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
if cnt[sandwiches[0]] == 0:
cnt.pop(sandwiches[0])
sandwiches.pop(0)
return len(students) | 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)
if len(sandwiches) == 0:
break
return len(students) | 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:
if student_counters[sandwich] == 0:
break
while students_queue[0] != sandwich:
students_queue.append(students_queue.popleft())
student_counters[sandwich] -= 1
students_queue.popleft()
return len(students_queue) | 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)
sandwiches.pop(0)
rot_cnt = len(students)
else:
temp = students[0]
students.pop(0)
students.append(temp)
rot_cnt -= 1
return total_length - eat | 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 len(sandwiches)-i | 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 sandwiches[0] in students:
students.append(students.pop(0))
else:
break
return a - x | 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
return students_counter[0]+students_counter[1] | 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.append(first)
return len(students)
def helper(self,students, sandwiches):#The while loop should stop if there are no students or if all the students want
#the same shape and the shape the want is different that the first sandwich in the stack
summ = sum(students)
if summ == 0:#It means that all students are 0
if sandwiches[0] != 0:
return False
else:
True
if summ == len(students):#It means that all students are 1
if sandwiches[0] != 1:
return False
else:
True
return True | 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)
return sum(arr) / len(arr) | 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 + overhead
end_prev_order = max(end_prev_order, arrival) + time
return waits_total / len(customers) | 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:
ans+=customers[i][1]
c = customers[i][1]+customers[i][0]
return ans/len(customers) | 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+=(count-start)
else:
s+=(time)
count=start+time
print(count, s)
return s/len(customers) | 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:
time += free-a + b
free += b
return time/n | 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_time / len(customers) | 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 = customers[i][1]
if arr < next_wt:
wt = (next_wt-arr)+(arr+dep)
else:
wt = arr+dep
time.append(wt - arr)
next_wt = wt
#print(time)
return sum(time)/len(time) | 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":
c+=1
for i in range (0,len(s)):
#finding the ist 0
if s[i]=="0":
lst[i+c-1]="0"
return "".join(lst)
return s | 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 '1' (not including leading '1's)
return f"{binary[:zero_idx]}{'1'*(zero-1)}{'0'*min(zero, 1)}{'1'*one}" | 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 index of 0, num_zeros => Total number of zeros in the input binary string.
2. If there are no zeros, directly return the binary string (Already in the desired form).
3. Put all the ones(1) till first 0 occurence as it is and covert remaining num_zeros - 1 zeros to 1.
3. Put the last remaining 0 now as we can't do anything about it.
4. Now put the remaining 1s => (total length of binary string) - (count of zeros) - (count of ones at the beginning)
'''
def maximumBinaryString(self, binary: str) -> str:
first_zero, num_zeros = binary.find('0'), binary.count('0')
return ('1' * ( first_zero + num_zeros - 1 )) + '0' + ('1' * (len(binary) - num_zeros - first_zero)) if zeros else binary | 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', first_zero)
return '1' * (first_zero) + '1' * (count_zeroes-1) + '0' + '1' * (len(binary) - first_zero - count_zeroes) | 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+=1
if count_zero > 1:
return '1'*(leading_one+count_zero-1)+'0'+'1'*(count_one-leading_one)
else:
return binary | 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 (ones + zeros - 1)*"1" + "0" + (len(binary) - zeros)*"1" | 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] - loc[m] - (len(loc)-ii)//2 # adding right
if len(loc) - ii > k:
m = (ii + len(loc))//2 # updated median
val -= loc[m] - loc[ii] - (len(loc)-ii)//2 # removing left
ii += 1
if len(loc)-ii == k: ans = min(ans, val) # len(ones) - ii effective length
return ans | 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)//2]) - (prefix[i+k//2] - prefix[i]))
return ans - (k//2)*((k+1)//2) | 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 middle position of the list
# counts the vowels in first and back half
for x in first:
if x.lower() in vowels:
firstc += 1
for y in back:
if y.lower() in vowels:
backc += 1
# returns whether the counts are equal to each other
return firstc == backc | 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
return a==b | 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==c2 | 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:
c2 += 1
return c1 == c2 | 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:
count-=1
if count == 0:
return True
else:
return False | 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
return left == right | 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:
count += 1
return count == 0 | 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 == temp):
return True
return False
``` | 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']:
x2+=1
return x1==x2 | 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
for char in right:
if char in vowels:
vowels_count -= 1
return vowels_count == 0 | 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:
if ch in vowels:
count_v_in_a += 1
for ch in b:
if ch in vowels:
count_v_in_b += 1
return count_v_in_a == count_v_in_b | 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
return lv == rv | 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[char] if not i else -half[char]
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 | 24,653 |
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 = len(s)//2
while i<=length and j>=length:
if dt.get(s[i],False):
count1+=1
i+=1
if dt.get(s[j],False):
count2+=1
j-=1
return count1==count2 | 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(firstHalf)):
if (firstHalf[i] == "a" or firstHalf[i] == "A"):
countFirstHalf+=1
if (firstHalf[i] == "e" or firstHalf[i] == "E"):
countFirstHalf+=1
if (firstHalf[i] == "i" or firstHalf[i] == "I"):
countFirstHalf+=1
if (firstHalf[i] == "o" or firstHalf[i] == "O"):
countFirstHalf+=1
if (firstHalf[i] == "u" or firstHalf[i] == "U"):
countFirstHalf+=1
for i in range(len(secondHalf)):
if (secondHalf[i] == "a" or secondHalf[i] == "A"):
countSecondHalf+=1
if (secondHalf[i] == "e" or secondHalf[i] == "E"):
countSecondHalf+=1
if (secondHalf[i] == "i" or secondHalf[i] == "I"):
countSecondHalf+=1
if (secondHalf[i] == "o" or secondHalf[i] == "O"):
countSecondHalf+=1
if (secondHalf[i] == "u" or secondHalf[i] == "U"):
countSecondHalf+=1
if countFirstHalf == countSecondHalf:
return True
else:
return False
# print(firstHalf)
# print(secondHalf) | 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 index < middle:
left_vowels_counter += 1
else:
right_vowels_counter += 1
return left_vowels_counter == right_vowels_counter | 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:
b+=1
return a==b | 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:
return 1
return 0 | 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:
if c2 in vowels:
cnt2 +=1
return cnt1 == cnt2 | 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 for each of the two halves.
first_count = 0
second_count = 0
# Start counting the two halves both in each iteration of the for-loop.
for i in range(halfway):
if s[i] in vowels:
first_count += 1
if s[i + halfway] in vowels:
second_count += 1
# Return the result.
return first_count == second_count | 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)
return i == j | 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:
cnt2 += 1
i += 1
j += 1
if cnt1 == cnt2:
return True
else:
return False | 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 s1 == s2 | 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+=1 #count number of vowels in second half
return c==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 sum([1 for c in s if self.isVowal(c)]) | 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
if s2_half[i] in vow:
d = d+1
if c == d:
return True
else:
return False | 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 in "aeiouAEIOU"]
if len(vowels_first_half) == len(vowels_second_half):
return True
else:
return False | 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):
if s[i] in d:
a-=1
return False if a else True | 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 False | 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
if s[end] in temp:
rc += 1
start += 1
end -= 1
return lc == rc | 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 be self-explainable, to avoid the need of having additional comments.
def is_vowel(letter):
return a in "aeiouAEIOU"
a,b = 0,0
L = len(s)
for i in range(L // 2):
if is_vowel(s[i]):
a += 1
if is_vowel(s[L - 1 - i]):
b += 1
return a == b | 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
for i in range(int(len(s)/2),len(s)):
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
if count==0:
return 1
return 0 | 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_vows_len | 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 a_c in vowels:
a += 1
if b_c in vowels:
b += 1
return a == b
# sol = Solution()
# print(sol.halvesAreAlike("MerryChristmas")) | 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:
ii, x = heappop(pq)
if x-1: heappush(pq, (ii, x-1))
ans += 1
i += 1
while pq:
ii, x = heappop(pq)
x = min(x, ii-i)
ans += x
i += x
return ans | 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
res=0
i=0
while(i<le):
A=l[i][0]
D=l[i][1]
if day<D:
diff_days=D-day
mn=min(diff_days,A)
day+=mn
res+=mn
i+=1
return res | 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)
elif grid[r][c] == -1 and 0 <= c-1 and grid[r][c-1] == -1:
return helper(r+1, c-1)
else:
return -1
return [helper(0, j) for j in range(n)] | 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 collumn to get the new column
new_col = col+grid[row][col]
### if the new column is already out of the box
### or the neighbor cell doesn't equal to grid[row][col]
### the ball will get stuck and we just return -1
if new_col==n or new_col==-1 or grid[row][new_col]!=grid[row][col]:
return -1
else:
return check(row+1,new_col)
res = []
for i in range(n):
res.append(check(0,i))
return res | 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][kk] < 0: break
k = kk
else: ans[j] = k # no break
return ans | 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
c_nxt = c + grid[r][c]
if c_nxt < 0 or c_nxt >= c_len or grid[r][c_nxt] == -grid[r][c]:
output[i] = -1
continue
output[i] += grid[r][c]
return output | 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:
return self.dfs(x+1,y+grid[x][y],m,n,grid,visited)
else:
return visited[x][y]
def findBall(self, grid: List[List[int]]) -> List[int]:
m=len(grid)
n=len(grid[0])
visited=[[None]*n for _ in range(m)]
result=[]
for i in range(n):
x=self.dfs(0,i,m,n,grid,visited)
result.append(x)
return result | 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]:
return c + grid[r][c]
return -1
elif grid[r][c] == 1 and c+1 < cols and grid[r][c] == grid[r][c+1]:
return dfs(r+1, c+1)
elif grid[r][c] == -1 and c-1 > -1 and grid[r][c] == grid[r][c-1]:
return dfs(r+1, c-1)
else:
return -1
ans = []
for c in range(cols):
ans.append(dfs(0, c))
return ans | 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, c = p[k]
if r==m:
ans[k] = c
else:
if grid[r][c]==1 and c+1<=n-1 and grid[r][c+1]==1:
p1[k] = (r+1, c+1)
elif grid[r][c]==-1 and c-1>=0 and grid[r][c-1]==-1:
p1[k] = (r+1, c-1)
p = p1
print("ans: ", ans)
print("=" * 20)
return ans
# print = lambda *a, **aa: () | 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: #ball goes right in next row
if col + 1 < cols:
balls[ball] += 1
else: #ball is out of boundary from right side
balls[ball] = -1
elif grid[row][col] == -1: # ball goes left in next row
if col - 1 >= 0:
balls[ball] -= 1
else: # ball is out of boundary from left side
balls[ball] = -1
else: #ball is in grid where it will get stuck
balls[ball] = -1
helper(row+1, balls)
#mark all cells where ball will get stuck as 0
for r, row in enumerate(grid):
for c in range(1, len(grid[0])):
if grid[r][c-1] == 1 and grid[r][c] == -1:
grid[r][c-1] = 0
grid[r][c] = 0
cols = len(grid[0])
ans = [i for i in range(cols)]
helper(0, ans)
return ans | 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][col] == 1 and (col == len_col-1 or grid[row][col+1] == -1)) or \
(grid[row][col] == -1 and (col == 0 or grid[row][col-1] == 1)):
answer.append(-1)
break
# Last Row
if row == len_row-1:
if grid[row][col] == 1: answer.append(col+1)
else: answer.append(col-1)
# Change Row, Column
if grid[row][col] == 1: col += 1
else: col -= 1
row += 1
return answer | 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][j-1] == 1)):
return -1
else:
return fall_pos(i, j+1, "below") if grid[i][j] == 1 else fall_pos(i, j-1, "below")
else:
return j if i == max_row-1 else fall_pos(i+1, j, "above")
return [fall_pos(0, j, "above") for j in range(max_col)] | 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] == 1):
return -1
if grid[i][j] == 1:
return dfs(i+1,j+1)
if grid[i][j] == -1:
return dfs(i+1,j-1)
return [dfs(0,i) for i in range(n)] | 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.