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/rotting-oranges/discuss/2771844/Chinese-Explanation-%2B-Python | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
# bfs
# 1. 定位rotten orange的坐标，并且save in visited and queue
# 2. start bfs
m,n = len(grid), len(grid[0])
queue = collections.deque()
# locate rotten orange
for i in range(m):
... | rotting-oranges | Chinese Explanation + Python | Michael_Songru | 0 | 2 | rotting oranges | 994 | 0.525 | Medium | 16,200 |
https://leetcode.com/problems/rotting-oranges/discuss/2749731/Efficient-Python-BFS-with-comments | class Solution:
'''
BFS approach
'''
def orangesRotting(self, grid: List[List[int]]) -> int:
queue = deque()
minutes, fresh = 0, 0
# Populate the initial queue with coordinates of 2's
# Count 1's as fresh
for i, row in enumerate(grid):
for j, v... | rotting-oranges | Efficient Python BFS with comments | decsery | 0 | 12 | rotting oranges | 994 | 0.525 | Medium | 16,201 |
https://leetcode.com/problems/rotting-oranges/discuss/2735774/Python-Easy-Solution-Time%3A-O(n)-Space%3A-O(n) | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
# record rotten and fresh oranges
rottens = []
freshs = []
for r in range(len(grid)):
for c in range(len(grid[0])):
if grid[r][c] == 1:
freshs.append((r,c))
... | rotting-oranges | Python Easy Solution Time: O(n) Space: O(n) | chienhsiang-hung | 0 | 9 | rotting oranges | 994 | 0.525 | Medium | 16,202 |
https://leetcode.com/problems/rotting-oranges/discuss/2707117/Simple-python-oror-beats-99-memory | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
mins=0 # store result
dirs = [(0,1),(1,0),(-1,0), (0,-1)] # movement directions
rotten_set = set()
total_r = 0 # total rotten in any minute
total_o = 0 # total oranges
m = len(grid)
n = ... | rotting-oranges | Simple python || beats 99% memory | user1090g | 0 | 6 | rotting oranges | 994 | 0.525 | Medium | 16,203 |
https://leetcode.com/problems/rotting-oranges/discuss/2687990/Simple-DFS-like-Solution-with-explanation | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
def Rotting(i, j): # function to make oranges rotten
if i >= 0 and i < len(grid) and j >= 0 and j < len(grid[0]) and grid[i][j] == 1:
grid[i][j] = 2
count = 0
prev = copy.deepcopy(grid) # us... | rotting-oranges | Simple DFS-like Solution with explanation | AustinHuang823 | 0 | 30 | rotting oranges | 994 | 0.525 | Medium | 16,204 |
https://leetcode.com/problems/rotting-oranges/discuss/2673232/python-! | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
q = deque()
m,n = len(grid), len(grid[0])
oranges = set()
for i in range(m):
for j in range(n):
if grid[i][j]==2:
q.append((i,j,0))
... | rotting-oranges | python ! | sanjeevpathak | 0 | 6 | rotting oranges | 994 | 0.525 | Medium | 16,205 |
https://leetcode.com/problems/rotting-oranges/discuss/2666239/python-BFS-with-readable-explanation | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
rows = len(grid)
cols = len(grid[0])
minutes = 0
directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
rotten = set()
fresh_count = 0
infected = set()
# go thru each cell to find the po... | rotting-oranges | python BFS with readable explanation | deezeey | 0 | 15 | rotting oranges | 994 | 0.525 | Medium | 16,206 |
https://leetcode.com/problems/rotting-oranges/discuss/2622386/python3-oror-easy-oror-bfs-solution | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
visited=[[0]*len(grid[0]) for i in range(len(grid))]
rowSize=len(grid)
colSize=len(grid[0])
q=collections.deque()
for r in range(rowSize):
... | rotting-oranges | python3 || easy || bfs solution | _soninirav | 0 | 11 | rotting oranges | 994 | 0.525 | Medium | 16,207 |
https://leetcode.com/problems/rotting-oranges/discuss/2610530/Simple-Python-solution-(faster-than-90)-with-detailed-explanation.-easy-to-understand. | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
M = len(grid) # num rows
N = len(grid[0]) # num cols
q = [] # queue
visited = [ [0]*N for _ in range(M)] #2D list
n_fresh = [0] # global variable (pointer)
def... | rotting-oranges | Simple Python solution (faster than 90%) with detailed explanation. easy to understand. | alexion1 | 0 | 37 | rotting oranges | 994 | 0.525 | Medium | 16,208 |
https://leetcode.com/problems/rotting-oranges/discuss/2600535/BREADTH-FIRST-SEARCH-APPROACH | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
"""
Each cell can have one of 3 values
0 - empty cell
1 - fresh orange
2 - rotten orange
output - time to turn all fresh oranges to rotten if possible
else return -1
... | rotting-oranges | BREADTH FIRST SEARCH APPROACH | leomensah | 0 | 42 | rotting oranges | 994 | 0.525 | Medium | 16,209 |
https://leetcode.com/problems/rotting-oranges/discuss/2541032/Python3-Solution-or-BFS | class Solution:
def orangesRotting(self, grid):
n, m = len(grid), len(grid[0])
q, ans = collections.deque(), -1
count = sum(row.count(1) for row in grid)
for i in range(n):
for j in range(m):
if grid[i][j] == 2:
q.append((i, j))
... | rotting-oranges | ✔ Python3 Solution | BFS | satyam2001 | 0 | 44 | rotting oranges | 994 | 0.525 | Medium | 16,210 |
https://leetcode.com/problems/rotting-oranges/discuss/2537000/Python-oror-BFS-oror-Queue-oror-97-fast | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
q = []
m = len(grid)
n = len(grid[0])
seen = set()
def makerot(i,j,q):
if (i,j) in seen:
return
seen.add((i,j))
if i+1 < m and (i+1,j) not in seen:
... | rotting-oranges | Python || BFS || Queue || 97% fast | 1md3nd | 0 | 70 | rotting oranges | 994 | 0.525 | Medium | 16,211 |
https://leetcode.com/problems/rotting-oranges/discuss/2470825/Clean-Fast-Python3-or-BFS | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
# for each fresh orange, bfs to nearest rotten one. Take max of these distances
rows, cols = len(grid), len(grid[0])
dirs = [(0, -1), (-1, 0), (0, 1), (1, 0)]
def bfs(start_row, start_col):
n... | rotting-oranges | Clean, Fast Python3 | BFS | ryangrayson | 0 | 47 | rotting oranges | 994 | 0.525 | Medium | 16,212 |
https://leetcode.com/problems/rotting-oranges/discuss/2417509/Python3-9792-Simple-Solution-w-Explanation | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
m = len(grid)
n = len(grid[0])
changing = True
infected = 1
# Loop through mxn changing grid entries until no entries are changed on a loop
while changing:
infected += 1
... | rotting-oranges | [Python3] 97%/92% Simple Solution w Explanation | connorthecrowe | 0 | 56 | rotting oranges | 994 | 0.525 | Medium | 16,213 |
https://leetcode.com/problems/rotting-oranges/discuss/2337831/Python3-knapsack-solution-with-mega-comprehension | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
rotten = frozenset((x, y) for y, row in enumerate(grid) for x, cell in enumerate(row) if cell == 2)
length = len(grid)
width = len(grid[0])
minutes = 0
while True:
neighbors = (cell for x, y i... | rotting-oranges | Python3 knapsack solution with mega comprehension | SkookumChoocher | 0 | 38 | rotting oranges | 994 | 0.525 | Medium | 16,214 |
https://leetcode.com/problems/rotting-oranges/discuss/2255567/Python-Readable-and-easy-to-understand-solution-with-explanation-using-only-a-queue | class Solution:
EMPTY = 0
FRESH = 1
ROTTEN = 2
ADJACENT_DIRECTIONS = [(-1, 0), (+1, 0), (0, -1), (0, +1)]
def orangesRotting(self, grid: List[List[int]]) -> int:
self.grid = grid
self.rows, self.columns = len(grid), len(grid[0])
rotten_coordinates_queue = self._get_rott... | rotting-oranges | [Python] Readable and easy to understand solution with explanation, using only a queue | julenn | 0 | 67 | rotting oranges | 994 | 0.525 | Medium | 16,215 |
https://leetcode.com/problems/rotting-oranges/discuss/2254223/Python-Basic-BFS-91-Less-Memory | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
def searchFreshAndRotten(grid):
freshes = 0
rots = []
for row in range(len(grid)):
for col in range(len(grid[row])):
if grid[row][col] == 1:
... | rotting-oranges | Python Basic BFS 91% Less Memory | codeee5141 | 0 | 54 | rotting oranges | 994 | 0.525 | Medium | 16,216 |
https://leetcode.com/problems/rotting-oranges/discuss/2116219/BFS-Solution-Python-(Time-87-Space-99) | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
# Breadth-first search
m, n = len(grid), len(grid[0])
minute = 0
qRotten, freshCount = [], 0
for i in range(m):
for j in range(n):
# We store rotten ... | rotting-oranges | BFS Solution - Python (Time 87%, Space 99%) | tylerpruitt | 0 | 92 | rotting oranges | 994 | 0.525 | Medium | 16,217 |
https://leetcode.com/problems/rotting-oranges/discuss/1988876/ororPYTHON-SOL-oror-VERY-EASY-oror-WELL-EXPLAINED-oror-JUST-AS-QUESTION-DEMANDS-oror-BFS-oror | class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
# 0 = means empty
# 1 = fresh orange
# 2 = rotten orange
# every minute any every rotten orange infects its adjacent fresh orange
# min no of minutes to make all rotten else -1
... | rotting-oranges | ||PYTHON SOL || VERY EASY || WELL EXPLAINED || JUST AS QUESTION DEMANDS || BFS || | reaper_27 | 0 | 77 | rotting oranges | 994 | 0.525 | Medium | 16,218 |
https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/discuss/2122927/Python-O(N)-S(N)-Queue-solution | class Solution:
def minKBitFlips(self, nums: List[int], k: int) -> int:
ans = 0
q = []
for i in range(len(nums)):
if len(q) % 2 == 0:
if nums[i] == 0:
if i+k-1 <= len(nums)-1:
ans += 1
q.append(i+... | minimum-number-of-k-consecutive-bit-flips | Python O(N) S(N) Queue solution | DietCoke777 | 0 | 64 | minimum number of k consecutive bit flips | 995 | 0.512 | Hard | 16,219 |
https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/discuss/1266069/Python3-greedy | class Solution:
def minKBitFlips(self, nums: List[int], k: int) -> int:
ans = flip = 0
queue = deque()
for i, x in enumerate(nums):
if queue and i == queue[0]:
flip ^= 1
queue.popleft()
if x == flip:
if len(nums) - i ... | minimum-number-of-k-consecutive-bit-flips | [Python3] greedy | ye15 | 0 | 186 | minimum number of k consecutive bit flips | 995 | 0.512 | Hard | 16,220 |
https://leetcode.com/problems/number-of-squareful-arrays/discuss/1375586/python-simple-backtracking.-20ms | class Solution(object):
def numSquarefulPerms(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
def dfs(temp,num,count = 0):
if len(num)==0:
return count+1
for i in xrange(len(num)):
if (i>0 and num[i]==num[i-1]) or... | number-of-squareful-arrays | python simple backtracking. 20ms | leah123 | 1 | 305 | number of squareful arrays | 996 | 0.492 | Hard | 16,221 |
https://leetcode.com/problems/number-of-squareful-arrays/discuss/1314226/Python3-TSP | class Solution:
def numSquarefulPerms(self, nums: List[int]) -> int:
@cache
def fn(v, mask):
"""Return squareful arrays given prev value and mask."""
if not mask: return 1
ans = 0
seen = set()
for i, x in enumerate(nums):
... | number-of-squareful-arrays | [Python3] TSP | ye15 | 1 | 152 | number of squareful arrays | 996 | 0.492 | Hard | 16,222 |
https://leetcode.com/problems/number-of-squareful-arrays/discuss/2674701/Python | class Solution:
def numSquarefulPerms(self, nums: List[int]) -> int:
n = len(nums)
nums.sort()
def dfs(prev,rem):
if not rem:
return 1
ans = 0
for i,a in enumerate(rem):
if (i > 0 and a == rem[i-1]) or (prev != -1 and mat... | number-of-squareful-arrays | Python | Akhil_krish_na | 0 | 4 | number of squareful arrays | 996 | 0.492 | Hard | 16,223 |
https://leetcode.com/problems/number-of-squareful-arrays/discuss/2447950/Python-3-Backtrack | class Solution:
def numSquarefulPerms(self, nums: List[int]) -> int:
def is_perfect(v):
k=int(math.sqrt(v))
return k*k==v
nums.sort()
def perm(A,prev):
if len(A)==0:
self.res+=1
return
for j in range(len(A)):
if j>0 and A[j]==A[j-1]:continue
if prev is None:
perm(A[:j]+A[j+1:],A... | number-of-squareful-arrays | [Python 3] Backtrack | gabhay | 0 | 40 | number of squareful arrays | 996 | 0.492 | Hard | 16,224 |
https://leetcode.com/problems/number-of-squareful-arrays/discuss/1991455/orPYTHON-SOL-or-BACKTRACKING-%2B-HASHMAP-or-SIMPLE-SOLUTION-or-WELL-EXPLAINED-or | class Solution:
def isSquare(self,num):
return int(num**0.5)**2 == num
def makePermutation(self,used,vis,prev,n):
if used == n:
#we reached the end
self.ans += 1
return
tmp = {}
for i in range(n):
if vis[i] == False and self.nums[i... | number-of-squareful-arrays | |PYTHON SOL | BACKTRACKING + HASHMAP | SIMPLE SOLUTION | WELL EXPLAINED | | reaper_27 | 0 | 96 | number of squareful arrays | 996 | 0.492 | Hard | 16,225 |
https://leetcode.com/problems/number-of-squareful-arrays/discuss/1691089/Python-optimal-backtrackingdfs-solution-(clean-code) | class Solution:
def numSquarefulPerms(self, nums: List[int]) -> int:
@lru_cache
def square(m):
left, right = 1, m
while left < right:
mid = (left + right) // 2
if mid**2 >= m:
right = mid
else:
... | number-of-squareful-arrays | Python optimal backtracking/dfs solution (clean code) | byuns9334 | 0 | 125 | number of squareful arrays | 996 | 0.492 | Hard | 16,226 |
https://leetcode.com/problems/number-of-squareful-arrays/discuss/831137/Similar-as-problem-47-and-use-DFS | class Solution:
def numSquarefulPerms(self, A: List[int]) -> int:
res = []
visited = [0] * len(A)
A.sort()
def helper(nums,out, res):
if len(out) == len(nums):
res.append(out[:])
return
else:
for i in range(len(A)):
... | number-of-squareful-arrays | Similar as problem 47 and use DFS | jppooo888 | 0 | 116 | number of squareful arrays | 996 | 0.492 | Hard | 16,227 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663344/C%2B%2BJavaPython3Javascript-Everything-you-need-to-know-from-start-to-end-. | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
Trusted = [0] * (N+1)
for (a, b) in trust:
Trusted[a] -= 1
Trusted[b] += 1
for i in range(1, len(Trusted)):
if Trusted[i] == N-1:
return i
retu... | find-the-town-judge | [C++/Java/Python3/Javascript] Everything you need to know from start to end . | Cosmic_Phantom | 144 | 9,500 | find the town judge | 997 | 0.493 | Easy | 16,228 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663192/Python3-EASY-TO-UNDERSTAND-CODE-Explained | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
trust_to, trusted = defaultdict(int), defaultdict(int)
for a, b in trust:
trust_to[a] += 1
trusted[b] += 1
for i in range(1, n+1):
if trust_to[i] == 0 and trusted[i] == n... | find-the-town-judge | ✔️ [Python3] EASY TO UNDERSTAND CODE, Explained | artod | 7 | 864 | find the town judge | 997 | 0.493 | Easy | 16,229 |
https://leetcode.com/problems/find-the-town-judge/discuss/1621150/O(n)-solution-in-Python | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
dg = [0] * (n + 1)
for a, b in trust:
dg[a] -= 1 # out
dg[b] += 1 # in
return next((i for i in range(1, n + 1) if dg[i] == n - 1), -1) | find-the-town-judge | O(n) solution in Python | mousun224 | 3 | 266 | find the town judge | 997 | 0.493 | Easy | 16,230 |
https://leetcode.com/problems/find-the-town-judge/discuss/404001/Python-Logical-solution.-No-hashing-required. | class Solution(object):
def findJudge(self, N, trust):
if trust==[] and N==1:
return 1
x1 = [x[1] for x in trust]
x0 = [x[0] for x in trust]
for i in range(1, N+1):
if i in x1:
if x1.count(i)==(N-1):
if i not in x0:
return i
return -1 | find-the-town-judge | Python Logical solution. No hashing required. | saffi | 3 | 877 | find the town judge | 997 | 0.493 | Easy | 16,231 |
https://leetcode.com/problems/find-the-town-judge/discuss/1664882/Python-easy-and-clean-solution-with-full-explanation | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
no_of_trust = [0] * (n+1) #because in trust numbers starts from 1 to N
for a,b in trust:
no_of_trust[a] -= 1 # a trusts b so a will become less
no_of_trust[b] += 1 # a trusts b so b will become more t... | find-the-town-judge | Python easy and clean solution with full explanation | yashitanamdeo | 2 | 201 | find the town judge | 997 | 0.493 | Easy | 16,232 |
https://leetcode.com/problems/find-the-town-judge/discuss/382225/Two-Short-Solutions-in-Python-3 | class Solution:
def findJudge(self, n: int, t: List[List[int]]) -> int:
N = set(range(1,n+1))
for i in t: N.discard(i[0])
if len(N) == 0: return -1
a = list(N)[0]
return a if sum(i[1] == a for i in t) == n-1 else -1 | find-the-town-judge | Two Short Solutions in Python 3 | junaidmansuri | 2 | 479 | find the town judge | 997 | 0.493 | Easy | 16,233 |
https://leetcode.com/problems/find-the-town-judge/discuss/382225/Two-Short-Solutions-in-Python-3 | class Solution:
def findJudge(self, n: int, t: List[List[int]]) -> int:
N = set(range(1,n+1))
for i in t: N.discard(i[0])
return (lambda x: x if sum(i[1] == x for i in t) == n-1 and len(N) == 1 else -1)(list(N)[0] if len(N) != 0 else -1)
- Junaid Mansuri
(LeetCode ID)@hotmail.com | find-the-town-judge | Two Short Solutions in Python 3 | junaidmansuri | 2 | 479 | find the town judge | 997 | 0.493 | Easy | 16,234 |
https://leetcode.com/problems/find-the-town-judge/discuss/242937/Python3-List-O(N)-space-O(N)-time | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
a = [0] * (N + 1)
for l in trust:
a[l[1]] += 1
a[l[0]] -= 1
for i in range(1, len(a)):
if a[i] == N - 1:
return i
return -1 | find-the-town-judge | Python3 List O(N) space, O(N) time | jimmyyentran | 2 | 319 | find the town judge | 997 | 0.493 | Easy | 16,235 |
https://leetcode.com/problems/find-the-town-judge/discuss/2515433/Efficient-Python-Solution-or-Memory-less-than-91.50 | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if not trust and n == 1:
return 1
degree = [0 for i in range(0,n+1)]
for u, v in trust:
degree[u] -= 1 #indegree = -1 for that node
degree[v] += 1 #outdegree = +1 for that node
for i in degree:
if i == (n - 1):
return... | find-the-town-judge | Efficient Python Solution | Memory less than 91.50% | nikhitamore | 1 | 73 | find the town judge | 997 | 0.493 | Easy | 16,236 |
https://leetcode.com/problems/find-the-town-judge/discuss/1810855/Python-3-hashmap-solution | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
trustCount = collections.Counter()
trustedCount = collections.Counter()
for a, b in trust:
trustCount[a] += 1
trustedCount[b] += 1
for i in range(1, n + 1):
... | find-the-town-judge | Python 3, hashmap solution | dereky4 | 1 | 195 | find the town judge | 997 | 0.493 | Easy | 16,237 |
https://leetcode.com/problems/find-the-town-judge/discuss/1664607/Python3-Explanation-and-Intuition-of-complete-solution. | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
trusted_by = [0] * n
for a, b in trust:
[a - 1] -= 1
trusted_by[b - 1] += 1
for i in range(n):
if trusted_by[i] == n - 1:
return i + 1
return -1 | find-the-town-judge | [Python3] Explanation and Intuition of complete solution. | Crimsoncad3 | 1 | 66 | find the town judge | 997 | 0.493 | Easy | 16,238 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663944/Python3-2-liner-and-one-liner | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
counts= collections.Counter([edge for p1,p2 in trust for edge in ((p1,0),(0,p2))])
return next(itertools.chain((p for p in range(1,n+1) if counts[(0,p)]-counts[(p,0)] == n-1),[-1])) | find-the-town-judge | Python3 2-liner and one-liner | pknoe3lh | 1 | 80 | find the town judge | 997 | 0.493 | Easy | 16,239 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663944/Python3-2-liner-and-one-liner | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
return (lambda counts: next(itertools.chain((p for p in range(1,n+1) if counts[(0,p)]-counts[(p,0)] == n-1),[-1])))(collections.Counter([edge for p1,p2 in trust for edge in ((p1,0),(0,p2))])) | find-the-town-judge | Python3 2-liner and one-liner | pknoe3lh | 1 | 80 | find the town judge | 997 | 0.493 | Easy | 16,240 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663227/Python3-O(V-%2B-E)-or-Clean-%2B-Simple-Solution-or-In-Degree-and-Out-Degree | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
inDegree = [0] * n
outDegree = [0] * n
for node, neighb in trust:
inDegree[neighb - 1] += 1
outDegree[node - 1] += 1
for i, (inD, outD) in enumerate(zip(inDegree, outDegree)):... | find-the-town-judge | ✅ [Python3] O(V + E) | Clean + Simple Solution | In-Degree & Out-Degree | PatrickOweijane | 1 | 207 | find the town judge | 997 | 0.493 | Easy | 16,241 |
https://leetcode.com/problems/find-the-town-judge/discuss/1258626/Python3-solution-faster-100 | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
first = []
second = []
if n==1 and len(trust)==0:
return 1
for i in trust:
first.append(i[0])
second.append(i[1])
x = list((set(second)-set(first)))
if len(... | find-the-town-judge | Python3 solution faster 100% | Sanyamx1x | 1 | 253 | find the town judge | 997 | 0.493 | Easy | 16,242 |
https://leetcode.com/problems/find-the-town-judge/discuss/1219967/Python3-simple-solution-using-two-lists | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if n == 1:
return 1
l1 = list(range(1,n+1))
l2 = []
for i in trust:
if i[0] in l1:
l1.remove(i[0])
if i[1] in l1:
l2.append(i[1])
fo... | find-the-town-judge | Python3 simple solution using two lists | EklavyaJoshi | 1 | 97 | find the town judge | 997 | 0.493 | Easy | 16,243 |
https://leetcode.com/problems/find-the-town-judge/discuss/891198/Python3-simple-using-iteration-and-degree | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
inDegree = [0]*N
outDegree = [0]*N
for a,b in trust:
outDegree[a-1] += 1
inDegree[b-1] += 1
for i in range(N):
if outDegree[i] == 0 and inDegree[i] == N-1... | find-the-town-judge | Python3 - simple using iteration and degree | gargprat | 1 | 112 | find the town judge | 997 | 0.493 | Easy | 16,244 |
https://leetcode.com/problems/find-the-town-judge/discuss/760354/Python-3Find-the-Town-Judge. | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
if N==1:
return 1
# Since it is a Directed Graph
# if -> income degree +=1
# if -> outgoing degree -=1
degree = [0]*(N+1)
for i,j in trust:
... | find-the-town-judge | [Python 3]Find the Town Judge. | tilak_ | 1 | 186 | find the town judge | 997 | 0.493 | Easy | 16,245 |
https://leetcode.com/problems/find-the-town-judge/discuss/624957/Python3-indeg-and-outdeg | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
degree = [0]*n
for u, v in trust:
degree[v-1] += 1
degree[u-1] -= 1
return next((i+1 for i, x in enumerate(degree) if x == n-1), -1) | find-the-town-judge | [Python3] indeg & outdeg | ye15 | 1 | 42 | find the town judge | 997 | 0.493 | Easy | 16,246 |
https://leetcode.com/problems/find-the-town-judge/discuss/2757946/Python3-81-faster-with-explanation | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
trustMap = {}
for i in range(1, n + 1):
trustMap[i] = [0, 0]
for tPath in trust:
trustMap[tPath[0]][1] += 1
trustMap[tPath[1]][0] += 1
for person in t... | find-the-town-judge | Python3, 81% faster with explanation | cvelazquez322 | 0 | 9 | find the town judge | 997 | 0.493 | Easy | 16,247 |
https://leetcode.com/problems/find-the-town-judge/discuss/2733453/Simple-Python-Solution | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
tracker = [0] * n
for a, b in trust:
tracker[a-1] -= 1
tracker[b-1] += 1
for i in range(0, n):
if tracker[i] == n - 1:
return i + 1
return -1 | find-the-town-judge | Simple Python Solution | ekomboy012 | 0 | 3 | find the town judge | 997 | 0.493 | Easy | 16,248 |
https://leetcode.com/problems/find-the-town-judge/discuss/2708890/Graph-or-Python-or-O(n) | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
t = defaultdict(int)
for i in range(n):
t[i] = 0
for a, b in trust:
t[a - 1] -= 1
t[b - 1] += 1
for k, _ in t.items():
if t[k... | find-the-town-judge | Graph | Python | O(n) | Kiyomi_ | 0 | 12 | find the town judge | 997 | 0.493 | Easy | 16,249 |
https://leetcode.com/problems/find-the-town-judge/discuss/1959649/Python-3-or-faster-than-99.91 | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if not trust and n == 1:
return 1
elif not trust:
return -1
judgeCnt = Counter([ y for x, y in trust]).most_common()[0]
if judgeCnt[1] != n - 1 or judgeCnt[0] in [ x ... | find-the-town-judge | Python 3 | faster than 99.91% | anels | 0 | 194 | find the town judge | 997 | 0.493 | Easy | 16,250 |
https://leetcode.com/problems/find-the-town-judge/discuss/1806350/3-Lines-Python-Solution-oror-slow-oror-Memory-less-than-60 | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
for i in range(1,n+1):
if sorted([trus[0] for trus in trust if trus[1]==i])==[x for x in range(1,n+1) if x!=i] and i not in [trus[0] for trus in trust]: return i
return -1 | find-the-town-judge | 3-Lines Python Solution || slow || Memory less than 60% | Taha-C | 0 | 107 | find the town judge | 997 | 0.493 | Easy | 16,251 |
https://leetcode.com/problems/find-the-town-judge/discuss/1664970/Using-sets-and-intersection-python-O(n)-solution | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if trust == []:
return 1 if n == 1 else -1
memory = {}
for t in trust:
memory.setdefault(t[0], set()).add(t[1])
trusted = set.intersection(*memory.val... | find-the-town-judge | Using sets & intersection python O(n) solution | Sima24 | 0 | 50 | find the town judge | 997 | 0.493 | Easy | 16,252 |
https://leetcode.com/problems/find-the-town-judge/discuss/1664611/Python-3-Easy-Solution | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if n==1:
return 1
ct={}
t={}
for i in trust:
if i[1] in ct:
ct[i[1]]+=1
else:
ct[i[1]]=1
for i in trust:
t[i[0]]=i[1]
... | find-the-town-judge | Python 3 Easy Solution | aryanagrawal2310 | 0 | 74 | find the town judge | 997 | 0.493 | Easy | 16,253 |
https://leetcode.com/problems/find-the-town-judge/discuss/1664288/Unique-Approach-Python3-Solution-O(1)-Memory | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
trust = list(sorted(trust, key=lambda x: x[1])) # Sort list based on the trustee
if n == 1: # Only one person condition
return 1
if not trust:
return -1
... | find-the-town-judge | [Unique Approach] Python3 Solution O(1) Memory | Sparkles4 | 0 | 73 | find the town judge | 997 | 0.493 | Easy | 16,254 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663815/Python3-hashmap-solution-or-easy-understanding | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if n == 1: return 1
if not trust: return -1
potential_judge = dict()
normal_people = []
for i in trust:
if i[1] in potential_judge.keys():
potential_judge[i[1... | find-the-town-judge | Python3 hashmap solution | easy-understanding | Janetcxy | 0 | 52 | find the town judge | 997 | 0.493 | Easy | 16,255 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663483/Python-Solution-SImple-to-understand | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if n == 1 and len(trust) == 0:
return 1
if len(trust) == 0:
return -1
persons = []
for i in range(n+1):
persons.append({"id": i, "trusted_by": 0, "trusts": 0})
... | find-the-town-judge | Python Solution SImple to understand | pradeep288 | 0 | 156 | find the town judge | 997 | 0.493 | Easy | 16,256 |
https://leetcode.com/problems/find-the-town-judge/discuss/1663177/python3-Simple-O(n)-Solution | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
votes = [0] * n
# track the current most popular candidate
c = 0
for a, b in trust:
# the judge trusts noone, so anyone that votes cannot possibly be in the running
votes[a-1] = -inf
... | find-the-town-judge | python3 Simple O(n) Solution | zldobbs | 0 | 51 | find the town judge | 997 | 0.493 | Easy | 16,257 |
https://leetcode.com/problems/find-the-town-judge/discuss/1342456/Easy-Python-Solution | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if not trust and n!=1:
return -1
s=set()
j=0
for i in (trust):
s.add(i[0])
for i in range(1,n+1):
if(i not in s):
j=i
break
... | find-the-town-judge | Easy Python Solution | Sneh17029 | 0 | 316 | find the town judge | 997 | 0.493 | Easy | 16,258 |
https://leetcode.com/problems/find-the-town-judge/discuss/1092248/Python-Solution-using-a-Dictionary | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
"""
Uses a hash table to store the valid candidates and who trusted this candidate.
Time complexity: O(N). Space complexity: O(N)
"""
# A dictionary maps a candidate to who trusted t... | find-the-town-judge | Python Solution using a Dictionary | QizhangJia | 0 | 263 | find the town judge | 997 | 0.493 | Easy | 16,259 |
https://leetcode.com/problems/find-the-town-judge/discuss/1080127/a-Python-solution-based-on-%22277.-Find-the-Celebrity%22 | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
# main ideas:
# 1. if a trusts b -> a is not a judge
# 2. if a doesn't trust b -> b is not a judge
candidate = 1
for i in range(2, N+1): # 1~N
if [candidate, i] in trust:
... | find-the-town-judge | a Python solution based on "277. Find the Celebrity" | kylu | 0 | 146 | find the town judge | 997 | 0.493 | Easy | 16,260 |
https://leetcode.com/problems/find-the-town-judge/discuss/991887/Python-easy-to-understand | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
if N==1:
return 1
d = defaultdict(list)
for n in trust:
d[n[0]].append(1)
d[n[1]].append(2)
for k,v in d.items():
cond = [True if a==2 els... | find-the-town-judge | Python easy to understand | vimoxshah | 0 | 207 | find the town judge | 997 | 0.493 | Easy | 16,261 |
https://leetcode.com/problems/find-the-town-judge/discuss/959284/Python3-O(n)-with-explanation | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
seen = {t[0] for t in trust}
j = None
for i in range(1, N+1):
if i in seen:
continue
if j:
return -1
j = i
if not j:
return -1
... | find-the-town-judge | [Python3] O(n) with explanation | gdm | 0 | 201 | find the town judge | 997 | 0.493 | Easy | 16,262 |
https://leetcode.com/problems/find-the-town-judge/discuss/624177/Python3-Beautiful-and-detailed-solution-with-O(N)-Time-and-O(N)-Extra-Space | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
all_people = set(range(1, N + 1))
people_who_trust = set([x[0] for x in trust])
people_who_dont_trust = all_people - people_who_trust
if len(people_who_dont_trust) != 1:
return - 1
judge =... | find-the-town-judge | [Python3] Beautiful & detailed solution with O(N) Time & O(N) Extra Space | timetoai | 0 | 46 | find the town judge | 997 | 0.493 | Easy | 16,263 |
https://leetcode.com/problems/find-the-town-judge/discuss/243955/Celebrity-question-detailed-explanation | class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
a_trust_b = set()
for a, b in trust:
a_trust_b.add((a, b))
if N == 1:
return 1
# find candidate judge as b
# each round elimnate 1 candidate
# after a... | find-the-town-judge | Celebrity question detailed explanation | leonmak | 0 | 158 | find the town judge | 997 | 0.493 | Easy | 16,264 |
https://leetcode.com/problems/find-the-town-judge/discuss/1520554/Python3-Two-kind-of-solutions | class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
if n == 1:
return 1
d = {}
trusted = set()
for t in trust:
if t[1] not in d:
d[t[1]] = []
d[t[1]].append(t[0])
trusted.add(t[0... | find-the-town-judge | [Python3] Two kind of solutions | maosipov11 | -1 | 111 | find the town judge | 997 | 0.493 | Easy | 16,265 |
https://leetcode.com/problems/maximum-binary-tree-ii/discuss/2709985/Python-short-python-solution | class Solution:
def insertIntoMaxTree(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root: return TreeNode(val)
if val > root.val: return TreeNode(val, root)
root.right = self.insertIntoMaxTree(root.right, val)
return root | maximum-binary-tree-ii | [Python] short python solution | scrptgeek | 0 | 6 | maximum binary tree ii | 998 | 0.665 | Medium | 16,266 |
https://leetcode.com/problems/maximum-binary-tree-ii/discuss/2533846/Python-Commented-and-simple-DFS-solution | class Solution:
def insertIntoMaxTree(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
# that can be solved using DFS, as it is quite easy to
# keep track of the parent node there
# take care of the edge case that there is no root
if not root:
... | maximum-binary-tree-ii | [Python] - Commented and simple DFS solution | Lucew | 0 | 18 | maximum binary tree ii | 998 | 0.665 | Medium | 16,267 |
https://leetcode.com/problems/maximum-binary-tree-ii/discuss/2533846/Python-Commented-and-simple-DFS-solution | class Solution:
def insertIntoMaxTree(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root:
return TreeNode(val=val)
if root.val < val:
return TreeNode(val=val, left=root)
dfs(root.right, root, val)
return root
def dfs(node, pa... | maximum-binary-tree-ii | [Python] - Commented and simple DFS solution | Lucew | 0 | 18 | maximum binary tree ii | 998 | 0.665 | Medium | 16,268 |
https://leetcode.com/problems/maximum-binary-tree-ii/discuss/985315/Python3-move-down-the-tree-O(logN) | class Solution:
def insertIntoMaxTree(self, root: TreeNode, val: int) -> TreeNode:
prev, node = None, root
while node and val < node.val: prev, node = node, node.right
temp = TreeNode(val, left=node)
if prev: prev.right = temp
else: root = temp
return root | maximum-binary-tree-ii | [Python3] move down the tree O(logN) | ye15 | 0 | 76 | maximum binary tree ii | 998 | 0.665 | Medium | 16,269 |
https://leetcode.com/problems/available-captures-for-rook/discuss/356593/Solution-in-Python-3-(beats-~97)-(three-lines) | class Solution:
def numRookCaptures(self, b: List[List[str]]) -> int:
I, J = divmod(sum(b,[]).index('R'),8)
C = "".join([i for i in [b[I]+['B']+[b[i][J] for i in range(8)]][0] if i != '.'])
return C.count('Rp') + C.count('pR')
- Junaid Mansuri
(LeetCode ID)@hotmail.com | available-captures-for-rook | Solution in Python 3 (beats ~97%) (three lines) | junaidmansuri | 8 | 839 | available captures for rook | 999 | 0.679 | Easy | 16,270 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1601678/Python-3-easy-to-understand-faster-than-96 | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
n = 8
for i in range(n): # find rook location
for j in range(n):
if board[i][j] == 'R':
x, y = i, j
break
res = 0
for i in range(x-1... | available-captures-for-rook | Python 3 easy to understand, faster than 96% | dereky4 | 3 | 159 | available captures for rook | 999 | 0.679 | Easy | 16,271 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1112858/Easiest-Recursive-solution-or-97.6-time-98.2-space | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
def find(rx, ry, direction, count):
if rx == 8 or ry == 8 or rx == -1 or ry == -1: return count
if board[rx][ry] == "B": return 0
if board[rx][ry] == "p": return count + 1
if di... | available-captures-for-rook | Easiest Recursive solution | 97.6% time, 98.2% space | vanigupta20024 | 3 | 243 | available captures for rook | 999 | 0.679 | Easy | 16,272 |
https://leetcode.com/problems/available-captures-for-rook/discuss/500938/Python3%3A-not-pretty-but-straight-forward | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
for i in range(len(board)):
for j in range(len(board[0])):
if board[i][j] == 'R':
count = 0;
l, r = j - 1, j + 1
while l >= 0:
... | available-captures-for-rook | Python3: not pretty, but straight forward | andnik | 1 | 157 | available captures for rook | 999 | 0.679 | Easy | 16,273 |
https://leetcode.com/problems/available-captures-for-rook/discuss/379640/Simon's-Note-Python3 | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
res=0
n_row=len(board)
n_col=len(board[0])
dirs=[[0,1],[0,-1],[-1,0],[1,0]]
for i in range(n_row):
for j in range(n_col):
if board[i][j]=="R":
for dir in ... | available-captures-for-rook | [🎈Simon's Note🎈] Python3 | SunTX | 1 | 101 | available captures for rook | 999 | 0.679 | Easy | 16,274 |
https://leetcode.com/problems/available-captures-for-rook/discuss/2843995/Python3-Solution-using-DFS | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
def dfs(r, c, i, j):
ans = 0
while 0 <= r < 8 and 0 <= c < 8:
if board[r][c] == 'p':
ans += 1
break
if board[r][c] == 'B':
... | available-captures-for-rook | [Python3] Solution using DFS | BLOCKS | 0 | 3 | available captures for rook | 999 | 0.679 | Easy | 16,275 |
https://leetcode.com/problems/available-captures-for-rook/discuss/2716302/Python-Easy-to-follow-with-comments | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
def find_pawn(board_slice):
for square in board_slice:
if square == 'B':
return 0
if square == 'p':
return 1
return 0
output = ... | available-captures-for-rook | Python - Easy to follow with comments | ptegan | 0 | 10 | available captures for rook | 999 | 0.679 | Easy | 16,276 |
https://leetcode.com/problems/available-captures-for-rook/discuss/2711816/Python-!-Simple-Solution | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
bod = board[::]
rows = [[i for i in ro if i != "."] for ro in bod]
cols = [[i for i in list(co) if i != "."]for co in list(zip(*bod))]
count = 0
rows = rows + cols
for row in ... | available-captures-for-rook | Python ! Simple Solution | w7Pratham | 0 | 11 | available captures for rook | 999 | 0.679 | Easy | 16,277 |
https://leetcode.com/problems/available-captures-for-rook/discuss/2681421/Python-Solution-Fast-and-Easy | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
ans = 0
boardT = list(zip(*board))
for i in range(8):
if "R" in board[i]:
j = board[i].index("R")
rookIndex = (i, j)
if "p" in board[i][:j]:
... | available-captures-for-rook | Python Solution - Fast and Easy | scifigurmeet | 0 | 2 | available captures for rook | 999 | 0.679 | Easy | 16,278 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1991449/easy-soln | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
rookcoord=[]
i=j=0
while i < len(board):
j=0
while j < len(board[0]):
if board[i][j] == "R":
#print("Rook found", i,j)
r = i
... | available-captures-for-rook | easy soln | golden-eagle | 0 | 18 | available captures for rook | 999 | 0.679 | Easy | 16,279 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1980496/python3-easy-solution-for-rook | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
counter=0
number=0
number2=0
for i in range(len(board)):
temp=board[i]
for j in range(len(temp)):
if temp[j] == 'R':
index_rook = j
... | available-captures-for-rook | python3 easy solution for rook | vishwahiren16 | 0 | 44 | available captures for rook | 999 | 0.679 | Easy | 16,280 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1650582/Simple-Python-Solution | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
x=y=0
for i in range(len(board)):
flag=0
for j in range(len(board[0])):
if board[i][j]=='R':
x,y=i,j
# print(x, y)
count=0
for i in range(x, -... | available-captures-for-rook | Simple Python Solution | Siddharth_singh | 0 | 83 | available captures for rook | 999 | 0.679 | Easy | 16,281 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1398741/Python3-Simulation-Faster-Than-95.27-Memory-Less-Than-63.43 | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
for i in range(8):
for j in range(8):
if board[i][j] == 'R':
x, y = i, j
break
cap = 0
flag1, flag2, flag3, flag4 = False, False, False, False... | available-captures-for-rook | Python3 Simulation Faster Than 95.27%, Memory Less Than 63.43% | Hejita | 0 | 48 | available captures for rook | 999 | 0.679 | Easy | 16,282 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1268386/Simple-Python-Solution-Recursive-Approach-(DFS-like) | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
# to store result
res = [0]
# recursive function to find number of available captures
def find(i,j,dirn):
# checking if the position is still out of bound or is 'B' meaning it ... | available-captures-for-rook | Simple Python Solution - Recursive Approach (DFS-like) | nagashekar | 0 | 77 | available captures for rook | 999 | 0.679 | Easy | 16,283 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1060944/Python3-simple-solution | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
def check(board, row, col):
res = 0
a = [[1,0],[-1,0],[0,1],[0,-1]]
for n in a:
x,y = row,col
i = n[0]
j = n[1]
while 0<=x<=7 and 0<=y... | available-captures-for-rook | Python3 simple solution | EklavyaJoshi | 0 | 78 | available captures for rook | 999 | 0.679 | Easy | 16,284 |
https://leetcode.com/problems/available-captures-for-rook/discuss/1004621/Python-Faster-than-99.49-20-ms-Search-and-Capture | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
def total_captures(i,j):
res = 0
# here we are searching for a pawn in top,
# bottom, left, and right directions
# If we find a pawn first, we can capture it
# If we find a bishop, then we can't ca... | available-captures-for-rook | Python Faster than 99.49% 20 ms - Search and Capture | prashantsengar | 0 | 181 | available captures for rook | 999 | 0.679 | Easy | 16,285 |
https://leetcode.com/problems/available-captures-for-rook/discuss/777981/Intuitive-approach-by-searching-four-directions | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
R, C = len(board), len(board[0])
# 1) Search for rock
rock_r = rock_c = 0
for i in range(R):
for j in range(C):
if board[i][j] == 'R':
rock_r, rock_c = i, j
... | available-captures-for-rook | Intuitive approach by searching four directions | puremonkey2001 | 0 | 52 | available captures for rook | 999 | 0.679 | Easy | 16,286 |
https://leetcode.com/problems/available-captures-for-rook/discuss/512642/Python3-94.24-extremely-easy-to-write-using-too-many-'break'-though...... | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
# find R
for j in range(8):
for i in range(8):
if board[j][i] == 'R':
count = 0
# find if there is any p vertically above R
for n in range(j,-1,-1):
... | available-captures-for-rook | Python3 94.24% - extremely easy to write using too many 'break' though...... | Bannbuuu | 0 | 111 | available captures for rook | 999 | 0.679 | Easy | 16,287 |
https://leetcode.com/problems/available-captures-for-rook/discuss/248407/Faster-Than-100-Python-Solution | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
row,column=self.findRook(board)
if(row is None and column is None):
return 0
count=0
#Above Rook
for i in range(row,0,-1):
if(board[i][column]=='B'):
bre... | available-captures-for-rook | Faster Than 100% Python Solution | bismeet | 0 | 117 | available captures for rook | 999 | 0.679 | Easy | 16,288 |
https://leetcode.com/problems/available-captures-for-rook/discuss/244982/Easy-to-read-and-understand-Python-16ms | class Solution(object):
def numRookCaptures(self, board):
"""
:type board: List[List[str]]
:rtype: int
"""
found, i, j = self.findWhiteRook(board)
if not found:
print ('Rook not found.')
return False
p = 0
p = self.blackPondCoun... | available-captures-for-rook | Easy to read and understand Python 16ms | jujbates | 0 | 113 | available captures for rook | 999 | 0.679 | Easy | 16,289 |
https://leetcode.com/problems/available-captures-for-rook/discuss/251469/Python-3-100-faster-100-memory | class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
# At first find position of rook and save in 'iR' and 'jR' variables
for i in range(len(board)):
for j in range(len(board[i])):
if board[i][j] == 'R':
iR = i
... | available-captures-for-rook | Python 3, 100% faster, 100% memory | astepano | -1 | 174 | available captures for rook | 999 | 0.679 | Easy | 16,290 |
https://leetcode.com/problems/minimum-cost-to-merge-stones/discuss/1465680/Python3-dp | class Solution:
def mergeStones(self, stones: List[int], k: int) -> int:
if (len(stones)-1) % (k-1): return -1 # impossible
prefix = [0]
for x in stones: prefix.append(prefix[-1] + x)
@cache
def fn(lo, hi):
"""Return min cost of merging stones[l... | minimum-cost-to-merge-stones | [Python3] dp | ye15 | 1 | 654 | minimum cost to merge stones | 1,000 | 0.423 | Hard | 16,291 |
https://leetcode.com/problems/minimum-cost-to-merge-stones/discuss/781323/python-Top-Down-solution | class Solution:
def minCost(self, n: int, cuts: List[int]) -> int:
_cuts = [0] + sorted(cuts) + [n]
N = len(_cuts)
@lru_cache(None)
def helper(lp,rp):
nonlocal _cuts
if rp-lp==1:
return 0
return _cuts[rp]-_cuts[lp] + min([h... | minimum-cost-to-merge-stones | python Top Down solution | e-yi | 1 | 564 | minimum cost to merge stones | 1,000 | 0.423 | Hard | 16,292 |
https://leetcode.com/problems/minimum-cost-to-merge-stones/discuss/2633516/Top-down-dynamic-programming-in-concise-Python | class Solution:
@cache
def dp(self, l, r, piles) -> int:
if r - l < piles:
return inf
if r - l == piles:
return 0
if piles == 1:
return self.dp(l, r, self.k) + self.prefix_sum[r] - self.prefix_sum[l]
return min(self.dp(l, m, i) + self.dp(m, r, ... | minimum-cost-to-merge-stones | Top-down dynamic programming in concise Python | metaphysicalist | 0 | 36 | minimum cost to merge stones | 1,000 | 0.423 | Hard | 16,293 |
https://leetcode.com/problems/grid-illumination/discuss/1233528/Python-or-HashMap-or-O(L%2BQ)-or-928ms | class Solution:
def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
lamps = {(r, c) for r, c in lamps}
row, col, left, right = dict(), dict(), dict(), dict()
for r, c in lamps:
row[r] = row.get(r, 0) + 1
col[c] =... | grid-illumination | Python | HashMap | O(L+Q) | 928ms | PuneethaPai | 2 | 103 | grid illumination | 1,001 | 0.362 | Hard | 16,294 |
https://leetcode.com/problems/grid-illumination/discuss/2181052/python-3-or-simple-4-hash-map-solution | class Solution:
def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
rows = collections.Counter()
cols = collections.Counter()
diags1 = collections.Counter()
diags2 = collections.Counter()
lamps = {tuple(lamp) for lamp in lamps}
... | grid-illumination | python 3 | simple 4 hash map solution | dereky4 | 1 | 142 | grid illumination | 1,001 | 0.362 | Hard | 16,295 |
https://leetcode.com/problems/grid-illumination/discuss/2115766/Python3-solution-or-Hashmap-or-Explained | class Solution:
def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
def check(i, j, dRow, dCol, dDiagS, dDiagP):
if (i in dRow and dRow[i] > 0) or (j in dCol and dCol[j] > 0) or (
i + j in dDiagS and dDiagS[i + j] > 0) or (
... | grid-illumination | Python3 solution | Hashmap | Explained | FlorinnC1 | 1 | 51 | grid illumination | 1,001 | 0.362 | Hard | 16,296 |
https://leetcode.com/problems/grid-illumination/discuss/1998042/PYTHON-SOL-oror-WELL-EXPLAINED-oror-HASHMAP-BASED-oror-SIMPLE-oror-EFFICIENT-oror | class Solution:
def checkIsOn(self,row,col):
return 1 if (self.rows[row] > 0 or self.cols[col] > 0 \
or self.digonal1[row-col] > 0 or self.digonal2[row+col] > 0) else 0
def TurnOff(self,row,col):
adj = ((row,col),(row+1,col),(row-1,col),(row,col-1),(row,col+1),\
(row+1... | grid-illumination | PYTHON SOL || WELL EXPLAINED || HASHMAP BASED || SIMPLE || EFFICIENT || | reaper_27 | 1 | 51 | grid illumination | 1,001 | 0.362 | Hard | 16,297 |
https://leetcode.com/problems/grid-illumination/discuss/1638153/python-solution-with-tables-tracking-rows-cols-and-diags-lit | class Solution:
from collections import defaultdict
from itertools import product
def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
rows = defaultdict(int)
cols = defaultdict(int)
downright = defaultdict(int)
downleft = default... | grid-illumination | python solution with tables tracking rows, cols, and diags lit | PsyKosh | 1 | 95 | grid illumination | 1,001 | 0.362 | Hard | 16,298 |
https://leetcode.com/problems/grid-illumination/discuss/1521789/Python3-freq-table | class Solution:
def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
lamps = {(i, j) for i, j in lamps}
rows = defaultdict(int)
cols = defaultdict(int)
anti = defaultdict(int)
diag = defaultdict(int)
for i, j in lamps:
... | grid-illumination | [Python3] freq table | ye15 | 0 | 92 | grid illumination | 1,001 | 0.362 | Hard | 16,299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.