598 | 0.644 | Easy | 23,457 |
https://leetcode.com/problems/crawler-log-folder/discuss/897949/Intuitive-approach-by-ifelse | class Solution:
def minOperations(self, logs: List[str]) -> int:
depth = 0
for a in logs:
if a == './':
continue
elif a == '../':
depth = max(0, depth-1)
else:
depth += 1
return depth | crawler-log-folder | Intuitive approach by if/else | puremonkey2001 | 0 | 25 | crawler log folder | 1,598 | 0.644 | Easy | 23,458 |
https://leetcode.com/problems/crawler-log-folder/discuss/1196008/python3-solution-with-stack | class Solution:
def minOperations(self, logs: List[str]) -> int:
res=[]
for word in logs:
if word=="../" and len(res)>0:
res.pop()
elif word=="./":
continue
elif word!="../" and word!="./":
res.append(word)
r... | crawler-log-folder | python3 solution with stack | janhaviborde23 | -1 | 26 | crawler log folder | 1,598 | 0.644 | Easy | 23,459 |
https://leetcode.com/problems/maximum-profit-of-operating-a-centennial-wheel/discuss/866356/Python3-simulation | class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = -1
most = pnl = waiting = 0
for i, x in enumerate(customers):
waiting += x # more people waiting in line
waiting -= (chg := min(4, waiting)) # b... | maximum-profit-of-operating-a-centennial-wheel | [Python3] simulation | ye15 | 5 | 381 | maximum profit of operating a centennial wheel | 1,599 | 0.436 | Medium | 23,460 |
https://leetcode.com/problems/maximum-profit-of-operating-a-centennial-wheel/discuss/866482/Python3-O(n) | class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
n=len(customers)
dp=[]
reserved=0
on_board=0
rotation=0
for i in range(n):
if reserved!=0:
if reserved>=4... | maximum-profit-of-operating-a-centennial-wheel | Python3 O(n) | harshitCode13 | 0 | 46 | maximum profit of operating a centennial wheel | 1,599 | 0.436 | Medium | 23,461 |
https://leetcode.com/problems/maximum-profit-of-operating-a-centennial-wheel/discuss/866454/Python3-simple-self-explanatory-code | class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
i=0
count=0
profit=0
rem=0
onBoard=0
max_profit_rounds=-1
max_profit=0
while True:
if i>=len(customers)-1 and rem==0: ... | maximum-profit-of-operating-a-centennial-wheel | Python3 simple self explanatory code | tharun_99 | 0 | 58 | maximum profit of operating a centennial wheel | 1,599 | 0.436 | Medium | 23,462 |
https://leetcode.com/problems/maximum-number-of-achievable-transfer-requests/discuss/866369/Python3-10-Lines-Bitmasking-or-Combinations-or-Easy-Explanation | class Solution:
def maximumRequests(self, n: int, req: List[List[int]]) -> int:
tot = len(req)
for i in range(tot, 0, -1):
comb = list(itertools.combinations([j for j in range(tot)], i))
for c in comb:
net = [0 for j in range(n)]
for idx in c:
... | maximum-number-of-achievable-transfer-requests | [Python3] 10 Lines Bitmasking | Combinations | Easy Explanation | uds5501 | 17 | 1,000 | maximum number of achievable transfer requests | 1,601 | 0.513 | Hard | 23,463 |
https://leetcode.com/problems/maximum-number-of-achievable-transfer-requests/discuss/866390/Python3-dfs-w-bitmask | class Solution:
def maximumRequests(self, n: int, requests: List[List[int]]) -> int:
def fn(k, mask):
"""Return maximum number of achievable transfer requests."""
if k == len(requests):
net = [0]*n
for i, (u, v) in enumerate(requests):
... | maximum-number-of-achievable-transfer-requests | [Python3] dfs w/ bitmask | ye15 | 1 | 95 | maximum number of achievable transfer requests | 1,601 | 0.513 | Hard | 23,464 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/1284866/Python3-or-Dict-%2B-Sort | class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
key_time = {}
for index, name in enumerate(keyName):
key_time[name] = key_time.get(name, [])
key_time[name].append(int(keyTime[index].replace(":", "")))
ans = []
for nam... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | Python3 | Dict + Sort | Sanjaychandak95 | 3 | 314 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,465 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/876966/Python-Simple-python-solution | class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
mapp = {}
for i in range(len(keyName)):
name = keyName[i]
if(name not in mapp):
mapp[name] = [keyTime[i]]
else:
mapp[name].append(keyTime[i])... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | [Python] Simple python solution | AnupBS | 3 | 426 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,466 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/1868311/Python-3-or-Default-Dict-or-Times-to-Minutes | class Solution:
def get_minute(self, time):
hour, minute = map(int, time.split(':'))
return hour * 60 + minute
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
lookup = defaultdict(list)
for name, time in zip(keyName, keyTime):
look... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | Python 3 | Default Dict | Times to Minutes | leeteatsleep | 1 | 146 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,467 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/2794500/Python3-Simply-Solution-by-casting-time-to-index | class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
def cast_to_index(time):
hour, minutes = int(time[:2]), int(time[3:])
return hour*60 + minutes
dicts = defaultdict(list)
for i in range(len(keyName)):
name, time = keyNa... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | Python3 Simply Solution by casting time to index | xxHRxx | 0 | 7 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,468 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/2560509/Python3-solution-with-dictmap-and-sliding-window | class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
hmap = {}
for i in range(len(keyName)):
hmap[keyName[i]] = hmap.get(keyName[i], []) + [int("".join(keyTime[i].split(":")))]
results = []
for ids, times in hmap.ite... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | Python3 solution with dict/map and sliding window | WizardOfGryfindor | 0 | 69 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,469 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/2077820/Python3%3A-O(nlogn)-Solution-using-defaultdict-(Dictionary)-and-Sort | class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
"""
1. use dict to record each name's min
2. for each name in dict, check if within 60 min
"""
def getMin(time):
"""
turn hr:min to min
"""
... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | Python3: O(nlogn) Solution using defaultdict (Dictionary) and Sort | yunglinchang | 0 | 71 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,470 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/1732535/Python-Using-Deque | class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
d = defaultdict(collections.deque)
items = list(zip(keyName, iter(datetime.datetime.strptime(item, "%H:%M") for item in keyTime)))
items.sort(key=lambda l: l[1])
res = set()
for name, t... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | Python Using Deque | etherealoptimist | 0 | 66 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,471 |
https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/discuss/1088753/Python3-dict-of-deques | class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
ans = set()
seen = {}
for key, time in sorted(zip(keyName, keyTime)):
if key not in ans:
h, m = time.split(":")
time = int(h) * 60 + int(m)
... | alert-using-same-key-card-three-or-more-times-in-a-one-hour-period | [Python3] dict of deques | ye15 | 0 | 80 | alert using same key card three or more times in a one hour period | 1,604 | 0.473 | Medium | 23,472 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/1734833/Python-or-Backtracking | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
def backtrack(y, x):
choice = min(rowSum[y], colSum[x])
result[y][x] = choice
rowSum[y] -= choice
colSum[x] -= choice
if y == 0 and x == 0:
... | find-valid-matrix-given-row-and-column-sums | Python | Backtracking | holdenkold | 1 | 102 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,473 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/2792270/Python3-or-Moving-Extra-Sum-to-next-column-or-Explanation | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
n,m=len(rowSum),len(colSum)
matrix = [[0 for i in range(m)] for j in range(n)]
total = 0
for i in range(n):
matrix[i][0] = rowSum[i]
total += rowSum[i]
fo... | find-valid-matrix-given-row-and-column-sums | [Python3] | Moving Extra Sum to next column | Explanation | swapnilsingh421 | 0 | 2 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,474 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/2769890/Simple-Approach | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
R = sorted([r, i] for i, r in enumerate(rowSum))
C = sorted([c, j] for j, c in enumerate(colSum))
# print(R, C)
i, j, m, n = 0, 0, len(rowSum), len(colSum)
ans = [[0]*n for _ in ... | find-valid-matrix-given-row-and-column-sums | Simple Approach | Mencibi | 0 | 6 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,475 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/2270771/Python!-Explained-and-easy-to-understand! | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
"""
We can firstly push the number in smallest rowSum and colSum, and push the remain number.
Time comlexity: O(n)(sort use O(nlogn)), space comlexity: O(1)(if ignore the space of answer and sor... | find-valid-matrix-given-row-and-column-sums | Python! Explained and easy to understand! | XRFXRF | 0 | 66 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,476 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/2263157/Python-Solution | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m=len(rowSum)
n=len(colSum)
arr=[[0]*n for _ in range(m)]
i,j=0,0
while(i<m and j<n):
arr[i][j]=min(rowSum[i],colSum[j])
rowSum[i]-=arr[i][j]
colSum[j]-=arr[i][j]
j+=1
... | find-valid-matrix-given-row-and-column-sums | Python Solution | SakshiMore22 | 0 | 28 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,477 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/1265281/python-oror-clean-and-concise-oror-easy | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
n= len(rowSum)
m = len(colSum)
t=[ [0]*(m) for j in range(n)]
for i in range(n):
for j in range(m):
... | find-valid-matrix-given-row-and-column-sums | python || clean and concise || easy | chikushen99 | 0 | 134 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,478 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/1182626/Easy-and-very-simple-Approach | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
row=len(rowSum)
col=len(colSum)
res=[[0 for _ in range(col)] for _ in range(row)]
for i in range(row):
for j in range(col):
res[i][... | find-valid-matrix-given-row-and-column-sums | Easy and very simple Approach | jaipoo | 0 | 83 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,479 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/1088759/Python3-greedy | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m, n = len(rowSum), len(colSum) # dimensions
ans = [[0]*n for _ in range(m)]
pq = [(-y, j) for j, y in enumerate(colSum)] # min-heap
heapify(pq)
for i, x in e... | find-valid-matrix-given-row-and-column-sums | [Python3] greedy | ye15 | 0 | 126 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,480 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/1088759/Python3-greedy | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m, n = len(rowSum), len(colSum) # dimensions
ans = [[0]*n for _ in range(m)]
i = j = 0
while i < len(rowSum) and j < len(colSum):
ans[i][j] = min(rowSum[i], colSum... | find-valid-matrix-given-row-and-column-sums | [Python3] greedy | ye15 | 0 | 126 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,481 |
https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/discuss/877972/Intuitive-approach-by-keeping-all-row-sum-in-first-column-and-make-deduction-from-them | class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
# 0) Initialization
ROW_SIZE = len(rowSum)
COL_SIZE = len(colSum)
mtx = [[0] * COL_SIZE for ri in range(ROW_SIZE)]
for i, rs in enumerate(rowSum):
mtx[i][0] = rs
... | find-valid-matrix-given-row-and-column-sums | Intuitive approach by keeping all row sum in first column and make deduction from them | puremonkey2001 | 0 | 38 | find valid matrix given row and column sums | 1,605 | 0.78 | Medium | 23,482 |
https://leetcode.com/problems/find-servers-that-handled-most-number-of-requests/discuss/1089184/Python3-summarizing-3-approaches | class Solution:
def busiestServers(self, k: int, arrival: List[int], load: List[int]) -> List[int]:
busy = [] # min-heap
free = list(range(k)) # min-heap
freq = [0]*k
for i, (ta, tl) in enumerate(zip(arrival, load)):
while busy and busy[0][0] <= ta:
... | find-servers-that-handled-most-number-of-requests | [Python3] summarizing 3 approaches | ye15 | 13 | 641 | find servers that handled most number of requests | 1,606 | 0.429 | Hard | 23,483 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/1527669/Python-or-Faster-than-94-or-2-methods-or-O(nlogn) | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
if n<=nums[0]:
return n
for i in range(1,n):
count = n-i #counts number of elements in nums greater than equal i
if nums[i]>=(count) and (coun... | special-array-with-x-elements-greater-than-or-equal-x | Python | Faster than 94% | 2 methods | O(nlogn) | ana_2kacer | 5 | 531 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,484 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/1527669/Python-or-Faster-than-94-or-2-methods-or-O(nlogn) | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
if n<=nums[0]:
return n
#binary search
start,end = 0,n
while(start<=end):
mid = (start+end)//2
#index of middle ele... | special-array-with-x-elements-greater-than-or-equal-x | Python | Faster than 94% | 2 methods | O(nlogn) | ana_2kacer | 5 | 531 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,485 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2330480/Python-Simple-Python-Solution-Using-Two-Approach | class Solution:
def specialArray(self, nums: List[int]) -> int:
low = 0
high = 1000
while low <= high:
mid = ( low + high ) //2
count = 0
for current_number in nums:
if current_number >= mid:
count = count + 1
if mid == count:
return mid
elif mid < count:
low = mid + 1
... | special-array-with-x-elements-greater-than-or-equal-x | [ Python ] ✅✅ Simple Python Solution Using Two Approach 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 3 | 148 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,486 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2330480/Python-Simple-Python-Solution-Using-Two-Approach | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums = sorted(nums)[::-1]
for num in range(len(nums) + 1):
count = 0
for current_number in nums:
if current_number >= num:
count = count + 1
if count == num:
return num
return -1 | special-array-with-x-elements-greater-than-or-equal-x | [ Python ] ✅✅ Simple Python Solution Using Two Approach 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 3 | 148 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,487 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2709600/Python-O(n)-Solution | class Solution:
def specialArray(self, nums: List[int]) -> int:
freq=[0 for _ in range(max(nums)+1)]
for i in nums:
freq[i]+=1
suff=[freq[-1]]
for i in freq[::-1][1:]:
suff.append(suff[-1]+i)
suff=suff[::-1]
for i in range(max(nums)+1):
if suff[i]==i:
return i
return -1 | special-array-with-x-elements-greater-than-or-equal-x | Python - O(n) Solution | prateekgoel7248 | 2 | 179 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,488 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2392444/Python-oror-Faster-than-97.27-oror-Sort-and-Binary-Search | class Solution:
def specialArray(self, nums: List[int]) -> int:
#Sorting nums array
nums.sort()
for i in range(len(nums), 0, -1):
#Binary Search for i
low, high = 0, len(nums) - 1
while low <= high:
mid = (low + high) // 2... | special-array-with-x-elements-greater-than-or-equal-x | Python || Faster than 97.27% || Sort and Binary Search | dhruva3223 | 1 | 89 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,489 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/1932987/orRuntime%3A-faster-than-92.71-or-or-Memory-Usage%3A-less-than-98.29-or | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
for x in range(1,len(nums)+1):
s = 0
e = len(nums) - 1
while s <=e:
mid = (s + e)//2
if nums[mid] <x:
s= mi... | special-array-with-x-elements-greater-than-or-equal-x | |Runtime: faster than 92.71% | | Memory Usage: less than 98.29% | | Ashi_garg | 1 | 91 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,490 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/1787793/Python3-accepted-solution | class Solution:
def specialArray(self, nums: List[int]) -> int:
for i in range(1,len(nums)+1):
if(len([nums[j] for j in range(len(nums)) if(nums[j]>=i)]) == i):
return i
return -1 | special-array-with-x-elements-greater-than-or-equal-x | Python3 accepted solution | sreeleetcode19 | 1 | 41 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,491 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/1569042/python-sorting-solution | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.sort()
nums.insert(0, -1)
n = len(nums)
for i in range(1, n):
if nums[i-1] < n - i <= nums[i]:
return n - i
return -1 | special-array-with-x-elements-greater-than-or-equal-x | python sorting solution | dereky4 | 1 | 149 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,492 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/1473375/Sort-and-pass-once-98-speed | class Solution:
def specialArray(self, nums: List[int]) -> int:
len_nums = len(nums)
nums.sort(reverse=True)
if nums[0] == 1:
return 1
for i in range(1, len_nums):
if nums[i - 1] >= i > nums[i]:
return i
if nums[-1] >= len_nums:
... | special-array-with-x-elements-greater-than-or-equal-x | Sort and pass once, 98% speed | EvgenySH | 1 | 150 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,493 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/1084572/Python3-simple-solution | class Solution:
def specialArray(self, nums: List[int]) -> int:
k = -1
while k <= len(nums):
count = 0
for i in nums:
if i >= k:
count += 1
if k == count:
return k
k += 1
return -1 | special-array-with-x-elements-greater-than-or-equal-x | Python3 simple solution | EklavyaJoshi | 1 | 87 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,494 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2829483/Special-Array-With-X-Elements-Greater-Than-or-Equal-X-or-Python | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.sort(reverse=True)
left, right = 0, len(nums)
while left < right:
mid = left + (right - left) // 2
if mid < nums[mid]:
left = mid + 1
else:
right = mid ... | special-array-with-x-elements-greater-than-or-equal-x | Special Array With X Elements Greater Than or Equal X | Python | jashii96 | 0 | 1 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,495 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2813355/simple-python-n(log-n) | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.sort()
for i in range(nums[-1]+1):
low = 0
high = len(nums)
while(low<high):
mid = (low+high)//2
if nums[mid]<i:
low = mid+1
el... | special-array-with-x-elements-greater-than-or-equal-x | simple python n(log n) | sudharsan1000m | 0 | 1 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,496 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2801013/Help-needed-why-two-seemingly-identical-codes-give-different-results | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.append(-1)
nums = sorted(nums,reverse = True)
for i,x in enumerate(nums[:len(nums)-1],start=1):
if i <= x and i > nums[i]:
return i
return -1 | special-array-with-x-elements-greater-than-or-equal-x | Help needed - why two seemingly identical codes give different results? | piotr_swicarz | 0 | 10 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,497 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2801013/Help-needed-why-two-seemingly-identical-codes-give-different-results | class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.append(-1)
for i,x in enumerate(sorted(nums,reverse = True)[:len(nums)-1],start=1):
if i <= x and i > nums[i]:
return i
return -1 | special-array-with-x-elements-greater-than-or-equal-x | Help needed - why two seemingly identical codes give different results? | piotr_swicarz | 0 | 10 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,498 |
https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/discuss/2689789/Easy-To-Understand-Python-Solution | class Solution:
def specialArray(self, nums: List[int]) -> int:
n = len(nums)
while n != len([x for x in nums if x >= n]):
n -= 1
if n == 0:
return -1
return n | special-array-with-x-elements-greater-than-or-equal-x | Easy To Understand Python Solution | scifigurmeet | 0 | 1 | special array with x elements greater than or equal x | 1,608 | 0.601 | Easy | 23,499 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.