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 |
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https://leetcode.com/problems/toeplitz-matrix/discuss/2763470/Python!-As-short-as-it-gets! | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
diag = defaultdict(lambda: None)
for x in range(len(matrix)):
for y in range(len(matrix[x])):
if diag[y - x] is None:
diag[y - x] = matrix[x][y]
elif diag[y - x] != matrix[x][y]:
return False
return True | toeplitz-matrix | 😎Python! As short as it gets! | aminjun | 0 | 4 | toeplitz matrix | 766 | 0.688 | Easy | 12,500 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2763262/Python-Solution | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
rows = len(matrix)
cols = len(matrix[0])
for r in range(1, rows):
for c in range(1, cols):
if matrix[r][c] != matrix[r - 1][c - 1]:
return False
return True | toeplitz-matrix | Python Solution | mansoorafzal | 0 | 3 | toeplitz matrix | 766 | 0.688 | Easy | 12,501 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762914/Python-oror-C%2B%2B-oror-Easily-Understood-oror-Fast-oror-Simple | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
n = len(matrix)-1
m = len(matrix[0])-1
for i in range(n):
for j in range(m):
if matrix[i][j] != matrix[i+1][j+1]:
return False
return True | toeplitz-matrix | ✅ Python || C++ || Easily Understood || Fast || Simple 🔥 | Marie-99 | 0 | 9 | toeplitz matrix | 766 | 0.688 | Easy | 12,502 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762878/Superfast-easy-and-mem-efficient-or-Explained-or-Python3 | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
for r in range(len(matrix)):
for c in range(len(matrix[0])):
i,j=r,c
while i+1<len(matrix) and j+1<len(matrix[0]): # go through the diagonal from current element
i+=1 # if yes, move to the next element
j+=1
if matrix[i][j] != matrix[r][c]: return False # if the next element is not equal to the current element, return False
return True | toeplitz-matrix | ✅ Superfast, easy and mem-efficient | Explained | Python3 | arnavjaiswal149 | 0 | 3 | toeplitz matrix | 766 | 0.688 | Easy | 12,503 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762772/Simple-Python-Solution-or-Easy-to-understand | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
r, c = len(matrix), len(matrix[0])
for i in range(r):
for j in range(c):
if 0<=i+1<r and 0<=j+1<c:
if matrix[i+1][j+1] != matrix[i][j]:
return False
return True | toeplitz-matrix | Simple Python Solution | Easy to understand | shreyans | 0 | 6 | toeplitz matrix | 766 | 0.688 | Easy | 12,504 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762688/Python-solution-(double-loops)-less-memory-than-98.45 | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
for i in range(1,len(matrix)):
for j in range(1,len(matrix[0])):
if matrix[i][j] != matrix[i-1][j-1]: return False
return True | toeplitz-matrix | Python solution (double loops) less memory than 98.45% | anandanshul001 | 0 | 4 | toeplitz matrix | 766 | 0.688 | Easy | 12,505 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762674/Easy-Python-Solution-with-Comments-or-Faster-than-93-Solutions-or-Like-if-you-find-it-beneficial-or | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
#let initial status be True
status = True
#iterating through the matrix leaving the last elements and last row
#top rightmost and bottom leftmost element will always satisfy the solution
#rest all the elements unexplored by loop had been already explored by their diagonal elements
for i in range(len(matrix)-1):
for j in range(len(matrix[0])-1):
#set status to False if the diagonal element is not equal
if matrix[i][j]==matrix[i+1][j+1]:
continue
else:
status = False
break
#check if status is false to break out of the loop and prevent any further proceedings
if status == False:
break
#return the status whether it is Toeplitz Matrix or not
return status | toeplitz-matrix | Easy Python Solution with Comments | Faster than 93% Solutions | Like if you find it beneficial | | Yash_A | 0 | 7 | toeplitz matrix | 766 | 0.688 | Easy | 12,506 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762659/Python-Simple-Python-Solution | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
rows = len(matrix)
cols = len(matrix[0])
all_diagonals = []
for row in range(rows):
r,c = row,0
array = []
while r < rows and c < cols:
array.append(matrix[r][c])
r = r + 1
c = c + 1
all_diagonals.append(array)
for col in range(1, cols):
r,c = 0,col
array = []
while r < rows and c < cols:
array.append(matrix[r][c])
r = r + 1
c = c + 1
all_diagonals.append(array)
for diagonal in all_diagonals:
if len(diagonal) != diagonal.count(diagonal[0]):
return False
return True | toeplitz-matrix | [ Python ] ✅✅ Simple Python Solution 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 0 | 8 | toeplitz matrix | 766 | 0.688 | Easy | 12,507 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762601/Python-Soln-using-Dictionary-ofSet | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
diagSet = {} #diagonal:set()
for i in range(len(matrix)):
for j in range(len(matrix[0])):
diagSet[i-j] = set()
for i in range(len(matrix)):
for j in range(len(matrix[0])):
diagSet[i-j].add(matrix[i][j])
for val in diagSet.values():
if len(val) > 1:
return False | toeplitz-matrix | Python Soln - using Dictionary ofSet | logeshsrinivasans | 0 | 5 | toeplitz matrix | 766 | 0.688 | Easy | 12,508 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762566/Python-easy | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
def helper(i,j,ele,matrix):
if(j >= len(matrix[0]) or i >= len(matrix)):
return True
return (matrix[i][j]==ele and helper(i+1,j+1,ele,matrix))
for i in range(len(matrix)):
if(not(helper(i,0,matrix[i][0],matrix))):
return False
for i in range(len(matrix[0])):
if(not(helper(0,i,matrix[0][i],matrix))):
return False
return True | toeplitz-matrix | Python easy | Pradyumankannan | 0 | 3 | toeplitz matrix | 766 | 0.688 | Easy | 12,509 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762510/Python3-or-Row-by-Row-check | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
m, n = len(matrix), len(matrix[0])
for i in range(1, m):
for j in range(1, n):
if matrix[i][j] != matrix[i - 1][j - 1]:
return False
return True | toeplitz-matrix | Python3 | Row by Row check | joshua_mur | 0 | 9 | toeplitz matrix | 766 | 0.688 | Easy | 12,510 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762445/Simple-Python-solution-in-one-loop. | class Solution:
def isToeplitzMatrix(self, matrix: list[list[int]]) -> bool:
height=len(matrix) # 3
width=len(matrix[0]) # 4
result=True
for i in range(1,height):
if matrix[i-1][:width-1] != matrix[i][1:width]:
result=False
return result | toeplitz-matrix | Simple Python solution in one loop. | hbr199320xy | 0 | 3 | toeplitz matrix | 766 | 0.688 | Easy | 12,511 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762236/Python-oror-Easy-oror-Beginners-solution | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
for i in range(len(matrix) - 1):
for j in range(len(matrix[0]) - 1):
if matrix[i][j] != matrix[i + 1][j + 1]:
return False
return True | toeplitz-matrix | Python || Easy || Beginners solution | its_iterator | 0 | 4 | toeplitz matrix | 766 | 0.688 | Easy | 12,512 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762178/Python3-easy-solution-Daily-solutions-in-python3 | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]])->bool:
rows,colms = len(matrix),len(matrix[0])
for r in range (1,rows):
for c in range (1,colms):
if matrix[r][c]!=matrix[r-1][c-1]:
return 0
return 1 | toeplitz-matrix | Python3 easy solution Daily solutions in python3 | rupamkarmakarcr7 | 0 | 2 | toeplitz matrix | 766 | 0.688 | Easy | 12,513 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762127/python-code-with-explanation | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
n=len(matrix) #no. of rows
m=len(matrix[0]) #no. of elements in a row,i.e.,no. of columns
for i in range(n-1):
for j in range(m-1):
if not(matrix[i][j]==matrix[i+1][j+1]):
return False
return True | toeplitz-matrix | python code with explanation | ayushigupta2409 | 0 | 4 | toeplitz matrix | 766 | 0.688 | Easy | 12,514 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2762020/Easy-Matrix-Traversal-or-Python-or-99-Faster | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
n, m = len(matrix), len(matrix[0])
i, j = n-1, 0
while i!=0 or j!=m-1:
i1, j1 = i, j
res = matrix[i1][j1]
while j1>=0 and i1>=0:
if matrix[i1][j1]!=res:
return False
j1-=1; i1-=1
if j<m-1: j+=1
else: i-=1
return True | toeplitz-matrix | Easy Matrix Traversal | Python | 99% Faster | RajatGanguly | 0 | 5 | toeplitz matrix | 766 | 0.688 | Easy | 12,515 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761979/Python-only-loop-through-top-left-point | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
def checkTopLeftToBottomRight(x, y):
i, j = x, y
while i+1 < len(matrix) and j+1 < len(matrix[0]):
i += 1
j += 1
if matrix[i][j] != matrix[x][y]:
return False
return True
for i in range(len(matrix)-1)[::-1]:
if not checkTopLeftToBottomRight(i,0):
return False
for j in range(1, len(matrix[0])-1):
if not checkTopLeftToBottomRight(0,j):
return False
return True | toeplitz-matrix | Python only loop through top left point | zxia545 | 0 | 5 | toeplitz matrix | 766 | 0.688 | Easy | 12,516 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761973/Beat-91-solution-east-to-understand-using-example | class Solution:
def solve(self , matrix ):
count = 0
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if i -1 > -1 and j - 1>-1 and matrix[i - 1][j-1]!=matrix[i][j]:
return False
return True
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
return self.solve(matrix) | toeplitz-matrix | Beat 91% solution , east to understand using example | darkgod | 0 | 4 | toeplitz matrix | 766 | 0.688 | Easy | 12,517 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761948/Easy-Python-or-Faster-or-4-Liner-or-O(n*m)-Time-or-O(1)-Space | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
for i in range(1, len(matrix)):
for j in range(1, len(matrix[0])):
if matrix[i][j]!=matrix[i-1][j-1]:
return False
return True | toeplitz-matrix | Easy Python | Faster | 4 Liner | O(n*m) Time | O(1) Space | coolakash10 | 0 | 1 | toeplitz matrix | 766 | 0.688 | Easy | 12,518 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761875/Python-simple-and-most-readable-code | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
diff = {}
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if(i-j not in diff):
diff[i-j] = matrix[i][j]
else:
if(diff[i-j]!=matrix[i][j]):
return False
else:
continue
return True | toeplitz-matrix | Python simple and most readable code | Jayu79 | 0 | 5 | toeplitz matrix | 766 | 0.688 | Easy | 12,519 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761809/easy-brute-force-solution | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
#calculating diagonals from the first row
for i in range(len(matrix[0])):
cur = matrix[0][i]
j = i+1
t = 1
while j < len(matrix[0]) and t < len(matrix) :
print(matrix[t][j])
if matrix[t][j] != cur:
return False
j += 1
t += 1
#calculating diagonals from the first column
for j in range(1,len(matrix)):
cur = matrix[j][0]
print(cur)
k = j+1
print(k)
t = 1
print(t)
while k < len(matrix) and t < len(matrix[0]) :
if matrix[k][t] != cur:
return False
k += 1
t += 1
return True | toeplitz-matrix | easy brute force solution | neeshumaini55 | 0 | 4 | toeplitz matrix | 766 | 0.688 | Easy | 12,520 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761776/easy-python-solutionoror-easy-to-understandororTime-complexityO(n2) | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
d = {}
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if (i-j) not in d:
d[(i-j)] = matrix[i][j]
else:
if matrix[i][j]!=d[(i-j)]:
return False
return True | toeplitz-matrix | ✅✅easy python solution|| easy to understand||Time complexityO(n2) | chessman_1 | 0 | 4 | toeplitz matrix | 766 | 0.688 | Easy | 12,521 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761766/Python3-Visually-Explained-Beats-91 | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
nrows,ncols = len(matrix),len(matrix[0])
#given the start of the diagonal, check if its univalue diagonal
def unival(start):
x,y = start
while x<nrows and y<ncols:
if matrix[x][y]==matrix[start[0]][start[1]]:
x+=1
y+=1
else:
return False
return True
#find all starting points for diagonals
#along first row and first column
for j in range(ncols):
if not unival((0,j)):
return False
for i in range(nrows):
if not unival((i,0)):
return False
return True | toeplitz-matrix | Python3 Visually Explained Beats 91% | user9611y | 0 | 2 | toeplitz matrix | 766 | 0.688 | Easy | 12,522 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761738/FASTER-THAN-99.79-PYTHON-SOLUTION | class Solution:
def help(self,matrix,i,j,x,n,m):
if i+1>n-1 or j+1>m-1:
return True
if matrix[i+1][j+1]!=x:
return False
else:
return self.help(matrix,i+1,j+1,x,n,m)
return True
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
n=len(matrix)
m=len(matrix[0])
i=0
for j in range(0,m):
x=matrix[i][j]
if self.help(matrix,i,j,x,n,m):
continue
else:
return False
j=0
for i in range(0,n):
x=matrix[i][j]
if self.help(matrix,i,j,x,n,m):
continue
else:
return False
return True | toeplitz-matrix | FASTER THAN 99.79% PYTHON SOLUTION | DG-Problemsolver | 0 | 2 | toeplitz matrix | 766 | 0.688 | Easy | 12,523 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761728/Easy-Python-Solution-or-Loops | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
def answer(r, c):
# print(matrix[r][c])
curr = matrix[r][c]
while r < len(matrix) and c < len(matrix[0]):
if curr != matrix[r][c]:
return False
r += 1
c += 1
return True
r = 0
c = len(matrix[0])-1
while c >= 0:
if not answer(r, c):
return False
c -= 1
c = 0
r += 1
while r < len(matrix):
if not answer(r, 0):
return False
r += 1
return True | toeplitz-matrix | Easy Python Solution | Loops | atharva77 | 0 | 1 | toeplitz matrix | 766 | 0.688 | Easy | 12,524 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761713/Python3-or-Straight-Forward | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
n, m = len(matrix), len(matrix[0])
def traverseDiagonal(x, y):
i, j = x, y
curEle = matrix[i][j]
while i < n and j < m:
if curEle != matrix[i][j]:
return False
i += 1
j += 1
return True
# top row
for r in range(len(matrix[0])):
if not traverseDiagonal(0, r):
return False
# left col
for c in range(1, len(matrix)):
if not traverseDiagonal(c, 0):
return False
return True | toeplitz-matrix | Python3 | Straight Forward | vikinam97 | 0 | 1 | toeplitz matrix | 766 | 0.688 | Easy | 12,525 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761669/Python3-Oneliner | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
return all(matrix[i][:len(matrix[0])-1] == matrix[i+1][1:] for i in range(len(matrix)-1)) | toeplitz-matrix | Python3 Oneliner | anels | 0 | 1 | toeplitz matrix | 766 | 0.688 | Easy | 12,526 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761660/Python-or-super-simple-iteration-O(n)-time-O(1)-space | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
width = len(matrix[0])
height = len(matrix)
r, c = height-1, 0
while r < height and c < width:
current = matrix[r][c]
outer_r, outer_c = r, c
r += 1
c += 1
while r < height and c < width:
if matrix[r][c] != current:
return False
r += 1
c += 1
c = outer_c+1 if outer_r==0 else 0
r = outer_r-1 if outer_r-1>=0 else 0
return True | toeplitz-matrix | Python | super simple iteration O(n) time, O(1) space | chienhsiang-hung | 0 | 6 | toeplitz matrix | 766 | 0.688 | Easy | 12,527 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761631/Easy-to-understand-Python-code-(not-a-one-liner) | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
for i in range(len(matrix)-1):
for j in range(len(matrix[0])-1):
if matrix[i][j]==matrix[i+1][j+1]:
continue
else:
return False
else:
return True | toeplitz-matrix | Easy to understand Python code (not a one-liner) | sarveshebs | 0 | 2 | toeplitz matrix | 766 | 0.688 | Easy | 12,528 |
https://leetcode.com/problems/toeplitz-matrix/discuss/2761592/Fast-and-Simple-Solution | class Solution:
def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
for i in range(1 ,len(matrix)):
for j in range(1, len(matrix[0])):
if matrix[i - 1][j - 1] != matrix[i][j]:
return False
return True | toeplitz-matrix | Fast and Simple Solution | user6770yv | 0 | 1 | toeplitz matrix | 766 | 0.688 | Easy | 12,529 |
https://leetcode.com/problems/reorganize-string/discuss/488325/Python-8-Liner-Memory-usage-less-than-100 | class Solution:
def reorganizeString(self, S: str) -> str:
counter = collections.Counter(S)
i, res, n = 0, [None] * len(S), len(S)
for k in sorted(counter, key = counter.get, reverse = True):
if counter[k] > n // 2 + (n % 2): return ""
for j in range(counter[k]):
if i >= n: i = 1
res[i] = k; i += 2
return "".join(res) | reorganize-string | Python - 8 Liner - Memory usage less than 100% | mmbhatk | 9 | 1,700 | reorganize string | 767 | 0.528 | Medium | 12,530 |
https://leetcode.com/problems/reorganize-string/discuss/741623/Python-heap-easy-solution-with-comments | class Solution:
def reorganizeString(self, S: str) -> str:
if not S: return ''
heap, last, ans = [], None, ''
counts = collections.Counter(S)
for ch in counts:
heapq.heappush(heap, (-counts[ch], ch))
while heap:
count, ch = heapq.heappop(heap)
ans += ch
if last:
heapq.heappush(heap, last)
last = (count+1, ch) if count != -1 else None
return ans if not last else '' | reorganize-string | Python heap easy solution with comments | sexylol | 4 | 229 | reorganize string | 767 | 0.528 | Medium | 12,531 |
https://leetcode.com/problems/reorganize-string/discuss/1930450/Python-Easy-Solution-Max-Heap-and-Counter-Hashmap-with-Comments | class Solution:
def reorganizeString(self, s: str) -> str:
counter = {}
for ch in s:
if ch in counter:
counter[ch] +=1
else:
counter[ch] = 1
queue = []
for elem in counter:
heapq.heappush(queue, (-1*counter[elem], elem))
result = []
prev_char = None
prev_freq = None
while(queue):
freq, elem = heapq.heappop(queue)
result.append(elem)
if prev_freq:
# it makes sure that character whihc is popped just now is not popped next, i.e. maintains non- adjacent condition
heapq.heappush(queue, (prev_freq, prev_char))
prev_char, prev_freq = elem, freq + 1 # +1 is equivalent to -1 as we used max_heap
if len(result) == len(s):
return "".join(result)
return "" | reorganize-string | Python Easy Solution Max Heap and Counter Hashmap with Comments | emerald19 | 2 | 175 | reorganize string | 767 | 0.528 | Medium | 12,532 |
https://leetcode.com/problems/reorganize-string/discuss/1918126/Python-No-Sort-O(N) | class Solution:
def reorganizeString(self, s: str) -> str:
count = Counter(s) # O(N)
most_common_c = max(count.items(), key=itemgetter(1))[0] # O(N)
if count[most_common_c] > (len(s)+1)//2:
return ""
output = ['']*len(s)
i = 0
for _ in range(count[most_common_c]):
output[i] = most_common_c
i += 2
count[most_common_c] = 0
for k, v in count.items():
for _ in range(v):
if i >= len(s):
i = 1
output[i] = k
i += 2
return "".join(output) | reorganize-string | Python, No Sort, O(N) | caseychen008 | 2 | 270 | reorganize string | 767 | 0.528 | Medium | 12,533 |
https://leetcode.com/problems/reorganize-string/discuss/1900082/Python-heap-solution-O(n-log-n) | class Solution:
def reorganizeString(self, s: str) -> str:
last, amountLast = None, None
count = Counter(s)
maxHeap = [(-count[char], char) for char in count.keys()]
heapq.heapify(maxHeap)
res = ""
while maxHeap:
tmp = heapq.heappop(maxHeap)
res += tmp[1]
if amountLast:
heapq.heappush(maxHeap, (amountLast, last))
last, amountLast = tmp[1], tmp[0] + 1
return res if len(res) == len(s) else "" | reorganize-string | Python, heap solution O(n log n) | bchong123 | 2 | 185 | reorganize string | 767 | 0.528 | Medium | 12,534 |
https://leetcode.com/problems/reorganize-string/discuss/2369932/simple-python-heap | class Solution:
def reorganizeString(self, S):
h, res, c = [], [], Counter(S)
for i in c:
heappush(h, (-c[i], i))
p_a, p_b = 0, ''
while h:
a, b = heapq.heappop(h)
res += [b]
if p_a < 0:
heapq.heappush(h, (p_a, p_b))
a += 1
p_a, p_b = a, b
res = ''.join(res)
if len(res) != len(S): return ""
return res | reorganize-string | simple python heap | gasohel336 | 1 | 95 | reorganize string | 767 | 0.528 | Medium | 12,535 |
https://leetcode.com/problems/reorganize-string/discuss/1843491/Python-easy-to-read-and-understand-or-max-heap | class Solution:
def reorganizeString(self, s: str) -> str:
d = {}
for i in s:
d[i] = d.get(i, 0) + 1
pq = []
for key in d:
heapq.heappush(pq, (-d[key], key))
res = ''
cnt, ch = heapq.heappop(pq)
res += ch
block = (cnt+1, ch)
while len(pq) > 0:
cnt, ch = heapq.heappop(pq)
res += ch
if block[0] < 0:
heapq.heappush(pq, block)
block = (cnt+1, ch)
if len(res) != len(s):
return ''
else:
return res | reorganize-string | Python easy to read and understand | max-heap | sanial2001 | 1 | 262 | reorganize string | 767 | 0.528 | Medium | 12,536 |
https://leetcode.com/problems/reorganize-string/discuss/1437851/Python-solution-98-Using-Dictionary-O(n) | class Solution:
def reorganizeString(self, s: str) -> str:
arr,res = Counter(s),""
arr = {k:v for k,v in sorted(arr.items(),key = lambda v:v[1], reverse=True)}
an = ""
m = max(arr.values())
for k,v in arr.items():
an+=k*v
#print(an,m)
if len(s)%2==0:
if m <=len(s)//2:
for i,j in zip(an[:len(s)//2],an[len(s)//2:]):
res+=i
res+=j
else:
if m <= (len(s)//2) + 1:
for i,j in zip(an[:(len(s)//2) + 1],an[(len(s)//2)+1:]):
res+=i
res+=j
res+=an[len(s)//2]
#print(res)
return "".join(res)
# solution By Rajeev Mohan | reorganize-string | Python solution 98% Using Dictionary, O(n) | rstudy211 | 1 | 483 | reorganize string | 767 | 0.528 | Medium | 12,537 |
https://leetcode.com/problems/reorganize-string/discuss/1399898/Python-16ms-faster-than-100 | class Solution:
def reorganizeString(self, s: str) -> str:
# count the character in the string
charMap = {}
for i in range(len(s)):
if s[i] in charMap:
charMap[s[i]] += 1
else:
charMap[s[i]] = 1
# sort the map by value - the character appears the most is at index 0
charMap = {k: v for k, v in sorted(charMap.items(), key=lambda x: x[1], reverse=True)}
firstMap = list(charMap.items())[0] # character appears the most
if len(s) <= (firstMap[1]-1)*2:
return ""
else:
# create a string contains of different "phrase"
# each character will be distinct in each phrase
newstr = [firstMap[0] for _ in range(firstMap[1])]
index = 0
charMap.pop(firstMap[0]) # the initialize value already contains the most appeared character
for k, v in charMap.items():
for _ in range(v):
if index == len(newstr):
index = 0 # reset index counter
newstr[index] += str(k)
index += 1
return "".join(newstr) | reorganize-string | Python 16ms faster than 100% | percy_pham | 1 | 297 | reorganize string | 767 | 0.528 | Medium | 12,538 |
https://leetcode.com/problems/reorganize-string/discuss/2845254/super-easy-solution-for-beginners-using-hashmap-in-**python** | class Solution:
def reorganizeString(self, s: str) -> str:
wc = {}
lenn = 0
for letter in s:
wc[letter] = wc.get(letter,0)+1
lenn += 1
if 2 * max(wc.values()) > sum(wc.values()) + 1:
return ""
else:
# sorting the word counter based on frequency in decending order
wc = dict(sorted(wc.items(),key=lambda x:x[1],reverse=True))
bs = ["" for i in range(lenn)]
# filling the odd places
for i in range(0,lenn,2):
for key in wc.keys():
if wc[key]:
bs[i] = key
wc[key] -= 1
break
# filling the even places
for i in range(1,lenn,2):
for key in wc.keys():
if wc[key]:
bs[i] = key
wc[key] -= 1
break
return "".join(bs) | reorganize-string | super easy solution for beginners using hashmap in **python** | experimentallyf | 0 | 4 | reorganize string | 767 | 0.528 | Medium | 12,539 |
https://leetcode.com/problems/reorganize-string/discuss/2830013/Python-hash-map-and-max-heap | class Solution:
def reorganizeString(self, s: str) -> str:
count = Counter(s)
maxHeap = [[-freq, c] for c, freq in count.items()]
heapq.heapify(maxHeap)
prev = None
res = ""
while maxHeap or prev:
if prev and not maxHeap:
return ""
freq, c = heapq.heappop(maxHeap)
res += c
freq += 1
if prev:
heapq.heappush(maxHeap, prev)
prev = None
if freq != 0:
prev = [freq, c]
return res | reorganize-string | Python hash map and max heap | zananpech9 | 0 | 4 | reorganize string | 767 | 0.528 | Medium | 12,540 |
https://leetcode.com/problems/reorganize-string/discuss/2777897/Python%3A-Heap-Sort-(Prority-Queue)-%2B-Hashmap-oror-Time-Complexity-O(nlogn) | class Solution:
def reorganizeString(self, s: str) -> str:
d={}
for i in range(len(s)):
curr=s[i]
if curr in d.keys():
d[curr]+=1
else:
d[curr]=1
m=[]
for i,j in d.items():
heapq.heappush(m,[-j,i])
s1=""
while (len(m)>=2):
a=heapq.heappop(m)
b=heapq.heappop(m)
s1=s1+a[1]+b[1]
a[0]=a[0]+1
b[0]=b[0]+1
if a[0]!=0:
heapq.heappush(m,a)
if b[0]!=0:
heapq.heappush(m,b)
if len(m)==1:
a=heapq.heappop(m)
if (a[0]*-1)>1:
return ""
else:
s1=s1+a[1]
return s1 | reorganize-string | Python: Heap Sort (Prority Queue) + Hashmap || Time Complexity= O(nlogn) | utsa_gupta | 0 | 13 | reorganize string | 767 | 0.528 | Medium | 12,541 |
https://leetcode.com/problems/reorganize-string/discuss/2764818/Python3-using-a-dictionary-for-character-frequency-(99) | class Solution:
def reorganizeString(self, s: str) -> str:
char_dict = {}
for c in s:
if c not in char_dict:
char_dict[c] = 1
else:
char_dict[c] += 1
if 2*max(char_dict.values()) > len(s) + 1:
return ''
char_list = []
max_char_val = 0
old = ''
new = ''
for _ in s:
new = list(char_dict.keys())[list(char_dict.values()).index(max(char_dict.values()))] #get char with max no. of occurrences from dict
if old:
char_dict[old] = max_char_val - 1 #reset the previous max char's value back to what it should be, minus one for using it
max_char_val = char_dict[new] #store the no. occurrences of the char with max occurrences
char_dict[new] = 0 #temporary devaluation to prevent repetition
char_list.append(new) #add the character to the list
old = new
rearranged_string = ''.join(char_list)
return rearranged_string | reorganize-string | Python3 using a dictionary for character frequency (99%) | Kenpari | 0 | 9 | reorganize string | 767 | 0.528 | Medium | 12,542 |
https://leetcode.com/problems/reorganize-string/discuss/2704330/Python-O(NlogN)-O(1) | class Solution:
def reorganizeString(self, s: str) -> str:
counter = collections.Counter(s)
heap = [(-counter[c], c) for c in counter]
heapq.heapify(heap)
res = []
lastOcc, lastChar = 0, ""
while heap:
tempOcc, tempChar = lastOcc, lastChar
lastOcc, lastChar = heapq.heappop(heap)
if tempOcc < 0:
heapq.heappush(heap, (tempOcc, tempChar))
res.append(lastChar)
lastOcc += 1
if lastOcc < 0:
return ""
return "".join(res) | reorganize-string | Python - O(NlogN), O(1) | Teecha13 | 0 | 19 | reorganize string | 767 | 0.528 | Medium | 12,543 |
https://leetcode.com/problems/reorganize-string/discuss/1825832/Two-methods%3A-python-heapq-and-greedy | class Solution:
def reorganizeString(self, s: str) -> str:
n = len(s)
s = Counter(s)
max_l = s.most_common(1)[0][1]
# if the number of any letter larger then the haft length (or plus1 when it is even) of the string
# impossible to reorganize
if (n+1)//2<max_l:
return ""
# using odd/even insert
oddIndex=0
evenIndex=1
out=[""]*n # create slot
for k,v in s.most_common():
while v and oddIndex<n: # fill the most letter in all odd places
out[oddIndex] = k
oddIndex+=2
v-=1
while v: # then fill the remain in evenIndex
out[evenIndex] = k
evenIndex+=2
v-=1
return "".join(out)
# using heapq
q = []
out=""
for k,v in s.items():
heapq.heappush(q,(-v,k))
while len(q)>1:
v,l=heapq.heappop(q)
if not len(out):
out+=l
if v+1<0:
heapq.heappush(q,(v+1,l))
elif l!=out[-1]:
out+=l
if v+1<0:
heapq.heappush(q,(v+1,l))
else:
v2,l2=heapq.heappop(q)
out+=l2
if v2+1<0:
heapq.heappush(q,(v2+1,l2))
heapq.heappush(q,(v,l))
if q[0][0]<-1:
return ""
else:
return out+q[0][1] | reorganize-string | Two methods: python heapq and greedy | cooldog6026 | 0 | 116 | reorganize string | 767 | 0.528 | Medium | 12,544 |
https://leetcode.com/problems/reorganize-string/discuss/1771697/Python-no-sort.-Time%3A-O(N).-Space%3A-O(N) | class Solution:
def reorganizeString(self, s: str) -> str:
h = [(-count, char) for char, count in Counter(s).items()]
heapq.heapify(h)
if -h[0][0] > (len(s) + 1) // 2:
return ""
idx = 0
result = [None] * len(s)
for count, char in h:
for _ in range(-count):
result[idx] = char
idx += 2
if idx >= len(s):
idx = 1
return "".join(result) | reorganize-string | Python, no sort. Time: O(N). Space: O(N) | blue_sky5 | 0 | 240 | reorganize string | 767 | 0.528 | Medium | 12,545 |
https://leetcode.com/problems/reorganize-string/discuss/1687901/Simple-python-solution-with-Counter-API | class Solution:
def reorganizeString(self, s: str) -> str:
s_counter = Counter(s)
def remove(s_counter, char):
s_counter[char] -= 1
if s_counter[char] == 0:
s_counter.pop(char)
def take(s_counter, adj_char):
for ch in s_counter.most_common(2):
if ch[0][0] == adj_char:
continue
return ch[0][0]
return ""
new_s = s_counter.most_common(1)[0][0]
remove(s_counter, new_s)
for i in range(1, len(s)):
new_char = take(s_counter, new_s[-1])
if not new_char: return ""
new_s += new_char
remove(s_counter, new_char)
return new_s | reorganize-string | Simple python solution with Counter API | Arvindn | 0 | 143 | reorganize string | 767 | 0.528 | Medium | 12,546 |
https://leetcode.com/problems/reorganize-string/discuss/1670521/Using-Max-heap-in-python-with-example | class Solution:
import heapq
def reorganizeString(self, s: str) -> str:
d={}
for i in s:
if i in d:
d[i]+=1
else:
d[i]=1
l=[]
for i in d:
# -ive sign is used beacuse there in no max-heap in python
x=[-d[i],i]
l.append(x)
# l will look like this l=[[-2, a],[-3, b],[-1, c]]
heapq.heapify(l)
ans=""
while len(l)>1:
m=heapq.heappop(l)
n=heapq.heappop(l)
ans+=m[1]+n[1]
# I am adding here is same as decreasing the count because the count here is -ive
# If the count is 0 than no need to push
if m[0]+1!=0:
heapq.heappush(l,[m[0]+1,m[1]])
if n[0]+1!=0:
heapq.heappush(l,[n[0]+1,n[1]])
if l==[]:
return ans
# if any character count is more than 1 left than its a not possible case
if (l[0][0]*-1)==1:
x=l.pop()
ans+=x[1]
return ans
else:
return "" | reorganize-string | Using Max-heap in python with example | gamitejpratapsingh998 | 0 | 180 | reorganize string | 767 | 0.528 | Medium | 12,547 |
https://leetcode.com/problems/reorganize-string/discuss/1512760/Python3-Solution-with-using-heap | class Solution:
def reorganizeString(self, s: str) -> str:
counter = collections.Counter(s)
l = [""] * len(s)
heap = []
for key in counter:
heapq.heappush(heap, [-counter[key], key])
idx = 0
while len(heap) > 0:
symb_count, symb = heapq.heappop(heap)
symb_count *= -1
while symb_count > 0:
l[idx] = symb
symb_count -= 1
if idx > 0 and l[idx] == l[idx - 1]:
return ""
if idx + 2 >= len(l):
idx = -1
idx += 2
return "".join(l) | reorganize-string | [Python3] Solution with using heap | maosipov11 | 0 | 127 | reorganize string | 767 | 0.528 | Medium | 12,548 |
https://leetcode.com/problems/reorganize-string/discuss/1193641/queue-based-python-sol | class Solution:
def reorganizeString(self, string: str) -> str:
n=len(string)
dict=Counter(string)
dict=sorted(dict.items(),key=lambda x:-x[1])
print(dict)
new=[]
for i,j in dict:
for k in range(j):
new.append(i)
ans=[0]*n
queue=[]
for i in range(0,n):
queue.append(i)
for i in new:
l=queue[0]
if l==0 and ans[l+1]!=i:
ans[queue.pop(0)]=i
if queue:
queue.append(queue.pop(0))
elif l==n-1 and ans[l-1]!=i:
ans[queue.pop(0)]=i
if queue:
queue.append(queue.pop(0))
elif 0<l<n-1 and ans[l-1]!=i and ans[l+1]!=i:
ans[queue.pop(0)]=i
if queue:
queue.append(queue.pop(0))
else:
return ""
return "".join(ans) | reorganize-string | queue based python sol | heisenbarg | 0 | 142 | reorganize string | 767 | 0.528 | Medium | 12,549 |
https://leetcode.com/problems/reorganize-string/discuss/776668/Python-Max-Heap | class Solution:
def reorganizeString(self, S: str) -> str:
if not S:
return ""
elif len(S) == 1:
return S
# Initialize max heap
counts, res, heap = Counter(S), [], []
for char in counts:
heapq.heappush(heap, (-counts[char], char))
while len(heap) > 1:
# Get next two chars...
count1, char1 = heapq.heappop(heap)
count2, char2 = heapq.heappop(heap)
# Update counts
count1 += 1
count2 += 1
# Add to result
res.extend([char1, char2])
# Add chars back to heap if we have not used them all...
if count1 < 0:
heapq.heappush(heap, (count1, char1))
if count2 < 0:
heapq.heappush(heap, (count2, char2))
# If we cannot alternate this char...
if len(heap) > 0 and abs(heap[0][0]) > 1:
return ""
# Otherwise add last char to result string
elif len(heap) > 0:
res.append(heap[0][1])
return "".join(res) | reorganize-string | Python Max Heap | zuu | 0 | 108 | reorganize string | 767 | 0.528 | Medium | 12,550 |
https://leetcode.com/problems/reorganize-string/discuss/760848/Python3-via-heap | class Solution:
def reorganizeString(self, S: str) -> str:
#frequency table
freq = dict()
for c in S: freq[c] = 1 + freq.get(c, 0)
#max heap
hp = [(-v, k) for k, v in freq.items()]
heapify(hp)
i = 0
ans = [""]*len(S)
while hp:
v, k = heappop(hp)
v = -v
if 2*v - 1 > len(S): return "" #impossible
for _ in range(v):
ans[i] = k
i = i+2 if i+2 < len(S) else 1
return "".join(ans) | reorganize-string | [Python3] via heap | ye15 | 0 | 72 | reorganize string | 767 | 0.528 | Medium | 12,551 |
https://leetcode.com/problems/reorganize-string/discuss/760848/Python3-via-heap | class Solution:
def reorganizeString(self, S: str) -> str:
freq = {}
for c in S: freq[c] = 1 +freq.get(c, 0) # frequency table
ans = [""]*len(S)
i = 0
for k in sorted(freq, reverse=True, key=freq.get):
if 2*freq[k] - 1 > len(S): return "" # impossible
for _ in range(freq[k]):
ans[i] = k
i = i+2 if i+2 < len(S) else 1 # reset to 1
return "".join(ans) | reorganize-string | [Python3] via heap | ye15 | 0 | 72 | reorganize string | 767 | 0.528 | Medium | 12,552 |
https://leetcode.com/problems/reorganize-string/discuss/449996/Easy-to-understand-Python3-solution | class Solution:
def reorganizeString(self, S: str) -> str:
from collections import Counter
c=Counter(S)
if(len(c.keys())==1):
return ''
else:
l=list(c.keys())
l.sort(key=lambda x:c[x])
v=[c[i] for i in l]
s=''
for i in range(len(l)-1):
x=l[i+1]+l[i]
s+=x*v[i]
v[i+1]-=v[i]
if(v[-1]>0):
s+=l[-1]
v[-1]-=1
ans=''
if(v[-1]>0):
for i in range(len(s)):
if(s[i]!=l[-1] and v[-1]>0):
ans+=(l[-1]+s[i])
v[-1]-=1
continue
elif(v[-1]==0):
ans+=s[i:]
return ans
elif(s[i]==l[-1] and v[-1]!=0):
return ''
return s | reorganize-string | Easy to understand Python3 solution | Banra123 | 0 | 138 | reorganize string | 767 | 0.528 | Medium | 12,553 |
https://leetcode.com/problems/reorganize-string/discuss/425376/Python3-heap-solution-24ms-beat-99 | class Solution:
def reorganizeString(self, S: str) -> str:
counted=collections.Counter(S)
heap=[]
result=[]
last=None #record the last placed node
for key in counted:
heapq.heappush(heap,[-1*counted[key],key]) #biggest heap with characters ranked by occuring times
while(heap):
node=heapq.heappop(heap) #take the most occuring cha out and hidding it for one time to avoid be adjacent to itself
result.append(node[1])
node[0]+=1 #the occuring times -1
if last and last[0]<0: #put hidded cha back to the heap
heapq.heappush(heap,last)
last=node # the hidded cha
if len(result)==len(S):
return "".join(result)
else:
return "" | reorganize-string | Python3 heap solution 24ms beat 99% | wangzi100 | 0 | 127 | reorganize string | 767 | 0.528 | Medium | 12,554 |
https://leetcode.com/problems/reorganize-string/discuss/405960/Python-3-(beats-~100)-(seven-lines) | class Solution:
def reorganizeString(self, S: str) -> str:
L, A, m, B = len(S), [0]*len(S), 1-len(S) % 2, []
for k,v in collections.Counter(S).most_common(): B += k*v
for i,c in enumerate(B):
I = (2*i + (2*i >= L)*m) % L
A[I] = c
if I != 0 and A[I-1] == c: return ''
return "".join(A)
- Junaid Mansuri | reorganize-string | Python 3 (beats ~100%) (seven lines) | junaidmansuri | -1 | 684 | reorganize string | 767 | 0.528 | Medium | 12,555 |
https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/1498733/Easy-to-Understand-oror-98-faster-oror-Well-Explained | class Solution:
def maxChunksToSorted(self, nums: List[int]) -> int:
st = []
for n in nums:
if len(st)==0 or st[-1]<=n:
st.append(n)
else:
ma = st[-1]
while st and st[-1]>n:
ma = max(ma,st.pop())
st.append(ma)
return len(st) | max-chunks-to-make-sorted-ii | 📌📌 Easy-to-Understand || 98% faster || Well-Explained 🐍 | abhi9Rai | 14 | 659 | max chunks to make sorted ii | 768 | 0.528 | Hard | 12,556 |
https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/1419866/Python-Solution-Monotonic-queue-O(N) | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
S = []
for a in arr:
min_i, max_i = a, a
while S and a < S[-1][1]:
start, end = S.pop()
min_i, max_i = min(start, min_i), max(end, max_i)
S.append([min_i, max_i])
return len(S) | max-chunks-to-make-sorted-ii | Python Solution - Monotonic queue O(N) | ahujara | 1 | 138 | max chunks to make sorted ii | 768 | 0.528 | Hard | 12,557 |
https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/1342227/Python3-greedy | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
mn = [inf]*(1 + len(arr))
for i in reversed(range(len(arr))): mn[i] = min(arr[i], mn[i+1])
ans = mx = 0
for i, x in enumerate(arr):
mx = max(mx, x)
if mx <= mn[i+1]: ans += 1
return ans | max-chunks-to-make-sorted-ii | [Python3] greedy | ye15 | 1 | 115 | max chunks to make sorted ii | 768 | 0.528 | Hard | 12,558 |
https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/1342227/Python3-greedy | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
stack = []
for i, x in enumerate(arr):
most = x
while stack and stack[-1] > x: most = max(most, stack.pop())
stack.append(most)
return len(stack) | max-chunks-to-make-sorted-ii | [Python3] greedy | ye15 | 1 | 115 | max chunks to make sorted ii | 768 | 0.528 | Hard | 12,559 |
https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/2833062/Python-solution-using-array-O(n) | class Solution:
def maxChunksToSorted(self, arr: list[int]) -> int:
max_chunk = 0
min_left_i = [float('inf')]*(len(arr)+1)
curr_min = math.inf
# find the
for i in range(len(arr)-1, -1, -1):
if arr[i] < curr_min:
curr_min = arr[i]
min_left_i[i] = curr_min
max_till_i = arr[0]
for i in range(len(arr)):
if arr[i] > max_till_i:
max_till_i = arr[i]
# check to verify if there is no element to the right which is lower than the current.
if max_till_i <= min_left_i[i+1]:
max_chunk += 1
return max_chunk | max-chunks-to-make-sorted-ii | Python solution using array O(n) | runForrest | 0 | 3 | max chunks to make sorted ii | 768 | 0.528 | Hard | 12,560 |
https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/2515259/Python3-or-Stack-Approach | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
stack=[]
for v in arr:
maxVal=-float('inf')
while stack and stack[-1]>v:
maxVal=max(maxVal,stack.pop())
stack.append(max(v,maxVal))
return len(stack) | max-chunks-to-make-sorted-ii | [Python3] | Stack Approach | swapnilsingh421 | 0 | 59 | max chunks to make sorted ii | 768 | 0.528 | Hard | 12,561 |
https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/2040797/python3-iterate-backwards | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
n=len(arr)
que=[]
res=0
for i in range(n-1,-1,-1):
if not que or arr[i]<=que[0]:
que.insert(0,arr[i])
res+=1
else:
a=bisect_left(que,arr[i])
if len(que)-1-a<0:
que=que[0:1]
res=1
elif a-1!=0:
res=len(que)-a+1
que=que[0:1]+que[a:]
return res | max-chunks-to-make-sorted-ii | python3 iterate backwards | appleface | 0 | 23 | max chunks to make sorted ii | 768 | 0.528 | Hard | 12,562 |
https://leetcode.com/problems/max-chunks-to-make-sorted/discuss/1579999/Python3-Solution-with-using-stack | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
stack = []
for num in arr:
lagest = num
while stack and num < stack[-1]:
lagest = max(lagest, stack.pop())
stack.append(lagest)
return len(stack) | max-chunks-to-make-sorted | [Python3] Solution with using stack | maosipov11 | 1 | 95 | max chunks to make sorted | 769 | 0.582 | Medium | 12,563 |
https://leetcode.com/problems/max-chunks-to-make-sorted/discuss/1407579/Python-3-easy-solution | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
temp_sum, block = 0, 0
for i in range(len(arr)):
temp_sum += arr[i] - i
if temp_sum == 0:
block += 1
return block | max-chunks-to-make-sorted | Python 3 easy solution | Andy_Feng97 | 1 | 64 | max chunks to make sorted | 769 | 0.582 | Medium | 12,564 |
https://leetcode.com/problems/max-chunks-to-make-sorted/discuss/2046530/python-3-oror-greedy-solution-oror-O(n)(1) | class Solution:
def maxChunksToSorted(self, nums: List[int]) -> int:
chunks = 0
left = 0
n = len(nums)
smallest, largest = n, -1
for right, num in enumerate(nums):
smallest = min(smallest, num)
largest = max(largest, num)
if left <= smallest and largest <= right:
chunks += 1
left = right + 1
smallest, largest = n, -1
return chunks | max-chunks-to-make-sorted | python 3 || greedy solution || O(n)/(1) | dereky4 | 0 | 82 | max chunks to make sorted | 769 | 0.582 | Medium | 12,565 |
https://leetcode.com/problems/max-chunks-to-make-sorted/discuss/923484/Python3-greedy | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
ans = prefix = 0
for i, x in enumerate(arr):
prefix = max(prefix, x)
if i == prefix: ans += 1
return ans | max-chunks-to-make-sorted | [Python3] greedy | ye15 | 0 | 71 | max chunks to make sorted | 769 | 0.582 | Medium | 12,566 |
https://leetcode.com/problems/max-chunks-to-make-sorted/discuss/526573/Python3-20ms-96-solution | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
start, max_value, chunks = 0, 0, 0
for index, value in enumerate(arr):
max_value = max(value, max_value)
if max_value - start == index - start:
chunks += 1
start = index + 1
return chunks | max-chunks-to-make-sorted | Python3 20ms 96% solution | tjucoder | 0 | 61 | max chunks to make sorted | 769 | 0.582 | Medium | 12,567 |
https://leetcode.com/problems/max-chunks-to-make-sorted/discuss/1410648/Python-3-oror-Easy-understanding-oror-self-understandable | class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
largest_chunk=0
start=0
for i in range(len(arr)):
visited=[False]*len(arr)
for j in range(start,i+1):
visited[arr[j]]=True
if all(visited[start:i+1]):
largest_chunk+=1
start=i+1
return largest_chunk | max-chunks-to-make-sorted | Python 3 || Easy-understanding || self-understandable | bug_buster | -2 | 186 | max chunks to make sorted | 769 | 0.582 | Medium | 12,568 |
https://leetcode.com/problems/jewels-and-stones/discuss/362196/Solution-in-Python-3-(beats-~94)-(one-line)-(three-solutions) | class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
return sum(i in J for i in S)
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
return sum(S.count(i) for i in J)
from collections import Counter
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
return sum(Counter(S)[i] for i in J)
- Junaid Mansuri
(LeetCode ID)@hotmail.com | jewels-and-stones | Solution in Python 3 (beats ~94%) (one line) (three solutions) | junaidmansuri | 28 | 4,500 | jewels and stones | 771 | 0.881 | Easy | 12,569 |
https://leetcode.com/problems/jewels-and-stones/discuss/1016607/Easy-and-Clear-Solution-Python-3 | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
occ=dict()
for i in stones:
if i in occ.keys():
occ[i]+=1
else:
occ.update({i:1})
res=0
for i in jewels:
if i in occ.keys():
res+=occ[i]
return res | jewels-and-stones | Easy & Clear Solution Python 3 | moazmar | 5 | 432 | jewels and stones | 771 | 0.881 | Easy | 12,570 |
https://leetcode.com/problems/jewels-and-stones/discuss/1016607/Easy-and-Clear-Solution-Python-3 | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
res=0
for i in stones:
if i in set(jewels):
res+=1
return res | jewels-and-stones | Easy & Clear Solution Python 3 | moazmar | 5 | 432 | jewels and stones | 771 | 0.881 | Easy | 12,571 |
https://leetcode.com/problems/jewels-and-stones/discuss/1804792/Python-3-Simple-solution | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
stones = list(stones)
jewels = list(jewels)
match = len([i for i in stones if i in jewels])
return match | jewels-and-stones | [Python 3] Simple solution | IvanTsukei | 4 | 147 | jewels and stones | 771 | 0.881 | Easy | 12,572 |
https://leetcode.com/problems/jewels-and-stones/discuss/1804792/Python-3-Simple-solution | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
return len([i for i in list(stones) if i in list(jewels)]) | jewels-and-stones | [Python 3] Simple solution | IvanTsukei | 4 | 147 | jewels and stones | 771 | 0.881 | Easy | 12,573 |
https://leetcode.com/problems/jewels-and-stones/discuss/1047701/Python-1-liner | class Solution:
def numJewelsInStones(self, j: str, s: str) -> int:
return sum(i in j for i in s) | jewels-and-stones | Python 1-liner | lokeshsenthilkumar | 4 | 385 | jewels and stones | 771 | 0.881 | Easy | 12,574 |
https://leetcode.com/problems/jewels-and-stones/discuss/2165889/Python3-O(j%2Bs)-oror-O(1)-Runtime%3A-27ms-97.69-memory%3A-10mb-10.87 | class Solution:
# O(j+s) where j is jewels and s is stones
# memory: O(1) we will english letters.
# Runtime: 27ms 97.69% memory: 10mb 10.87%
def numJewelsInStones(self, jewels: str, stones: str) -> int:
jewelsMap = dict()
for i in jewels:
jewelsMap[i] = jewelsMap.get(i, 0) + 1
oldSum = sum(jewelsMap.values())
for i in stones:
if i in jewelsMap:
jewelsMap[i] = jewelsMap.get(i, 0) + 1
return sum(jewelsMap.values()) - oldSum | jewels-and-stones | Python3 O(j+s) || O(1) Runtime: 27ms 97.69% memory: 10mb 10.87% | arshergon | 3 | 124 | jewels and stones | 771 | 0.881 | Easy | 12,575 |
https://leetcode.com/problems/jewels-and-stones/discuss/1273148/or-One-line-or-Python-3-or-95.69-faster-or-Using-count()-or | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
return sum([stones.count(l) for l in jewels if l in stones]) | jewels-and-stones | | One line | Python 3 | 95.69% faster | Using count() | | anotherprogramer | 2 | 125 | jewels and stones | 771 | 0.881 | Easy | 12,576 |
https://leetcode.com/problems/jewels-and-stones/discuss/1171051/Simple-Python-Solutions%3A-Multiple-Approaches_O(n%2Bm) | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
if len(jewels)==0 or len(stones)==0:
return 0
jewels = set(jewels)
j_count = 0
for s in stones:
if s in jewels:
j_count+=1
return j_count | jewels-and-stones | Simple Python Solutions: Multiple Approaches_O(n+m) | smaranjitghose | 2 | 181 | jewels and stones | 771 | 0.881 | Easy | 12,577 |
https://leetcode.com/problems/jewels-and-stones/discuss/1171051/Simple-Python-Solutions%3A-Multiple-Approaches_O(n%2Bm) | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
if len(jewels)==0 or len(stones)==0:
return 0
j_count = 0
for j in jewels:
j_count+=stones.count(j)
return j_count | jewels-and-stones | Simple Python Solutions: Multiple Approaches_O(n+m) | smaranjitghose | 2 | 181 | jewels and stones | 771 | 0.881 | Easy | 12,578 |
https://leetcode.com/problems/jewels-and-stones/discuss/1171051/Simple-Python-Solutions%3A-Multiple-Approaches_O(n%2Bm) | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
if len(jewels)==0 or len(stones)==0:
return 0
return sum([1 for s in stones if s in jewels]) | jewels-and-stones | Simple Python Solutions: Multiple Approaches_O(n+m) | smaranjitghose | 2 | 181 | jewels and stones | 771 | 0.881 | Easy | 12,579 |
https://leetcode.com/problems/jewels-and-stones/discuss/2752400/Python-or-Runtime-beats-99.00-of-submitted-solutions-or-Explained | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
target = 0
jewels = sorted(jewels)
for jewel in jewels:
if jewel in stones:
target += stones.count(jewel)
return target | jewels-and-stones | Python | Runtime beats 99.00 % of submitted solutions | Explained | sahil193101 | 1 | 16 | jewels and stones | 771 | 0.881 | Easy | 12,580 |
https://leetcode.com/problems/jewels-and-stones/discuss/2582286/Pythonoror2-line-codeororList-comprehension | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
st=Counter(stones)
return sum([st[j] for j in jewels if j in st]) | jewels-and-stones | Python||2-line code||List comprehension | shersam999 | 1 | 58 | jewels and stones | 771 | 0.881 | Easy | 12,581 |
https://leetcode.com/problems/jewels-and-stones/discuss/2148998/hashmap | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
# use a hashmap to map stones to their freq
# iterate jewels as keys
# add the key loopup to a result
# use get to void a KeyError
# Time: O(n + m) Space: O(n)
d = Counter(stones)
res = 0
for k in jewels:
res += d.get(k, 0)
return res | jewels-and-stones | hashmap | andrewnerdimo | 1 | 50 | jewels and stones | 771 | 0.881 | Easy | 12,582 |
https://leetcode.com/problems/jewels-and-stones/discuss/1863207/Python-easy-solution-faster-than-95 | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
count = 0
for i in jewels:
count += stones.count(i)
return count | jewels-and-stones | Python easy solution faster than 95% | alishak1999 | 1 | 122 | jewels and stones | 771 | 0.881 | Easy | 12,583 |
https://leetcode.com/problems/jewels-and-stones/discuss/1796626/python-3-or-simple-solution | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
c = 0
for i in stones:
if i in jewels:
c += 1
return c | jewels-and-stones | ✔python 3 | simple solution | Coding_Tan3 | 1 | 47 | jewels and stones | 771 | 0.881 | Easy | 12,584 |
https://leetcode.com/problems/jewels-and-stones/discuss/1733625/771-JEWELS-and-STONES-PYTHON | class Solution(object):
def numJewelsInStones(self, jewels, stones):
jewel_found = 0
for each in jewels:
for stone in stones:
if each == stone:
jewel_found += 1
return jewel_found | jewels-and-stones | 771 JEWELS & STONES PYTHON | ankit61d | 1 | 64 | jewels and stones | 771 | 0.881 | Easy | 12,585 |
https://leetcode.com/problems/jewels-and-stones/discuss/1212973/Python-Beating-99-in-Time | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
d = {}
for i in stones:
if i not in d:
d[i] = 1
else:
d[i]+=1
ans = 0
for i in jewels:
if i in d:
ans+=d[i]
return ans | jewels-and-stones | Python Beating 99% in Time | iamkshitij77 | 1 | 95 | jewels and stones | 771 | 0.881 | Easy | 12,586 |
https://leetcode.com/problems/jewels-and-stones/discuss/1122545/one-liner-28ms-easy | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
return sum([stones.count(i) for i in [j for j in jewels]]) | jewels-and-stones | one liner 28ms easy | nuke01 | 1 | 100 | jewels and stones | 771 | 0.881 | Easy | 12,587 |
https://leetcode.com/problems/jewels-and-stones/discuss/1011101/python3-solution-with-list-comprehension | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
return sum([stones.count(item) for item in jewels]) | jewels-and-stones | python3 solution with list comprehension | MartenMink | 1 | 133 | jewels and stones | 771 | 0.881 | Easy | 12,588 |
https://leetcode.com/problems/jewels-and-stones/discuss/998322/Python-or-3-Methods-or-Efficient-Solution | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
res = 0
for i in stones:
if i in jewels:
res+=1
return res | jewels-and-stones | Python | 3 Methods | Efficient Solution | hritik5102 | 1 | 80 | jewels and stones | 771 | 0.881 | Easy | 12,589 |
https://leetcode.com/problems/jewels-and-stones/discuss/998322/Python-or-3-Methods-or-Efficient-Solution | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
res = 0
d = {}
for i in stones:
if i in d:
d[i]+=1
else:
d[i] = 1
for i in jewels:
if i in d:
res+= d[i]
return res | jewels-and-stones | Python | 3 Methods | Efficient Solution | hritik5102 | 1 | 80 | jewels and stones | 771 | 0.881 | Easy | 12,590 |
https://leetcode.com/problems/jewels-and-stones/discuss/998322/Python-or-3-Methods-or-Efficient-Solution | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
res = 0
for i in list(jewels):
res+= stones.count(i)
return res | jewels-and-stones | Python | 3 Methods | Efficient Solution | hritik5102 | 1 | 80 | jewels and stones | 771 | 0.881 | Easy | 12,591 |
https://leetcode.com/problems/jewels-and-stones/discuss/941811/Python-Easy-Solution | class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
return sum(S.count(i) for i in J) | jewels-and-stones | Python Easy Solution | lokeshsenthilkumar | 1 | 212 | jewels and stones | 771 | 0.881 | Easy | 12,592 |
https://leetcode.com/problems/jewels-and-stones/discuss/941811/Python-Easy-Solution | class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
return sum(i in J for i in S) | jewels-and-stones | Python Easy Solution | lokeshsenthilkumar | 1 | 212 | jewels and stones | 771 | 0.881 | Easy | 12,593 |
https://leetcode.com/problems/jewels-and-stones/discuss/262418/Python3-99-Explanation | class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
ss = {}
for s in S:
ss[s] = ss.get(s, 0) + 1 # if the key is not there, return a default value of 0, then increment by 1
counter = 0
for j in J:
counter += ss.get(j, 0) # if that character is not in the dictionary, return 0
return counter | jewels-and-stones | Python3 99% Explanation | Khris | 1 | 177 | jewels and stones | 771 | 0.881 | Easy | 12,594 |
https://leetcode.com/problems/jewels-and-stones/discuss/2844360/Easy-Python | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
count1 = 0
for i in range(0,len(stones)):
if stones[i] in jewels:
count1+=1
return count1 | jewels-and-stones | Easy Python | khanismail_1 | 0 | 1 | jewels and stones | 771 | 0.881 | Easy | 12,595 |
https://leetcode.com/problems/jewels-and-stones/discuss/2837464/Beginners-PythonCode | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
count = ""
for i in range (0, len(jewels)):
for j in range(0, len(stones)):
if jewels[i]==stones[j]:
count += jewels[i]
return len(count) | jewels-and-stones | Beginners - PythonCode | bharatvishwa | 0 | 1 | jewels and stones | 771 | 0.881 | Easy | 12,596 |
https://leetcode.com/problems/jewels-and-stones/discuss/2830949/Basic-Solution-For-Beginners | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
count=0
for i in stones:
if i in jewels:
count+=1
return count | jewels-and-stones | Basic Solution For Beginners | VidishaPandey | 0 | 1 | jewels and stones | 771 | 0.881 | Easy | 12,597 |
https://leetcode.com/problems/jewels-and-stones/discuss/2830357/fastest-python-solution | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
jewels = set(jewels)
return sum(x in jewels for x in stones) | jewels-and-stones | fastest python solution | TrickyUnicorn | 0 | 2 | jewels and stones | 771 | 0.881 | Easy | 12,598 |
https://leetcode.com/problems/jewels-and-stones/discuss/2815478/Multiple-Fast-and-Simple-Solutions-Python-(One-Liner-Included) | class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
count = 0
for i in stones:
if i in jewels:
count += 1
return count | jewels-and-stones | Multiple Fast and Simple Solutions - Python (One-Liner Included) | PranavBhatt | 0 | 2 | jewels and stones | 771 | 0.881 | Easy | 12,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.