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/house-robber/discuss/2821702/DYNAMIC-PROGRAMMING-oror-FASTEST-oror-BEATS-99-SUBMISSIONS-oror-EASIEST | class Solution:
def rob(self, nums: List[int]) -> int:
r1,r2=0,0
for i in nums:
t=max(r1+i,r2)
r1=r2
r2=t
return r2 | house-robber | DYNAMIC PROGRAMMING || FASTEST || BEATS 99% SUBMISSIONS || EASIEST | Pritz10 | 0 | 7 | house robber | 198 | 0.488 | Medium | 3,200 |
https://leetcode.com/problems/house-robber/discuss/2814165/Fully-Optimized-Code-oror-Python-oror-Easy-to-Understand | class Solution(object):
def rob(self, nums):
prev = 0
prev2 = 0
for i in range(len(nums)):
#Pick the number
take = nums[i]
if i > 1: take += prev2
#Not pick the number
notTake = prev
curr = max... | house-robber | Fully Optimized Code💯 || Python || Easy to Understand✅ | __avinash_007 | 0 | 5 | house robber | 198 | 0.488 | Medium | 3,201 |
https://leetcode.com/problems/house-robber/discuss/2791948/Backtracking-%2B-DP | class Solution:
def rob(self, nums: List[int]) -> int:
n = len(nums)
dp = {}
def dfs(index, steal):
if (index, steal) in dp:
return dp[(index, steal)]
if index == n:
return 0
result = 0
if steal:
... | house-robber | Backtracking + DP | ayhamhashesh | 0 | 2 | house robber | 198 | 0.488 | Medium | 3,202 |
https://leetcode.com/problems/house-robber/discuss/2780327/Simple-and-Clear-O(N)-Method | class Solution:
def rob(self, nums: List[int]) -> int:
if len(nums) <= 2:
return max(nums)
dp = [0 for _ in range(len(nums))]
dp[0] = nums[0]
dp[1] = nums[1]
prev_max = -inf
for i in range(2, len(nums)):
prev_max = max(prev_max, dp[i-2])
... | house-robber | Simple and Clear O(N) Method | SnoopyLovesCoding | 0 | 3 | house robber | 198 | 0.488 | Medium | 3,203 |
https://leetcode.com/problems/house-robber/discuss/2764118/House-Robber-Python-solution | class Solution:
def rob(self, nums: List[int]) -> int:
rob1, rob2 = 0,0
for i in nums:
ans = max (rob1 + i , rob2)
rob1,rob2 = rob2,ans
return rob2 | house-robber | House Robber Python solution | Sujit773 | 0 | 1 | house robber | 198 | 0.488 | Medium | 3,204 |
https://leetcode.com/problems/house-robber/discuss/2730842/Python-Bottom-up | class Solution:
def rob(self, nums: List[int]) -> int:
n = len(nums)
dp = [0]*n
dp[0] = nums[0]
for i in range(1 , n):
take = nums[i]
if i > 1:
take += dp[i-2]
notTake = dp[i-1]
dp[i] = max(take , notTake)
retur... | house-robber | Python Bottom - up | rajitkumarchauhan99 | 0 | 7 | house robber | 198 | 0.488 | Medium | 3,205 |
https://leetcode.com/problems/house-robber/discuss/2730327/Python-or-DP-or-Top-Down-or-Bottom-Up-or-3-Approaches | class Solution:
def rob(self, nums: List[int]) -> int:
def bottom_up():
'''
Bottom up 1d solution
dp[i]: maximum amount of money you can rob up till house[i]
Time: O(N) => 1 iteration
Space: O(N) => dp array
'''
dp = [0] * l... | house-robber | Python | DP | Top-Down | Bottom-Up | 3 Approaches | seangohck | 0 | 8 | house robber | 198 | 0.488 | Medium | 3,206 |
https://leetcode.com/problems/house-robber/discuss/2706053/Python-recursion-DP-beats-97 | class Solution:
def rob(self, arr: List[int]) -> int:
d = [-1] * len(arr)
def dp(cur):
if d[cur] != -1:
return d[cur]
if cur >= len(arr) - 2:
d[cur] = arr[cur]
elif cur < len(arr) - 3:
d[cur] = arr[cur] + max(dp(cur ... | house-robber | Python recursion DP beats 97% | JSTM2022 | 0 | 3 | house robber | 198 | 0.488 | Medium | 3,207 |
https://leetcode.com/problems/house-robber/discuss/2694612/Binary-recursion-on-Python | class Solution:
def rob(self, nums: List[int]) -> int:
n = len(nums)
if n == 0:
return 0
if n <= 2:
return max(nums)
if n == 3:
return max(nums[0] + nums[2], nums[1])
middle = n // 2
left_with_middle = self.rob(nums[:middle - 1])
... | house-robber | Binary recursion on Python | vadim_vadim | 0 | 5 | house robber | 198 | 0.488 | Medium | 3,208 |
https://leetcode.com/problems/house-robber/discuss/2675550/Python-oror-DP-solution | class Solution:
def rob(self, nums: List[int]) -> int:
dp= [0] * len(nums)
if len(nums) == 1:
return nums[0]
for i in range(len(nums)):
dp[i] = max(dp[i-2]+nums[i],dp[i-1])
return max(dp) | house-robber | Python || DP solution | sinjan_singh | 0 | 8 | house robber | 198 | 0.488 | Medium | 3,209 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2266055/C%2B%2B-oror-PYTHON-oror-EXPLAINED-oror | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
def solve(root, lvl):
if root:
if len(res)==lvl:
res.append(root.val)
solve(root.right, lvl + 1)
solve(root.left, lvl + 1)
return
res = []
sol... | binary-tree-right-side-view | ✔️ C++ || PYTHON || EXPLAINED || ; ] | karan_8082 | 80 | 4,400 | binary tree right side view | 199 | 0.612 | Medium | 3,210 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2265611/Python3-oror-dfs-recursion-10-lines-oror-TM%3A-98-70 | class Solution:
# The plan here is to dfs the tree, right-first
# (opposite of the usual left-first method), and
# keeping track of the tree levels as we proceed. The
# first node we visit on each level is the right-side v... | binary-tree-right-side-view | Python3 || dfs, recursion, 10 lines || T/M: 98%/ 70% | warrenruud | 7 | 563 | binary tree right side view | 199 | 0.612 | Medium | 3,211 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/327960/Python-solution-with-no-additional-space-and-96.61-faster | class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
res, max_level = [], -1
def traverse_tree(node, level):
nonlocal res, max_level
if not node:
return
if max_level < level:
res.append(node.val)
... | binary-tree-right-side-view | Python solution with no additional space and 96.61% faster | cyrilbeschi | 5 | 400 | binary tree right side view | 199 | 0.612 | Medium | 3,212 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2680849/python-3-simple-BFS-solution | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
ans = []
if not root: return ans
q = [root]
while q:
lv = []
ans.append(q[-1].val)
for node in q:
i... | binary-tree-right-side-view | python 3 simple BFS solution | yfwong | 3 | 318 | binary tree right side view | 199 | 0.612 | Medium | 3,213 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2094856/Python3-O(N)-Time-Easy-to-Understand-Solution-with-Comments | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
res = []
if not root: return res
q = collections.deque()
q.append(root)
while q:
res.append(q[-1].val) # the top element of q is the right-most
n = len(q... | binary-tree-right-side-view | [Python3] O(N) Time Easy to Understand Solution with Comments | samirpaul1 | 3 | 122 | binary tree right side view | 199 | 0.612 | Medium | 3,214 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/1052999/Python.-Cool-and-easy-solution.-O(n). | class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
Heights = {}
def sol(root: TreeNode, height: int):
if not root:
return
sol( root.right, height + 1)
if height not in Heights:
Heights[height] = root.val
sol( root.left, height + 1)
sol(root, 0)
return [Heights[index] ... | binary-tree-right-side-view | Python. Cool & easy solution. O(n). | m-d-f | 3 | 185 | binary tree right side view | 199 | 0.612 | Medium | 3,215 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/454028/Python-3-Four-liner-Memory-usage-less-than-100 | class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
levels = [[root]]
while any(levels[-1]):
levels.append([x for node in levels[-1] for x in [node.left, node.right] if x])
return [level[-1].val for level in levels[:-1]] | binary-tree-right-side-view | Python 3 - Four liner - Memory usage less than 100% | mmbhatk | 3 | 454 | binary tree right side view | 199 | 0.612 | Medium | 3,216 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/1123139/Python-BFS-100-Space-95-Time | class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root:
return
ans = []
q = [root]
subq = []
while q:
element = q.pop(0)
if element.left:
subq.append(element.left... | binary-tree-right-side-view | Python BFS 100% Space, 95% Time | rooky1905 | 2 | 203 | binary tree right side view | 199 | 0.612 | Medium | 3,217 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2266545/python3-or-easy-or-explained-or-dictionary-approach | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if not root:
return [] # if root is None
self.d = {} # used to store right most element of each level
self.trav(root, 0) # function calli... | binary-tree-right-side-view | python3 | easy | explained | dictionary approach | H-R-S | 1 | 35 | binary tree right side view | 199 | 0.612 | Medium | 3,218 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2265795/C%2B%2B-and-Python3-Easy-Solution-using-level-order-traversal | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
ans = list()
if not root: return []
q = [root]
while q:
for i in range(len(q)):
node = q.pop(0)
if node.left: q.append(node.left)
if node.right:... | binary-tree-right-side-view | C++ & Python3 Easy Solution using level order traversal | Akash3502 | 1 | 53 | binary tree right side view | 199 | 0.612 | Medium | 3,219 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2265779/Python3-solution-using-preorder-traversal | class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
result = []
def get_result(root,level):
if not root:
return
if level==len(result):
result.append(root.val)
get_result(root.right,level+1)
ge... | binary-tree-right-side-view | 📌 Python3 solution using preorder traversal | Dark_wolf_jss | 1 | 5 | binary tree right side view | 199 | 0.612 | Medium | 3,220 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2265581/Python-BFS-solution | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if root is None:
return None
queue = deque()
queue.append(root)
res = []
while queue:
temp = []
for _ in range(len(queue)):
no... | binary-tree-right-side-view | Python BFS solution | PythonicLava | 1 | 156 | binary tree right side view | 199 | 0.612 | Medium | 3,221 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/1597806/Python-space-complexity-O(n)-time-complexity-O(n) | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
res = {}
if not root: return res
def dfs(node,level):
if level not in res:
res[level] = node.val
if node.right:
dfs(node.right,level+1)
... | binary-tree-right-side-view | Python, space complexity O(n), time complexity O(n) | sineman | 1 | 139 | binary tree right side view | 199 | 0.612 | Medium | 3,222 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/1473833/Python-Clean-BFS-and-DFS | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
queue, view = deque([(root, 0)]), []
while queue:
node, level = queue.popleft()
if not node:
continue
if level >= len(view):
view.append(node.val)
... | binary-tree-right-side-view | [Python] Clean BFS & DFS | soma28 | 1 | 148 | binary tree right side view | 199 | 0.612 | Medium | 3,223 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/1473833/Python-Clean-BFS-and-DFS | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
queue, view = deque([(root, 0)]), []
while queue:
node, level = queue.pop()
if not node:
continue
if level >= len(view):
view.append(node.val)
... | binary-tree-right-side-view | [Python] Clean BFS & DFS | soma28 | 1 | 148 | binary tree right side view | 199 | 0.612 | Medium | 3,224 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/1414651/Row-by-row-89-speed | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if not root:
return []
row = [root]
values = []
while row:
values.append(row[-1].val)
new_row = []
for node in row:
if node.left:
... | binary-tree-right-side-view | Row by row, 89% speed | EvgenySH | 1 | 100 | binary tree right side view | 199 | 0.612 | Medium | 3,225 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/863385/Python3-Array-Filling-and-BFS-O(n)-Space-and-Time | class Solution:
# Array filling - O(n) space and time
def helper(self, node: TreeNode, height: int, res: List[int]):
if not node:
return
res[height] = node.val
self.helper(node.left, height + 1, res)
self.helper(node.right, height + 1, res)
def recursive_array_fi... | binary-tree-right-side-view | [Python3] Array Filling and BFS - O(n) Space & Time | nachiketsd | 1 | 106 | binary tree right side view | 199 | 0.612 | Medium | 3,226 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2844165/python3-oror-Easy-way | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
res =[]
def risidview(root, level):
if not root:
return res
if len(res) < level:
res.append(root.val)
l = root.left
r = root.right
... | binary-tree-right-side-view | python3 || Easy way | dsai | 0 | 1 | binary tree right side view | 199 | 0.612 | Medium | 3,227 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2775267/Python3-or-Simple-Explanation-or-Beats-96.20-or-32-ms-or-Recursion-or-DFS-or-O(n) | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
def recursive(node: TreeNode, depth: Optional[int] = 0) -> None:
if len(right_side_values) > depth:
right_side_values[depth] = node.val
else:
right_side_values.append(node... | binary-tree-right-side-view | Python3 | Simple Explanation | Beats 96.20% | 32 ms | Recursion | DFS | O(n) | thomwebb | 0 | 2 | binary tree right side view | 199 | 0.612 | Medium | 3,228 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2726411/Python3-simple-and-fast-Sol. | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if root is None:
return []
q=deque([root])
ans=[]
while q:
l=len(q)
ans.append(q[-1].val)
for i in range(l):
node=q.popleft()
... | binary-tree-right-side-view | Python3 simple and fast Sol. | pranjalmishra334 | 0 | 4 | binary tree right side view | 199 | 0.612 | Medium | 3,229 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2483923/BFS-Python-solution-98-faster.-Use-a-level-flag-to-store-nodule's-depth | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if root is None:
return None
level = 0
queue = [[root,level]]
output = []
cur_level = 0
while len(queue)>0:
node,level = queue.pop(0)
cur_level = level
... | binary-tree-right-side-view | BFS Python solution 98% faster. Use a level flag to store nodule's depth | Arana | 0 | 38 | binary tree right side view | 199 | 0.612 | Medium | 3,230 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2407531/Using-level-order-traversal-Runtime%3A-34-ms-faster-than-92.88-of-Python3 | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if not root:
return None
queue = [root]
res = []
next_level = []
while queue:
node = queue.pop(0)
if node.left:
next_level.append(node.left)
... | binary-tree-right-side-view | Using level order traversal - Runtime: 34 ms, faster than 92.88% of Python3 | krithika117 | 0 | 30 | binary tree right side view | 199 | 0.612 | Medium | 3,231 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2283631/Fast-and-extremely-simple-solution | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
def recur(root,depth):
nonlocal valueDepth
if not root:
return None
if depth>=valueDepth:
res.append(root.val)
valueDepth+=1
recur(r... | binary-tree-right-side-view | Fast and extremely simple solution | HaoChenNus | 0 | 23 | binary tree right side view | 199 | 0.612 | Medium | 3,232 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2282881/Python-Easy-to-understand-DFS-solution | class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if not root:
return []
level_value_map = {}
def helper(node, level):
if not node:
return
if level not in level_value_map:
... | binary-tree-right-side-view | [Python] Easy to understand DFS solution | julenn | 0 | 21 | binary tree right side view | 199 | 0.612 | Medium | 3,233 |
https://leetcode.com/problems/binary-tree-right-side-view/discuss/2270119/Simple-solution | class Solution:
def rightSideView(self, root: TreeNode):
rightside = {}
self.seeTree(root, rightside, 1)
numbers = [value for value in rightside.values()]
return numbers
def seeTree(self, root: TreeNode, rightside: dict, nodo: int):
if root == None:
... | binary-tree-right-side-view | Simple solution | jivanbasurtom | 0 | 3 | binary tree right side view | 199 | 0.612 | Medium | 3,234 |
https://leetcode.com/problems/number-of-islands/discuss/863366/Python-3-or-DFS-BFS-Union-Find-All-3-methods-or-Explanation | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid: return 0
m, n = len(grid), len(grid[0])
ans = 0
def dfs(i, j):
grid[i][j] = '2'
for di, dj in (0, 1), (0, -1), (1, 0), (-1, 0):
ii, jj = i+di, j+dj
... | number-of-islands | Python 3 | DFS, BFS, Union Find, All 3 methods | Explanation | idontknoooo | 50 | 5,200 | number of islands | 200 | 0.564 | Medium | 3,235 |
https://leetcode.com/problems/number-of-islands/discuss/863366/Python-3-or-DFS-BFS-Union-Find-All-3-methods-or-Explanation | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid: return 0
m, n = len(grid), len(grid[0])
ans = 0
for i in range(m):
for j in range(n):
if grid[i][j] == '1':
q = collections.deque([(i, j)])
... | number-of-islands | Python 3 | DFS, BFS, Union Find, All 3 methods | Explanation | idontknoooo | 50 | 5,200 | number of islands | 200 | 0.564 | Medium | 3,236 |
https://leetcode.com/problems/number-of-islands/discuss/863366/Python-3-or-DFS-BFS-Union-Find-All-3-methods-or-Explanation | class UF:
def __init__(self, n):
self.p = [i for i in range(n)]
self.n = n
self.size = n
def union(self, i, j):
pi, pj = self.find(i), self.find(j)
if pi != pj:
self.size -= 1
self.p[pj] = pi
def find(self, i):
if i != self.p[i]:
... | number-of-islands | Python 3 | DFS, BFS, Union Find, All 3 methods | Explanation | idontknoooo | 50 | 5,200 | number of islands | 200 | 0.564 | Medium | 3,237 |
https://leetcode.com/problems/number-of-islands/discuss/479692/Intuitive-Disjoint-Set-Union-Find-in-JavaPython3 | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if grid == None or len(grid) == 0: return 0
r, c = len(grid), len(grid[0])
dsu = DSU(r * c)
# union an island with its adjacent islands
for i in range(r):
for j in range(c):
... | number-of-islands | Intuitive Disjoint Set Union Find in [Java/Python3] | BryanBoCao | 7 | 1,200 | number of islands | 200 | 0.564 | Medium | 3,238 |
https://leetcode.com/problems/number-of-islands/discuss/1446732/Flood-Fill-approach.-(733). | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
count = 0
R, C = len(grid), len(grid[0])
def dfs(r, c):
grid[r][c] = '0'
if r-1 >= 0 and grid[r-1][c] == '1':
dfs(r-1, c)
if r+1 < R and grid[r+1][c] == '... | number-of-islands | Flood Fill approach. (733). | AmrinderKaur1 | 5 | 273 | number of islands | 200 | 0.564 | Medium | 3,239 |
https://leetcode.com/problems/number-of-islands/discuss/673201/Python3-union-find-(better-than-83) | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if len(grid) == 0:
return 0
uf = {}
N = len(grid)
M = len(grid[0])
for i in range(N):
for j in range(M):
if grid[i][j] == '1':
uf[... | number-of-islands | Python3 union find (better than 83%) | sird0d0 | 4 | 294 | number of islands | 200 | 0.564 | Medium | 3,240 |
https://leetcode.com/problems/number-of-islands/discuss/2336379/Iterative-solution-using-Stack-Python | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
# m = len(grid)
# n = len(grid[0])
# def mark(r, c):
# if r < 0 or c < 0 or r > m-1 or c > n-1 or grid[r][c] != '1' :
# return
# grid[r][c] = '2'
# mark(r + 1, c)
# ... | number-of-islands | Iterative solution using Stack Python | glebbs | 3 | 270 | number of islands | 200 | 0.564 | Medium | 3,241 |
https://leetcode.com/problems/number-of-islands/discuss/2160946/Python3-Very-Concise-AC-Solution-Both-DFS-BFS | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
row = len(grid); col = len(grid[0])
res = 0
def dfs(i, j): # make surrounded '1' to '0'
if ( not 0 <= i < row) or (not 0 <= j < col) or (grid[i][j] == '0'): return
grid[i][j] = '0'
... | number-of-islands | [Python3] Very Concise AC Solution Both DFS, BFS | samirpaul1 | 3 | 287 | number of islands | 200 | 0.564 | Medium | 3,242 |
https://leetcode.com/problems/number-of-islands/discuss/2160946/Python3-Very-Concise-AC-Solution-Both-DFS-BFS | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
row = len(grid); col = len(grid[0])
res = 0
def bfs(i, j):
q = collections.deque()
q.append((i, j))
while q:
r, c = q.popleft()
if ( not 0 <= r < r... | number-of-islands | [Python3] Very Concise AC Solution Both DFS, BFS | samirpaul1 | 3 | 287 | number of islands | 200 | 0.564 | Medium | 3,243 |
https://leetcode.com/problems/number-of-islands/discuss/1658537/Python-Simple-recursive-DFS-explained | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
res = 0
row = len(grid)
col = len(grid[0])
visited = [[0 for x in range(col)] for y in range(row)]
def perm(i, j):
# check if we're on a land because if yes, we need
... | number-of-islands | [Python] Simple recursive DFS explained | buccatini | 3 | 252 | number of islands | 200 | 0.564 | Medium | 3,244 |
https://leetcode.com/problems/number-of-islands/discuss/2737580/Number-of-Islands-using-BFS-faster-then-96 | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
from collections import deque
n=len(grid)
m=len(grid[0])
visited=[[0 for j in range(m)]for i in range(n)]
c=0
l=[[-1,0],[1,0],[0,-1],[0,1]]
for i in range(n):
for j in range(m):
... | number-of-islands | Number of Islands using BFS faster then 96% | ravinuthalavamsikrishna | 2 | 388 | number of islands | 200 | 0.564 | Medium | 3,245 |
https://leetcode.com/problems/number-of-islands/discuss/2498797/Python-Elegant-and-Short-or-99.90-faster | class Solution:
"""
Time: O(n*m)
Memory: O(n*m)
"""
LAND = '1'
WATER = '0'
def numIslands(self, grid: List[List[str]]) -> int:
n, m = len(grid), len(grid[0])
islands = 0
for row in range(n):
for col in range(m):
if grid[row][col] == self.LAND:
self._visit(row, col, grid)
islands += 1
... | number-of-islands | Python Elegant & Short | 99.90% faster | Kyrylo-Ktl | 2 | 218 | number of islands | 200 | 0.564 | Medium | 3,246 |
https://leetcode.com/problems/number-of-islands/discuss/1292751/Be-careful-and-avoid-this-mistake! | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
rows, cols = len(grid), len(grid[0]) if grid else 0
islands = 0
dirs = [(-1, 0), (0, 1), (1, 0), (0, -1)]
for row in range(rows):
for col in range(cols):
cell = grid[row][col]
... | number-of-islands | Be careful and avoid this mistake! | v21 | 2 | 260 | number of islands | 200 | 0.564 | Medium | 3,247 |
https://leetcode.com/problems/number-of-islands/discuss/1258779/Python-simple-13-lines-solution-easy-to-understand | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
c=0
for i in range(len(grid)):
for j in range(len(grid[0])):
if(grid[i][j]=='1'):
c=c+1
grid=Solution.cut(grid,i,j)
return c
def cut(grid,i,j):
... | number-of-islands | Python simple 13 lines solution easy to understand | Rajashekar_Booreddy | 2 | 749 | number of islands | 200 | 0.564 | Medium | 3,248 |
https://leetcode.com/problems/number-of-islands/discuss/583740/Python-sol-by-DFS-traversal.-w-Comment | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
self.h = len(grid)
if self.h == 0:
# Quick response for empty grid
return 0
self.w = len( grid[0] )
if self.h * self.w == 1:
# Quick response for 1x1 gr... | number-of-islands | Python sol by DFS traversal. [w/ Comment] | brianchiang_tw | 2 | 659 | number of islands | 200 | 0.564 | Medium | 3,249 |
https://leetcode.com/problems/number-of-islands/discuss/583740/Python-sol-by-DFS-traversal.-w-Comment | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
is_land = lambda i,j : grid[i][j] == "1"
h, w = len(grid), len(grid[0])
class DisjointSet():
def __init__(self):
self.pare... | number-of-islands | Python sol by DFS traversal. [w/ Comment] | brianchiang_tw | 2 | 659 | number of islands | 200 | 0.564 | Medium | 3,250 |
https://leetcode.com/problems/number-of-islands/discuss/2500891/Python-or-C%2B%2B-%3A-DFS-with-recursion | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m, n = len(grid), len(grid[0])
ans, count = 0, 0
visited = [[0] * n for i in range(m)]
def search(i, j):
if grid[i][j] == '1' and not visited[i][j]:
visited[i][j] = 1
... | number-of-islands | Python | C++ : DFS with recursion | joshua_mur | 1 | 19 | number of islands | 200 | 0.564 | Medium | 3,251 |
https://leetcode.com/problems/number-of-islands/discuss/2498192/python3-or-explained-with-comment-or-easy-understanding | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
n, m = len(grid), len(grid[0])
visited = [[False for _ in range(m)] for _ in range(n)]
num_island = 0
for i in range(n):
for j in range(m):
if grid[i][j] == '1' and not visited[i][j]: #... | number-of-islands | python3 | explained with comment | easy-understanding | H-R-S | 1 | 27 | number of islands | 200 | 0.564 | Medium | 3,252 |
https://leetcode.com/problems/number-of-islands/discuss/2378109/python3-or-with-comments | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
count = 0
m = len(grid)
n = len(grid[0])
# here marking all the islands as false visited
visited = [[False for i in range(n)] for j in range(m)]
# This method will mark the island cell as visited as well... | number-of-islands | python3 | with comments | jayeshmaheshwari555 | 1 | 125 | number of islands | 200 | 0.564 | Medium | 3,253 |
https://leetcode.com/problems/number-of-islands/discuss/2268125/Python-DFS-Solution-(Faster-than-95) | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m=len(grid)
n=len(grid[0])
self.directions=[(0,1),(0,-1),(1,0),(-1,0)]
res=0
for i in range(m):
for j in range(n):
if grid[i][j]=="1":
self.dfs(i,j,grid,m,n)
... | number-of-islands | Python DFS Solution (Faster than 95%) | shatheesh | 1 | 166 | number of islands | 200 | 0.564 | Medium | 3,254 |
https://leetcode.com/problems/number-of-islands/discuss/2266815/Python3-Solution-with-using-dfs | class Solution:
def dfs(self, grid, i, j):
if i < 0 or i > len(grid) - 1 or j < 0 or j > len(grid[i]) - 1 or grid[i][j] == '0':
return
grid[i][j] = '0'
self.dfs(grid, i - 1, j)
self.dfs(grid, i, j - 1)
self.dfs(grid, i + 1, j)
self.dfs(grid, i, j... | number-of-islands | [Python3] Solution with using dfs | maosipov11 | 1 | 45 | number of islands | 200 | 0.564 | Medium | 3,255 |
https://leetcode.com/problems/number-of-islands/discuss/2265786/Python3-DFS-solution | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
visited = {}
rowLength = len(grid)
columnLength = len(grid[0])
def dfs(i,j):
if(i<0 or i>=rowLength or j<0 or j>=columnLength or (i,j) in visited):
return
if(... | number-of-islands | 📌 Python3 DFS solution | Dark_wolf_jss | 1 | 30 | number of islands | 200 | 0.564 | Medium | 3,256 |
https://leetcode.com/problems/number-of-islands/discuss/2184036/Python-DFSBFS-with-full-working-explanation | class Solution:
def numIslands(self, grid: List[List[str]]) -> int: # Time: O(n*n) and Space: O(n)
if not grid: return 0
rows, cols = len(grid), len(grid[0])
visit = set()
islands = 0
def bfs(r, c): # using bfs to search for adjacent 1's in a breadth first or row manner
... | number-of-islands | Python DFS/BFS with full working explanation | DanishKhanbx | 1 | 152 | number of islands | 200 | 0.564 | Medium | 3,257 |
https://leetcode.com/problems/number-of-islands/discuss/1903920/Passing-most-cases.-But-off-by-1-for-one-case-on-submission | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
X = len(grid)
Y = len(grid[0])
count = 0
seen = set()
def search(x,y):
if f'{x}{y}' in seen:
return False
el = grid[x][y]
if el == "0":
return False
s... | number-of-islands | Passing most cases. But off by 1 for one case on submission | yathi | 1 | 45 | number of islands | 200 | 0.564 | Medium | 3,258 |
https://leetcode.com/problems/number-of-islands/discuss/1862578/python-dfs-solution-super-simple | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
numOfIslands = 0
self.grid = grid
for r in range(len(grid)):
for c in range(len(grid[r])):
if grid[r][c] == "1":
numOfIslands+=1
self.callDFS(r,c)
... | number-of-islands | python dfs solution super simple | karanbhandari | 1 | 129 | number of islands | 200 | 0.564 | Medium | 3,259 |
https://leetcode.com/problems/number-of-islands/discuss/1851696/Python-DFS | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
num_islands = 0
num_rows, num_cols = len(grid), len(grid[0])
neighbor_directions = [(0,1), (0,-1), (1,0), (-1,0)]
def dfs(i, j):
if grid[i][j] == "1":
grid[i][j] = "-1"
... | number-of-islands | Python DFS | doubleimpostor | 1 | 112 | number of islands | 200 | 0.564 | Medium | 3,260 |
https://leetcode.com/problems/number-of-islands/discuss/1755076/Python-Easy-and-clean-DFS-solution | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m, n = len(grid), len(grid[0])
def dfs(x, y):
for i, j in [(1,0),(-1,0),(0,1),(0,-1)]:
if 0 <= x+i < m and 0 <= y+j < n and grid[x+i][y+j] == "1":
grid[x+i][y+j] = "0"
... | number-of-islands | [Python] Easy and clean DFS solution | nomofika | 1 | 296 | number of islands | 200 | 0.564 | Medium | 3,261 |
https://leetcode.com/problems/number-of-islands/discuss/1737834/Python-(DFS)-Easy-In-Place-Solution-%2B-Explanation-Beats-96 | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
# Convert to grid of ints from grid of strings
grid = [list( map(int,i) ) for i in grid]
counter = 0
# 1 = not visited 0 = visited
def dfs(i,j):
grid[i][j] = 0
if i > 0 and grid[i-1][j] == 1:... | number-of-islands | Python (DFS) Easy In-Place Solution + Explanation Beats 96% | Meerxn | 1 | 306 | number of islands | 200 | 0.564 | Medium | 3,262 |
https://leetcode.com/problems/number-of-islands/discuss/1684705/Python3-easy-to-understand-with-explanation | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m = len(grid)
n = len(grid[0])
res = 0
def count(i, j, res):
if i < 0 or j < 0 or i >= m or j >= n: # Neglect out-of-range case.
return
if grid[i][j] == '1': ... | number-of-islands | Python3 easy to understand with explanation | byroncharly3 | 1 | 182 | number of islands | 200 | 0.564 | Medium | 3,263 |
https://leetcode.com/problems/number-of-islands/discuss/1628042/Python-DFS-faster-than-98 | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
res = 0
def find_island(i,j):
grid[i][j] = '0'
#search up:
if i-1 >= 0 and grid[i-1][j] == '1':
find_island(i-1,j)
# down
if i... | number-of-islands | Python DFS faster than 98% | jcm497 | 1 | 407 | number of islands | 200 | 0.564 | Medium | 3,264 |
https://leetcode.com/problems/number-of-islands/discuss/1613343/DFS-Python-fastest-or-Latest | class Solution:
def dfs(self,graph,i,j):
if i<0 or j<0 or i>=len(graph) or j>=len(graph[0]):
return
if graph[i][j]=="0":
return
graph[i][j]="0"
self.dfs(graph,i-1,j)
self.dfs(graph,i+1,j)
self.dfs(graph,i,j-1)
self.df... | number-of-islands | DFS Python fastest | Latest | Brillianttyagi | 1 | 295 | number of islands | 200 | 0.564 | Medium | 3,265 |
https://leetcode.com/problems/number-of-islands/discuss/1607189/python-dfs | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m, n = len(grid), len(grid[0])
def dfs(grid, r, c):
if r < 0 or c < 0 or r > m - 1 or c > n - 1 or grid[r][c] == '0':
return
grid[r][c] = '0'
dfs... | number-of-islands | python dfs | alextoqc | 1 | 117 | number of islands | 200 | 0.564 | Medium | 3,266 |
https://leetcode.com/problems/number-of-islands/discuss/1579494/Better-then-97-iterative-python | class Solution:
def deleteIsland(self, grid, x, y):
island_field = [] # array of fields I want to delete
island_field.append((x,y)) # adding first one
grid[y][x] = '0'
while len(island_field) != 0:
curr = island_field[0] # gettin... | number-of-islands | Better then 97% iterative python | Pladq | 1 | 359 | number of islands | 200 | 0.564 | Medium | 3,267 |
https://leetcode.com/problems/number-of-islands/discuss/1401860/Python-easy-to-understand-solution.-Mark-and-count. | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
islandCount = 0
def creep(i, j, matrix):
# if already explored, then return
if matrix[i][j] == "0" or matrix[i][j] == "x" :
return
# mark 1 to x so it wont interfere agai... | number-of-islands | Python easy to understand solution. Mark and count. | vineeth_moturu | 1 | 197 | number of islands | 200 | 0.564 | Medium | 3,268 |
https://leetcode.com/problems/number-of-islands/discuss/1189760/Short-and-Simple-oror-93.6-faster-oror-Python-DFS-oror | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
def dfs(i,j):
if i>=m or i<0 or j>=n or j<0 or grid[i][j]=="0":
return
grid[i][j]="0"
dfs(i+1,j)
dfs(i-1,j)
dfs(i,j+1)
dfs(i,j-1)
m=len(grid)
n=len(grid[0])
f=0
... | number-of-islands | Short and Simple || 93.6% faster || Python DFS || | abhi9Rai | 1 | 215 | number of islands | 200 | 0.564 | Medium | 3,269 |
https://leetcode.com/problems/number-of-islands/discuss/1134032/WEEB-DOES-PYTHON-BFS | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
row, col = len(grid), len(grid[0])
queue = deque([])
count = 0
for x in range(row):
for y in range(col):
if grid[x][y] == "1":
count+=1
queue.append((x,y))
self.bfs(gri... | number-of-islands | WEEB DOES PYTHON BFS | Skywalker5423 | 1 | 353 | number of islands | 200 | 0.564 | Medium | 3,270 |
https://leetcode.com/problems/number-of-islands/discuss/1103937/Python3-BFS-and-DFS(Recursion)-Solution | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
'''
Breath-first Search:
'''
num_rows = len(grid)
num_cols = len(grid[0])
total_island = 0
for row in range(num_rows):
for col in range(num_cols):
if grid[row][col] == "1":
# Start doing Breath-first search:
total_isl... | number-of-islands | Python3 BFS & DFS(Recursion) Solution | Keyuan_Huang | 1 | 184 | number of islands | 200 | 0.564 | Medium | 3,271 |
https://leetcode.com/problems/number-of-islands/discuss/584270/Python3-dfs | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid: return 0
m, n = len(grid), len(grid[0])
def fn(i, j):
"""Flood fill cell with "0"."""
grid[i][j] = "0"
for ii, jj in (i-1, j), (i, j-1), (i, j+1), (i+1, j):
... | number-of-islands | [Python3] dfs | ye15 | 1 | 110 | number of islands | 200 | 0.564 | Medium | 3,272 |
https://leetcode.com/problems/number-of-islands/discuss/584270/Python3-dfs | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m, n = len(grid), len(grid[0])
ans = 0
for r in range(m):
for c in range(n):
if grid[r][c] == '1':
ans += 1
grid[r][c] = '0'
stack = [(... | number-of-islands | [Python3] dfs | ye15 | 1 | 110 | number of islands | 200 | 0.564 | Medium | 3,273 |
https://leetcode.com/problems/number-of-islands/discuss/452352/Python3-Simple-%2B-Fast-Implementation | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
count = 0
for i in range(len(grid)):
for j in range(len(grid[i])):
if(grid[i][j] == "1"):
count += 1
self.callBFS(grid, i, j)
return count
def call... | number-of-islands | Python3 Simple + Fast Implementation | iamyatin | 1 | 255 | number of islands | 200 | 0.564 | Medium | 3,274 |
https://leetcode.com/problems/number-of-islands/discuss/384878/Python-BFS-solution | class Solution:
def bfs(self, grid, queue):
while queue:
i, j = queue.pop(0)
for x in (-1, +1):
if i+x>=0 and i+x<len(grid) and grid[i+x][j] == '1':
grid[i+x][j] = '#'
queue.append((i+x, j))
... | number-of-islands | Python BFS solution | ahisa | 1 | 536 | number of islands | 200 | 0.564 | Medium | 3,275 |
https://leetcode.com/problems/number-of-islands/discuss/337907/Python-solution-using-dictionary-and-stack | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
d={}
res=0
for i in range(len(grid)):
for j in range(len(grid[i])):
if grid[i][j]=='1':d[(i,j)]=1
print(d)
while d:
l=list(d.keys())
stack=[l[0]]
... | number-of-islands | Python solution using dictionary and stack | ketan35 | 1 | 183 | number of islands | 200 | 0.564 | Medium | 3,276 |
https://leetcode.com/problems/number-of-islands/discuss/292868/Python-DFS-solution | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
num_islands = 0
for row in range(len(grid)):
for column in range(len(grid[0])):
if grid[row][column] == '1':
num_islands += 1
self.island_dfs(grid, row, column... | number-of-islands | Python DFS solution | AdyD | 1 | 574 | number of islands | 200 | 0.564 | Medium | 3,277 |
https://leetcode.com/problems/number-of-islands/discuss/2842030/Python-DFS-Solution | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
X = len(grid[0]) - 1
Y = len(grid) - 1
cnt = 0
def explore_island(v):
x, y = v
if grid[y][x] == '1':
grid[y][x] = 'land'
if x + 1 <= X: explore_island((x + 1, ... | number-of-islands | Python DFS Solution | LaggerKrd | 0 | 4 | number of islands | 200 | 0.564 | Medium | 3,278 |
https://leetcode.com/problems/number-of-islands/discuss/2835299/Easy-approachor-O(1)-space-or-Violent-imagination-or-Python3 | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
def sink(x,y):
# index out of range so return
if x<0 or x>=len(grid) or y < 0 or y >=len(grid[0]):return
# no land to sink so return
if grid[x][y] == '0':return
... | number-of-islands | Easy approach| O(1) space | Violent imagination | Python3 | pardunmeplz | 0 | 4 | number of islands | 200 | 0.564 | Medium | 3,279 |
https://leetcode.com/problems/number-of-islands/discuss/2834271/Intuitive-Easy-To-Understand-PYTHON-Sol | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m = len(grid)
n = len(grid[0])
islands=0
def dfs(row, col):
if grid[row][col]=="0":
return
grid[row][col]="0"
for increment in (1,-1):
if 0<=row+i... | number-of-islands | Intuitive Easy To Understand PYTHON Sol | taabish_khan22 | 0 | 5 | number of islands | 200 | 0.564 | Medium | 3,280 |
https://leetcode.com/problems/number-of-islands/discuss/2827024/Number-of-Islands | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
rows = len(grid)
cols = len(grid[0])
def isValid(i,j):
if i < 0 or j < 0 or i >= rows or j >= cols:
return False
else:
return True
... | number-of-islands | Number of Islands | yashchells | 0 | 3 | number of islands | 200 | 0.564 | Medium | 3,281 |
https://leetcode.com/problems/number-of-islands/discuss/2823646/Easy-Python-BFS-Approach | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
#BFS Approach
if not grid:
return 0
count = 0
r, c = len(grid), len(grid[0])
visited = [[False for _ in range(c)] for _ in range(r)]
queue = []
for i in range(r):
... | number-of-islands | Easy Python BFS Approach | dipupaul | 0 | 5 | number of islands | 200 | 0.564 | Medium | 3,282 |
https://leetcode.com/problems/number-of-islands/discuss/2815355/Python-solution | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
ans = 0
directions = [(-1,0),(1,0),(0,1),(0,-1)]
row = len(grid)
col = len(grid[0])
def dfs(i,j):
for x, y in directions:
dx = i + x
dy = j + y
... | number-of-islands | Python solution | maomao1010 | 0 | 6 | number of islands | 200 | 0.564 | Medium | 3,283 |
https://leetcode.com/problems/number-of-islands/discuss/2812473/simple-solution-with-deep-first-search | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
q=[]
m=len(grid)
n=len(grid[0])
visited=set([])
direction=[[0,1],[0,-1],[1,0],[-1,0]]
count=0
for i in range(m):
for j in range(n):
if grid[i][j]=="1" and (i,j) not... | number-of-islands | simple solution with deep first search | althrun | 0 | 1 | number of islands | 200 | 0.564 | Medium | 3,284 |
https://leetcode.com/problems/number-of-islands/discuss/2797091/Python-oror-DFS-Recursive-oror-DFS-Iterative-oror-BFS | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
R, C = len(grid), len(grid[0])
visited = set()
count = 0
dirs = [[-1,0],[1,0],[0,-1],[0,1]]
W, L = "0", "1"
def is_valid(r,c):
return (0 <= r < R and 0 <= c < C
and (... | number-of-islands | Python || DFS Recursive || DFS Iterative || BFS | sc1233 | 0 | 17 | number of islands | 200 | 0.564 | Medium | 3,285 |
https://leetcode.com/problems/number-of-islands/discuss/2793724/python-3-solution-using-DFS-recursive-method-Please-upvote-if-u-like-it | class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid:
return 0
rows,cols = len(grid),len(grid[0])
islands = 0
visit = set()
# using dfs in 2 methods
# firstone
def dfs(r,c):
if r <0 or r >= rows or c <0 or c >... | number-of-islands | python 3 solution using DFS recursive method Please upvote if u like it | mohamedsheded606 | 0 | 12 | number of islands | 200 | 0.564 | Medium | 3,286 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/469130/Python-iterative-sol.-based-on-bit-manipulation | class Solution:
def rangeBitwiseAnd(self, m: int, n: int) -> int:
shift = 0
# find the common MSB bits.
while m != n:
m = m >> 1
n = n >> 1
shift += 1
return m << shift | bitwise-and-of-numbers-range | Python iterative sol. based on bit-manipulation | brianchiang_tw | 9 | 723 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,287 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/2740509/Python-Solution-or-bit-manipulation | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
leftb = "{0:b}".format(left)
rightb = "{0:b}".format(right)
if len(rightb) > len(leftb):
return 0
res = left
for i in range(left + 1, right + 1):
res = res &... | bitwise-and-of-numbers-range | Python Solution | bit manipulation | maomao1010 | 1 | 72 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,288 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/1514406/Bitwise-AND-of-Numbers-Range-or-Beginner-friendly-or-Python-Solution | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
if left == 0 and right == 0: #To check if both inputs are 0
return 0
elif len(bin(left)) != len(bin(right)): #To check if both inputs have unequal binary number length
... | bitwise-and-of-numbers-range | Bitwise AND of Numbers Range | Beginner friendly | Python Solution | Shreya19595 | 1 | 217 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,289 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/1514129/Solution-With-Intuition-Explained | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
def check(left, right, p):
d = right - left
x = 1 << p
if d >= x:
return 0
else:
return 1 if right & x != 0 else 0
ans = left... | bitwise-and-of-numbers-range | Solution With Intuition Explained 🏞 | yasir991925 | 1 | 97 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,290 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/2742371/Python-3-line-solution-with-time-less-O(1's-in-bin(right)) | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
while right > left:
right = right & (right - 1)
return right | bitwise-and-of-numbers-range | Python 3-line solution with time < O(#1's in bin(right)) | Fredrick_LI | 0 | 3 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,291 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/2689537/Easy-Fast-Basic-Python-Solution | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
#Optimal Solution
if left==0 or right==0:
return 0
left_limit = math.floor(log(left,2))
right_limit = int(log(right,2))
if right_limit - left_limit:
return 0
return reduce... | bitwise-and-of-numbers-range | Easy Fast Basic Python Solution | RajatGanguly | 0 | 6 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,292 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/1992017/Python-5-lines-or-Easy-to-understand-O(1)-Space-O(1)-Time | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
# check is going to be our result, dist is the distance between the left and the right values
check, dist = right & left, right-left
# for ease, we just start with 31 instead of highest set bit
for ... | bitwise-and-of-numbers-range | Python 5 lines | Easy to understand O(1) Space O(1) Time | Apurv_Moroney | 0 | 206 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,293 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/1974912/Python-bit-shifting-solution-(heavily-commented) | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
"""
Given
11010 &
11011 &
11100 &
11101 &
11110 &
-----
11000 result
The AND operator keeps those bits whi... | bitwise-and-of-numbers-range | Python bit-shifting solution (heavily commented) | Zikker | 0 | 67 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,294 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/1698884/easy-to-understand-solution-counting-trailing-zeros | class Solution:
# if num = 600000000 next_num = 600000512, since all the numbers between them will return the same number "num"
def get_next_num(self, num):
trailing_zeros = 0
offset = 1
while num&(num+offset) == num:
trailing_zeros+=1
offset=(2*offset)+1... | bitwise-and-of-numbers-range | easy to understand solution - counting trailing zeros | SN009006 | 0 | 100 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,295 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/1622474/Mechanical-boring-non-genius-ideas-based-on-binary-strings... | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
right_binary = ['0']+list(bin(right)[2:])
left_binary = list(bin(left)[2:].zfill(len(right_binary)))
res = ['0']*len(right_binary)
last_zero = 0
for i in range(len(left_binary)):
if left_binary[i] == '0':
last_zero = i
elif left_bina... | bitwise-and-of-numbers-range | Mechanical, boring, non-genius ideas based on binary strings... | throwawayleetcoder19843 | 0 | 68 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,296 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/594583/Python3-Faster-than-92.22-100-memory-efficient | class Solution:
def rangeBitwiseAnd(self, m, n):
a = m
b = n
while(a < b):
b -= (b & -b)
return b | bitwise-and-of-numbers-range | Python3 Faster than 92.22%, 100% memory efficient | Hannibal404 | 0 | 189 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,297 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/593941/Python3-binary-common-prefix | class Solution:
def rangeBitwiseAnd(self, m: int, n: int) -> int:
k = 0
while m != n and (k:=k+1):
m, n = m >> 1, n >> 1
return m << k | bitwise-and-of-numbers-range | [Python3] binary common prefix | ye15 | 0 | 46 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,298 |
https://leetcode.com/problems/bitwise-and-of-numbers-range/discuss/593941/Python3-binary-common-prefix | class Solution:
def rangeBitwiseAnd(self, left: int, right: int) -> int:
while left < right: right &= right - 1
return right | bitwise-and-of-numbers-range | [Python3] binary common prefix | ye15 | 0 | 46 | bitwise and of numbers range | 201 | 0.423 | Medium | 3,299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.