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/verifying-an-alien-dictionary/discuss/1448977/Python3Python-Simple-solution-using-dictionary-w-comments | class Solution:
def isAlienSorted(self, words: List[str], order: str) -> bool:
# '∅' is defined as the blank character which
# is less than any other character
order = {c: i for i,c in enumerate(order)}
# Get num of words
m = len(words)
# l... | verifying-an-alien-dictionary | [Python3/Python] Simple solution using dictionary w/ comments | ssshukla26 | 0 | 91 | verifying an alien dictionary | 953 | 0.527 | Easy | 15,500 |
https://leetcode.com/problems/verifying-an-alien-dictionary/discuss/1426071/Python3-Faster-Than-85.26-Memory-Less-Than-98.32 | class Solution:
def isAlienSorted(self, words: List[str], order: str) -> bool:
d, i = {}, 0
for index, letter in enumerate(order):
d[letter] = index
while(i < len(words) - 1):
equal = True
s1, s2, j = words[i], words[i + 1... | verifying-an-alien-dictionary | Python3 Faster Than 85.26%, Memory Less Than 98.32% | Hejita | 0 | 96 | verifying an alien dictionary | 953 | 0.527 | Easy | 15,501 |
https://leetcode.com/problems/verifying-an-alien-dictionary/discuss/1265077/Python-Non-One-Liner-With-Explanation | class Solution(object):
def isAlienSorted(self, words, order):
"""
:type words: List[str]
:type order: str
:rtype: bool
"""
# 1. Create a dict which assigns a value to each letter in dictionary
# 2. Iterate through words, checking letter by letter for if value... | verifying-an-alien-dictionary | Python Non-One Liner With Explanation | jmwyds | 0 | 128 | verifying an alien dictionary | 953 | 0.527 | Easy | 15,502 |
https://leetcode.com/problems/verifying-an-alien-dictionary/discuss/1149951/Simple-solution-in-Python-using-a-Hashmap-and-sorting | class Solution:
def isAlienSorted(self, words: List[str], order: str) -> bool:
hashmap = dict()
for i in range(len(order)):
hashmap[order[i]] = i
score = [[hashmap[i] for i in word] for word in words]
return sorted(score) == score | verifying-an-alien-dictionary | Simple solution in Python using a Hashmap and sorting | amoghrajesh1999 | 0 | 39 | verifying an alien dictionary | 953 | 0.527 | Easy | 15,503 |
https://leetcode.com/problems/verifying-an-alien-dictionary/discuss/541049/Python-sol-by-comparison.-90%2B-wHint | class Solution:
def isAlienSorted(self, words: List[str], order: str) -> bool:
# alphabet dictionary with alien order
dictionary = { letter : idx for idx, letter in enumerate(order) }
def chk_in_order( word1: str, word2: str):
for i in range( min(len(word... | verifying-an-alien-dictionary | Python sol by comparison. 90%+ [w/Hint] | brianchiang_tw | 0 | 243 | verifying an alien dictionary | 953 | 0.527 | Easy | 15,504 |
https://leetcode.com/problems/verifying-an-alien-dictionary/discuss/382166/Solution-in-Python-3-(one-line) | class Solution:
def isAlienSorted(self, w: List[str], a: str) -> bool:
return (lambda x: all([w[i].translate(x) < w[i+1].translate(x) for i in range(len(w)-1)]))(str.maketrans(a,string.ascii_lowercase))
- Junaid Mansuri
(LeetCode ID)@hotmail.com | verifying-an-alien-dictionary | Solution in Python 3 (one line) | junaidmansuri | -2 | 559 | verifying an alien dictionary | 953 | 0.527 | Easy | 15,505 |
https://leetcode.com/problems/verifying-an-alien-dictionary/discuss/1378617/Best-code-Explaination-or-Any-programmer-can-understand-or-Python-Code-or-Very-easy-to-understand | class Solution:
def compare(self, cache, word1, word2):
i, j = 0,0
while i < len(word1) and j < len(word2):
if cache[word1[i]] < cache[word2[j]]:
return True
elif cache[word1[i]] > cache[word2[j]]:
return False
else: #letters e... | verifying-an-alien-dictionary | Best code Explaination | Any programmer can understand | Python Code | Very easy to understand | sathwickreddy | -3 | 404 | verifying an alien dictionary | 953 | 0.527 | Easy | 15,506 |
https://leetcode.com/problems/array-of-doubled-pairs/discuss/1840844/python-3-oror-O(nlogn) | class Solution:
def canReorderDoubled(self, arr: List[int]) -> bool:
count = collections.Counter(arr)
for n in sorted(arr, key=abs):
if count[n] == 0:
continue
if count[n * 2] == 0:
return False
count[n] -= 1
count[n * 2... | array-of-doubled-pairs | python 3 || O(nlogn) | dereky4 | 1 | 81 | array of doubled pairs | 954 | 0.391 | Medium | 15,507 |
https://leetcode.com/problems/array-of-doubled-pairs/discuss/1810465/Python-Optimised-Counter-Solution-Explained | class Solution:
def canReorderDoubled(self, arr: List[int]) -> bool:
arr.sort(key=abs) # Sort the array based on the absolute value
cnt = Counter(arr) # Count the number of each element in arr
while cnt:
x = next(iter(cnt.keys())) # Get the next unique element x
# Handle co... | array-of-doubled-pairs | [Python] Optimised Counter Solution Explained | zayne-siew | 1 | 69 | array of doubled pairs | 954 | 0.391 | Medium | 15,508 |
https://leetcode.com/problems/array-of-doubled-pairs/discuss/1398727/Greedy-oror-O(NlogN)-oror-Clean-and-Concise-oror-93-faster | class Solution:
def canReorderDoubled(self, arr: List[int]) -> bool:
dic = Counter(arr)
res = []
for n in sorted(arr):
if dic[n]>0:
dic[n]-=1 # This is to check 0 if it present take e.g. : [4,0]
if 2*n in dic and dic[2*n]>0:
dic[2*n]-=1
... | array-of-doubled-pairs | 🐍 Greedy || O(NlogN) || Clean & Concise || 93% faster 📌📌 | abhi9Rai | 0 | 73 | array of doubled pairs | 954 | 0.391 | Medium | 15,509 |
https://leetcode.com/problems/array-of-doubled-pairs/discuss/1397824/python3-oror-clean-oror-simple-approach | class Solution:
def canReorderDoubled(self, arr: List[int]) -> bool:
d={}
arr.sort()
print(arr)
for i in range (0,len(arr)):
if arr[i] in d:
d[arr[i]] +=1
else:
d[arr[i]]=1
print(d)
for key in d:... | array-of-doubled-pairs | python3 || clean || simple approach | minato_namikaze | 0 | 93 | array of doubled pairs | 954 | 0.391 | Medium | 15,510 |
https://leetcode.com/problems/array-of-doubled-pairs/discuss/1397426/Just-did-what-I-know-! | class Solution:
def canReorderDoubled(self, arr: List[int]) -> bool:
def helper(nums) :
stack = []
while nums :
top = nums.pop(0)
if stack and 2*stack[0] == top :
stack.pop(0)
else:
stac... | array-of-doubled-pairs | Just did what I know ! | abhijeetgupto | 0 | 47 | array of doubled pairs | 954 | 0.391 | Medium | 15,511 |
https://leetcode.com/problems/array-of-doubled-pairs/discuss/975041/Python3-freq-table | class Solution:
def canReorderDoubled(self, arr: List[int]) -> bool:
freq = Counter(arr)
for x in sorted(freq, key=abs):
if freq[2*x] < freq[x]: return False
freq[2*x] -= freq[x]
return True | array-of-doubled-pairs | [Python3] freq table | ye15 | 0 | 81 | array of doubled pairs | 954 | 0.391 | Medium | 15,512 |
https://leetcode.com/problems/array-of-doubled-pairs/discuss/838398/Python-3-or-Hash-Table-or-Explanations | class Solution:
def canReorderDoubled(self, A: List[int]) -> bool:
A = sorted([-i if i < 0 else i for i in A], reverse=True)
n, c, cnt = len(A), collections.Counter(A), 0
for i in range(n):
if c[2*A[i]] > 0:
c[2*A[i]] -= 1
c[A[i]] -= 1
... | array-of-doubled-pairs | Python 3 | Hash Table | Explanations | idontknoooo | -2 | 156 | array of doubled pairs | 954 | 0.391 | Medium | 15,513 |
https://leetcode.com/problems/delete-columns-to-make-sorted-ii/discuss/844457/Python-3-or-Greedy-DP-(28-ms)-or-Explanation | class Solution:
def minDeletionSize(self, A: List[str]) -> int:
m, n = len(A), len(A[0])
ans, in_order = 0, [False] * (m-1)
for j in range(n):
tmp_in_order = in_order[:]
for i in range(m-1):
# previous step, rows are not in order; and current step rows are not in ... | delete-columns-to-make-sorted-ii | Python 3 | Greedy, DP (28 ms) | Explanation | idontknoooo | 6 | 550 | delete columns to make sorted ii | 955 | 0.346 | Medium | 15,514 |
https://leetcode.com/problems/delete-columns-to-make-sorted-ii/discuss/1906030/PYTHON-SOL-oror-EXPLAINED-oror-WELL-COMMENTED-SOLUTION-oror-GREEDY-oror | class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
n = len(strs)
col_size = len(strs[0])
# a b c d e f g h i j k l m n o p q r s t u v w x y z
i = 0
ans = 0
def getRemoved(idx):
# removing the idx column
for ... | delete-columns-to-make-sorted-ii | PYTHON SOL || EXPLAINED || WELL COMMENTED SOLUTION || GREEDY || | reaper_27 | 0 | 124 | delete columns to make sorted ii | 955 | 0.346 | Medium | 15,515 |
https://leetcode.com/problems/delete-columns-to-make-sorted-ii/discuss/1294831/Python3-greedy | class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
m, n = len(strs), len(strs[0]) # dimensions
ans, grp = 0, [0]*m
for j in range(n):
for i in range(1, m):
if grp[i-1] == grp[i] and strs[i-1][j] > strs[i][j]:
ans += 1
... | delete-columns-to-make-sorted-ii | [Python3] greedy | ye15 | -1 | 105 | delete columns to make sorted ii | 955 | 0.346 | Medium | 15,516 |
https://leetcode.com/problems/tallest-billboard/discuss/1561795/Python3-dp-and-binary-search | class Solution:
def tallestBillboard(self, rods: List[int]) -> int:
dp = {0: 0}
for x in rods:
for k, v in dp.copy().items():
dp[k+x] = max(dp.get(k+x, 0), v)
if k >= x: dp[k-x] = max(dp.get(k-x, 0), v+x)
else: dp[x-k] = max(dp.get(x-k, 0... | tallest-billboard | [Python3] dp & binary search | ye15 | 0 | 273 | tallest billboard | 956 | 0.399 | Hard | 15,517 |
https://leetcode.com/problems/tallest-billboard/discuss/1561795/Python3-dp-and-binary-search | class Solution:
def tallestBillboard(self, rods: List[int]) -> int:
def fn(arr):
"""Possible a mapping from diff to length"""
mp = defaultdict(int)
for t in range(1, len(arr)+1):
for total in combinations(arr, t):
tt = sum(t... | tallest-billboard | [Python3] dp & binary search | ye15 | 0 | 273 | tallest billboard | 956 | 0.399 | Hard | 15,518 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/347500/Python3-Prison-Cells-After-N-days%3A-dictionary-to-store-pattern | class Solution:
def prisonAfterNDays(self, cells: List[int], N: int) -> List[int]:
def nextday(cells):
next_day_cells = [0] *len(cells)
for i in range(1,len(cells)-1):
if cells[i-1] == cells[i+1]:
next_day_cells[i] = 1
else:
... | prison-cells-after-n-days | [Python3] Prison Cells After N days: dictionary to store pattern | zhanweiting | 33 | 3,800 | prison cells after n days | 957 | 0.391 | Medium | 15,519 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/744326/Very-Intuitive-Python-Detailed-Solution-with-Example | class Solution(object):
def prisonAfterNDays(self, cells, N):
"""
:type cells: List[int]
:type N: int
:rtype: List[int]
"""
def next(state):
return tuple([1 if i>0 and i<len(state)-1 and state[i-1] == state[i+1] else 0 for i in range(len(state))])
... | prison-cells-after-n-days | Very Intuitive Python Detailed Solution with Example | ivankatrump | 11 | 841 | prison cells after n days | 957 | 0.391 | Medium | 15,520 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/2077309/EXPLAINED-by-picture | class Solution:
def prisonAfterNDays(self, c, n) :
return self.prisonAfterNDays([0, c[0]==c[2], c[1]==c[3], c[2]==c[4], c[3]==c[5], c[4]==c[6], c[5]==c[7], 0], (n-1)%14) if n else [int(i) for i in c] | prison-cells-after-n-days | 🚩 EXPLAINED by picture ^^ | andrii_khlevniuk | 3 | 199 | prison cells after n days | 957 | 0.391 | Medium | 15,521 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/2077309/EXPLAINED-by-picture | class Solution:
def prisonAfterNDays(self, c, n) :
for _ in range(1+(n-1)%14) :
p=-1
for i in range(8) :
p, c[i] = c[i], int(i<7 and p==c[i+1])
return c | prison-cells-after-n-days | 🚩 EXPLAINED by picture ^^ | andrii_khlevniuk | 3 | 199 | prison cells after n days | 957 | 0.391 | Medium | 15,522 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/2518634/Python-Detecting-repeated-patterns-No-Dictionary-With-Intuition | class Solution:
def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
# since the first and the last cell will always be zero after the
# first day, the array only has 2**(8-2) = 64 states
# therefore the cells will be repeated after more than 64 days
#
... | prison-cells-after-n-days | [Python] - Detecting repeated patterns - No Dictionary - With Intuition | Lucew | 2 | 140 | prison cells after n days | 957 | 0.391 | Medium | 15,523 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/1918973/PYTHON-SOL-oror-FAST-oror-MATHS-oror-WELL-EXPLAINED-oror-PATTERN-oror-LOGIC-EXPLAINED-oror | class Solution:
def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
n = n%14 if n%14!=0 else 14
for _ in range(n):
new = [0]*8
for i in range(1,7):
if cells[i-1]==cells[i+1]: new[i] = 1
cells = new
return cells | prison-cells-after-n-days | PYTHON SOL || FAST || MATHS || WELL EXPLAINED || PATTERN || LOGIC EXPLAINED || | reaper_27 | 2 | 207 | prison cells after n days | 957 | 0.391 | Medium | 15,524 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/2129910/Easy-Python-Solution-Faster-than-95-(36ms) | class Solution:
def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
new_cells = [0] * len(cells)
if n % 14 != 0:
n = n % 14
else:
n = 14
while n > 0:
for i in range(1, len(cells)-1):
if cells[i-1... | prison-cells-after-n-days | Easy Python Solution - Faster than 95% (36ms) | kn_vardhan | 1 | 87 | prison cells after n days | 957 | 0.391 | Medium | 15,525 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/1743890/Python-Dynamic-and-Fix-Loop | class Solution:
def prisonAfterNDays(self, nums: List[int], n: int) -> List[int]:
n=(n-1)%14+1
for i in range(1,n+1):
temp=list(nums)
temp[0],temp[7] = 0,0
for j in range(1,7):
if nums[j-1]==nums[j+1]:
temp[j]=1
... | prison-cells-after-n-days | Python Dynamic and Fix Loop | shandilayasujay | 1 | 204 | prison cells after n days | 957 | 0.391 | Medium | 15,526 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/1743890/Python-Dynamic-and-Fix-Loop | class Solution:
def prisonAfterNDays(self, nums: List[int], n: int) -> List[int]:
s=[]
for i in range(1,n+1):
temp=list(nums)
temp[0],temp[7] = 0,0
for j in range(1,7):
if nums[j-1]==nums[j+1]:
temp[j]=1
else: te... | prison-cells-after-n-days | Python Dynamic and Fix Loop | shandilayasujay | 1 | 204 | prison cells after n days | 957 | 0.391 | Medium | 15,527 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/864181/Python-3-or-Cycle-Hash-Table-or-Explanation | class Solution:
def prisonAfterNDays(self, cells: List[int], N: int) -> List[int]:
id_cells, cells_id = dict(), dict()
prev, loop = cells[:], None
for k in range(N):
for i in range(1, 7):
cells[i] = 1 if prev[i+1] == prev[i-1] else 0
cells[0] = cells[7... | prison-cells-after-n-days | Python 3 | Cycle, Hash Table | Explanation | idontknoooo | 1 | 464 | prison cells after n days | 957 | 0.391 | Medium | 15,528 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/717747/Python-3-solution-with-concise-explanation | class Solution:
def prisonAfterNDays(self, cells: List[int], N: int) -> List[int]:
"""
1. The first and the last element will always become 0 after one
transformation step.
2. So the actual loop happened between the combinations of the 6
numbers in the middle.
... | prison-cells-after-n-days | Python 3 solution with concise explanation | eroneko | 1 | 56 | prison cells after n days | 957 | 0.391 | Medium | 15,529 |
https://leetcode.com/problems/prison-cells-after-n-days/discuss/1304180/Python-Bit-manipulation-and-memory-table | class Solution:
def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
counter = 0
bitmap = 0
# cells to bitmap
for b in cells:
bitmap = (bitmap<<1)|b
# memory table
idToCount = {bitmap:counter}
countToCells = {counter:bitmap}
for j in rang... | prison-cells-after-n-days | Python Bit manipulation and memory table | johnnylu305 | 0 | 130 | prison cells after n days | 957 | 0.391 | Medium | 15,530 |
https://leetcode.com/problems/check-completeness-of-a-binary-tree/discuss/2287813/Python3-oror-bfs-8-lines-w-explanation-oror-TM%3A-9797 | class Solution:
def isCompleteTree(self, root: TreeNode) -> bool:
# The criteria for an n-level complete tree:
#
# • The first n-1 rows have no null nodes.
#
# • The nth row has no non-null no... | check-completeness-of-a-binary-tree | Python3 || bfs, 8 lines, w/ explanation || T/M: 97%/97% | warrenruud | 3 | 111 | check completeness of a binary tree | 958 | 0.538 | Medium | 15,531 |
https://leetcode.com/problems/check-completeness-of-a-binary-tree/discuss/1775415/Easy-python3-solution-using-BFS-and-indexes | class Solution:
def isCompleteTree(self, root: Optional[TreeNode]) -> bool:
res=[]
from collections import deque
q=deque()
q.append((root,1))
while q:
node,ind=q.popleft()
res.append(ind)
if node.left:
q.ap... | check-completeness-of-a-binary-tree | Easy python3 solution using BFS and indexes | Karna61814 | 1 | 74 | check completeness of a binary tree | 958 | 0.538 | Medium | 15,532 |
https://leetcode.com/problems/check-completeness-of-a-binary-tree/discuss/2234169/Python-BFS | class Solution:
def isCompleteTree(self, root: Optional[TreeNode]) -> bool:
q = deque([root])
missing = False
while q:
length = len(q)
for _ in range(length):
node = q.popleft()
if node.left:
if missing:
... | check-completeness-of-a-binary-tree | Python, BFS | blue_sky5 | 0 | 24 | check completeness of a binary tree | 958 | 0.538 | Medium | 15,533 |
https://leetcode.com/problems/check-completeness-of-a-binary-tree/discuss/977153/Python3-BFS-O(N) | class Solution:
def isCompleteTree(self, root: TreeNode) -> bool:
i = -1
prev = 1/2
queue = [root]
while queue:
newq = []
flag = False
if prev != 2**i: return False # check previous level is full
for node in queue:
fo... | check-completeness-of-a-binary-tree | [Python3] BFS O(N) | ye15 | 0 | 90 | check completeness of a binary tree | 958 | 0.538 | Medium | 15,534 |
https://leetcode.com/problems/check-completeness-of-a-binary-tree/discuss/860671/python-queue-appending-check-final-array-for-non-Nones | class Solution:
def isCompleteTree(self, root: TreeNode) -> bool:
queue = [root]
while queue[0]:
node = queue[0]
queue = queue[1:]
queue.append(node.left)
queue.append(node.right)
return not any(queue) | check-completeness-of-a-binary-tree | python, queue appending, check final array for non-Nones | anthcor | 0 | 68 | check completeness of a binary tree | 958 | 0.538 | Medium | 15,535 |
https://leetcode.com/problems/regions-cut-by-slashes/discuss/205674/DFS-on-upscaled-grid | class Solution:
def regionsBySlashes(self, grid: List[str]) -> int:
def dfs(i: int, j: int) -> int:
if min(i, j) < 0 or max(i, j) >= len(g) or g[i][j] != 0:
return 0
g[i][j] = 1
return 1 + dfs(i - 1, j) + dfs(i + 1, j) + dfs(i, j - 1) + dfs(i, j + 1)
... | regions-cut-by-slashes | DFS on upscaled grid | votrubac | 733 | 23,400 | regions cut by slashes | 959 | 0.691 | Medium | 15,536 |
https://leetcode.com/problems/regions-cut-by-slashes/discuss/1311948/DFS-PYTHON-CLEAN | class Solution:
def regionsBySlashes(self, grid: List[str]) -> int:
def check(x,y,t):
if x>=0 and y>=0 and x<row and y<col and (x,y,t) not in visited:
return True
return False
row,col=len(grid),len(grid[0])
visited=set()
... | regions-cut-by-slashes | DFS-PYTHON-CLEAN | manmohan1105 | 2 | 221 | regions cut by slashes | 959 | 0.691 | Medium | 15,537 |
https://leetcode.com/problems/regions-cut-by-slashes/discuss/2129956/Python-or-Flood-Fill-like-DFS-solution-Not-short-But-Clear~ | class Solution:
def regionsBySlashes(self, grid):
return self.flood_fill(grid)
def flood_fill(self, grid):
''' |\0/|
set area of cell like: |3X1|
|/2\|
dye and count like flood fill
'''
... | regions-cut-by-slashes | Python | Flood-Fill-like DFS solution, Not short But Clear~ | steve-jokes | 0 | 92 | regions cut by slashes | 959 | 0.691 | Medium | 15,538 |
https://leetcode.com/problems/regions-cut-by-slashes/discuss/1920789/Python3-or-3x3-Matrix-or-DFS-or-Faster-than-81 | class Solution:
def regionsBySlashes(self, grid: List[str]) -> int:
n, n3 = len(grid), len(grid) * 3
mat = [[0] * n3 for _ in range(n3)]
for i in range(n):
for j in range(n):
if grid[i][j] == "/":
mat[i * 3][j * 3 + 2] = 1;
... | regions-cut-by-slashes | Python3 | 3x3 Matrix | DFS | Faster than 81% | milannzz | 0 | 79 | regions cut by slashes | 959 | 0.691 | Medium | 15,539 |
https://leetcode.com/problems/regions-cut-by-slashes/discuss/815325/Python-3-better-than-98-(DSU-method) | class Solution:
def regionsBySlashes(self, grid: List[str]) -> int:
n,m=len(grid),len(grid[0])
parent=[i for i in range((n+1)*(m+1))]
rank=[1 for i in range((n+1)*(m+1))]
matrix=[]
cc=0
for i in range(n+1):
li=[]
for j in range(m+1):
... | regions-cut-by-slashes | Python 3 better than 98% (DSU method) | codeenrrun | 0 | 153 | regions cut by slashes | 959 | 0.691 | Medium | 15,540 |
https://leetcode.com/problems/regions-cut-by-slashes/discuss/730936/Python-3-Easy-understanding-Union-Find-with-explanation | class Solution:
def regionsBySlashes(self, grid: List[str]) -> int:
'''
--------
| \ 0 / |
| 3\ / |
| /\ 1 |
|/ 2 \ |
|-------|
'''
self.N = len(grid)
# m_ = [i for i in range(self.N * self.N * 4)]
m_ = list(range(self.N * self.N * 4))
self.count = self.N * self.N * 4
... | regions-cut-by-slashes | Python 3 - Easy understanding Union Find -- with explanation | nbismoi | 0 | 282 | regions cut by slashes | 959 | 0.691 | Medium | 15,541 |
https://leetcode.com/problems/delete-columns-to-make-sorted-iii/discuss/1258211/Python3-top-down-dp | class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
m, n = len(strs), len(strs[0]) # dimensions
@cache
def fn(k, prev):
"""Return min deleted columns to make sorted."""
if k == n: return 0
ans = 1 + fn(k+1, prev) # delete kth colu... | delete-columns-to-make-sorted-iii | [Python3] top-down dp | ye15 | 2 | 139 | delete columns to make sorted iii | 960 | 0.571 | Hard | 15,542 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1337509/PYTHON-3-%3A-SUPER-EASY-99.52-FASTER | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
list1 = []
for i in nums :
if i in list1 :
return i
else :
list1.append(i) | n-repeated-element-in-size-2n-array | PYTHON 3 : SUPER EASY 99.52% FASTER | rohitkhairnar | 11 | 510 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,543 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1337509/PYTHON-3-%3A-SUPER-EASY-99.52-FASTER | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
set1 = set()
for i in nums :
if i in set1 :
return i
else :
set1.add(i) | n-repeated-element-in-size-2n-array | PYTHON 3 : SUPER EASY 99.52% FASTER | rohitkhairnar | 11 | 510 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,544 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1337509/PYTHON-3-%3A-SUPER-EASY-99.52-FASTER | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
for i in nums :
if nums.count(i) == len(nums)/2 :
return i | n-repeated-element-in-size-2n-array | PYTHON 3 : SUPER EASY 99.52% FASTER | rohitkhairnar | 11 | 510 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,545 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1337509/PYTHON-3-%3A-SUPER-EASY-99.52-FASTER | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
dic = {}
for i in nums :
if i in dic :
dic[i] += 1
if dic[i] == len(nums)/2 :
return i
else :
dic[i] = 1 | n-repeated-element-in-size-2n-array | PYTHON 3 : SUPER EASY 99.52% FASTER | rohitkhairnar | 11 | 510 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,546 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1209477/Python-Explained-Simple-one-liner-Counter | class Solution(object):
def repeatedNTimes(self, nums):
return Counter(nums).most_common(1)[0][0] | n-repeated-element-in-size-2n-array | [Python Explained] Simple one-liner Counter | akashadhikari | 2 | 81 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,547 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1182101/python-or-faster-than-97.20 | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
d = {}
for i in A:
if i not in d:
d[i] = 1
else:
return i | n-repeated-element-in-size-2n-array | python | faster than 97.20% | prachijpatel | 2 | 133 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,548 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1562650/Python3-one-line-solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
return (sum(nums) - sum(set(nums))) // (len(nums) // 2 - 1) | n-repeated-element-in-size-2n-array | Python3 one line solution | sirenescx | 1 | 78 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,549 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1376630/Python-3-%3A-FASTER-THAN-99.66-O(N)-TIME-COMPLEXITY | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
dic = {}
for i in nums:
if i in dic:
return i
if i not in dic:
dic[i] = 1 | n-repeated-element-in-size-2n-array | Python 3 : FASTER THAN 99.66% , O(N) TIME COMPLEXITY | iron_man_365 | 1 | 90 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,550 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2821341/Simple-One-Line-Python3 | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
return Counter(nums).most_common(1)[0][0] | n-repeated-element-in-size-2n-array | Simple One Line Python3 | Jlonerawesome | 0 | 3 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,551 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2809876/Easiest-Python-Solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
s = set()
for num in nums:
if num in s:
return num
s.add(num) | n-repeated-element-in-size-2n-array | Easiest Python Solution | namashin | 0 | 4 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,552 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2770013/Simple-Python-Solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
for i in nums :
if nums.count(i) == len(nums)/2 :
return i | n-repeated-element-in-size-2n-array | Simple Python Solution | dnvavinash | 0 | 4 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,553 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2711402/3-lines-Python3-SolutionororO(n) | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
for i in nums:
if nums.count(i)!=1:
return i | n-repeated-element-in-size-2n-array | 3 lines Python3 Solution||O(n) | sowmika_chaluvadi | 0 | 6 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,554 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2659808/Easy-to-Understand-or-Beginner's-Friendly-or-Python | class Solution(object):
def repeatedNTimes(self, nums):
nums.sort()
n = len(nums) // 2
if nums[n + 1] == nums[n]: return nums[n]
return nums[n - 1] | n-repeated-element-in-size-2n-array | Easy to Understand | Beginner's Friendly | Python | its_krish_here | 0 | 8 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,555 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2619751/O(1)-space-python3 | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
n =len(nums)
for i in range(2,n):
if nums[i]==nums[i-1] or nums[i]==nums[i-2]:
return nums[i]
return nums[0] | n-repeated-element-in-size-2n-array | O(1) space python3 | abhayCodes | 0 | 14 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,556 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2619698/return-first-element-to-occur-2-times-python | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
s=set()
for i in nums:
if i in s:
return i
s.add(i) | n-repeated-element-in-size-2n-array | return first element to occur 2 times python | abhayCodes | 0 | 7 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,557 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2466512/Python-solution-using-dictionary | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
num_dict = {}
for i in nums:
if i in num_dict:
num_dict[i] += 1
else:
num_dict[i] = 1
for k, v in num_dict.items():
if v > 1:
return ... | n-repeated-element-in-size-2n-array | Python solution using dictionary | samanehghafouri | 0 | 10 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,558 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2414318/Python3-Simple-Set-Solution-beats-95 | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
seen = set()
for num in nums:
if num in seen:
return num
seen.add(num) | n-repeated-element-in-size-2n-array | [Python3] Simple Set Solution, beats 95% | ivnvalex | 0 | 39 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,559 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2328821/Simple-Python3-Solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
unique = len(set(nums))
cnt = Counter(nums)
for k, v in cnt.items():
if v == unique - 1:
return k | n-repeated-element-in-size-2n-array | Simple Python3 Solution | vem5688 | 0 | 26 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,560 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2309975/My-Solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
for element in nums:
if nums.count(element) == len(nums) // 2:
return element | n-repeated-element-in-size-2n-array | My Solution 😀 | ibrahimbayburtlu5 | 0 | 7 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,561 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2181996/Simple-Logic | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
freq = Counter(nums)
return freq.most_common(1)[0][0] | n-repeated-element-in-size-2n-array | Simple Logic | writemeom | 0 | 42 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,562 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/2095052/PYTHON-or-Simple-python-solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
hashMap = {}
for i in nums:
hashMap[i] = 1 + hashMap.get(i, 0)
n = len(nums) // 2
for i in hashMap:
if hashMap[i] == n:
return i | n-repeated-element-in-size-2n-array | PYTHON | Simple python solution | shreeruparel | 0 | 50 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,563 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1928296/easy-python-code | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
x = {}
for i in nums:
if i in x:
return i
else:
x[i] = 0 | n-repeated-element-in-size-2n-array | easy python code | dakash682 | 0 | 32 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,564 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1882796/easy-simple-python-solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
n = len(nums)//2
d={}
for num in nums:
if num in d:
d[num]+=1
else:
d[num]=1
for key in d:
if d[key] == n :
return key | n-repeated-element-in-size-2n-array | easy simple python solution | Buyanjargal | 0 | 35 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,565 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1807550/1-Line-Python-Solution-oror-80-Faster-oror-Memory-less-than-75 | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
return Counter(nums).most_common(1)[0][0] | n-repeated-element-in-size-2n-array | 1-Line Python Solution || 80% Faster || Memory less than 75% | Taha-C | 0 | 46 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,566 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1805255/Python-Easy-to-Understand-Hashmap | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
dic = {}
for n in nums:
if n in dic:
dic[n]+=1
else:
dic[n]=1
for k in dic:
# print(f'Ans is {k} -> {dic[k]}') k -> d[k] # 1->1,2->1,3->2 # {1:1,2:1,3:2}
... | n-repeated-element-in-size-2n-array | Python Easy to Understand [Hashmap] | Ron99 | 0 | 37 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,567 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1725426/(JavaScript%2BPython3)-solution-using-Frequency-counter | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
dic = dict()
for item in nums:
if dic.get(item):
dic[item] += 1
else:
dic[item] = 1
items = dic.items()
ab = sorted(dic.items(), key=lambda x: x[... | n-repeated-element-in-size-2n-array | (JavaScript+Python3) solution using - Frequency-counter | shakilbabu | 0 | 28 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,568 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1491902/Python-dollarolution-(95-Faster) | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
for i in range(len(nums)):
if nums[i] in nums[i+1:]:
return nums[i] | n-repeated-element-in-size-2n-array | Python $olution (95% Faster) | AakRay | 0 | 87 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,569 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1476997/Python-O(n)-time-O(1)-space-solution | class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
n = len(nums)
if n == 4:
return sum(nums)-sum(list(set(nums)))
for i in range(n-2):
a, b, c = nums[i], nums[i+1], nums[i+2]
if a== b or b ==c or c ==a:
return a+b+c-max(a,b,c... | n-repeated-element-in-size-2n-array | Python O(n) time, O(1) space solution | byuns9334 | 0 | 62 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,570 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1082166/ONE-LINER-PYTHON-CODE | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
return ([a for a,c in Counter(A).items() if c==len(A)//2][0]) | n-repeated-element-in-size-2n-array | ONE LINER PYTHON CODE | yashwanthreddz | 0 | 47 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,571 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/1030407/Python3-easy-solution | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
d = {}
for i in A:
d[i] = d.get(i,0)+1
return list(d.keys())[list(d.values()).index(len(A)//2)] | n-repeated-element-in-size-2n-array | Python3 easy solution | EklavyaJoshi | 0 | 75 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,572 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/985049/O(1)-Space-and-O(n)-time-solution | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
a,b = A[0], A[1]
for i in range(2,len(A)):
if A[i]==a or A[i]==b:
return A[i]
else:
a = b
b = A[i]
return b | n-repeated-element-in-size-2n-array | O(1) Space and O(n) time solution | majinlion | 0 | 81 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,573 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/679849/Python-Easy-to-Understand | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
l=[]
A.sort()
for i in A:
if i not in l:
l.append(i)
else:
return i | n-repeated-element-in-size-2n-array | Python Easy to Understand | sulabh98 | 0 | 42 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,574 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/580070/Python-One-Liner-(Slow) | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
return max(collections.Counter(A),key = collections.Counter(A).get) | n-repeated-element-in-size-2n-array | Python One Liner (Slow) | pratushah | 0 | 89 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,575 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/382179/Three-Solutions-in-Python-3 | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
return (lambda x: max(x, key = lambda y: x[y]))(collections.Counter(A)) | n-repeated-element-in-size-2n-array | Three Solutions in Python 3 | junaidmansuri | 0 | 365 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,576 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/382179/Three-Solutions-in-Python-3 | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
S = set()
for a in A:
if a in S: return a
else: S.add(a) | n-repeated-element-in-size-2n-array | Three Solutions in Python 3 | junaidmansuri | 0 | 365 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,577 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/382179/Three-Solutions-in-Python-3 | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
for a in set(A):
if A.count(a) != 1: return a
- Junaid Mansuri
(LeetCode ID)@hotmail.com | n-repeated-element-in-size-2n-array | Three Solutions in Python 3 | junaidmansuri | 0 | 365 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,578 |
https://leetcode.com/problems/n-repeated-element-in-size-2n-array/discuss/318920/Easy-way-by-python | class Solution:
def repeatedNTimes(self, A: List[int]) -> int:
length = len(A)
for i in A:
if A.count(i)>1:
return i | n-repeated-element-in-size-2n-array | Easy way by python | PPdog520 | 0 | 68 | n repeated element in size 2n array | 961 | 0.759 | Easy | 15,579 |
https://leetcode.com/problems/maximum-width-ramp/discuss/977244/Python3-binary-search-O(NlogN)-and-stack-O(N) | class Solution:
def maxWidthRamp(self, A: List[int]) -> int:
ans = 0
stack = []
for i in range(len(A)):
if not stack or A[stack[-1]] > A[i]: stack.append(i)
else:
lo, hi = 0, len(stack)
while lo < hi:
mid = lo + h... | maximum-width-ramp | [Python3] binary search O(NlogN) & stack O(N) | ye15 | 8 | 299 | maximum width ramp | 962 | 0.49 | Medium | 15,580 |
https://leetcode.com/problems/maximum-width-ramp/discuss/977244/Python3-binary-search-O(NlogN)-and-stack-O(N) | class Solution:
def maxWidthRamp(self, A: List[int]) -> int:
ans = 0
stack = []
for i in range(len(A)):
if not stack or A[stack[-1]] > A[i]: stack.append(i)
for i in reversed(range(len(A))):
while stack and A[stack[-1]] <= A[i]:
ans... | maximum-width-ramp | [Python3] binary search O(NlogN) & stack O(N) | ye15 | 8 | 299 | maximum width ramp | 962 | 0.49 | Medium | 15,581 |
https://leetcode.com/problems/maximum-width-ramp/discuss/1938716/Python-easy-to-read-and-understand-or-stack | class Solution:
def maxWidthRamp(self, nums: List[int]) -> int:
stack = []
n = len(nums)
for i in range(n):
if not stack or nums[stack[-1]] > nums[i]:
stack.append(i)
ans = 0
for i in range(n-1, -1, -1):
while stack and nums[i] >= nums[... | maximum-width-ramp | Python easy to read and understand | stack | sanial2001 | 2 | 183 | maximum width ramp | 962 | 0.49 | Medium | 15,582 |
https://leetcode.com/problems/maximum-width-ramp/discuss/1592965/Simple-and-Readable%3A-O(N)-Time-O(N)-Space | class Solution:
def maxWidthRamp(self, nums: List[int]) -> int:
if not nums: return 0
max_vals = self.create_max_vals(nums)
max_width = 0
i = 0
while i + max_width < len(nums):
j = i + max_width
while j < len(nums) and nu... | maximum-width-ramp | Simple and Readable: O(N) Time, O(N) Space | owenc343 | 0 | 243 | maximum width ramp | 962 | 0.49 | Medium | 15,583 |
https://leetcode.com/problems/minimum-area-rectangle-ii/discuss/980956/Python3-center-point-O(N2) | class Solution:
def minAreaFreeRect(self, points: List[List[int]]) -> float:
ans = inf
seen = {}
for i, (x0, y0) in enumerate(points):
for x1, y1 in points[i+1:]:
cx = (x0 + x1)/2
cy = (y0 + y1)/2
d2 = (x0 - x1)**2 + (y0 - y1)**2
... | minimum-area-rectangle-ii | [Python3] center point O(N^2) | ye15 | 28 | 1,200 | minimum area rectangle ii | 963 | 0.547 | Medium | 15,584 |
https://leetcode.com/problems/minimum-area-rectangle-ii/discuss/2340767/Runtime%3A-96-ms-faster-than-97.40-of-Python3-oror-Python-solution | class Solution:
def minAreaFreeRect(self, points: List[List[int]]) -> float:
N = len(points)
seen = set()
for point in points:
seen.add(tuple(point))
# length^2
def length2(a, b):
return (a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] -... | minimum-area-rectangle-ii | Runtime: 96 ms, faster than 97.40% of Python3 || Python solution | vimla_kushwaha | 1 | 82 | minimum area rectangle ii | 963 | 0.547 | Medium | 15,585 |
https://leetcode.com/problems/least-operators-to-express-number/discuss/1367268/Python3-top-down-dp | class Solution:
def leastOpsExpressTarget(self, x: int, target: int) -> int:
@cache
def fn(val):
"""Return min ops to express val."""
if val < x: return min(2*val-1, 2*(x-val))
k = int(log(val)//log(x))
ans = k + fn(val - x**k)
if... | least-operators-to-express-number | [Python3] top-down dp | ye15 | 4 | 277 | least operators to express number | 964 | 0.48 | Hard | 15,586 |
https://leetcode.com/problems/least-operators-to-express-number/discuss/1925389/PYTHON-SOL-oror-RECURSION-%2B-MEMOIZATION-oror-WELL-EXPLAINED-oror-APPROACH-EXPLAINED-oror-WELL-COMMENTED-oror | class Solution:
def solve(self,x,target):
if target in self.dp : return self.dp[target]
# when target == 1 we can solve just by doing x/x
if target == 1: return 1
# current value = x and operations performed = 0
cur = x
op = 0
# i... | least-operators-to-express-number | PYTHON SOL || RECURSION + MEMOIZATION || WELL EXPLAINED || APPROACH EXPLAINED || WELL COMMENTED || | reaper_27 | 1 | 88 | least operators to express number | 964 | 0.48 | Hard | 15,587 |
https://leetcode.com/problems/univalued-binary-tree/discuss/1569046/python-dfs-recursion-faster-than-97 | class Solution:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
val = root.val
def helper(root):
return root is None or (root.val == val and helper(root.left) and helper(root.right))
return helper(root) | univalued-binary-tree | python dfs recursion faster than 97% | dereky4 | 2 | 89 | univalued binary tree | 965 | 0.693 | Easy | 15,588 |
https://leetcode.com/problems/univalued-binary-tree/discuss/469876/Python-3-(five-lines)-(beats-~93)-(Simple-Recursion) | class Solution:
def isUnivalTree(self, R: TreeNode) -> bool:
def IUT(n):
if n == None: return True
if n.val != R.val: return False
return IUT(n.left) and IUT(n.right)
return IUT(R)
- Junaid Mansuri
- Chicago, IL | univalued-binary-tree | Python 3 (five lines) (beats ~93%) (Simple Recursion) | junaidmansuri | 2 | 259 | univalued binary tree | 965 | 0.693 | Easy | 15,589 |
https://leetcode.com/problems/univalued-binary-tree/discuss/1600243/Python-short-recursive-solution-or-faster-than-96.1-submission | class Solution:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
self.value=root.val
self.flag=True
def helper(node):
if node is None:
return True
if node.val!=self.value:
return False
flag=helper(node.left) and hel... | univalued-binary-tree | Python short recursive solution | faster than 96.1% submission | diksha_choudhary | 1 | 30 | univalued binary tree | 965 | 0.693 | Easy | 15,590 |
https://leetcode.com/problems/univalued-binary-tree/discuss/1167940/WEEB-DOES-PYTHON-BFS | class Solution:
def isUnivalTree(self, root: TreeNode) -> bool:
queue, Unival = deque([root]), root.val
while queue:
curNode = queue.popleft()
if curNode.val != Unival: return False
if curNode.left:
queue.append(curNode.left)
if curNode.right:
queue.append(curNode.right)
return True | univalued-binary-tree | WEEB DOES PYTHON BFS | Skywalker5423 | 1 | 48 | univalued binary tree | 965 | 0.693 | Easy | 15,591 |
https://leetcode.com/problems/univalued-binary-tree/discuss/2823623/Faster-than-98-python-solution | class Solution:
def isUnivalTree(self, root) -> bool:
x=root.val
q=[root]
while q:
node=q.pop(0)
if node.val!=x:
return False
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
... | univalued-binary-tree | Faster than 98% python solution | sushants007 | 0 | 3 | univalued binary tree | 965 | 0.693 | Easy | 15,592 |
https://leetcode.com/problems/univalued-binary-tree/discuss/2823623/Faster-than-98-python-solution | class Solution:
def isUnivalTree(self, root) -> bool:
if not root:
return True
if root.left:
if root.val!=root.left.val :
return False
if root.right:
if root.val!=root.right.val:
return False
return self... | univalued-binary-tree | Faster than 98% python solution | sushants007 | 0 | 3 | univalued binary tree | 965 | 0.693 | Easy | 15,593 |
https://leetcode.com/problems/univalued-binary-tree/discuss/2789350/Explained-recursive-Python-solution | class Solution:
def isUnivalTree(self, root, value=float('inf')):
if not(root):
return True
if value==float('inf'):
return self.isUnivalTree(root.left, root.val) and self.isUnivalTree(root.right, root.val)
else:
return root.val==value and self.isUnivalTree... | univalued-binary-tree | Explained recursive Python solution | Pirmil | 0 | 1 | univalued binary tree | 965 | 0.693 | Easy | 15,594 |
https://leetcode.com/problems/univalued-binary-tree/discuss/2654035/DFS-Solution | class Solution:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
i = set()
def inorderTraversal(root):
if root:
inorderTraversal(root.left)
i.add(root.val)
inorderTraversal(root.right)
inorderTraversal(root)
return ... | univalued-binary-tree | DFS Solution | jaisalShah | 0 | 2 | univalued binary tree | 965 | 0.693 | Easy | 15,595 |
https://leetcode.com/problems/univalued-binary-tree/discuss/2334018/Python-greater98-Time | class Solution:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
uni = []
def search(root):
if root == None:
return
uni.append(root.val)
search(root.left)
search(root.right)
search(root)
return len(set(uni)) ==... | univalued-binary-tree | Python >98% Time | tq326 | 0 | 12 | univalued binary tree | 965 | 0.693 | Easy | 15,596 |
https://leetcode.com/problems/univalued-binary-tree/discuss/2073728/Python-Easy-Solution | class Solution:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
res=[]
def tree(node):
if not node: return
tree(node.left)
res.append(node.val)
tree(node.right)
tree(root)
print(res)
v=res[0]
for x in range(1,l... | univalued-binary-tree | [Python] Easy Solution | Perindhavallipadem | 0 | 25 | univalued binary tree | 965 | 0.693 | Easy | 15,597 |
https://leetcode.com/problems/univalued-binary-tree/discuss/1913648/Python3-Recursion-Solution-With-Prev-Variable-No-Memory-Needed-Faster-Than-88.92 | class Solution:
def isUnivalTree(self, root: Optional[TreeNode]) -> bool:
flag = [True]
def dfs(tree, prev):
if not tree:
return
if tree.val != prev:
flag[0] = False
dfs(tree.l... | univalued-binary-tree | Python3 Recursion Solution With Prev Variable, No Memory Needed, Faster Than 88.92% | Hejita | 0 | 27 | univalued binary tree | 965 | 0.693 | Easy | 15,598 |
https://leetcode.com/problems/univalued-binary-tree/discuss/1878371/Python-Simple-and-Elegant!-Recursive | class Solution(object):
def isUnivalTree(self, root):
if root.left and root.right:
return root.val == root.left.val == root.right.val and self.isUnivalTree(root.left) and self.isUnivalTree(root.right)
elif root.left:
return root.val == root.left.val and self.isUnivalTree(root... | univalued-binary-tree | Python - Simple and Elegant! Recursive | domthedeveloper | 0 | 58 | univalued binary tree | 965 | 0.693 | Easy | 15,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.