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/next-greater-element-i/discuss/2765312/Next-greater-element-python-O(nlogn) | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
nums2_sorted = sorted(range(len(nums2)), key = lambda x: nums2[x])
# [0, 3, 1, 2]
nums1_sorted = sorted(range(len(nums1)), key = lambda x: nums1[x])
# [1, 2, 0]
result = [-1]*len(nu... | next-greater-element-i | Next greater element - python O(nlogn) | DavidCastillo | 0 | 4 | next greater element i | 496 | 0.714 | Easy | 8,700 |
https://leetcode.com/problems/next-greater-element-i/discuss/2764611/Python-Solution-beats-99.6-users-with-explanation-Fastest-solution | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
if not nums2:
return None
mapping = {}
result = []
stack = []
stack.append(nums2[0])
for i in range(1, len(nums2)):
while stack and nums2[i] > stack[-1]: # if stack is not... | next-greater-element-i | Python Solution beats 99.6% users with explanation Fastest solution | mritunjayyy | 0 | 4 | next greater element i | 496 | 0.714 | Easy | 8,701 |
https://leetcode.com/problems/next-greater-element-i/discuss/2746312/easy-method-using-python | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
l=[]
s=-1
for i in nums1:
p=nums2.index(i)
d=nums2[p:]
i=d[0]
u=0
for j in d:
if(j>i):
u=j
... | next-greater-element-i | easy method using python | sindhu_300 | 0 | 9 | next greater element i | 496 | 0.714 | Easy | 8,702 |
https://leetcode.com/problems/next-greater-element-i/discuss/2743106/python-Explanation | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
s=len(nums2)
l1=[]
l,stk=[],[]
for i in range(s-1,-1,-1):
if len(stk)==0:
stk.append(nums2[i])
l.append(-1)
elif len(stk)>0 and stk[... | next-greater-element-i | python Explanation | Kevin7777777 | 0 | 5 | next greater element i | 496 | 0.714 | Easy | 8,703 |
https://leetcode.com/problems/next-greater-element-i/discuss/2738322/Beats-97.66-with-Stacks-with-explanation | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
pairs = {}
result = []
stack = []
stack.append(nums2[0])
def makePairs(num, greaterValue):
pairs[num] = greaterValue
for i in range(1, len(nums2)):
... | next-greater-element-i | Beats 97.66 % with Stacks with explanation | karanvirsagar98 | 0 | 8 | next greater element i | 496 | 0.714 | Easy | 8,704 |
https://leetcode.com/problems/next-greater-element-i/discuss/2734653/Python-34ms-Easy-Understanding | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
k=[]
for i in nums1:
for j in range(nums2.index(i),len(nums2)):
if nums2[j]>i and len(k)<nums1.index(i)+1:
k+=nums2[j],
break
... | next-greater-element-i | Python 34ms Easy Understanding | Jlonerawesome | 0 | 10 | next greater element i | 496 | 0.714 | Easy | 8,705 |
https://leetcode.com/problems/next-greater-element-i/discuss/2730520/Next-Greater-Element-I-or-Python-Solution | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
def findGreater(num):
length = len(nums2)
temp = nums2[nums2.index(num)+1:]
for i in temp:
if i > num:
return i
return -1
... | next-greater-element-i | Next Greater Element I | Python Solution | ygygupta0 | 0 | 2 | next greater element i | 496 | 0.714 | Easy | 8,706 |
https://leetcode.com/problems/next-greater-element-i/discuss/2723590/Python-solution | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
hm={}
for i in range(len(nums1)):
for j in range(nums2.index(nums1[i]),len(nums2)):
hm[i]=[]
if nums2[j]>nums1[i]:
hm[i].append(nums2[j])
... | next-greater-element-i | Python solution | annazhengzhu | 0 | 3 | next greater element i | 496 | 0.714 | Easy | 8,707 |
https://leetcode.com/problems/next-greater-element-i/discuss/2721317/Python-%3A-O(n2)-and-O(n)-solution-with-explaination | class Solution:
#Order n^2 solution
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
nums1IndexMap = {v:i for i,v in enumerate(nums1)} #Searching array
output = [-1]*len(nums1)
for i,v in enumerate(nums2):
if v not in nums1IndexMap: continue
... | next-greater-element-i | Python : O(n^2) and O(n) solution with explaination | abrarjahin | 0 | 6 | next greater element i | 496 | 0.714 | Easy | 8,708 |
https://leetcode.com/problems/next-greater-element-i/discuss/2721317/Python-%3A-O(n2)-and-O(n)-solution-with-explaination | class Solution:
#Order n solution using monotonic stack
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
nums1IndexMap = {v:i for i,v in enumerate(nums1)} #Searching array
output = [-1]*len(nums1)
searchingElementStack = [] #Should store only values if-... | next-greater-element-i | Python : O(n^2) and O(n) solution with explaination | abrarjahin | 0 | 6 | next greater element i | 496 | 0.714 | Easy | 8,709 |
https://leetcode.com/problems/next-greater-element-i/discuss/2717244/Python3-64-faster-with-explanation | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
rlist = []
for num in nums1:
i = nums2.index(num)
if nums2[i:]:
check = 0
for item in nums2[i:]:
if item > nums2[i]:
... | next-greater-element-i | Python3, 64% faster with explanation | cvelazquez322 | 0 | 2 | next greater element i | 496 | 0.714 | Easy | 8,710 |
https://leetcode.com/problems/next-greater-element-i/discuss/2715303/Python-O(m%2Bn)-using-stack-with-easy-to-understand-approach | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
#Time-- O(n+m)
#Space-- O(m)
hset={n:i for i,n in enumerate(nums1)}
stack =[]
res=[-1]*len(nums1)
for i in range(len(nums2)):
while stack and nums2[... | next-greater-element-i | Python O(m+n) using stack with easy to understand approach | kartikchoudhary96 | 0 | 2 | next greater element i | 496 | 0.714 | Easy | 8,711 |
https://leetcode.com/problems/next-greater-element-i/discuss/2709432/O(nums1.length-%2B-nums2.length)-time-python3-solution | class Solution:
# O(nums1.length + nums2.length) time,
# O(nums2.length) space,
# Approach: monotonic stack, hashmap
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
stack = []
next_greater = {}
for index, num in enumerate(nums2):
... | next-greater-element-i | O(nums1.length + nums2.length) time python3 solution | destifo | 0 | 3 | next greater element i | 496 | 0.714 | Easy | 8,712 |
https://leetcode.com/problems/next-greater-element-i/discuss/2697288/O(N)-solution-with-monotonic-stack | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
stack = []
hashmap = {}
for n in nums2:
while stack and n > stack[-1]:
hashmap[stack.pop()] = n
stack.append(n)
while stack:
hashmap[st... | next-greater-element-i | O(N) solution with monotonic stack | michaelniki | 0 | 9 | next greater element i | 496 | 0.714 | Easy | 8,713 |
https://leetcode.com/problems/next-greater-element-i/discuss/2674219/Python-or-Stack-solution | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
arr = [-1 for i in range(10**4 + 1)]
st = [nums2[0]]
for i in range(1, len(nums2)):
while st and nums2[i] > st[-1]:
arr[st.pop()] = nums2[i]
st.append(nums2[... | next-greater-element-i | Python | Stack solution | LordVader1 | 0 | 76 | next greater element i | 496 | 0.714 | Easy | 8,714 |
https://leetcode.com/problems/next-greater-element-i/discuss/2673430/Python-Solution | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
output=[]
for i in range(len(nums1)):
index=nums2.index(nums1[i])
lock=False
for j in range(index,len(nums2)):
if(nums2[j]>nums1[i] and lock==False):
... | next-greater-element-i | Python Solution | Phoenix_18 | 0 | 1 | next greater element i | 496 | 0.714 | Easy | 8,715 |
https://leetcode.com/problems/next-greater-element-i/discuss/2654012/Next-Greater-Element-using-stacks(python) | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
n=len(nums2)
m=len(nums1)
rge=[-1]*n
stack=[]
stack.append(nums2[-1])
for i in range(n-2,-1,-1):
while stack and nums2[i]>=stack[-1]:
stack.pop()... | next-greater-element-i | Next Greater Element using stacks(python) | kaushik555 | 0 | 5 | next greater element i | 496 | 0.714 | Easy | 8,716 |
https://leetcode.com/problems/next-greater-element-i/discuss/2643343/python-easy-solution-using-monotonic-stack-and-dictionary | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
n=len(nums1)
m=len(nums2)
ans=[0]*n
d={}
for i in range(n):
d[nums1[i]]=i
#print(d)
stack=[]
for i in range(m-1,-1,-1):
while len(sta... | next-greater-element-i | python easy solution using monotonic stack and dictionary | tush18 | 0 | 70 | next greater element i | 496 | 0.714 | Easy | 8,717 |
https://leetcode.com/problems/next-greater-element-i/discuss/2605639/Python-Solution-or-2-solutions-or-Brute-Force-or-Stack-%2B-Hashmap-or-O(n%2Bm) | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
# Brute Force: Accepted, TC: O(len(nums1)*len(nums2))
# ans=[-1]*len(nums1)
# for i in range(len(nums1)):
# index=nums2.index(nums1[i])
# for j in range(index+1, le... | next-greater-element-i | Python Solution | 2 solutions | Brute Force | Stack + Hashmap | O(n+m) | Siddharth_singh | 0 | 47 | next greater element i | 496 | 0.714 | Easy | 8,718 |
https://leetcode.com/problems/next-greater-element-i/discuss/2575428/Simple-dictonary-solution | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
dic = {}
for i in nums1:
if i in nums2:
dic[i] = nums2.index(i)
res = []
for k in dic.keys():
j = dic[k]
while j < len(nums2):
... | next-greater-element-i | Simple dictonary solution | aruj900 | 0 | 39 | next greater element i | 496 | 0.714 | Easy | 8,719 |
https://leetcode.com/problems/next-greater-element-i/discuss/2572371/Python3-or-3-Different-Solutions-or-Optimal-Complexity | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
# Method 1: Brute Force T.C: O(nˆ3) S.C: O(1)
ans = [-1] * len(nums1)
for i in range(len(nums1)):
for j in range(len(nums2)):
if nums1[i] == nums2[j]:
... | next-greater-element-i | Python3 | 3 Different Solutions | Optimal Complexity | chawlashivansh | 0 | 58 | next greater element i | 496 | 0.714 | Easy | 8,720 |
https://leetcode.com/problems/next-greater-element-i/discuss/2564311/python-solution-using-stack | class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
dict_greater = {}
stack = [nums2[0]]
for num in nums2[1:]:
if num < stack[-1]:
stack.append(num)
while len(stack) > 0 and num > stack[-1]:
... | next-greater-element-i | python solution using stack | samanehghafouri | 0 | 44 | next greater element i | 496 | 0.714 | Easy | 8,721 |
https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/1453111/Python-Binary-Search | class Solution:
def __init__(self, rects: List[List[int]]):
self.rects = rects
self.search_space = []
for i, rect in enumerate(rects):
a, b, c, d = rect
self.search_space.append((d - b + 1) * (c - a + 1))
if i != 0:
self.search_space[i] +=... | random-point-in-non-overlapping-rectangles | Python Binary Search | ypatel38 | 1 | 155 | random point in non overlapping rectangles | 497 | 0.393 | Medium | 8,722 |
https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/1923494/Python-easy-understanding-solution-with-comment | class Solution:
def __init__(self, rects: List[List[int]]):
self.rects = rects
self.weights = [] # self.weights record the "points" one rectangle have respectively
for i in range(len(rects)):
num_points = (rects[i][2] - rects[i][0] + ... | random-point-in-non-overlapping-rectangles | Python easy - understanding solution with comment | byroncharly3 | 0 | 93 | random point in non overlapping rectangles | 497 | 0.393 | Medium | 8,723 |
https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/808568/Python3-sampling-by-weight | class Solution:
def __init__(self, rects: List[List[int]]):
self.rects = rects #store rectangle
self.wt = [0]
for x1, y1, x2, y2 in rects:
wt = (x2 - x1 + 1) * (y2 - y1 + 1) # number of points
self.wt.append(self.wt[-1] + wt)
def pick(self) -> List[int]:
... | random-point-in-non-overlapping-rectangles | [Python3] sampling by weight | ye15 | 0 | 51 | random point in non overlapping rectangles | 497 | 0.393 | Medium | 8,724 |
https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/808568/Python3-sampling-by-weight | class Solution:
def __init__(self, rects: List[List[int]]):
self.rects = rects # store rectangles
self.wt = [0] # weighted sampling
for x1, y1, x2, y2 in rects:
wt = (x2 - x1 + 1) * (y2 - y1 + 1) # wt ~ number of points
self.wt.append(self.wt[-1] + wt)
def p... | random-point-in-non-overlapping-rectangles | [Python3] sampling by weight | ye15 | 0 | 51 | random point in non overlapping rectangles | 497 | 0.393 | Medium | 8,725 |
https://leetcode.com/problems/diagonal-traverse/discuss/272114/Simple-Python-Solution-(with-comments) | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
res = []
if not matrix:
return res
# group values in matrix by the sum of their indices in a map
map = {}
for i in range(len(matrix) + len(matrix[0]) - 1):
map[i] = []
... | diagonal-traverse | Simple Python Solution (with comments) | AnthonyChao | 5 | 612 | diagonal traverse | 498 | 0.581 | Medium | 8,726 |
https://leetcode.com/problems/diagonal-traverse/discuss/1508963/Python-solution-or-dictionary-or-sum-of-index | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
# Making a dictionary of keys = (row + col) as in a Diagonal treversal line (index of row + index of col) = constant. And values of Dictionary are elements of mat
myDict = {}
for i in range(len(mat)):
... | diagonal-traverse | [Python] solution | dictionary | sum of index | samirpaul1 | 3 | 97 | diagonal traverse | 498 | 0.581 | Medium | 8,727 |
https://leetcode.com/problems/diagonal-traverse/discuss/2406808/Python-solution | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
M,N = len(mat), len(mat[0])
diagonals = []
# traverse first column
for i in range(len(mat)):
idx_i = i
idx_j = 0
diagonals.append([])
while idx_i >... | diagonal-traverse | Python solution | pivovar3al | 2 | 84 | diagonal traverse | 498 | 0.581 | Medium | 8,728 |
https://leetcode.com/problems/diagonal-traverse/discuss/1362507/Intuitive-Python-solution-with-debugging-statements-to-run-through. | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
rows = len(mat)-1 # index upperbound
columns = len(mat[0])-1 # index upperbound
up = False # Next iteration flag
ans = [mat[0][0]]
lastPoint = [0,0]
while True:
r,c ... | diagonal-traverse | Intuitive Python solution with debugging statements to run through. | SathvikPN | 2 | 74 | diagonal traverse | 498 | 0.581 | Medium | 8,729 |
https://leetcode.com/problems/diagonal-traverse/discuss/2226442/Python3-Simple-Solution | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
flip=0
res=[]
R=len(mat)
C=len(mat[0])
def checker(r,c):
if r<0 or c<0 or r>=R or c>=C:
return False
return True
def runner(r,c):
temp=[]... | diagonal-traverse | Python3 Simple Solution | sonikartik2021 | 1 | 79 | diagonal traverse | 498 | 0.581 | Medium | 8,730 |
https://leetcode.com/problems/diagonal-traverse/discuss/2038679/Python-detailed-explanation-Tree-based-implementation | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
# special cases
if len(mat) == 1:
return mat[0]
elif len(mat[0]) == 1:
return [m[0] for m in mat]
def neighb(x, y): # gives left and right child
for dirx, diry in [(1,... | diagonal-traverse | Python detailed explanation [Tree based implementation] | 96sayak | 1 | 55 | diagonal traverse | 498 | 0.581 | Medium | 8,731 |
https://leetcode.com/problems/diagonal-traverse/discuss/1651119/Python%3AUsing-sum-of-indexes | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
m = len(mat)
n = len(mat[0])
res = list()
sum_dict = defaultdict(list)
# Segregate elements as per the sum if indexes (i,j)
for i in range(m):
for j in range(n):
... | diagonal-traverse | Python:Using sum of indexes | jnaik | 1 | 98 | diagonal traverse | 498 | 0.581 | Medium | 8,732 |
https://leetcode.com/problems/diagonal-traverse/discuss/861312/Python3-two-solutions | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix: return [] # edge case
m, n = len(matrix), len(matrix[0]) # dimension
ans = []
i = j = 0
stride = 1
for _ in range(m+n-1):
val = []
ii,... | diagonal-traverse | [Python3] two solutions | ye15 | 1 | 82 | diagonal traverse | 498 | 0.581 | Medium | 8,733 |
https://leetcode.com/problems/diagonal-traverse/discuss/861312/Python3-two-solutions | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix: return [] # edge case
ans = []
m, n = len(matrix), len(matrix[0]) # dimension
i = j = 0
for _ in range(m*n):
ans.append(matrix[i][j])
if (i+j)... | diagonal-traverse | [Python3] two solutions | ye15 | 1 | 82 | diagonal traverse | 498 | 0.581 | Medium | 8,734 |
https://leetcode.com/problems/diagonal-traverse/discuss/861312/Python3-two-solutions | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
ans = []
if matrix:
m, n = len(matrix), len(matrix[0]) # dimensions
i, j, di, dj = 0, 0, -1, 1
for _ in range(m*n):
ans.append(matrix[i][j])
if 0 ... | diagonal-traverse | [Python3] two solutions | ye15 | 1 | 82 | diagonal traverse | 498 | 0.581 | Medium | 8,735 |
https://leetcode.com/problems/diagonal-traverse/discuss/2848652/python3-solution | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
d = defaultdict(list)
for i in range(len(mat)):
for j in range(len(mat[0])):
d[i+j].append(mat[i][j])
arr = []
for i in d.items():
if i[0]%2 == 0:
... | diagonal-traverse | python3 solution | Cosmodude | 0 | 1 | diagonal traverse | 498 | 0.581 | Medium | 8,736 |
https://leetcode.com/problems/diagonal-traverse/discuss/2816617/Python-Simple-and-Easy-Solution-With-Diagram | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
# get no of rows and cols
row = len(mat)
col = len(mat[0])
# get no of diagonals needs
no_diagonals = [[] for _ in range(row+col -1)]
# put every element of mat in coresp... | diagonal-traverse | Python Simple and Easy Solution With Diagram | devzohaib | 0 | 7 | diagonal traverse | 498 | 0.581 | Medium | 8,737 |
https://leetcode.com/problems/diagonal-traverse/discuss/2796650/Beats-91.54-in-Memory | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
res=[]
r=c=0
going_up=True
rmax,cmax=len(mat),len(mat[0])
while len(res)< rmax*cmax:
if going_up:
while r>=0 and c<cmax:
... | diagonal-traverse | Beats 91.54% in Memory | avinash_konduri | 0 | 3 | diagonal traverse | 498 | 0.581 | Medium | 8,738 |
https://leetcode.com/problems/diagonal-traverse/discuss/2785748/python-traversing-optimised-solution | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
# O(n), O(n) where n = len(rows) * len(cols)
rows, cols = len(mat), len(mat[0])
res = []
cur_row = cur_col = 0
going_up = True
while len(res) != rows * cols:
if going_up:
... | diagonal-traverse | python traversing optimised solution | sahilkumar158 | 0 | 6 | diagonal traverse | 498 | 0.581 | Medium | 8,739 |
https://leetcode.com/problems/diagonal-traverse/discuss/2743450/Python-98.33-or-O-(M%2BN) | class Solution:
def findDiagonalOrder(self, mat) :
n=len(mat)
m=len(mat[0])
ans=[]
row,col=0,0
for d in range(m+n-1):
temp=[]
if d < n:
row=d
col=0
else:
row= n-1
col= d-row
... | diagonal-traverse | Python 98.33 % | O (M+N) | user3904Q | 0 | 4 | diagonal traverse | 498 | 0.581 | Medium | 8,740 |
https://leetcode.com/problems/diagonal-traverse/discuss/2683848/O(N)-Intuitive-Python-BFS-Solution | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
m,n = len(mat), len(mat[0])
lvls = []
added = set()
q = []
def add(i,j):
if (i,j) not in added:
added.add((i,j))
q.append((i,j))
add(0,0)
... | diagonal-traverse | O(N) Intuitive Python BFS Solution | Dylan_Yifan | 0 | 10 | diagonal traverse | 498 | 0.581 | Medium | 8,741 |
https://leetcode.com/problems/diagonal-traverse/discuss/2678714/Python3-or-Simple-or-Intuitive-or-Neat | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
m, n, r, c = len(mat), len(mat[0]), 0, 0
diag = []
def getNext(r, c):
# go Northeast
if (r+c)%2 == 0 and r > 0 and c < n-1:
return r-1, c+1
... | diagonal-traverse | Python3 | Simple | Intuitive | Neat | aashi111989 | 0 | 22 | diagonal traverse | 498 | 0.581 | Medium | 8,742 |
https://leetcode.com/problems/diagonal-traverse/discuss/2558455/Python3-or-Using-Stack-or-O(m*n) | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
stack_1,stack_2=[],[]
stack_1.append((0,0))
ans=[mat[0][0]]
r,c=len(mat),len(mat[0])
mat[0][0]=float('inf')
while stack_1 or stack_2:
while stack_1:
x,y=stack_1... | diagonal-traverse | [Python3] | Using Stack | O(m*n) | swapnilsingh421 | 0 | 23 | diagonal traverse | 498 | 0.581 | Medium | 8,743 |
https://leetcode.com/problems/diagonal-traverse/discuss/2455706/Python-3-or-Iterative-or-Direction-Check | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
m, n = len(mat), len(mat[0])
self.res = []
q = deque()
q.append((0,0,True))
while q:
i, j, goesUp = q.popleft()
if i < 0 or i >= m or j < 0 or j >= n:... | diagonal-traverse | Python 3 | Iterative | Direction Check | Ploypaphat | 0 | 46 | diagonal traverse | 498 | 0.581 | Medium | 8,744 |
https://leetcode.com/problems/diagonal-traverse/discuss/2408554/Python-Implementation-Using-Hashmap | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
ele = {}
for i in range(len(mat)):
for j in range(len(mat[i])):
if i+j not in ele:
ele[i+j] = [mat[i][j]]
else:
ele[i+j].append(mat[i][j... | diagonal-traverse | Python Implementation Using Hashmap | Abhi_-_- | 0 | 37 | diagonal traverse | 498 | 0.581 | Medium | 8,745 |
https://leetcode.com/problems/diagonal-traverse/discuss/2334066/Easy-Simple-Python-Defaultdict(List) | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
# Default dict for simpler code
d = defaultdict(list)
for i in range(len(mat)):
for j in range(len(mat[0])):
# Add values to diagonal dictionary
d[i+j].append(mat[i][j])
a... | diagonal-traverse | Easy Simple Python Defaultdict(List) | drblessing | 0 | 23 | diagonal traverse | 498 | 0.581 | Medium | 8,746 |
https://leetcode.com/problems/diagonal-traverse/discuss/2074320/Simple-solution-Python-O(NM)-time-and-O(1)-space | class Solution:
def change_direction(self):
self.isign *= -1
self.jsign *= -1
def i_overflow(self, i):
return i+self.isign < 0 or i+self.isign > self.m-1
def j_overflow(self, j):
return j+self.jsign < 0 or j+self.jsign > self.n-1
def next_cell(self, i, j):
... | diagonal-traverse | Simple solution, Python, O(NM) time and O(1) space | tushar-rishav | 0 | 43 | diagonal traverse | 498 | 0.581 | Medium | 8,747 |
https://leetcode.com/problems/diagonal-traverse/discuss/1872793/Simplest-Python-Solution-%2B-Visual-Explanation-Time%3A-O(M*N)-Space%3A-O(1)-or-grandma-can-understand | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
m,n=len(mat),len(mat[0])
res=[]
for k in range(m+n-1):
if k%2==0:
i=min(m-1,k)
j=k-i
while 0<=i<m and 0<=j<n:
res.append(mat[i][j])... | diagonal-traverse | Simplest Python Solution + Visual Explanation, Time: O(M*N), Space: O(1) | 🥴 grandma can understand | noobcoderbaby | 0 | 45 | diagonal traverse | 498 | 0.581 | Medium | 8,748 |
https://leetcode.com/problems/diagonal-traverse/discuss/1865220/Python-iterative-dfs | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
rows, cols = len(mat), len(mat[0])
stack = [(0,0,0)] # (row, column, direction), 0 is up direction and 1 is down direction
output = []
n = rows * cols
while len(output) != n:
... | diagonal-traverse | Python iterative dfs | Rush_P | 0 | 54 | diagonal traverse | 498 | 0.581 | Medium | 8,749 |
https://leetcode.com/problems/diagonal-traverse/discuss/1789445/Easy-money | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
m = len(mat)
n = len(mat[0])
res = []
for s in range(m+n-1):
if s%2 == 0: # even, go from left to right
for j in range(max(s-m+1,0), min(s+1,n)):
res.append... | diagonal-traverse | Easy money | justicesuker | 0 | 38 | diagonal traverse | 498 | 0.581 | Medium | 8,750 |
https://leetcode.com/problems/diagonal-traverse/discuss/1750946/quick-easy-to-understand-python | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
row_dict =defaultdict(list)
for row in range(len(mat)):
for col in range(len(mat[0])):
row_dict[row+col].append(mat[row][col])
flag = False
res = []
for i,val in ... | diagonal-traverse | quick, easy to understand, python | aaronat | 0 | 124 | diagonal traverse | 498 | 0.581 | Medium | 8,751 |
https://leetcode.com/problems/diagonal-traverse/discuss/1684420/Python-no-reverse-diagonal-iteration-simulation | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
n, m = len(mat), len(mat[0])
res = []
for s in range(0, n + m - 1):
if s % 2 == 0:
x = min(s, n - 1)
y = s - x
while x >= 0 and y < m:
... | diagonal-traverse | Python no-reverse diagonal iteration simulation | Krymore | 0 | 59 | diagonal traverse | 498 | 0.581 | Medium | 8,752 |
https://leetcode.com/problems/diagonal-traverse/discuss/1600012/Easy-Python-very-simple-cases | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
tr = True
row = col = 0
num_row = len(mat)
num_col = len(mat[0])
ordered = []
while row < num_row and col < num_col:
ordered.append(mat[row][col])
... | diagonal-traverse | Easy Python - very simple cases | AbitamimBharmal | 0 | 84 | diagonal traverse | 498 | 0.581 | Medium | 8,753 |
https://leetcode.com/problems/diagonal-traverse/discuss/1500518/Python3 | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
d = collections.defaultdict(list)
for i in range(len(mat)):
for j in range(len(mat[0])):
d[i+j].append(mat[i][j])
res = []
index = 0
for k, v ... | diagonal-traverse | Python3 | immutable_23 | 0 | 73 | diagonal traverse | 498 | 0.581 | Medium | 8,754 |
https://leetcode.com/problems/diagonal-traverse/discuss/1500449/Easy-Python-Solution-with-greater-Direction-Vector-greater | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
index = (1, -1)
direction = (-1, 1)
arr = []
while len(arr) != len(mat) * len(mat[0]):
index = (index[0] + direction[0], index[1] + direction[1])
if (index[0] < 0 or index... | diagonal-traverse | Easy Python Solution with -> Direction Vector -> | schedutron | 0 | 56 | diagonal traverse | 498 | 0.581 | Medium | 8,755 |
https://leetcode.com/problems/diagonal-traverse/discuss/1500448/Easy-Python-Solution-with-greater-Direction-Vector-greater | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
index = (1, -1)
direction = (-1, 1)
arr = []
while len(arr) != len(mat) * len(mat[0]):
index = (index[0] + direction[0], index[1] + direction[1])
if (index[0] < 0 or index... | diagonal-traverse | Easy Python Solution with -> Direction Vector -> | schedutron | 0 | 27 | diagonal traverse | 498 | 0.581 | Medium | 8,756 |
https://leetcode.com/problems/diagonal-traverse/discuss/1431210/Simple-Python | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
num_bins = len(mat) + len(mat[0]) - 1
bins = [[] for _ in range(num_bins)]
index = 0
for i in range(len(mat)):
index = i
for j in range(len(mat[0])):
bins[index]... | diagonal-traverse | Simple Python | envy7 | 0 | 83 | diagonal traverse | 498 | 0.581 | Medium | 8,757 |
https://leetcode.com/problems/diagonal-traverse/discuss/1386313/Python-Solution | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
diagonal_order = []
n = len(matrix)
m = len(matrix[0])
i = 0
j = 0
while len(diagonal_order) < n * m:
while i >= 0 and j < m: # we go upwards
diagonal_order... | diagonal-traverse | Python Solution | mariandanaila01 | 0 | 133 | diagonal traverse | 498 | 0.581 | Medium | 8,758 |
https://leetcode.com/problems/diagonal-traverse/discuss/1099305/Python-or-RuntimeSpace-Efficiency-99-or-Simple | class Solution:
def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
if mat==[]: return mat
r,c=len(mat),len(mat[0])
if r==1 or c==1:
return [mat[i][j] for i in range(r) for j in range(c)]
def getDiag(i,j,rev):
diag=[]
while j<c... | diagonal-traverse | Python | Runtime/Space Efficiency 99% | Simple | rajatrai1206 | 0 | 129 | diagonal traverse | 498 | 0.581 | Medium | 8,759 |
https://leetcode.com/problems/diagonal-traverse/discuss/1033801/Straight-forward-Python-solution | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix:
return []
M = len(matrix)
N = len(matrix[0])
result = {}
for i in range(M):
for j in range(N):
if i+j not in result:
... | diagonal-traverse | Straight forward Python solution | sirajali05 | 0 | 123 | diagonal traverse | 498 | 0.581 | Medium | 8,760 |
https://leetcode.com/problems/diagonal-traverse/discuss/1020147/Python-Clean-solution-based-on-coordinates-w-comments-(TC%3A-O(nm)-SC%3A-O(1)) | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
"""
Time Complexity: O(nm), n - number of rows, m - number of cols
Space Complexity: O(1), Note that the space occupied by the output array doesn't count towards the space complexity since that is a requiremen... | diagonal-traverse | Python Clean solution based on coordinates /w comments (TC: O(nm), SC: O(1)) | elzzz | 0 | 63 | diagonal traverse | 498 | 0.581 | Medium | 8,761 |
https://leetcode.com/problems/diagonal-traverse/discuss/914468/Python3-Easy-solution | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
d = defaultdict(list)
for i in range(len(matrix)):
for j in range(len(matrix[0])):
d[i+j].append(matrix[i][j])
res = []
... | diagonal-traverse | Python3 Easy solution | ermolushka2 | 0 | 127 | diagonal traverse | 498 | 0.581 | Medium | 8,762 |
https://leetcode.com/problems/diagonal-traverse/discuss/487810/Python3-super-simple-solution | class Solution:
def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix: return []
temp = []
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if len(temp)-1<i+j: temp.append([matrix[i][j]])
else: temp[i+j].appen... | diagonal-traverse | Python3 super simple solution | jb07 | 0 | 115 | diagonal traverse | 498 | 0.581 | Medium | 8,763 |
https://leetcode.com/problems/keyboard-row/discuss/1525751/Easy-Python-Solution-or-Faster-than-97-(24-ms) | class Solution:
def findWords(self, wds: List[str]) -> List[str]:
st = {'q': 1, 'w': 1, 'e': 1, 'r': 1, 't': 1, 'y': 1, 'u': 1, 'i': 1, 'o': 1, 'p': 1, 'a': 2, 's': 2, 'd': 2, 'f': 2, 'g': 2, 'h': 2, 'j': 2, 'k': 2, 'l': 2, 'z': 3, 'x': 3, 'c': 3, 'v': 3, 'b': 3, 'n': 3, 'm': 3}
ret = []
f... | keyboard-row | Easy Python Solution | Faster than 97% (24 ms) | the_sky_high | 5 | 462 | keyboard row | 500 | 0.692 | Easy | 8,764 |
https://leetcode.com/problems/keyboard-row/discuss/437422/C%2B%2BJavaPython-RegEx | class Solution:
def findWords(self, words):
letters_a, letters_b, letters_c = set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm')
row_words = []
for word in words:
unique_word = set(word.lower())
if unique_word <= letters_a or unique_word <= letters_b or unique_word <... | keyboard-row | [C++/Java/Python] RegEx | i-i | 5 | 366 | keyboard row | 500 | 0.692 | Easy | 8,765 |
https://leetcode.com/problems/keyboard-row/discuss/437422/C%2B%2BJavaPython-RegEx | class Solution:
def findWords(self, words):
keyboard = r'(?i)^(?:[qwertyuiop]+|[asdfghjkl]+|[zxcvbnm]+)$'
row_words = []
for word in words:
if word == "":
row_words.append("")
row_words.extend(re.findall(keyboard, word))
return row_words | keyboard-row | [C++/Java/Python] RegEx | i-i | 5 | 366 | keyboard row | 500 | 0.692 | Easy | 8,766 |
https://leetcode.com/problems/keyboard-row/discuss/1058707/Python3-simple-solution-using-%22set%22 | class Solution:
def findWords(self, words: List[str]) -> List[str]:
l = ["qwertyuiop", "asdfghjkl", "zxcvbnm"]
ans = []
for i in words:
a = i.lower()
if len(set(a).difference(set(l[0]))) == 0 or len(set(a).difference(set(l[1]))) == 0 or len(set(a).difference(set(l[2])... | keyboard-row | Python3 simple solution using "set" | EklavyaJoshi | 4 | 126 | keyboard row | 500 | 0.692 | Easy | 8,767 |
https://leetcode.com/problems/keyboard-row/discuss/230863/Python-simple-solution | class Solution(object):
def findWords(self, words):
"""
:type words: List[str]
:rtype: List[str]
"""
word_list=[]
top_row=set('qwertyuiop')
mid_row=set('asdfghjkl')
bottom_row=set('zxcvbnm')
for word in words:
if set(word.lower()).i... | keyboard-row | Python simple solution | kishoreravi97 | 4 | 748 | keyboard row | 500 | 0.692 | Easy | 8,768 |
https://leetcode.com/problems/keyboard-row/discuss/1474014/python-or-simple-or-Faster-than-96.17 | class Solution:
def findWords(self, words: List[str]) -> List[str]:
set1 = set('qwertyuiop')
set2 = set('asdfghjkl')
set3 = set('zxcvbnm')
res = list()
for word in words:
w = set(word.lower())
if len(w | set1) == len(set1) or len(w | set2) == len(set2)... | keyboard-row | python | simple | Faster than 96.17 % | deep765 | 3 | 239 | keyboard row | 500 | 0.692 | Easy | 8,769 |
https://leetcode.com/problems/keyboard-row/discuss/1353116/Python-Solution-using-Set-28ms | class Solution:
def findWords(self, words: List[str]) -> List[str]:
row1 = set('qwertyuiopQWERTYUIOP')
row2 = set('asdfghjklASDFGHJKL')
row3 = set('zxcvbnmZXCVBNM')
result = []
for word in words:
w = set(list(word))
if w.issubset(row1) or w.issubset(ro... | keyboard-row | Python Solution using Set - 28ms | _Mansiii_ | 3 | 174 | keyboard row | 500 | 0.692 | Easy | 8,770 |
https://leetcode.com/problems/keyboard-row/discuss/1879448/Python-solution-faster-than-95 | class Solution:
def findWords(self, words: List[str]) -> List[str]:
row1 = "qwertyuiop"
row2 = "asdfghjkl"
row3 = "zxcvbnm"
res = []
for i in words:
if i[0].lower() in row1:
if all(x in row1 for x in i.lower()):
res.append(i)
... | keyboard-row | Python solution faster than 95% | alishak1999 | 2 | 181 | keyboard row | 500 | 0.692 | Easy | 8,771 |
https://leetcode.com/problems/keyboard-row/discuss/2594367/Python-simple-solution-(using-set-to-find-common-letters) | class Solution:
def findWords(self, words: list[str]) -> list[str]:
def is_valid(s):
s, r1, r2, r3 = map(set, [s, 'qwertyuiop', 'asdfghjkl', 'zxcvbnm'])
return s <= r1 or s <= r2 or s <= r3
return [i for i in words if is_valid(i.lower())] | keyboard-row | Python simple solution (using set to find common letters) | Mark_computer | 1 | 27 | keyboard row | 500 | 0.692 | Easy | 8,772 |
https://leetcode.com/problems/keyboard-row/discuss/2408563/Python-faster-than-99.78-using-dict | class Solution:
def findWords(self, words: List[str]) -> List[str]:
d = {}
for c in "qwertyuiop":
d[c] = 1
for c in "asdfghjkl":
d[c] = 2
for c in "zxcvbnm":
d[c] = 3
result = []
for word in words:
if len(wo... | keyboard-row | Python faster than 99.78 using dict | Mohan01234 | 1 | 211 | keyboard row | 500 | 0.692 | Easy | 8,773 |
https://leetcode.com/problems/keyboard-row/discuss/1993013/My-Solution-Without-python3-built-ins | class Solution:
def findWords(self, words: List[str]) -> List[str]:
charMap = {'q': 1, 'w': 1, 'e': 1, 'r': 1, 't': 1, 'y': 1, 'u': 1, 'i': 1, 'o': 1, 'p': 1, 'a': 2, 's': 2, 'd': 2, 'f': 2, 'g': 2, 'h': 2, 'j': 2, 'k': 2, 'l': 2, 'z': 3, 'x': 3, 'c': 3, 'v': 3, 'b': 3, 'n': 3, 'm': 3}
result = []
... | keyboard-row | My Solution Without python3 built-ins | huyanguyen3695 | 1 | 67 | keyboard row | 500 | 0.692 | Easy | 8,774 |
https://leetcode.com/problems/keyboard-row/discuss/1746680/Python-solution-(Using-Subset) | class Solution:
def findWords(self, words: List[str]) -> List[str]:
first_row_keywords = set("qwertyuiop")
second_row_keywords = set("asdfghjkl")
third_row_keywords = set("zxcvbnm")
result_list = []
for word in words:
word_to_set = set(word.lower())
# Checking ... | keyboard-row | Python solution (Using Subset) | aakshay740 | 1 | 83 | keyboard row | 500 | 0.692 | Easy | 8,775 |
https://leetcode.com/problems/keyboard-row/discuss/1695414/**-Python-code%3A | class Solution:
def findWords(self, words: List[str]) -> List[str]:
row1 = set("QWERTYUIOPqwertyuiop")
row2 = set("ASDFGHJKLasdfghjkl")
row3 = set("ZXCVBNMzxcvbnm")
res=[]
for word in words:
check=True
if word[0] in row1:... | keyboard-row | ** Python code: | Anilchouhan181 | 1 | 68 | keyboard row | 500 | 0.692 | Easy | 8,776 |
https://leetcode.com/problems/keyboard-row/discuss/1468568/Easy-to-understand-Set-Operations-PythonPython3-Python3 | class Solution:
def findWords(self, words: List[str]) -> List[str]:
row1 = {'q','w','e','r','t','y','u','i','o','p'}
row2 = {'a','s','d','f','g','h','j','k','l'}
row3 = {'z','x','c','v','b','n','m'}
common = []
for word in words:
string = set(word.lower())
... | keyboard-row | Easy to understand, Set Operations, Python/Python3, Python3 | AshwinBalaji52 | 1 | 150 | keyboard row | 500 | 0.692 | Easy | 8,777 |
https://leetcode.com/problems/keyboard-row/discuss/1404858/Python-Easily-Understandable-or-Runtime%3A-28-ms | class Solution:
def findWords(self, words: List[str]) -> List[str]:
fr="qwertyuiop"
sr="asdfghjkl"
tr="zxcvbnm"
ans=[]
for j in words:
if all(i.lower() in fr for i in j) or all(i.lower() in sr for i in j) or all(i.lower() in tr for i in j):
ans.app... | keyboard-row | Python Easily Understandable | Runtime: 28 ms | satyamshrma | 1 | 103 | keyboard row | 500 | 0.692 | Easy | 8,778 |
https://leetcode.com/problems/keyboard-row/discuss/1292284/Python3-dollarolution(99-Faster) | class Solution:
def findWords(self, words: List[str]) -> List[str]:
r1 = 'qwertyuiop'
r2 = 'asdfghjkl'
r3 = 'zxcvbnm'
m = []
for i in range(len(words)):
f = 0
k = words[i].lower()
for j in k:
if j in r1:
... | keyboard-row | Python3 $olution(99% Faster) | AakRay | 1 | 296 | keyboard row | 500 | 0.692 | Easy | 8,779 |
https://leetcode.com/problems/keyboard-row/discuss/1217990/python-solution | class Solution:
def check1(self,s):
l = "qwertyuiopQWERTYUIOP"
for c in s:
if not c in l:
return False
return True
def check2(self,s):
l = "asdfghjklASDFGHJKL"
for c in s:
if not c in l:
return False
ret... | keyboard-row | python solution | ArnabCk | 1 | 75 | keyboard row | 500 | 0.692 | Easy | 8,780 |
https://leetcode.com/problems/keyboard-row/discuss/577650/Python-using-bit-mask | class Solution:
def bitset(self, word):
r = 0
for c in word:
r |= 1 << (ord(c)-97)
return r
def findWords(self, words: List[str]) -> List[str]:
r1 = self.bitset('qwertyuiop')
r2 = self.bitset('asdfghjkl')
r3 = self.bitset('zxcvbnm')
res = []
... | keyboard-row | Python, using bit mask | karbayev | 1 | 103 | keyboard row | 500 | 0.692 | Easy | 8,781 |
https://leetcode.com/problems/keyboard-row/discuss/2836433/Python-Simple-Solution | class Solution:
def findWords(self, words):
inSet = []
topSet = set(['q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p',
'Q', 'W', 'E', 'R', 'T', 'Y', 'U', 'I', 'O', 'P'])
middleSet = set(['a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l',
'A', 'S', 'D', 'F', 'G', 'H', 'J', 'K', ... | keyboard-row | Python - Simple Solution | BladeStew | 0 | 2 | keyboard row | 500 | 0.692 | Easy | 8,782 |
https://leetcode.com/problems/keyboard-row/discuss/2821282/Simple-solution-using-brute-force | class Solution:
def findWords(self, words: List[str]) -> List[str]:
return [z for z in words if sum(all(k.lower() in l for k in z)for l in['qwertyuiop','asdfghjkl','zxcvbnm']) > 0] | keyboard-row | Simple solution using brute force | alex41542 | 0 | 3 | keyboard row | 500 | 0.692 | Easy | 8,783 |
https://leetcode.com/problems/keyboard-row/discuss/2820595/Simple-for-loop | class Solution:
def findWords(self, words: List[str]) -> List[str]:
keyboard = ["qwertyuiop","asdfghjkl","zxcvbnm"]
result = []
for word in words:
if word[0].lower() in keyboard[0]:
idx = 0
elif word[0].lower() in keyboard[1]:
idx = 1
... | keyboard-row | Simple for loop | aruj900 | 0 | 5 | keyboard row | 500 | 0.692 | Easy | 8,784 |
https://leetcode.com/problems/keyboard-row/discuss/2761671/Simple-Python | class Solution:
def findWords(self, words: List[str]) -> List[str]:
row1 = set("qwertyuiop")
row2 = set("asdfghjkl")
row3 = set("zxcvbnm")
ans = []
for word in words:
w = word.lower()
if (all(ch in row1 for ch in w)
or all(ch in row2... | keyboard-row | Simple Python | on_danse_encore_on_rit_encore | 0 | 4 | keyboard row | 500 | 0.692 | Easy | 8,785 |
https://leetcode.com/problems/keyboard-row/discuss/2751887/One-liner-python-using-sets-Easy-solution | class Solution:
def findWords(self, words: List[str]) -> List[str]:
f = set('qwertyuiop')
s = set('asdfghjkl')
t = set('zxcvbnm')
return [word for word in words if len(set(word.lower()) - f) == 0 or len(set(word.lower()) - s) == 0 or len(set(word.lower()) - t) == 0] | keyboard-row | One liner python using sets - Easy solution | jacobsimonareickal | 0 | 3 | keyboard row | 500 | 0.692 | Easy | 8,786 |
https://leetcode.com/problems/keyboard-row/discuss/2744515/Solution-Using-Python | class Solution:
def findWords(self, words: List[str]) -> List[str]:
set1 = {'q','w','e','r','t','y','u','i','o','p'}
set2 = {'a','s','d','f','g','h','j','k','l'}
set3 = {'z','x','c','v','b','n','m'}
res = []
for i in words:
wordset = set(i.lower())
... | keyboard-row | Solution Using Python | dnvavinash | 0 | 5 | keyboard row | 500 | 0.692 | Easy | 8,787 |
https://leetcode.com/problems/keyboard-row/discuss/2520882/Python-oror-easy-and-well-explained-solution | class Solution(object):
def findWords(self, words):
"""
:type words: List[str]
:rtype: List[str]
"""
firstRow = "qwertyuiop"
secondRow = "asdfghjkl"
thirdRow = "zxcvbnm"
result = []
for i in range(len(words)):
add = words[i... | keyboard-row | Python || easy and well explained solution | ride-coder | 0 | 27 | keyboard row | 500 | 0.692 | Easy | 8,788 |
https://leetcode.com/problems/keyboard-row/discuss/2175152/Python-Faster-than-96 | class Solution:
def findWords(self, words: List[str]) -> List[str]:
first_row = set("q w e r t y u i o p Q W E R T Y U I O P".split())
second_row = set("a s d f g h j k l A S D F G H J K L".split())
third_row = set("z x c v b n m Z X C V B N M".split())
results = []
... | keyboard-row | [Python] Faster than 96% | julenn | 0 | 112 | keyboard row | 500 | 0.692 | Easy | 8,789 |
https://leetcode.com/problems/keyboard-row/discuss/2127129/Python-solution | class Solution:
def findWords(self, words: List[str]) -> List[str]:
row1 = list('qwertyuiop')
row2 = list('asdfghjkl')
row3 = list('zxcvbnm')
final_list = []
for word in words:
char_list = list(word.lower())
if all(i in row1 for i in char_list) or all(i in row2 for i in char_list) or... | keyboard-row | Python solution | NiketaM | 0 | 98 | keyboard row | 500 | 0.692 | Easy | 8,790 |
https://leetcode.com/problems/keyboard-row/discuss/2096300/Python3-brute-force | class Solution:
def findWords(self, words: List[str]) -> List[str]:
rows, output = ["qwertyuiop", "asdfghjkl", "zxcvbnm"], []
for word in words:
i, rowI, wordInRow, wLen = 0, 0, 0, len(word)
while True:
if i == wLen:
... | keyboard-row | [Python3] brute force | Shiyinq | 0 | 41 | keyboard row | 500 | 0.692 | Easy | 8,791 |
https://leetcode.com/problems/keyboard-row/discuss/2025722/Python-set-solution | class Solution:
def findWords(self, words: List[str]) -> List[str]:
f = set("qwertyuiop")
s = set("asdfghjkl")
t = set("zxcvbnm")
ans = []
for word in words:
if len(set(word.lower()) | f) == len(f) or len(set(word.lower()) | s) == len(s) or len(set(word.lower()) |... | keyboard-row | Python set solution | StikS32 | 0 | 105 | keyboard row | 500 | 0.692 | Easy | 8,792 |
https://leetcode.com/problems/keyboard-row/discuss/2010751/Easy-and-short-python-solution | class Solution:
def findWords(self, words: List[str]) -> List[str]:
keyRows = ['qwertyusiop', 'asdfghjkl', 'zxcvbnm']
printable = []
for word in words:
for keyRow in keyRows:
if set(word.lower()).issubset(set(keyRow)):
printable.append(word)
... | keyboard-row | Easy and short python solution | ssshekhu53 | 0 | 64 | keyboard row | 500 | 0.692 | Easy | 8,793 |
https://leetcode.com/problems/keyboard-row/discuss/1980907/Python-one-line-faster-than-88.11-(29)-Memory-usage-less-than-97.81-(13.8MB) | class Solution:
def findWords(self, words: list[str]) -> list[str]:
return [i for i in words if all(1 if x in tuple("qwertyuiop") else 0 for x in i.lower()) or all(1 if x in tuple("asdfghjkl") else 0 for x in i.lower()) or all(1 if x in tuple("zxcvbnm") else 0 for x in i.lower())] | keyboard-row | Python one-line faster than 88.11% (29), Memory usage less than 97.81% (13.8MB) | Minh4893IT | 0 | 63 | keyboard row | 500 | 0.692 | Easy | 8,794 |
https://leetcode.com/problems/keyboard-row/discuss/1940151/Python-Clean-and-Simple!-Solution-%2B-One-Liner | class Solution:
def findWords(self, words):
def helper(w, r): return all(c in r for c in set(w.lower()))
r1, r2, r3 = "qwertyuiop", "asdfghjkl", "zxcvbnm"
return [w for w in words if helper(w,r1) or helper(w,r2) or helper(w,r3)] | keyboard-row | Python - Clean and Simple! Solution + One-Liner | domthedeveloper | 0 | 76 | keyboard row | 500 | 0.692 | Easy | 8,795 |
https://leetcode.com/problems/keyboard-row/discuss/1940151/Python-Clean-and-Simple!-Solution-%2B-One-Liner | class Solution:
def findWords(self, words):
def helper(w, r): return all(c in r for c in set(w.lower()))
return [w for w in words if helper(w,"qwertyuiop") or helper(w,"asdfghjkl") or helper(w,"zxcvbnm")] | keyboard-row | Python - Clean and Simple! Solution + One-Liner | domthedeveloper | 0 | 76 | keyboard row | 500 | 0.692 | Easy | 8,796 |
https://leetcode.com/problems/keyboard-row/discuss/1940151/Python-Clean-and-Simple!-Solution-%2B-One-Liner | class Solution:
def findWords(self, words):
return (lambda f : [w for w in words if f(w,"qwertyuiop") or f(w,"asdfghjkl") or f(w,"zxcvbnm")])(lambda w,r : all(c in r for c in set(w.lower()))) | keyboard-row | Python - Clean and Simple! Solution + One-Liner | domthedeveloper | 0 | 76 | keyboard row | 500 | 0.692 | Easy | 8,797 |
https://leetcode.com/problems/keyboard-row/discuss/1826664/python-solution-Siimple | class Solution:
def findWords(self, words: List[str]) -> List[str]:
first_row = "qwertyuiop"
second_row = "asdfghjkl"
third_row = "zxcvbnm"
output = []
# iterate through
for word in words:
# use score to keep track of the different rows
score = [0,... | keyboard-row | python solution- Siimple | Tobi_Akin | 0 | 47 | keyboard row | 500 | 0.692 | Easy | 8,798 |
https://leetcode.com/problems/keyboard-row/discuss/1800960/6-Lines-Python-Solution-oror-60-Faster-(38-ms)-oror-Memory-less-than-90 | class Solution:
def findWords(self, words: List[str]) -> List[str]:
rows, ans = ["qwertyuiop", "asdfghjkl","zxcvbnm"], []
for word in words:
row = [row for row in rows if word[0].lower() in row][0]
for char in list(word):
if char.lower() not in row: ans.append... | keyboard-row | 6-Lines Python Solution || 60% Faster (38 ms) || Memory less than 90% | Taha-C | 0 | 68 | keyboard row | 500 | 0.692 | Easy | 8,799 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.