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/number-of-students-unable-to-eat-lunch/discuss/1751095/Easy-Python-solution | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
for i in sandwiches:
if i in students:
sandwiches = sandwiches[1:]
students.remove(i)
else:
break
return len(students) | number-of-students-unable-to-eat-lunch | Easy Python solution | MengyingLin | 0 | 86 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,600 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1545572/Python3-solution | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
cnt = Counter(students)
st = deque(sandwiches) # stack
while st and cnt[st[0]] > 0:
cnt[st[0]] -= 1
st.popleft()
return len(st) | number-of-students-unable-to-eat-lunch | Python3 solution | dalechoi | 0 | 66 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,601 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1485656/Counter-for-students-97-speed | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
cnt = Counter(students)
while sandwiches and sandwiches[0] in cnt:
i = students.index(sandwiches[0])
students = students[i + 1:] + students[:i]
cnt[sandwiches[0]] -= 1
... | number-of-students-unable-to-eat-lunch | Counter for students, 97% speed | EvgenySH | 0 | 136 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,602 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1332624/Python3-or-easy-understanding | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
while sandwiches[0] in students:
if students[0] == sandwiches[0]:
sandwiches.pop(0)
else:
students.append(students[0])
students.pop(0)
i... | number-of-students-unable-to-eat-lunch | Python3 | easy-understanding | Wyhever | 0 | 65 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,603 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1304432/Python-28-ms | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
student_counters = {1: sum(students)}
student_counters[0] = len(students) - student_counters[1]
students_queue = collections.deque(students)
for sandwich in sandwiches:
... | number-of-students-unable-to-eat-lunch | Python, 28 ms | MihailP | 0 | 107 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,604 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1169571/Python-approach-using-stack-operations | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
rot_cnt = len(sandwiches)
total_length = rot_cnt
eat = 0
while(rot_cnt):
if students[0] == sandwiches[0]:
eat += 1
students.pop(0)
s... | number-of-students-unable-to-eat-lunch | Python- approach using stack operations | ShivamBhirud | 0 | 112 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,605 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1092434/Python3-O(n)-faster-than-93 | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
count = Counter(students)
for i, s in enumerate(sandwiches):
if not count[s]:
break
count[s]-=1
else:
i = len(sandwiches)
return l... | number-of-students-unable-to-eat-lunch | Python3 O(n) faster than 93% | amol1729 | 0 | 61 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,606 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1040726/Python3-easy-solution | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
x = 0
a = len(students)
while len(students) > 0:
if students[0] == sandwiches[0]:
students.pop(0)
sandwiches.pop(0)
x += 1
elif ... | number-of-students-unable-to-eat-lunch | Python3 easy solution | EklavyaJoshi | 0 | 65 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,607 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/1003643/Python3-matches-students-with-sandwiches | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
ss = sum(students)
for i, x in enumerate(sandwiches):
if (x and not ss) or (not x and ss == len(sandwiches) - i): return len(sandwiches)-i
ss -= x
return 0 | number-of-students-unable-to-eat-lunch | [Python3] matches students with sandwiches | ye15 | 0 | 74 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,608 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/990290/Runtime-100-faster-and-Memory-100-less | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
students_counter = collections.Counter(students)
for sandwich in sandwiches:
if students_counter[sandwich]>0:
students_counter[sandwich]-=1
else:
break
... | number-of-students-unable-to-eat-lunch | Runtime 100 % faster and Memory 100% less | WiseLin | 0 | 102 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,609 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/987544/python-3-easy-to-understand-solution | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
while students and self.helper(students,sandwiches):
first = students.pop(0)
if first == sandwiches[0]:
sandwiches.pop(0)
else:
students.ap... | number-of-students-unable-to-eat-lunch | python 3 easy to understand solution | GiorgosMarga | 0 | 45 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,610 |
https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/discuss/987496/Python-no-dictionary.-Time%3A-O(N)-Space%3A-O(1)-or-100-100 | class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
counts = [len(students) - sum(students), sum(students)]
for s in sandwiches:
if not counts[s]:
return sum(counts)
counts[s] -= 1
return 0 | number-of-students-unable-to-eat-lunch | Python, no dictionary. Time: O(N), Space: O(1) | 100% / 100% | blue_sky5 | -1 | 59 | number of students unable to eat lunch | 1,700 | 0.679 | Easy | 24,611 |
https://leetcode.com/problems/average-waiting-time/discuss/1236349/Python3-Simple-And-Fast-Solution | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
arr = []
time = 0
for i , j in customers:
if(i > time):
time = i + j
else:
time += j
arr.append(time - i)
... | average-waiting-time | [Python3] Simple And Fast Solution | VoidCupboard | 6 | 194 | average waiting time | 1,701 | 0.624 | Medium | 24,612 |
https://leetcode.com/problems/average-waiting-time/discuss/2542272/Python3-or-Self-explanatory-1-pass-solution | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
waits_total = 0
end_prev_order = customers[0][0]
for arrival, time in customers:
overhead = end_prev_order - arrival if end_prev_order > arrival else 0
waits_total += time + ov... | average-waiting-time | Python3 | Self-explanatory 1-pass solution | Ploypaphat | 0 | 13 | average waiting time | 1,701 | 0.624 | Medium | 24,613 |
https://leetcode.com/problems/average-waiting-time/discuss/2524399/Python-or-Easy-to-solve | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
ans = 0
c = -1
for i in range(0,len(customers)):
if c>customers[i][0]:
c = c+customers[i][1]
ans+=(c)-customers[i][0]
else:
... | average-waiting-time | Python | Easy to solve | Brillianttyagi | 0 | 16 | average waiting time | 1,701 | 0.624 | Medium | 24,614 |
https://leetcode.com/problems/average-waiting-time/discuss/2402264/Simple-Python-Solution-or-Beats-74-in-TC-and-88-in-SC | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
s=0
count=0
flag=1
for start, time in customers:
if flag:
count+=start
flag=0
if count>=start:
count+=time
s+=(co... | average-waiting-time | Simple Python Solution | Beats 74% in TC & 88% in SC | Siddharth_singh | 0 | 7 | average waiting time | 1,701 | 0.624 | Medium | 24,615 |
https://leetcode.com/problems/average-waiting-time/discuss/2192265/Simplest-Intuition | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
totalTime = 0
current = 0
for arr, time in customers:
current = max(current, arr) + time
totalTime += current - arr
return totalTime/len(customers) | average-waiting-time | Simplest Intuition | Vaibhav7860 | 0 | 11 | average waiting time | 1,701 | 0.624 | Medium | 24,616 |
https://leetcode.com/problems/average-waiting-time/discuss/1777411/Python-3-very-easy-O(n)-O(1) | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
totalWait = curTime = 0
for arrival, time in customers:
curTime = max(curTime, arrival) + time
totalWait += curTime - arrival
return totalWait / len(customers) | average-waiting-time | Python 3, very easy, O(n) / O(1) | dereky4 | 0 | 49 | average waiting time | 1,701 | 0.624 | Medium | 24,617 |
https://leetcode.com/problems/average-waiting-time/discuss/1402183/Python3-simple-solution | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
time = 0
n = len(customers)
free = 0
for i in range(n):
a,b = customers[i]
if free <= a:
free += a - free + b
time += b
else:
... | average-waiting-time | Python3 simple solution | EklavyaJoshi | 0 | 52 | average waiting time | 1,701 | 0.624 | Medium | 24,618 |
https://leetcode.com/problems/average-waiting-time/discuss/1003331/Python3-sweep | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
ans = t = 0
for arrvl, tt in customers:
t = max(t, arrvl) + tt
ans += t - arrvl
return ans/len(customers) | average-waiting-time | [Python3] sweep | ye15 | 0 | 37 | average waiting time | 1,701 | 0.624 | Medium | 24,619 |
https://leetcode.com/problems/average-waiting-time/discuss/988689/PYTHON-oror-EASY-oror-O(N)-TIME-oror-O(1)-SPACE | class Solution(object):
def averageWaitingTime(self, customers):
current_time = 0
wait_time = 0
for arrival_time, duration in customers:
current_time = max(current_time, arrival_time) + duration
wait_time += (current_time - arrival_time)
return wait_... | average-waiting-time | PYTHON || EASY || O(N) TIME || O(1) SPACE | akashgkrishnan | 0 | 43 | average waiting time | 1,701 | 0.624 | Medium | 24,620 |
https://leetcode.com/problems/average-waiting-time/discuss/987569/Python-Solution | class Solution:
def averageWaitingTime(self, customers: List[List[int]]) -> float:
next_wt = customers[0][0]+customers[0][1]
time = []
time.append(next_wt-customers[0][0])
for i in range(1,len(customers)):
arr = customers[i][0]
dep = custome... | average-waiting-time | Python Solution | SaSha59 | 0 | 26 | average waiting time | 1,701 | 0.624 | Medium | 24,621 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/1382851/python-3-oror-clean-oror-easy-approach | lass Solution:
def maximumBinaryString(self, s: str) -> str:
#count of 0
c=0
#final ans string will contain only one zero.therefore shift the first 0 to c places.Initialize ans string with all 1s
lst=["1"]*len(s)
for i in range (0,len(s)):
if s[i]=="0":
... | maximum-binary-string-after-change | python 3 || clean || easy approach | minato_namikaze | 4 | 175 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,622 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/1492879/Python-3-or-Greedy-or-Explanation | class Solution:
def maximumBinaryString(self, binary: str) -> str:
zero = binary.count('0') # count number of '0'
zero_idx = binary.index('0') if zero > 0 else 0 # find the index of fist '0' if exists
one = len(binary) - zero_idx - zero # count number of... | maximum-binary-string-after-change | Python 3 | Greedy | Explanation | idontknoooo | 2 | 178 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,623 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/1173116/Easy-Python-2-lines-Beats-100 | class Solution:
'''
Runtime: 44 ms, faster than 100.00% of Python3 online submissions for Maximum Binary String After Change.
Memory Usage: 15.6 MB, less than 59.78% of Python3 online submissions for Maximum Binary String After Change.
'''
'''
1. Variable Description -> first_zero => first occurence in... | maximum-binary-string-after-change | Easy Python - 2 lines - Beats 100% | shubhsaxena | 1 | 122 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,624 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/2604358/Maximum-binary-string-oror-Python-oror-O(n)-Time-Complexity-oror-Solution-with-explanation-and-comments | class Solution:
def maximumBinaryString(self, binary: str) -> str:
# Find the first zero
first_zero = binary.find('0')
# If there are no zeroes then no optimization is needed
if(first_zero == -1):
return binary
# Counting number of zeros
count_zeroes = binary.count('0... | maximum-binary-string-after-change | Maximum binary string || Python || O(n) Time Complexity || Solution with explanation and comments | vanshika_2507 | 0 | 10 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,625 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/987983/Python3-One-liner | class Solution:
def maximumBinaryString(self, binary: str) -> str:
return '1'*binary.find('0')+reduce(lambda a,b:a+b if b=='1' else a[:-1]+'10', sorted(binary[max(binary.find('0'),0):])) | maximum-binary-string-after-change | [Python3] One-liner | vilchinsky | 0 | 73 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,626 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/987414/Python3-Detailed-explanation | class Solution:
def maximumBinaryString(self, binary: str) -> str:
leading_one, count_zero, count_one = 0, 0, 0
for x in binary:
if x == '1':
count_one+=1
if not count_zero:
leading_one+=1
else:
count_zero+=... | maximum-binary-string-after-change | [Python3] Detailed explanation | vilchinsky | 0 | 68 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,627 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/1003614/Python3-string-processing | class Solution:
def maximumBinaryString(self, binary: str) -> str:
n = len(binary)
binary = binary.lstrip("1")
ones = n - len(binary) # count of leading "1"s
zeros = binary.count("0")
if zeros <= 1: return ones*"1" + binary # nothing to do
return (o... | maximum-binary-string-after-change | [Python3] string processing | ye15 | -1 | 111 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,628 |
https://leetcode.com/problems/maximum-binary-string-after-change/discuss/1003614/Python3-string-processing | class Solution:
def maximumBinaryString(self, binary: str) -> str:
if binary.count("0") <= 1: return binary
ones = binary.count("1", binary.index("0"))
return (len(binary)-ones-1)*"1" + "0" + ones*"1" | maximum-binary-string-after-change | [Python3] string processing | ye15 | -1 | 111 | maximum binary string after change | 1,702 | 0.462 | Medium | 24,629 |
https://leetcode.com/problems/minimum-adjacent-swaps-for-k-consecutive-ones/discuss/1002574/Python3-1-pass-O(N) | class Solution:
def minMoves(self, nums: List[int], k: int) -> int:
ii = val = 0
ans = inf
loc = [] # location of 1s
for i, x in enumerate(nums):
if x:
loc.append(i)
m = (ii + len(loc) - 1)//2 # median
val += loc[-1] - l... | minimum-adjacent-swaps-for-k-consecutive-ones | [Python3] 1-pass O(N) | ye15 | 2 | 572 | minimum adjacent swaps for k consecutive ones | 1,703 | 0.423 | Hard | 24,630 |
https://leetcode.com/problems/minimum-adjacent-swaps-for-k-consecutive-ones/discuss/1002574/Python3-1-pass-O(N) | class Solution:
def minMoves(self, nums: List[int], k: int) -> int:
loc = [i for i, x in enumerate(nums) if x]
prefix = [0]
for x in loc: prefix.append(prefix[-1] + x)
ans = inf
for i in range(len(loc)-k+1):
ans = min(ans, (prefix[i+k] - prefix[i+(k+1)//... | minimum-adjacent-swaps-for-k-consecutive-ones | [Python3] 1-pass O(N) | ye15 | 2 | 572 | minimum adjacent swaps for k consecutive ones | 1,703 | 0.423 | Hard | 24,631 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/991430/Runtime-is-faster-than-98-and-the-memory-usage-is-less-than-90-Python-3-Accepted | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = set('aeiouAEIOU')
count = 0
for i in range(len(s)//2):
if s[i] in vowels:
count+=1
if s[-i-1] in vowels:
count-=1
return count == 0 | determine-if-string-halves-are-alike | Runtime is faster than 98% and the memory usage is less than 90% Python 3 [Accepted] | WiseLin | 5 | 341 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,632 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2669019/PYTHON-3-STRING-SLICING-or-EASY-or-EXPLANATION | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = ['a', 'e', 'i', 'o', 'u']
# initializing the first half of the word and the back half variables
first, firstc = [*s][:len(s)//2], 0
back, backc = [*s][len(s)//2:], 0
# [*s] creates a list
# [len(s)//2] finds the m... | determine-if-string-halves-are-alike | [PYTHON 3] STRING SLICING | EASY | EXPLANATION | omkarxpatel | 3 | 108 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,633 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1222548/python-or-98.51-or-easy | class Solution:
def halvesAreAlike(self, s: str) -> bool:
y=['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
a=b=0
for i in range(len(s)//2):
if s[i] in y:
a+=1
if s[len(s)//2 + i] in y:
b+=1
... | determine-if-string-halves-are-alike | python | 98.51% | easy | chikushen99 | 3 | 161 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,634 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1565203/simple-fast-python-solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = {'a', 'A', 'e', 'E', 'i', 'I', 'o', 'O', 'u', 'U'}
n = len(s)
m = n // 2
return sum(s[i] in vowels for i in range(m)) == sum(s[i] in vowels for i in range(m, n)) | determine-if-string-halves-are-alike | simple, fast python solution | dereky4 | 2 | 63 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,635 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2693513/Python-oror-Simple-oror-Runtime-57-ms-Beats-62.96-Memory-13.9-MB-Beats-78.52-oror-O(N) | class Solution:
def halvesAreAlike(self, s: str) -> bool:
n=len(s)
s=s.lower()
a,b=s[:n//2],s[n//2:]
vol="aeiou"
c1,c2=0,0
for i in a:
if i in vol:
c1+=1
for i in b:
if i in vol:
c2+=1
return c1==... | determine-if-string-halves-are-alike | Python || Simple || Runtime 57 ms Beats 62.96% Memory 13.9 MB Beats 78.52% || O(N) | Sneh713 | 1 | 49 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,636 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2429959/Python-Simple-Solution-(-Brute-Force-Approach-) | class Solution:
def halvesAreAlike(self, s: str) -> bool:
f1,f2 = s[:len(s)//2] , s[len(s)//2:]
c1 , c2 = 0,0
v = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
for i in f1:
if i in v:
c1 += 1
for j in f2:
if j in v:
... | determine-if-string-halves-are-alike | Python Simple Solution ( Brute Force Approach ) | SouravSingh49 | 1 | 25 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,637 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1483661/Set-of-vowels-100-speed | class Solution:
vowels = set("aeiouAEIOU")
def halvesAreAlike(self, s: str) -> bool:
half = len(s) // 2
return (sum(c in Solution.vowels for c in s[:half]) ==
sum(c in Solution.vowels for c in s[half:])) | determine-if-string-halves-are-alike | Set of vowels, 100% speed | EvgenySH | 1 | 65 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,638 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1385763/Easy-to-Understand-or-Faster-than-86 | class Solution:
def halvesAreAlike(self, s: str) -> bool:
count= 0
l = len(s)
d = l//2
s = s.lower()
lis = ['a', 'e', 'i', 'o', 'u']
for i in range(0,d):
if s[i] in lis:
count+=1
for j in range(d,l):
if s[j] in lis:
... | determine-if-string-halves-are-alike | Easy to Understand | Faster than 86% | kdevharsh2001 | 1 | 52 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,639 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1298413/Python3-weird-one-liner | class Solution:
def halvesAreAlike(self, s: str) -> bool:
return sum(map(s[:len(s)//2].lower().count, 'aeiou')) == sum(map(s[len(s)//2:].lower().count, 'aeiou')) | determine-if-string-halves-are-alike | [Python3] weird one-liner | denizen-ru | 1 | 37 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,640 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1173214/Python-Simple-Zip-Loop | class Solution:
def halvesAreAlike(self, s: str) -> bool:
left = right = 0
vowels = {'a','e','i','o','u','A', 'E','I','O','U'}
for (x,y) in zip(s[:len(s)//2], s[len(s)//2:]):
if x in vowels:
left+=1
if y in vowels:
right+=1
retu... | determine-if-string-halves-are-alike | Python Simple Zip Loop | X00X | 1 | 73 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,641 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1126182/Python-Faster-than-95-of-Submissions | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
count = 0
for i in range(floor(len(s)/2)):
if s[i] in vowels:
count -= 1
if s[-i - 1] in vowels:
... | determine-if-string-halves-are-alike | Python Faster than 95% of Submissions | APet99 | 1 | 114 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,642 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1120322/Runtime%3A-faster-than-95.81Memory-Usage%3A-less-than-97.18-or-Easy-Beginner | class Solution:
def halvesAreAlike(self, s: str) -> bool:
count =0
temp=0
arr= "AEIOUaeiou"
for x in range(len(s)//2):
if s[x] in arr:
count+=1
for x in range(len(s)//2,len(s)):
if s[x] in arr:
temp+=1
if (count ... | determine-if-string-halves-are-alike | Runtime: faster than 95.81%,Memory Usage: less than 97.18% | Easy, Beginner | ak-akkinapelli | 1 | 53 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,643 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/988509/Python-solution%3A-faster-than-100.00-submissions-in-runtime | class Solution:
def halvesAreAlike(self, s: str) -> bool:
a,b=s[:len(s)//2],s[len(s)//2:]
x1,x2=0,0
for i in a:
if i in ['a','e','i','o','u','A','E','I','O','U']:
x1+=1
for i in b:
if i in ['a','e','i','o','u','A','E','I','O','U']:
... | determine-if-string-halves-are-alike | Python solution: faster than 100.00% submissions in runtime | thisisakshat | 1 | 105 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,644 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/988430/Python3-count-O(N) | class Solution:
def halvesAreAlike(self, s: str) -> bool:
cnt = 0
for i, c in enumerate(s):
if c in "aeiouAEIOU": cnt += 1 if i < len(s)//2 else -1
return cnt == 0 | determine-if-string-halves-are-alike | [Python3] count O(N) | ye15 | 1 | 46 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,645 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2847085/Python-easy-Solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
mid = len(s) // 2
left = s[:mid]
right = s[mid:]
vowels_count = 0
for char in left:
if char in vowels:
vowels_count += 1... | determine-if-string-halves-are-alike | Python easy Solution | namashin | 0 | 1 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,646 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2811811/Python-Solution-oror-Simple-and-Fast-oror-Understandable | class Solution:
def halvesAreAlike(self, s):
cnt1 = 0
cnt2 = 0
for i in range(0, int(len(s)/2)):
if s[i].lower() in 'aeiou': cnt1 += 1
for i in range(int(len(s)/2), len(s)):
if s[i].lower() in 'aeiou': cnt2 += 1
return cnt1 == cnt2 | determine-if-string-halves-are-alike | Python Solution || Simple & Fast || Understandable | qiy2019 | 0 | 1 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,647 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2786283/Python-solution-95.97-faster | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
len_s = len(s)
half_index = len_s // 2
a = s[:half_index]
b = s[half_index:]
count_v_in_a = 0
count_v_in_b = 0
for ch in a:
... | determine-if-string-halves-are-alike | Python solution 95.97% faster | samanehghafouri | 0 | 5 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,648 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2780654/Python-3-O(n)-two-pointers-approach | class Solution:
def halvesAreAlike(self, s: str) -> bool:
l, r = 0, len(s) - 1
lv, rv = 0, 0
vowel = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
while l <= r:
if s[l] in vowel:
lv += 1
if s[r] in vowel:
rv += 1
l += 1
r -= 1
r... | determine-if-string-halves-are-alike | Python 3 O(n) two pointers approach | WJTTW | 0 | 1 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,649 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2591599/Python-5-Solutions-or-Easy-or-Clean-or-Straightforward | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels, count = 'aeiouAEIOU', 0
for i in range(len(s) // 2):
count += (s[i] in vowels) - (s[~i] in vowels)
return not count | determine-if-string-halves-are-alike | [Python] 5 Solutions | Easy | Clean | Straightforward | andy2167565 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,650 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2591599/Python-5-Solutions-or-Easy-or-Clean-or-Straightforward | class Solution:
def halvesAreAlike(self, s: str) -> bool:
return not sum((s[i] in 'aeiouAEIOU') - (s[~i] in 'aeiouAEIOU') for i in range(len(s) // 2)) | determine-if-string-halves-are-alike | [Python] 5 Solutions | Easy | Clean | Straightforward | andy2167565 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,651 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2591599/Python-5-Solutions-or-Easy-or-Clean-or-Straightforward | class Solution:
def halvesAreAlike(self, s: str) -> bool:
s, count = [s[:len(s) // 2], s[len(s) // 2:]], 0
for i, half in enumerate(s):
for vowel in set(half).intersection('aeiouAEIOU'):
count += half.count(vowel) if not i else -half.count(vowel)
return not count | determine-if-string-halves-are-alike | [Python] 5 Solutions | Easy | Clean | Straightforward | andy2167565 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,652 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2591599/Python-5-Solutions-or-Easy-or-Clean-or-Straightforward | class Solution:
def halvesAreAlike(self, s: str) -> bool:
from collections import Counter
C, count = list(map(Counter, (s[:len(s) // 2], s[len(s) // 2:]))), 0
for i, half in enumerate(C):
for char in half:
if char in 'aeiouAEIOU':
count += half... | determine-if-string-halves-are-alike | [Python] 5 Solutions | Easy | Clean | Straightforward | andy2167565 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,653 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2591599/Python-5-Solutions-or-Easy-or-Clean-or-Straightforward | class Solution:
def halvesAreAlike(self, s: str) -> bool:
return sum(char in 'aeiouAEIOU' for char in s[:len(s) // 2]) == sum(char in 'aeiouAEIOU' for char in s[len(s) // 2:]) | determine-if-string-halves-are-alike | [Python] 5 Solutions | Easy | Clean | Straightforward | andy2167565 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,654 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2591599/Python-5-Solutions-or-Easy-or-Clean-or-Straightforward | class Solution:
def halvesAreAlike(self, s: str) -> bool:
count = 0
for i, char in enumerate(s):
if char in 'aeiouAEIOU':
count += 1 if i < len(s) // 2 else -1
return not count | determine-if-string-halves-are-alike | [Python] 5 Solutions | Easy | Clean | Straightforward | andy2167565 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,655 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2591599/Python-5-Solutions-or-Easy-or-Clean-or-Straightforward | class Solution:
def halvesAreAlike(self, s: str) -> bool:
return not sum(((i < len(s) // 2) * 2 - 1) * (char in 'aeiouAEIOU') for i, char in enumerate(s)) | determine-if-string-halves-are-alike | [Python] 5 Solutions | Easy | Clean | Straightforward | andy2167565 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,656 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2572541/python3-oror-Optimized-solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
dt = {
'a' : 1,'A':1,
'e' : 1,'E':1,
'i' : 1,'I':1,
'o' : 1,'O':1,
'u' : 1,'U':1
}
i = 0
j= len(s)-1
count1 = count2 = 0
length = le... | determine-if-string-halves-are-alike | python3 || Optimized solution | shacid | 0 | 11 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,657 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2546279/Easy-python-solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
c1=0
c2=0
vowels=["a","e","i","o","u"]
s=s.lower()
s1=s[:len(s)//2]
s2=s[len(s)//2:]
for i in vowels:
c1=c1+s1.count(i)
c2=c2+s2.count(i)
return c1==c2 | determine-if-string-halves-are-alike | Easy python solution | keertika27 | 0 | 14 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,658 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2535914/Python-oror-Easy-and-well-explained-solution | class Solution(object):
def halvesAreAlike(self, s):
"""
:type s: str
:rtype: bool
"""
length = len(s)
firstHalf = s[:length/2]
secondHalf = s[length/2:]
countFirstHalf = 0
countSecondHalf = 0
for i in range(len(firstH... | determine-if-string-halves-are-alike | Python || Easy and well explained solution | ride-coder | 0 | 8 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,659 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2535194/Python-O(n)-solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = ['a', 'e', 'i', 'o', 'u']
left_vowels_counter: int = 0
right_vowels_counter: int = 0
middle: int = int(len(s) / 2)
for index, letter in enumerate(s):
if letter.lower() in vowels:
if... | determine-if-string-halves-are-alike | Python O(n) solution | anton_python | 0 | 10 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,660 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2409788/Python3-93-faster | class Solution:
def halvesAreAlike(self, s: str) -> bool:
k = len(s)//2
vowels=['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
a=0
b=0
for i in range(len(s)):
if(s[i] in vowels):
if(i<k):
a+=1
else:
... | determine-if-string-halves-are-alike | Python3 93% faster | morrismoppp | 0 | 7 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,661 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2299049/easy-python-solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
x = s[len(s)//2:]
y = s[:len(s)//2]
v = "aeiouAEIOU"
p, q = 0, 0
for i in x:
if i in v:
p += 1
for i in y:
if i in v:
q += 1
if p == q:
... | determine-if-string-halves-are-alike | easy python solution | rohansardar | 0 | 10 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,662 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2293229/Simple-Python3-Solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
length = len(s) // 2
a, b = s[:length], s[length:]
cnt1 = cnt2 = 0
for c1 in a:
if c1 in vowels:
cnt1 +=1
for c2 in b:
... | determine-if-string-halves-are-alike | Simple Python3 Solution | vem5688 | 0 | 11 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,663 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2242296/Beginner-Friendly-Solution-oror-41ms-Faster-Than-85-oror-Python | class Solution:
def halvesAreAlike(self, s: str) -> bool:
# Set up: - the half-way index number for s.
# - a list of vowels to perform property checks with.
halfway = len(s) // 2
vowels = ['a','e','i','o','u','A','E','I','O','U']
# Initialize vowel counters... | determine-if-string-halves-are-alike | Beginner Friendly Solution || 41ms, Faster Than 85% || Python | cool-huip | 0 | 18 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,664 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2077711/Basic-solution-O(N) | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels = "aeiou"
s = s.lower()
s1, s2 = s[:len(s) // 2], s[len(s) // 2:]
return sum([1 for c in s1 if c in vowels]) == sum([1 for c in s2 if c in vowels]) | determine-if-string-halves-are-alike | Basic solution O(N) | andrewnerdimo | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,665 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2030653/Runtime%3A-23-ms-93.06 | class Solution(object):
def halvesAreAlike(self, s):
"""
:type s: str
:rtype: bool
"""
vowels = ('a', 'e', 'i', 'o', 'u')
i, j = 0 , 0
s = s.lower()
l = len(s)//2
for v in vowels: i += s[:l].count(v)
for v in vowels: j += s[l:].count(v)... | determine-if-string-halves-are-alike | Runtime: 23 ms 93.06% ට වඩා වේගවත් | akilaocj | 0 | 17 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,666 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/2004839/python-solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels =('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U')
i = 0
cnt1 = cnt2 = 0
k = j = len(s) // 2
while(i != k):
if s[i] in vowels:
cnt1 += 1
if s[j] in vowels:
... | determine-if-string-halves-are-alike | python solution | dharmrajrathod98 | 0 | 19 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,667 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1938762/Python-Solution-or-Simple-Vowel-Counting-Approach-or-Over-90-Faster | class Solution:
def countVowels(self,s):
count = 0
for e in s:
if e in "aeiouAEIOU":
count += 1
return count
def halvesAreAlike(self, s: str) -> bool:
return self.countVowels(s[:len(s)//2]) == self.countVowels(s[len(s)//2:]) | determine-if-string-halves-are-alike | Python Solution | Simple Vowel Counting Approach | Over 90% Faster | Gautam_ProMax | 0 | 17 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,668 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1906932/Python-Solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
n = len(s) // 2
target = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
s1 = s2 = 0
for char1, char2 in zip(s[:n], s[n:]):
s1 += (0, 1)[char1 in target]
s2 += (0, 1)[char2 in target]
return ... | determine-if-string-halves-are-alike | Python Solution | hgalytoby | 0 | 25 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,669 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1906932/Python-Solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
n = len(s) // 2
target = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
return len(list(filter(lambda x: x in target, s[:n]))) == len(list(filter(lambda x: x in target, s[n:]))) | determine-if-string-halves-are-alike | Python Solution | hgalytoby | 0 | 25 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,670 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1808588/Easy-and-Fast-solution-95 | class Solution:
def halvesAreAlike(self, s: str) -> bool:
a=s[:len(s)//2]
b=s[len(s)//2:]
v='aeiouAEIOU'
c,d=0,0
for i in a:
if i in v:
c+=1 #count number of vowels in first half
for j in b:
if j in v:
d... | determine-if-string-halves-are-alike | Easy and Fast solution 95% | adityabaner | 0 | 48 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,671 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1700168/Python-with-functions | class Solution:
def halvesAreAlike(self, s: str) -> bool:
mid = len(s)//2
return self.sumVowal(s[:mid]) == self.sumVowal(s[mid:])
def isVowal(self, c) -> bool:
if c in "aeiouAEIOU":
return True
return False
def sumVowal(self, s) -> int:
return su... | determine-if-string-halves-are-alike | Python with functions | CleverUzbek | 0 | 41 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,672 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1690640/Python-Easy-Understanding | class Solution:
def halvesAreAlike(self, s: str) -> bool:
mid = len(s)//2
s1_half = s[0:mid]
s2_half = s[mid:]
vow = ['a','e','i','o','u','A','E','I','O','U']
c=0
d=0
for i in range(len(s1_half)):
if s1_half[i] in vow:
c = c+1
... | determine-if-string-halves-are-alike | [Python] Easy Understanding | Ron99 | 0 | 50 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,673 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1690329/**-Python-code%3A | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels="aeiouAEIOU"
l=len(s)//2
a,b=s[:l],s[l:]
aC,bC=0,0
for i in range(l):
if a[i] in vowels: aC+=1
if b[i] in vowels: bC+=1
return aC==bC | determine-if-string-halves-are-alike | ** Python code: | Anilchouhan181 | 0 | 25 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,674 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1659442/Super-simple-Python3-solution-faster-than-100-of-submissions | class Solution:
def halvesAreAlike(self, s: str) -> bool:
half_index = int(len(s)/2)
first_half = s[:half_index]
second_half = s[half_index:]
vowels_first_half = [i for i in first_half if i in "aeiouAEIOU"]
vowels_second_half = [i for i in second_half if i i... | determine-if-string-halves-are-alike | Super simple Python3 solution, faster than 100% of submissions | y-arjun-y | 0 | 59 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,675 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1412707/Runtime%3A-24-ms-faster-than-98.83-of-Python3-online-submissions-f | class Solution:
def halvesAreAlike(self, s: str) -> bool:
a=0
n=len(s)
if n%2!=0:
return False
d={"a","e","i","o","u","A","E","I","O","U"}
for i in range((n//2)):
if s[i] in d:
a+=1
for i in range((n//2),n):
... | determine-if-string-halves-are-alike | Runtime: 24 ms, faster than 98.83% of Python3 online submissions f | harshmalviya7 | 0 | 80 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,676 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1411064/Python-Best-Solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
lst = set("aeiouAEIOU")
n = len(s)//2
count = 0
for i in s[:n]:
if i in lst:
count +=1
for i in s[n:]:
if i in lst:
count -=1
return True if count ==0 else F... | determine-if-string-halves-are-alike | Python Best Solution | phuoctfx11137 | 0 | 45 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,677 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1148976/Python-3-or-Two-pointer-or-Set | class Solution:
def halvesAreAlike(self, s: str) -> bool:
lc = 0
rc = 0
temp = set(['a', 'e', 'i', 'o','u', 'A','E', 'I','O', 'U'])
start = 0
end = len(s) - 1
while start < end:
if s[start] in temp:
lc += 1
... | determine-if-string-halves-are-alike | Python 3 | Two pointer | Set | abhyasa | 0 | 35 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,678 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1147173/Python-3-count-vowel-one-pass-in-place | class Solution:
def halvesAreAlike(self, s: str) -> bool:
"""
Place pointer a at the index 0 and pointer b at last index. Traverse the string s up to half of its length. Count vowel with helper function
Time: O(N)
Space: O(1)
"""
# Function name and parameter itself should ... | determine-if-string-halves-are-alike | [Python 3] count vowel one pass in place | anonymous1127 | 0 | 23 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,679 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/1009743/Ultra-Simple-CppPython3-Solution-or-Suggestions-for-optimization-are-welcomed | class Solution:
def halvesAreAlike(self, s: str) -> bool:
count=0
for i in range(0,int(len(s)/2)):
if s[i]=='a' or s[i]=='e' or s[i]=='i' or s[i]=='o' or s[i]=='u' or s[i]=='A' or s[i]=='E' or s[i]=='I' or s[i]=='O' or s[i]=='U':
count=count+1
... | determine-if-string-halves-are-alike | Ultra Simple Cpp/Python3 Solution | Suggestions for optimization are welcomed | angiras_rohit | 0 | 17 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,680 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/994657/Python-One-line-Faster-than-96 | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowels,size = {"a","e","i","o","u","A","E","I", "O","U"}, len(s)
return sum([ch in vowels for ch in s[0:size//2]]) == sum([ch in vowels for ch in s[size//2:]]) | determine-if-string-halves-are-alike | Python One line, Faster than 96% | selimtanriverdien | 0 | 108 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,681 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/993723/Short-Python-Solution | class Solution:
def halvesAreAlike(self, s: str) -> bool:
a, b = s[:len(s)//2], s[len(s)//2:]
return True if sum(map(a.lower().count, "aeiou")) == sum(map(b.lower().count, "aeiou")) else False | determine-if-string-halves-are-alike | Short Python Solution | BirdLQ | 0 | 59 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,682 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/988713/Intuitive-approach-and-code-explain-itself-well | class Solution:
def halvesAreAlike(self, s: str) -> bool:
s, s_half_len = s.lower(), len(s) // 2
a, b = s[:s_half_len], s[s_half_len:]
a_vows_len = len(list(filter(lambda e: e in 'aeiou', a)))
b_vows_len = len(list(filter(lambda e: e in 'aeiou', b)))
return a_vows_len == b_v... | determine-if-string-halves-are-alike | Intuitive approach and code explain itself well | puremonkey2001 | 0 | 20 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,683 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/988685/PYTHON-oror-EASY-oror-O(N)-TIME-oror-O(1)-SPACE | class Solution:
def halvesAreAlike(self, string: str) -> bool:
j = len(string) // 2
a = 0
b = 0
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
for i in range(len(string) // 2):
a_c = string[i]
b_c = string[j + i]
if ... | determine-if-string-halves-are-alike | PYTHON || EASY || O(N) TIME || O(1) SPACE | akashgkrishnan | 0 | 46 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,684 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/988344/Python-one-for-loop | class Solution:
def halvesAreAlike(self, s: str) -> bool:
vowel = set('aeiouAEIOU')
count = 0
for i in range(len(s)//2):
if s[i] in vowel:
count += 1
if s[-i-1] in vowel:
count -= 1
return count == 0 | determine-if-string-halves-are-alike | Python, one for loop | blue_sky5 | 0 | 30 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,685 |
https://leetcode.com/problems/determine-if-string-halves-are-alike/discuss/988344/Python-one-for-loop | class Solution:
def halvesAreAlike(self, s: str) -> bool:
return sum(1 if i < len(s)//2 else -1 for i, c in enumerate(s) if c in 'aeiouAEIOU') == 0 | determine-if-string-halves-are-alike | Python, one for loop | blue_sky5 | 0 | 30 | determine if string halves are alike | 1,704 | 0.774 | Easy | 24,686 |
https://leetcode.com/problems/maximum-number-of-eaten-apples/discuss/988437/Python3-priority-queue | class Solution:
def eatenApples(self, apples: List[int], days: List[int]) -> int:
ans = 0
pq = [] # min-heap
for i, (x, d) in enumerate(zip(apples, days)):
while pq and pq[0][0] <= i: heappop(pq) # rotten
if x: heappush(pq, (i+d, x))
if pq:
... | maximum-number-of-eaten-apples | [Python3] priority queue | ye15 | 3 | 113 | maximum number of eaten apples | 1,705 | 0.381 | Medium | 24,687 |
https://leetcode.com/problems/maximum-number-of-eaten-apples/discuss/995033/Sorting-doesn't-work-here-(only-4463-test-cases-passed) | class Solution:
def eatenApples(self, apples: List[int], days: List[int]) -> int:
n=len(days)
l=[]
le=0
for i in range(n):
if days[i]!=0 or apples[i]!=0:
l.append([apples[i],days[i]+i])
le+=1
l.sort(key=lambda x:x[1])
day=0
... | maximum-number-of-eaten-apples | Sorting doesn't work here (only 44/63 test cases passed) | _Rehan12 | 1 | 78 | maximum number of eaten apples | 1,705 | 0.381 | Medium | 24,688 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/1443268/Python-3-or-DFS-Simulation-or-Explanation | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0])
@cache
def helper(r, c):
if r == m:
return c
elif grid[r][c] == 1 and c+1 < n and grid[r][c+1] == 1:
return helper(r+1, c+1)
... | where-will-the-ball-fall | Python 3 | DFS, Simulation | Explanation | idontknoooo | 28 | 1,400 | where will the ball fall | 1,706 | 0.716 | Medium | 24,689 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2765461/EASY-PYTHON-SOLUTON-oror-O(mxn)-oror-Comments | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
m,n = len(grid),len(grid[0])
def check(row,col):
### If a ball get out of the box, return col
if row==m:
return col
### note that since grid contains 1 and -1 representing to right and to left,
### we can just add the grid[row][col] to current co... | where-will-the-ball-fall | EASY PYTHON SOLUTON || O(mxn) || Comments | raghavdabra | 4 | 189 | where will the ball fall | 1,706 | 0.716 | Medium | 24,690 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/988447/Python3-simulation | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0]) # dimensions
ans = [-1]*n
for j in range(n):
k = j
for i in range(m):
kk = k + grid[i][k]
if not 0 <= kk < n or grid[i][k] * grid[i... | where-will-the-ball-fall | [Python3] simulation | ye15 | 2 | 166 | where will the ball fall | 1,706 | 0.716 | Medium | 24,691 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2765400/Python-Solution-with-Explanation-and-Diagram-or-97-Faster | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
r_len = len(grid)
c_len = len(grid[0])
output = list(range(c_len))
for r in range(r_len):
for i in range(c_len):
c = output[i]
if c == -1: continue
... | where-will-the-ball-fall | ✔️ Python Solution with Explanation and Diagram | 97% Faster 🔥 | pniraj657 | 1 | 109 | where will the ball fall | 1,706 | 0.716 | Medium | 24,692 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2765283/SIMPLE-SOLUTION-USING-DFS | class Solution:
def dfs(self,x,y,m,n,grid,visited):
if visited[x][y]!=None:
return visited[x][y]
if y+grid[x][y]<0 or y+grid[x][y]>=n or (grid[x][y]+grid[x][y+grid[x][y]]==0):
visited[x][y]=-1
return -1
visited[x][y]=y+grid[x][y]
if x+1<m:
... | where-will-the-ball-fall | SIMPLE SOLUTION USING DFS | beneath_ocean | 1 | 58 | where will the ball fall | 1,706 | 0.716 | Medium | 24,693 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2764916/Python-3-oror-DFS-solution | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
rows, cols = len(grid), len(grid[0])
def dfs(r, c):
if r == rows-1:
if grid[r][c] == 1 and c+1 < cols and grid[r][c] == grid[r][c+1]:
return c + grid[r][c]
if grid[r][c] == -1 and c-1 > -1 and grid[r][c] == grid[r][c-1]:
... | where-will-the-ball-fall | Python 3 || DFS solution | sagarhasan273 | 1 | 73 | where will the ball fall | 1,706 | 0.716 | Medium | 24,694 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2428020/O(n)-BFS-by-swapping-two-queues-and-moving-pattern-(easy-understanding-with-examples) | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
"""
"""
m, n = len(grid), len(grid[0])
p = {i: (0,i) for i in range(n)}
ans = [-1] * n
while len(p)>0:
print(p)
p1 = {}
for k in p:
r, ... | where-will-the-ball-fall | O(n) BFS by swapping two queues and moving pattern (easy understanding with examples) | dntai | 1 | 41 | where will the ball fall | 1,706 | 0.716 | Medium | 24,695 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2826586/Well-explained-with-commented-code-easy-to-follow-python-solution. | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
def helper(row, balls):
if row == len(grid):
return
for ball, col in enumerate(balls):
if col != -1: #the ball is not already stuck
if grid[row][col] == 1: #bal... | where-will-the-ball-fall | Well explained with commented code, easy to follow python solution. | dkashi | 0 | 1 | where will the ball fall | 1,706 | 0.716 | Medium | 24,696 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2824482/Python3-Easy-solution | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
len_row, len_col = len(grid),len(grid[0])
answer = []
for c in range(len_col):
row, col = 0, c
for _ in range(len_row):
# Stuck conditions
if (grid[row][c... | where-will-the-ball-fall | Python3 Easy solution | Coaspe | 0 | 1 | where will the ball fall | 1,706 | 0.716 | Medium | 24,697 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2816878/Python-short-recursive-solution-or-O(M-*-N)-or-O(M) | class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
max_row, max_col = len(grid), len(grid[0])
def fall_pos(i, j, layer):
if layer == "above":
if (grid[i][j] == 1 and (j == max_col-1 or grid[i][j+1] == -1)) or (grid[i][j] == -1 and (j == 0 or grid[i][... | where-will-the-ball-fall | Python short recursive solution | O(M * N) | O(M) | xyp7x | 0 | 2 | where will the ball fall | 1,706 | 0.716 | Medium | 24,698 |
https://leetcode.com/problems/where-will-the-ball-fall/discuss/2814138/Python-(Simple-DFS) | class Solution:
def findBall(self, grid):
m, n = len(grid), len(grid[0])
def dfs(i,j):
if i == m:
return j
if (j == 0 and grid[i][j] == -1) or (j == n-1 and grid[i][j] == 1) or (grid[i][j] == 1 and grid[i][j+1] == -1) or (grid[i][j] == -1 and grid[i][j-1] ==... | where-will-the-ball-fall | Python (Simple DFS) | rnotappl | 0 | 1 | where will the ball fall | 1,706 | 0.716 | Medium | 24,699 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.