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/max-area-of-island/discuss/2305261/Python-Optimized-DFS-Solution | class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
rows = len(grid)
cols = len(grid[0])
def dfs(i, j):
if grid[i][j] == 1:
grid[i][j] = 0
return 1 + goToLeft(i, j-1) + goToRight(i, j+1) + goToTop(i-1, j) + goToBottom(i+1, j)
... | max-area-of-island | [Python] Optimized DFS Solution | Buntynara | 0 | 39 | max area of island | 695 | 0.717 | Medium | 11,500 |
https://leetcode.com/problems/max-area-of-island/discuss/2296706/python3-bfs | class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
row = len(grid)
col = len(grid[0])
total = 0
for r in range(row):
for c in range(col):
if grid[r][c] == 1:
q = collections.deque()
land = 0 ... | max-area-of-island | python3 bfs | yhh1 | 0 | 25 | max area of island | 695 | 0.717 | Medium | 11,501 |
https://leetcode.com/problems/max-area-of-island/discuss/2287925/Basic-Python-BFS-solution-beats-90 | class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
if not grid:
return 0
len_row = len(grid)
len_col = len(grid[0])
directions = [(0,1),(1,0),(-1,0),(0,-1)]
def bfs(start_row,start_col):
area = 1
queue = [(st... | max-area-of-island | Basic Python BFS solution, beats 90% | prejudice23 | 0 | 20 | max area of island | 695 | 0.717 | Medium | 11,502 |
https://leetcode.com/problems/max-area-of-island/discuss/2287137/Python3-dfs-solution | class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
res=0
n=len(grid)
m=len(grid[0])
def isValid(x,y):
return n>x>=0 and m>y>=0
self.visited=[[0 for j in range(m)] for i in range(n)]
dirs=[[-1,0],[1,0],[0,-1],[0,1]]
... | max-area-of-island | Python3 dfs solution | atm1504 | 0 | 8 | max area of island | 695 | 0.717 | Medium | 11,503 |
https://leetcode.com/problems/max-area-of-island/discuss/2287118/Simple-intuitive-dfs-change-the-grid-value-move-updownrightleft-%3A) | class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
def dfs(i,j):
if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j]==0:
return 0
grid[i][j]=0
down = dfs(i+1,j)
up = dfs(i-1,j)
right = dfs(i,j+1)
... | max-area-of-island | Simple intuitive dfs - change the grid value - move up,down,right,left :) | ana_2kacer | 0 | 9 | max area of island | 695 | 0.717 | Medium | 11,504 |
https://leetcode.com/problems/count-binary-substrings/discuss/384054/Only-using-stack-with-one-iteration-logic-solution-in-Python-O(N) | class Solution:
def countBinarySubstrings(self, s: str) -> int:
stack = [[], []]
latest = int(s[0])
stack[latest].append(latest)
result = 0
for i in range(1,len(s)):
v = int(s[i])
if v != latest:
stack[v].clear()
latest ... | count-binary-substrings | Only using stack with one iteration, logic solution in Python O(N) | zouqiwu09 | 7 | 994 | count binary substrings | 696 | 0.656 | Easy | 11,505 |
https://leetcode.com/problems/count-binary-substrings/discuss/1849843/python-3-oror-O(n)-oror-O(1) | class Solution:
def countBinarySubstrings(self, s: str) -> int:
prev, cur = 0, 1
res = 0
for i in range(1, len(s)):
if s[i] == s[i - 1]:
cur += 1
else:
prev, cur = cur, 1
if cur <= prev:
res += 1
... | count-binary-substrings | python 3 || O(n) || O(1) | dereky4 | 2 | 462 | count binary substrings | 696 | 0.656 | Easy | 11,506 |
https://leetcode.com/problems/count-binary-substrings/discuss/2273332/Python-3-O(n)-solution-faster-then-maximum-submission | class Solution:
def countBinarySubstrings(self, s: str) -> int:
prev , curr , res = 0 , 1 , 0
for i in range(1,len(s)):
if s[i-1] == s[i]:
curr +=1
else:
prev = curr
curr = 1
if prev >= curr:
... | count-binary-substrings | Python 3 O(n) solution faster then maximum submission | ronipaul9972 | 1 | 283 | count binary substrings | 696 | 0.656 | Easy | 11,507 |
https://leetcode.com/problems/count-binary-substrings/discuss/380683/Solution-in-Python-3-(one-line) | class Solution:
def countBinarySubstrings(self, s: str) -> int:
L = len(s)
a = [-1]+[i for i in range(L-1) if s[i] != s[i+1]]+[L-1]
b = [a[i]-a[i-1] for i in range(1,len(a))]
c = [min(b[i-1],b[i]) for i in range(1,len(b))]
return sum(c) | count-binary-substrings | Solution in Python 3 (one line) | junaidmansuri | 1 | 731 | count binary substrings | 696 | 0.656 | Easy | 11,508 |
https://leetcode.com/problems/count-binary-substrings/discuss/380683/Solution-in-Python-3-(one-line) | class Solution:
def countBinarySubstrings(self, s: str) -> int:
return sum((lambda x: [min(x[i]-x[i-1],x[i+1]-x[i]) for i in range(1,len(x)-1)])([-1]+[i for i in range(len(s)-1) if s[i] != s[i+1]]+[len(s)-1]))
- Junaid Mansuri
(LeetCode ID)@hotmail.com | count-binary-substrings | Solution in Python 3 (one line) | junaidmansuri | 1 | 731 | count binary substrings | 696 | 0.656 | Easy | 11,509 |
https://leetcode.com/problems/count-binary-substrings/discuss/2180709/python-straght-forward-2-pointers | class Solution:
def countBinarySubstrings(self, s: str) -> int:
n = len(s)
left = right = 0
res = 0
for i in range(1, n):
if s[i] != s[i-1]:
left = i -1
right = i
while left >= 0 and right < n and s[left] == s[i-1] and s[right... | count-binary-substrings | python straght forward, 2 pointers | hardernharder | 0 | 237 | count binary substrings | 696 | 0.656 | Easy | 11,510 |
https://leetcode.com/problems/count-binary-substrings/discuss/2087348/Python3-Simple-approach-beats-96-in-time-and-98-in-space | class Solution:
def countBinarySubstrings(self, s: str) -> int:
ret_val = 0
ones = 0
zeros = 0
prev = s[0]
for bit in s:
if bit == "1":
if prev == "0":
ones = 0
ones += 1
if zeros > 0:
... | count-binary-substrings | Python3 Simple approach, beats 96% in time and 98% in space | kukamble | 0 | 222 | count binary substrings | 696 | 0.656 | Easy | 11,511 |
https://leetcode.com/problems/count-binary-substrings/discuss/1982595/Python-Solution | class Solution:
def countBinarySubstrings(self, s: str) -> int:
slen = len(s)
index = 0
interval = []
while index < slen:
cnt = 1
while index + 1 < slen and s[index + 1] == s[index]:
cnt += 1
index += 1
interval.appe... | count-binary-substrings | Python Solution | DietCoke777 | 0 | 103 | count binary substrings | 696 | 0.656 | Easy | 11,512 |
https://leetcode.com/problems/count-binary-substrings/discuss/1735320/Python3-solution-using-list-comprehensions | class Solution:
def countBinarySubstrings(self, s: str) -> int:
# Find the index where each change occurs, pick this apart by trying out `list(zip(s, s[1:]))` and then the enumeration
# Note we need to add index 0 and index len(s) for this to calculate correctly
indexes = [0] + [i for i, (l, r) in enume... | count-binary-substrings | Python3 solution using list comprehensions | mike72 | 0 | 164 | count binary substrings | 696 | 0.656 | Easy | 11,513 |
https://leetcode.com/problems/count-binary-substrings/discuss/1551757/O(n)-Time-Python-3-Solution | class Solution:
def countBinarySubstrings(self, s: str) -> int:
curr = s[0]
change_index_list = []
change_index_list.append(0)
for i in range(1, len(s)):
if curr != s[i]:
curr = s[i]
change_index_list.append(i)
change_index_list.app... | count-binary-substrings | O(n) Time Python 3 Solution | zzjharry | 0 | 442 | count binary substrings | 696 | 0.656 | Easy | 11,514 |
https://leetcode.com/problems/count-binary-substrings/discuss/1173477/Python-solution.-Expand-from-middle-of-'01'-or-'10' | class Solution:
def countBinarySubstrings(self, s: str) -> int:
def helper(left, right):
count = 1
while left - 1 >= 0 and right + 1 < len(s) and s[left] == s[left-1] and s[right] == s[right+1]:
count += 1
left -= 1
... | count-binary-substrings | Python solution. Expand from middle of '01' or '10' | pochy | 0 | 74 | count binary substrings | 696 | 0.656 | Easy | 11,515 |
https://leetcode.com/problems/count-binary-substrings/discuss/1173109/Python3-linear-sweep | class Solution:
def countBinarySubstrings(self, s: str) -> int:
ans = prev = curr = 0
for i in range(len(s)+1):
if i == len(s) or i and s[i-1] != s[i]:
ans += min(prev, curr)
prev = curr
curr = 1
else: curr += 1
return... | count-binary-substrings | [Python3] linear sweep | ye15 | 0 | 51 | count binary substrings | 696 | 0.656 | Easy | 11,516 |
https://leetcode.com/problems/count-binary-substrings/discuss/1173086/Simple-solution-using-a-binary-like-counter-for-both-types-of-characters-in-Python3 | class Solution:
def countBinarySubstrings(self, s: str) -> int:
c = [0, 0]
c[int(s[0])] += 1
ans = 0
for i in range(1, len(s)):
if s[i] == s[i - 1]:
c[int(s[i])] += 1
else:
c[int(s[i])] = 1
if c... | count-binary-substrings | Simple solution using a binary like counter for both types of characters in Python3 | amoghrajesh1999 | 0 | 124 | count binary substrings | 696 | 0.656 | Easy | 11,517 |
https://leetcode.com/problems/count-binary-substrings/discuss/1172942/Easy-Solution-Python-3 | class Solution:
def countBinarySubstrings(self, s: str) -> int:
occ,acc,res=[],1,0
for i in range(1,len(s)):
if s[i]==s[i-1]:
acc+=1
else:
occ.append(acc)
acc=1
occ.append(acc)
print(occ)
for i in range(1... | count-binary-substrings | Easy Solution Python 3 | moazmar | 0 | 170 | count binary substrings | 696 | 0.656 | Easy | 11,518 |
https://leetcode.com/problems/count-binary-substrings/discuss/1507982/Groupby-contiguous-blocks-86-speed | class Solution:
def countBinarySubstrings(self, s: str) -> int:
group_lens = [len(list(g)) for _, g in groupby(s)]
return sum(min(a, b) for a, b in zip(group_lens, group_lens[1:])) | count-binary-substrings | Groupby contiguous blocks, 86% speed | EvgenySH | -1 | 272 | count binary substrings | 696 | 0.656 | Easy | 11,519 |
https://leetcode.com/problems/degree-of-an-array/discuss/349801/Solution-in-Python-3-(beats-~98) | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
C = {}
for i, n in enumerate(nums):
if n in C: C[n].append(i)
else: C[n] = [i]
M = max([len(i) for i in C.values()])
return min([i[-1]-i[0] for i in C.values() if len(i) == M]) + 1
- Junaid Mansuri | degree-of-an-array | Solution in Python 3 (beats ~98%) | junaidmansuri | 43 | 3,000 | degree of an array | 697 | 0.559 | Easy | 11,520 |
https://leetcode.com/problems/degree-of-an-array/discuss/1179769/Simple-Python-Solution-with-explanation-O(n) | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
'''
step 1: find the degree
- create a hashmap of a number and value as list of occurance indices
- the largest indices array in the hashmap gives us the degree
step 2: find the minimum length sub... | degree-of-an-array | Simple Python Solution with explanation - O(n) | jitin11 | 10 | 888 | degree of an array | 697 | 0.559 | Easy | 11,521 |
https://leetcode.com/problems/degree-of-an-array/discuss/434794/Python3-solution-or-Beat-100 | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
if nums == []:
return 0
dic = {}
for n in nums:
if n not in dic:
dic[n] = 1
else:
dic[n] += 1
degree = max(dic.values())
if degree == 1:... | degree-of-an-array | Python3 solution | Beat 100% | YuanYao666 | 2 | 382 | degree of an array | 697 | 0.559 | Easy | 11,522 |
https://leetcode.com/problems/degree-of-an-array/discuss/1697602/Python-or-just-23-line-code-or-Dictionary-or-O(N)-time | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
frq = defaultdict(int) # frequency map for nums
fnl = {} # stores first and last index of each num
deg = 0 # degree
for i in range(len(nums)):
frq[nums[i]] += 1
deg = ma... | degree-of-an-array | [Python] | just 23 line code | Dictionary | O(N) time | Divyanshuk34 | 1 | 186 | degree of an array | 697 | 0.559 | Easy | 11,523 |
https://leetcode.com/problems/degree-of-an-array/discuss/1669876/Python3-Straight-forward-statistic | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
if not nums: return 0
stats = {}
for i, n in enumerate(nums):
if n not in stats:
stats[n] = {"start":i, "end":i, "count":1}
else:
stats[n]["end"] = i
... | degree-of-an-array | [Python3] Straight forward statistic | BigTailWolf | 1 | 82 | degree of an array | 697 | 0.559 | Easy | 11,524 |
https://leetcode.com/problems/degree-of-an-array/discuss/1375342/90.17-faster | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
maxv=0
d={}
count=1000000
for i in nums:
if i in d:
d[i]+=1
maxv=max(maxv, d[i])
else:
d[i]=1
if maxv<=1:
retur... | degree-of-an-array | 90.17% faster | prajwalahluwalia | 1 | 218 | degree of an array | 697 | 0.559 | Easy | 11,525 |
https://leetcode.com/problems/degree-of-an-array/discuss/2732712/Simple-Python-Solution-%3A | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
d = {}
min_len = len(nums)+1
max_deg = 0
for i in range(len(nums)):
if nums[i] not in d.keys():
d[nums[i]] = [ 1,i,1]
else:
d[nums[i]][0] += 1
... | degree-of-an-array | Simple Python Solution : | MaviOp | 0 | 5 | degree of an array | 697 | 0.559 | Easy | 11,526 |
https://leetcode.com/problems/degree-of-an-array/discuss/2319915/Fast-python-with-explanation | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
dict_ = {}
l=len(nums)
for i in range(l):
if nums[i] in dict_:
dict_[nums[i]][0]+=1
dict_[nums[i]][2]=i-dict_[nums[i]][1] #updating difference between first and last idx
... | degree-of-an-array | Fast python with explanation | sunakshi132 | 0 | 71 | degree of an array | 697 | 0.559 | Easy | 11,527 |
https://leetcode.com/problems/degree-of-an-array/discuss/2282461/Python-3-dicts-with-defaultdict-for-cleaner-code | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
maxFreq = 0
count = defaultdict(lambda: 0)
minPos = defaultdict(lambda: 100000)
maxPos = defaultdict(lambda: -1)
for i, n in enumerate(nums):
count[n] += 1
minPos[n] = min... | degree-of-an-array | Python, 3 dicts with defaultdict for cleaner code | boris17 | 0 | 40 | degree of an array | 697 | 0.559 | Easy | 11,528 |
https://leetcode.com/problems/degree-of-an-array/discuss/2158324/C%2B%2BPython-optimal-solution-with-comments | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
m = {} # dict to store freq, first index and last index
maxf,ans = 0,inf
for i, num in enumerate(nums):
# if this is the first time we encounter this number
# then simply initialize frequency as 1 and inde... | degree-of-an-array | [C++/Python optimal solution with comments | chaitanya_29 | 0 | 52 | degree of an array | 697 | 0.559 | Easy | 11,529 |
https://leetcode.com/problems/degree-of-an-array/discuss/1312521/Python3-dollarolution-(84-faster-and-98-better-memory-usage) | class Solution:
def findShortestSubArray(self, nums: List[int]) -> int:
d = {}
m, j, count = 1, [nums[0]], []
for i in nums:
if i not in d:
d[i] = 1
else:
d[i] += 1
if d[i] > m:
m = d[i]
... | degree-of-an-array | Python3 $olution (84% faster & 98% better memory usage) | AakRay | 0 | 439 | degree of an array | 697 | 0.559 | Easy | 11,530 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1627241/python-simple-with-detailed-explanation-or-96.13 | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
if k==1:
return True
total = sum(nums)
n = len(nums)
if total%k!=0:
return False
nums.sort(reverse=True)
average = total//k
if nums[0]>average:
... | partition-to-k-equal-sum-subsets | python simple with detailed explanation | 96.13% | 1579901970cg | 5 | 563 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,531 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1772609/Python3-Solution-with-DFS | class Solution(object):
def canPartitionKSubsets(self, nums, k):
target = sum(nums)
if target % k != 0: return False
target //= k
cur = [0] * k; nums.sort( reverse = True)
def foo( index):
if index == len( nums): return True
for i in range( k):
... | partition-to-k-equal-sum-subsets | Python3 Solution with DFS | JuboGe | 2 | 190 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,532 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2837264/Python-Backtrack-%2B-Exploiting-%40Cache | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
if sum(nums) % k != 0: return False
target = sum(nums) // k
if any(num > target for num in nums): return False
nums.sort()
while target in nums:
nums.remove(target)
k -=... | partition-to-k-equal-sum-subsets | [Python] Backtrack + Exploiting @Cache | Nezuko-NoBamboo | 1 | 43 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,533 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2054135/Python-DFS-Solution | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
n, expected_sum = len(nums), sum(nums) / k
nums.sort(reverse=True)
if expected_sum != int(expected_sum) or nums[0] > expected_sum:
return False
def btrack(pos, target, done):
if ... | partition-to-k-equal-sum-subsets | Python DFS Solution | TongHeartYes | 1 | 125 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,534 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1385240/Intuitive-recursive-solution-(less-15-lines)-WITHOUT-bitmask-or-visited-maps. | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
total = sum(nums)
if total%k:
return False
partitions = [total//k]*k
# Sorting was an after thought to get rid of the TLEs
nums.sort(reverse=True)
#... | partition-to-k-equal-sum-subsets | Intuitive recursive solution (< 15 lines) WITHOUT bitmask or visited maps. | worker-bee | 1 | 123 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,535 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2839608/Python-Backtrack | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
if sum(nums) % k != 0: return False
target = sum(nums) // k
if any(num > target for num in nums): return False
nums.sort()
while target in nums:
nums.remove(target)
k -=... | partition-to-k-equal-sum-subsets | Python Backtrack | Farawayy | 0 | 1 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,536 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2837456/backtrack-python-no-time-limit-exceeded | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
total = sum(nums)
if total%k != 0:
return False
target = total//k
n = len(nums)
buckets = [0 for _ in range(k)]
nums.sort(reverse=True)
def backtrack(buckets, index, t... | partition-to-k-equal-sum-subsets | backtrack python no time limit exceeded | ychhhen | 0 | 2 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,537 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2837400/Python-Solution-using-Backtracking | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
total = sum(nums)
if total % k:
return False
target = total // k
subSets = [0] * k
nums.sort(reverse = True)
used = [False] * len(nums)
def backtrack(i, k, ... | partition-to-k-equal-sum-subsets | Python Solution using Backtracking | taoxinyyyun | 0 | 1 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,538 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2752864/Python-backtracking-solution | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
total = sum(nums)
n = len(nums)
if total%k != 0:
return False
nums.sort(reverse=True)
target = total // k
used = [False] * n
memo = {} # Record the current state of us... | partition-to-k-equal-sum-subsets | Python backtracking solution | gcheng81 | 0 | 7 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,539 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2751927/python | # class Solution:
# def backtrack(self, used, bucket, start, k):
# if k == 0:
# return True
# if bucket == self.target:
# return self.backtrack(used[:], 0, 0, k-1)
# for i in range(start, self.n):
# if used[i]:
# continue
# ... | partition-to-k-equal-sum-subsets | python | lucy_sea | 0 | 8 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,540 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2750438/Python-Backtrack-Solution | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
target = sum(nums) / k
n = len(nums)
seen = [False for _ in range(n)]
memo = {}
nums.sort(reverse=True)
if target != int(target):
return False
# k bigger tha... | partition-to-k-equal-sum-subsets | Python Backtrack Solution | Rui_Liu_Rachel | 0 | 8 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,541 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2346059/Python3-Solution-or-Backtracking | class Solution:
def canPartitionKSubsets(self, A, k):
n, val = len(A), sum(A) / k
if val != floor(val):
return False
A.sort()
def btrack(i, space, k):
if k == 1: return True
for j in range(i, n):
if val >= A[j] > space: break
... | partition-to-k-equal-sum-subsets | Python3 Solution | Backtracking | satyam2001 | 0 | 189 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,542 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2292308/Python-backtracking-solution-with-memo | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
if k > len(nums): return False
if (sum(nums) % k != 0): return False
# hashmap: int -> bool
memo = {}
# binary used as bool array(used[]) to indicate if nums[i] has been placed in a subs... | partition-to-k-equal-sum-subsets | Python backtracking solution with memo | leqinancy | 0 | 51 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,543 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/2073591/Python3-DP-Solution | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
n, expected_sum = len(nums), sum(nums) / k
nums.sort(reverse=True)
if expected_sum != int(expected_sum) or nums[0] > expected_sum:
return False
def btrack(pos, target, done):
if ... | partition-to-k-equal-sum-subsets | Python3 DP Solution | TongHeartYes | 0 | 168 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,544 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1962631/Python3-Backtracking-%2B-Memoization-%2B-Bit-Masking-(Detailed-Explanation) | class Solution:
def canPartitionKSubsets(self, nums: List[int], K: int) -> bool:
T = sum(nums)
if T%K:
return False
N = len(nums)
T = T//K
taken = [False]*N
d={}
def helper(S,K,curr):
# Base case
if K==0:
ret... | partition-to-k-equal-sum-subsets | Python3 Backtracking + Memoization + Bit Masking (Detailed Explanation) | KiranRaghavendra | 0 | 159 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,545 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1886213/Python3-Backtracking-with-memo | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
if k > len(nums): return False
if (sum(nums) % k != 0): return False
memo = {}
used = 0
target = sum(nums)/k
def backtrack(k, currPath, nums, start, used, target):
... | partition-to-k-equal-sum-subsets | [Python3] Backtracking with memo | leqinancy | 0 | 60 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,546 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1813800/Python-or-Still-Confusing | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
# There are k buckets
bucket = [0] * k
# Target sum in each bucket
target = sum(nums) / k
nums.sort(reverse=True)
def backtrack(num, index, bucket, target):
# W... | partition-to-k-equal-sum-subsets | Python | Still Confusing | Fayeyf | 0 | 59 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,547 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1728529/python-using-backtracking-and-memoization | class Solution:
def canPartitionKSubsets(self, matchsticks: List[int], k: int) -> bool:
s = 0
n = 0
vis = []
for i in matchsticks:
s = s + i
n = n + 1
vis.append('0')
if(s%k):
return False
s = s//k
matchsticks.sort(reverse = True)
if(matchsticks[0] > s):
return False
h = {}
def solve... | partition-to-k-equal-sum-subsets | python using backtracking and memoization | jagdishpawar8105 | 0 | 245 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,548 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1705743/698-Partition-to-K-Equal-Sum-Subsets-via-Backtracking | class Solution:
def canPartitionKSubsets(self, nums, k):
if sum(nums) % k != 0:
return False
bucket = [0] * k
target = sum(nums) / k
nums.sort(reverse = True)
return self.dfs(nums, k, bucket, target, 0)
def dfs(self, nums, k, bucket, target, i):
if i == len(nums):
for ele in bucket:
if... | partition-to-k-equal-sum-subsets | 698 Partition to K Equal Sum Subsets via Backtracking | zwang198 | 0 | 104 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,549 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1705743/698-Partition-to-K-Equal-Sum-Subsets-via-Backtracking | class Solution:
def canPartitionKSubsets(self, nums, k):
if sum(nums) % k != 0:
return False
bucket = [0] * k
target = sum(nums) / k
nums.sort(reverse = True)
return self.dfs(nums, k, bucket, target, 0, 0)
def dfs(self, nums, k, bucket, target, b, iStart):
if b == k - 1:
return True
i... | partition-to-k-equal-sum-subsets | 698 Partition to K Equal Sum Subsets via Backtracking | zwang198 | 0 | 104 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,550 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1624376/Python-FAST-method-using-dictionary-explained | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
# basic check if answer is possible in this case
if sum(nums) % k > 0:
return False
target = sum(nums) / k
if max(nums) > target:
return False
# start ad... | partition-to-k-equal-sum-subsets | Python FAST method using dictionary explained | PoHandsome | 0 | 199 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,551 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1511412/Python-simple-backtracking-solution-(easy-to-understand) | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
s = sum(nums)
if s % k != 0:
return False
target = s // k
n = len(nums)
def dfs(m):
stack = [(m, 0, {m}, nums[m])]
while stack:... | partition-to-k-equal-sum-subsets | Python simple backtracking solution (easy to understand) | byuns9334 | 0 | 282 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,552 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/1364360/Python-Code-or-90-Faster | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
s=sum(nums)
if s%k !=0 or not nums or len(nums)==0:
return False
self.tar=s//k
nums.sort(reverse=True)
self.k=k
self.seen=[]
def dfs(nums,start,group,curr):
... | partition-to-k-equal-sum-subsets | Python Code | 90% Faster | rackle28 | 0 | 231 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,553 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/904900/Python3-backtracking | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
total = sum(nums)
if total % k: return False
avg = total // k
sm = [0]*k
nums.sort(reverse=True)
def fn(i):
"""Return True if possible to partition."""
... | partition-to-k-equal-sum-subsets | [Python3] backtracking | ye15 | 0 | 147 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,554 |
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/discuss/904900/Python3-backtracking | class Solution:
def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
total = sum(nums)
if total % k: return False
avg = total // k
@cache
def fn(x, mask):
"""Return True if available elements can be parititioned."""
if x ... | partition-to-k-equal-sum-subsets | [Python3] backtracking | ye15 | 0 | 147 | partition to k equal sum subsets | 698 | 0.408 | Medium | 11,555 |
https://leetcode.com/problems/falling-squares/discuss/2397036/faster-than-90.37-or-python-or-solution-or-explained | class Solution:
def fallingSquares(self, positions):
height, pos, max_h,res = [0],[0],0,[]
for left, side in positions:
i = bisect.bisect_right(pos, left)
j = bisect.bisect_left(pos, left + side)
high = max(height[i - 1:j] or [0]) + side
... | falling-squares | faster than 90.37% | python | solution | explained | vimla_kushwaha | 1 | 57 | falling squares | 699 | 0.444 | Hard | 11,556 |
https://leetcode.com/problems/falling-squares/discuss/1494448/Python3-brute-force | class Solution:
def fallingSquares(self, positions: List[List[int]]) -> List[int]:
ans = []
for i, (x, l) in enumerate(positions):
val = 0
for ii in range(i):
xx, ll = positions[ii]
if xx < x+l and x < xx+ll: val = max(val, ans[ii])
... | falling-squares | [Python3] brute-force | ye15 | 0 | 51 | falling squares | 699 | 0.444 | Hard | 11,557 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/943397/Python-Simple-Solution | class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
if not root:
return
if root.val==val:
return root
if root.val<val:
return self.searchBST(root.right,val)
else:
return self.searchBST(root.left,val) | search-in-a-binary-search-tree | Python Simple Solution | lokeshsenthilkumar | 20 | 1,300 | search in a binary search tree | 700 | 0.772 | Easy | 11,558 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1944884/Simple-5-Line-Python-Code | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root is None or root.val == val: # If end is reached or a node with a value of target is found found.
return root # Return that node.
# If target > current nodes value search in left side of node ... | search-in-a-binary-search-tree | Simple 5 Line Python Code | anCoderr | 12 | 991 | search in a binary search tree | 700 | 0.772 | Easy | 11,559 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1944884/Simple-5-Line-Python-Code | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
return root if not root or root.val == val else self.searchBST(root.left if root.val > val else root.right, val) | search-in-a-binary-search-tree | Simple 5 Line Python Code | anCoderr | 12 | 991 | search in a binary search tree | 700 | 0.772 | Easy | 11,560 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1046218/Python.-Super-simple-and-clear-solution.-iterative.-5-lines. | class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
while root:
if val < root.val: root = root.left
elif val > root.val: root = root.right
else: return root
return root | search-in-a-binary-search-tree | Python. Super simple & clear solution. iterative. 5 lines. | m-d-f | 5 | 299 | search in a binary search tree | 700 | 0.772 | Easy | 11,561 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2180667/Python3-Binary-search-faster-than-96 | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
def search(root):
if root == None:
return None
if root.val == val:
return root
if val < root.val:
return search(root.lef... | search-in-a-binary-search-tree | 📌 Python3 Binary search faster than 96% | Dark_wolf_jss | 4 | 50 | search in a binary search tree | 700 | 0.772 | Easy | 11,562 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1944579/Python3-oror-Simple-3-lines | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root or root.val == val:
return root
return self.searchBST(root.left, val) if val < root.val else self.searchBST(root.right, val) | search-in-a-binary-search-tree | ✔️ Python3 || Simple 3 lines | constantine786 | 4 | 289 | search in a binary search tree | 700 | 0.772 | Easy | 11,563 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1944579/Python3-oror-Simple-3-lines | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root and root.val != val:
root = root.left if val < root.val else root.right
return root | search-in-a-binary-search-tree | ✔️ Python3 || Simple 3 lines | constantine786 | 4 | 289 | search in a binary search tree | 700 | 0.772 | Easy | 11,564 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2508885/Python-Elegant-and-Short-or-Two-solutions-or-Iterative-and-Recursive-or-Three-lines | class Solution:
"""
Time: O(n)
Memory: O(1)
"""
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root is not None and root.val != val:
root = root.left if val < root.val else root.right
return root
class Solution:
"""
Time: O(n)
Memory: O(n)
"""
def searchBST... | search-in-a-binary-search-tree | Python Elegant & Short | Two solutions | Iterative & Recursive | Three lines | Kyrylo-Ktl | 2 | 144 | search in a binary search tree | 700 | 0.772 | Easy | 11,565 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2207822/Python-3-or-89ms-or-Recursion | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root:
return
elif root.val==val:
return root
elif root.val > val:
return self.searchBST(root.left, val)
else:
return self.searchBST(ro... | search-in-a-binary-search-tree | Python 3 | 89ms | Recursion | yashpurohit763 | 1 | 62 | search in a binary search tree | 700 | 0.772 | Easy | 11,566 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1520987/python3-intuitive-solution | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root or val == root.val: # base case
return root
# recurrence relation
return self.searchBST(root.left if val < root.val else root.right, val) | search-in-a-binary-search-tree | python3 intuitive solution | feexon | 1 | 68 | search in a binary search tree | 700 | 0.772 | Easy | 11,567 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1520987/python3-intuitive-solution | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root:
if val == root.val:
return root
root = root.left if val < root.val else root.right | search-in-a-binary-search-tree | python3 intuitive solution | feexon | 1 | 68 | search in a binary search tree | 700 | 0.772 | Easy | 11,568 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1032317/Python3-easy-solution | class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
bfs = [root]
while bfs:
node = bfs.pop(0)
if node.val == val:
return node
if node.left:
bfs.append(node.left)
if node.right:
bfs.... | search-in-a-binary-search-tree | Python3 easy solution | EklavyaJoshi | 1 | 111 | search in a binary search tree | 700 | 0.772 | Easy | 11,569 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1032317/Python3-easy-solution | class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
if not root:
return None
if root.val == val:
return root
elif root.val > val:
return self.searchBST(root.left,val)
elif root.val < val:
return self.searchBST(roo... | search-in-a-binary-search-tree | Python3 easy solution | EklavyaJoshi | 1 | 111 | search in a binary search tree | 700 | 0.772 | Easy | 11,570 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2834410/Python-or-Search-in-BST-or-simple-code-with-detailed-explanation | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root == None:
return None
if root.val == val:
return root
elif root.val > val:
return self.searchBST(root.left,val)
else:
return self.sear... | search-in-a-binary-search-tree | Python | Search in BST | simple code with detailed explanation | utkarshjain | 0 | 4 | search in a binary search tree | 700 | 0.772 | Easy | 11,571 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2794746/Super-easy-python-solutiopn | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root is None:
return None
if root.val == val:
return root
return self.searchBST(root.left,val) or self.searchBST(root.right,val) | search-in-a-binary-search-tree | Super easy python solutiopn | betaal | 0 | 1 | search in a binary search tree | 700 | 0.772 | Easy | 11,572 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2730523/Recursive-%2B-Iterative | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root:
return None
if root.val == val:
return root
if root.val < val:
return self.searchBST(root.right, val)
else:
return self.searchBS... | search-in-a-binary-search-tree | Recursive + Iterative | hacktheirlives | 0 | 1 | search in a binary search tree | 700 | 0.772 | Easy | 11,573 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2730523/Recursive-%2B-Iterative | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root is not None and root.val != val:
root = root.left if val < root.val else root.right
return root | search-in-a-binary-search-tree | Recursive + Iterative | hacktheirlives | 0 | 1 | search in a binary search tree | 700 | 0.772 | Easy | 11,574 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2707070/94-Accepted-Solution-or-Easy-to-Understand-or-Python | class Solution(object):
def searchBST(self, root, val):
q = deque()
q.append(root)
while q:
for _ in range(len(q)):
p = q.popleft()
if p.val == val: return p
if p.left: q.append(p.left)
if p.right: q.append(p.right)
... | search-in-a-binary-search-tree | 94% Accepted Solution | Easy to Understand | Python | its_krish_here | 0 | 3 | search in a binary search tree | 700 | 0.772 | Easy | 11,575 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2688447/python-or-easy-solution | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root:
return None
if root.val == val:
return root
if root.val < val:
return self.searchBST(root.right, val)
if root... | search-in-a-binary-search-tree | python | easy solution | MichelleZou | 0 | 10 | search in a binary search tree | 700 | 0.772 | Easy | 11,576 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2681514/Python-3-Solution-(Recursion) | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root:
if root.val == val:
return root
elif val > root.val:
return (self.searchBST(root.right,val))
elif val < root.val:
... | search-in-a-binary-search-tree | Python 3 Solution (Recursion) | rahulnakum | 0 | 16 | search in a binary search tree | 700 | 0.772 | Easy | 11,577 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2565867/Python-or-BFS-or-Faster-than-98.30 | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
q = deque()
q.append(root)
while q:
node = q.popleft()
# print(node.val)
if node.val == val:
return node
... | search-in-a-binary-search-tree | Python | BFS | Faster than 98.30% | ckayfok | 0 | 26 | search in a binary search tree | 700 | 0.772 | Easy | 11,578 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2336516/Python-Simple-Iterative-S-O(1)-and-Recursive-oror-Documented | class Solution:
# Recursive: T = O(log N) and S = O(log N) for stack created recursively
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root: return None
# if target value matched, return root
if root.val == val: return root
# if targe... | search-in-a-binary-search-tree | [Python] Simple Iterative S = O(1) and Recursive || Documented | Buntynara | 0 | 10 | search in a binary search tree | 700 | 0.772 | Easy | 11,579 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2276178/Python3-65ms-faster-than-99.52 | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root == None: return None
if root.val == val:
return root
return self.searchBST(root.left, val) or self.searchBST(root.right, val) | search-in-a-binary-search-tree | Python3 - 65ms - faster than 99.52% | spjoshis | 0 | 57 | search in a binary search tree | 700 | 0.772 | Easy | 11,580 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2222900/Python-simple-solution | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root:
if root.val == val:
return root
root = root.left if root.val > val else root.right | search-in-a-binary-search-tree | Python simple solution | meatcodex | 0 | 5 | search in a binary search tree | 700 | 0.772 | Easy | 11,581 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2153313/Python3-Solution | class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
if not root or root.val == val:
return root
if root.val < val: #要是目标值比根节点大 则在右子树中寻找
return self.searchBST(root.right, val)
if root.val > val: #要是目标值比根节点小 则在左子树中寻找
return self.searchBST(root.left,... | search-in-a-binary-search-tree | Python3 Solution | qywang | 0 | 33 | search in a binary search tree | 700 | 0.772 | Easy | 11,582 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/2033083/Python-Simple-readable-easy-to-understand-recursive-solution-(77-ms) | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root:
if root.val == val:
return root
elif root.val > val:
return self.searchBST(root.left, val)
else:
return self.searchBST(r... | search-in-a-binary-search-tree | [Python] Simple, readable, easy to understand, recursive solution (77 ms) | FedMartinez | 0 | 47 | search in a binary search tree | 700 | 0.772 | Easy | 11,583 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1960569/Python-iterative-3-lines-of-code.-No-recursion-O(1)-Space-O(nlogn)-average-time. | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root and root.val != val:
root = root.left if root.val > val else root.right
return root | search-in-a-binary-search-tree | Python iterative 3 lines of code. No recursion, O(1) Space, O(n·logn) average time. | sEzio | 0 | 42 | search in a binary search tree | 700 | 0.772 | Easy | 11,584 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1947577/python-3-oror-simple-iterative-solution | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root:
if val == root.val:
return root
elif val < root.val:
root = root.left
else:
root = root.right | search-in-a-binary-search-tree | python 3 || simple iterative solution | dereky4 | 0 | 12 | search in a binary search tree | 700 | 0.772 | Easy | 11,585 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1947357/Simple-python3-Solution | class Solution(object):
def searchBST(self, root, val):
trav = root
while trav:
if trav.val == val:
return trav
elif trav.val < val:
trav = trav.right
else:
trav = trav.left | search-in-a-binary-search-tree | Simple python3 Solution | nomanaasif9 | 0 | 11 | search in a binary search tree | 700 | 0.772 | Easy | 11,586 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1947127/Python-Recursive-%2B-Iterative-%2B-Bonus%3A-One-Liner! | class Solution:
def searchBST(self, root, val):
if not root: return None
elif root.val > val: return self.searchBST(root.left, val)
elif root.val < val: return self.searchBST(root.right, val)
else: return root | search-in-a-binary-search-tree | Python - Recursive + Iterative + Bonus: One-Liner! | domthedeveloper | 0 | 18 | search in a binary search tree | 700 | 0.772 | Easy | 11,587 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1947127/Python-Recursive-%2B-Iterative-%2B-Bonus%3A-One-Liner! | class Solution:
def searchBST(self, root, val):
return None if not root else self.searchBST(root.left, val) if root.val > val else self.searchBST(root.right, val) if root.val < val else root | search-in-a-binary-search-tree | Python - Recursive + Iterative + Bonus: One-Liner! | domthedeveloper | 0 | 18 | search in a binary search tree | 700 | 0.772 | Easy | 11,588 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1947127/Python-Recursive-%2B-Iterative-%2B-Bonus%3A-One-Liner! | class Solution:
def searchBST(self, root, val):
stack = [root]
while stack:
node = stack.pop()
if not node: return None
elif node.val > val: stack.append(node.left)
elif node.val < val: stack.append(node.right)
else: return node | search-in-a-binary-search-tree | Python - Recursive + Iterative + Bonus: One-Liner! | domthedeveloper | 0 | 18 | search in a binary search tree | 700 | 0.772 | Easy | 11,589 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1946834/Python-Fast-Recursive-Solution-No-Memory-Used | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
while root and root.val != val:
root = root.left if val < root.val else root.right
return root | search-in-a-binary-search-tree | Python Fast Recursive Solution, No Memory Used | Hejita | 0 | 14 | search in a binary search tree | 700 | 0.772 | Easy | 11,590 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1946088/Python-or-Recursion-or-Beats-99.7 | class Solution(object):
def searchBST(self, root, val):
"""
:type root: TreeNode
:type val: int
:rtype: TreeNode
"""
def search(node, val):
op = None
if node.val == val:
op = node
elif node.val > val and node.left:
... | search-in-a-binary-search-tree | Python | Recursion | Beats 99.7% | prajyotgurav | 0 | 13 | search in a binary search tree | 700 | 0.772 | Easy | 11,591 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1945657/Python-solution-92-faster | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[
TreeNode]:
while root:
if root.val == val:
return root
if root.val < val:
root = root.right
else:
root = root.left
retur... | search-in-a-binary-search-tree | Python solution 92% faster | pradeep288 | 0 | 14 | search in a binary search tree | 700 | 0.772 | Easy | 11,592 |
https://leetcode.com/problems/search-in-a-binary-search-tree/discuss/1945344/Python3-Solution-with-using-recursive-dfs | class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root:
return None
if root.val == val:
return root
return self.searchBST(root.left, val) or self.searchBST(root.right, val) | search-in-a-binary-search-tree | [Python3] Solution with using recursive dfs | maosipov11 | 0 | 7 | search in a binary search tree | 700 | 0.772 | Easy | 11,593 |
https://leetcode.com/problems/insert-into-a-binary-search-tree/discuss/1683883/Python3-ITERATIVE-(-)-Explained | class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root: return TreeNode(val)
cur, next = None, root
while next:
cur = next
next = cur.left if val < cur.val else cur.right
if val < cu... | insert-into-a-binary-search-tree | ✔️ [Python3] ITERATIVE (づ￣ ³￣)づ ❤, Explained | artod | 26 | 1,300 | insert into a binary search tree | 701 | 0.746 | Medium | 11,594 |
https://leetcode.com/problems/insert-into-a-binary-search-tree/discuss/1683964/Python-or-Simple-Recursive-or-Clean-or-O(N)-Timeor-O(1)-Space | class Solution:
def insertIntoBST(self, root, val):
if not root:
return TreeNode(val)
if val<root.val:
root.left = self.insertIntoBST(root.left, val)
else:
root.right = self.insertIntoBST(root.right, val)
return root | insert-into-a-binary-search-tree | [Python] | Simple Recursive | Clean | O(N) Time| O(1) Space | matthewlkey | 9 | 459 | insert into a binary search tree | 701 | 0.746 | Medium | 11,595 |
https://leetcode.com/problems/insert-into-a-binary-search-tree/discuss/2180670/Python3-recursive-solution | class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root == None:
return TreeNode(val)
def insert(root):
if root == None:
return TreeNode(val)
if val < root.val:
if root.lef... | insert-into-a-binary-search-tree | 📌 Python3 recursive solution | Dark_wolf_jss | 4 | 28 | insert into a binary search tree | 701 | 0.746 | Medium | 11,596 |
https://leetcode.com/problems/insert-into-a-binary-search-tree/discuss/305489/Python3-Iterative-solution | class Solution:
def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
cur_node = root
while True:
if val > cur_node.val:
if cur_node.right == None:
cur_node.right = TreeNode(val)
break
else:
... | insert-into-a-binary-search-tree | Python3 Iterative solution | decimalst | 2 | 109 | insert into a binary search tree | 701 | 0.746 | Medium | 11,597 |
https://leetcode.com/problems/insert-into-a-binary-search-tree/discuss/1994441/Python-oror-Simple-Iterative-Solution | class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
new, origin = TreeNode(val), root
if not root: return new
while root:
prev = root
if val > root.val: root = root.right
else: root = root.left
if val > prev.val: prev.right = new
else: prev.left = new... | insert-into-a-binary-search-tree | Python || Simple Iterative Solution | morpheusdurden | 1 | 56 | insert into a binary search tree | 701 | 0.746 | Medium | 11,598 |
https://leetcode.com/problems/insert-into-a-binary-search-tree/discuss/1684033/Python-3-Recursive-simple-O(n)-Explained | class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
def searchPlaceAndInsert(root,val):
if val < root.val:
if root.left:
searchPlaceAndInsert(root.left, val)
else:
... | insert-into-a-binary-search-tree | 👩💻 Python 3 Recursive simple O(n) Explained | letyrodri | 1 | 19 | insert into a binary search tree | 701 | 0.746 | Medium | 11,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.