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https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/2221822/Python-simple-solution | class Solution:
def sumOfThree(self, n: int) -> List[int]:
if n//3 == n/3:
return [n//3-1,n//3,n//3+1]
else:
return [] | find-three-consecutive-integers-that-sum-to-a-given-number | Python simple solution | StikS32 | 0 | 10 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,200 |
https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/2009445/python-3-oror-simple-O(1)-solution | class Solution:
def sumOfThree(self, num: int) -> List[int]:
if num % 3:
return []
middle = num // 3
return [middle - 1, middle, middle + 1] | find-three-consecutive-integers-that-sum-to-a-given-number | python 3 || simple O(1) solution | dereky4 | 0 | 26 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,201 |
https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/1806196/Python-easy-to-read-and-understand | class Solution:
def sumOfThree(self, num: int) -> List[int]:
ans = []
if num % 3 == 0:
ans = [num//3-1, num//3, num//3+1]
return ans | find-three-consecutive-integers-that-sum-to-a-given-number | Python easy to read and understand | sanial2001 | 0 | 37 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,202 |
https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/1799983/Python3-accepted-one-liner-solution | class Solution:
def sumOfThree(self, num: int) -> List[int]:
return [num//3 -1,num//3,num//3 +1] if(num%3==0) else [] | find-three-consecutive-integers-that-sum-to-a-given-number | Python3 accepted one-liner solution | sreeleetcode19 | 0 | 17 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,203 |
https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/1798731/Why-is-this-even-medium-level | class Solution:
def sumOfThree(self, num: int) -> List[int]:
if num % 3 != 0:
return []
k = num//3
return [k-1, k, k+1] | find-three-consecutive-integers-that-sum-to-a-given-number | Why is this even medium level? | byuns9334 | 0 | 26 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,204 |
https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/1785045/python3-number-theory-concept | class Solution:
def sumOfThree(self, num: int) -> List[int]:
if num%3!=0:
return []
return [num//3-1,num//3,num//3+1] | find-three-consecutive-integers-that-sum-to-a-given-number | python3 number theory concept | Karna61814 | 0 | 8 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,205 |
https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/1783860/Python-math | class Solution:
def sumOfThree(self, num: int) -> List[int]:
return [num //3 - 1, num // 3, num // 3 + 1] if num % 3 == 0 else [] | find-three-consecutive-integers-that-sum-to-a-given-number | Python, math | blue_sky5 | 0 | 14 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,206 |
https://leetcode.com/problems/find-three-consecutive-integers-that-sum-to-a-given-number/discuss/1783291/Python%3A-Very-Simple-Math-formula-SC%3A-O(1)-TC%3A-O(1) | class Solution:
def sumOfThree(self, num: int) -> List[int]:
rightHandSide = num - 3
if rightHandSide % 3 != 0:
return []
firstNumber = rightHandSide // 3
return [firstNumber, firstNumber + 1, firstNumber + 2] | find-three-consecutive-integers-that-sum-to-a-given-number | Python: Very Simple Math formula SC: O(1) TC: O(1) | tyrocoder | 0 | 33 | find three consecutive integers that sum to a given number | 2,177 | 0.637 | Medium | 30,207 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1783966/Simple-Python-Solution-with-Explanation-oror-O(sqrt(n))-Time-Complexity-oror-O(1)-Space-Complexity | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
l=set()
if finalSum%2!=0:
return l
else:
s=0
i=2 # even pointer 2, 4, 6, 8, 10, 12...........
while(s<finalSum):
s+=i #sum
l.add(i) # append the i in list
i+=2
if s==finalSum: #if sum s is equal to finalSum then no modidfication required
return l
else:
l.discard(s-finalSum) #Deleting the element which makes s greater than finalSum
return l | maximum-split-of-positive-even-integers | Simple Python Solution with Explanation || O(sqrt(n)) Time Complexity || O(1) Space Complexity | HimanshuGupta_p1 | 51 | 2,000 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,208 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1783248/Python-Solution-by-Splitting-and-Merging | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
arr = []
if finalSum % 2 == 0: # If finalSum is odd then we cannot ever divide it with the given conditions
a, i = finalSum // 2, 1 # a is the number of 2's and i is the number of 2's that we will use to form a even number in the current iteration
while i <= a: # Till we have sufficient number of 2's available
arr.append(2*i) # Join the i number of 2's to form a even number
a -= i # Number of 2's remaining reduces by i
i += 1 # Number of 2's required in next itertation increases by 1
s = sum(arr)
arr[-1] += finalSum - s # This is done if their were still some 2's remaining that could not form a number due to insufficient count, then we add the remaining 2's into the last number.
return arr | maximum-split-of-positive-even-integers | Python Solution by Splitting and Merging | anCoderr | 10 | 1,000 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,209 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1783481/Python-O(finalSum0.5)-Easy-Solution | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if(finalSum%2 != 0):
return []
finalSum = finalSum//2
result = []
total = 0
remove = None
for i in range(1, finalSum+1):
result.append(i)
total += i
if(total == finalSum):
break
elif(total > finalSum):
remove = total-finalSum
break
output = []
for num in result:
if(remove==None):
output.append(2*num)
else:
if(num!=remove):
output.append(2*num)
return output | maximum-split-of-positive-even-integers | Python - O(finalSum^0.5) Easy Solution | RoshanNaik | 7 | 282 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,210 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1783483/Well-Explained-oror-For-Beginners-oror-O(n) | class Solution:
def maximumEvenSplit(self, f: int) -> List[int]:
if f%2:
return []
res, i = [], 2
while i<=f:
res.append(i)
f -= i
i += 2
res[-1]+=f
return res | maximum-split-of-positive-even-integers | 📌📌 Well-Explained || For Beginners || O(n) 🐍 | abhi9Rai | 2 | 80 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,211 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1934288/python-3-oror-simple-greedy-solution | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum % 2:
return []
res = []
n = 2
while finalSum >= 2*n + 2:
res.append(n)
finalSum -= n
n += 2
res.append(finalSum)
return res | maximum-split-of-positive-even-integers | python 3 || simple greedy solution | dereky4 | 1 | 313 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,212 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1814930/Python-3-Math-and-Summations-O(MaxSplit)-or-Beats-95 | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum%2 != 0:
return []
total = 0
i=2
answer = []
while total+i <= finalSum:
total += i
answer.append(i)
i += 2
answer[-1] += finalSum-total
return answer | maximum-split-of-positive-even-integers | [Python 3] Math and Summations O(MaxSplit) | Beats 95% | hari19041 | 1 | 122 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,213 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1783428/Python3-simulation | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum % 2: return []
ans = []
x = 2
while finalSum >= x:
ans.append(x)
finalSum -= x
x += 2
ans[-1] += finalSum
return ans | maximum-split-of-positive-even-integers | [Python3] simulation | ye15 | 1 | 60 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,214 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/2542035/Python-simple-solution | class Solution:
'''
The idea is summing all even numbers starting from 2.
It is guaranteed to be larger than finalSum when we reach 2i
and i^2+i just larger than finalSum. The difference between
the sum and finalSum must also be an even number and is
contained in the sum. So we skip it.
'''
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum%2: return []
i = math.floor(sqrt(finalSum)) #To save some time
while i**2+i<finalSum:
i+=1
skip = i**2+i-finalSum
return [2*j for j in range(1,i+1) if 2*j != skip] | maximum-split-of-positive-even-integers | Python simple solution | li87o | 0 | 80 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,215 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/2366897/Easy-Python-solution-following-hints-provided-in-the-question | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum % 2 == 1: # odd number
return []
temp = 0
last_number = 2 # first even number
res = list()
while temp < finalSum:
temp += last_number
res.append(last_number)
last_number += 2
if temp == finalSum:
return res
else:
# remove the final element from list and subtract it from the running sum
temp -= res.pop()
diff = finalSum - temp
res[-1] += diff
return res | maximum-split-of-positive-even-integers | Easy Python solution following hints provided in the question | lau125 | 0 | 60 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,216 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/2039079/5-lines-simple-solution-beat-100-(with-explanation) | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
# (2 + 2 * n) * n / 2 <= finalSum
# math.floor(math.sqrt(finalSum))
# if n * (n + 1) > finalSum then n-=1
if finalSum % 2 != 0: return []
n = int(math.floor(math.sqrt(finalSum)))
if n * (n + 1) > finalSum: n -= 1
ret = [2 * i for i in range(1, n + 1)]
ret[-1] += finalSum - n * (n + 1)
return ret | maximum-split-of-positive-even-integers | 5 lines simple solution beat 100% (with explanation) | BichengWang | 0 | 159 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,217 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/2014843/Python3-Binary-Search-Solution | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum % 2 != 0:
return []
else:
#binary search
left, right = 1, finalSum//2
while left < right:
middle = ceil((left + right)/2)
diff_needed = (2 + (middle-1)*2)*(middle-1)//2
base = (finalSum - diff_needed)//middle
if base % 2 == 1:
base -= 1
if base >= 2:
left = middle
else:
right = middle-1
res = [2 for _ in range(left)]
current = 2
left = finalSum - sum(res)
for t in range(len(res)-2):
res[t+1] += current
left -= current
current += 2
res[-1] += left
return res | maximum-split-of-positive-even-integers | Python3 Binary Search Solution | xxHRxx | 0 | 116 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,218 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1960822/5-Lines-Python-Solution-oror-75-Faster | class Solution:
def maximumEvenSplit(self, num: int) -> List[int]:
if num%2: return []
ans=set() ; i=1 ; s=0
while s<num: ans.add(2*i) ; s+=2*i ; i+=1
ans.discard(s-num)
return ans | maximum-split-of-positive-even-integers | 5-Lines Python Solution || 75% Faster | Taha-C | 0 | 224 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,219 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1846473/Greedy-solution | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
# we have to split the sum as a maximum number of unique positive integers
if finalSum%2!=0:
return []
result=[]
count=0
final=0
while final<=finalSum:
result.append(count+2)
count+=2
final+=count
if (finalSum-final-(count+2))<=count+2:
if finalSum-final>result[-1]:
result.append(finalSum-final)
break
if sum(result)!=finalSum:
result=[finalSum]
return result | maximum-split-of-positive-even-integers | Greedy solution | g0urav | 0 | 91 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,220 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1785048/python3-solution | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum%2!=0:
return []
even=2
res=[]
while finalSum>=even:
finalSum-=even
res.append(even)
even+=2
res[-1]+=finalSum
return res | maximum-split-of-positive-even-integers | python3 solution | Karna61814 | 0 | 15 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,221 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1784492/Python-3-or-Math-or-Explanation | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum % 2: return []
value = int(sqrt(finalSum))
ans = [2 * i for i in range(1, value)]
diff = finalSum - sum(ans)
if ans and diff <= ans[-1]:
return ans[:-1] + [diff + ans[-1]]
else:
return ans + [diff] | maximum-split-of-positive-even-integers | Python 3 | Math | Explanation | idontknoooo | 0 | 93 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,222 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1784445/Python-3-or-Newbie | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum%2!=0: return []
res=[]
i=1
rest = finalSum
while rest >0:
add = 2*i
if rest<add:
temp = res.pop()
res.append(temp+add+(rest-add))
else:
res.append(add)
rest = rest - add
i+=1
return res | maximum-split-of-positive-even-integers | Python 3 | Newbie | hugsypenguin | 0 | 19 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,223 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1783853/Python-O(n)-Solution | class Solution:
def maximumEvenSplit(self, finalSum: int) -> List[int]:
if finalSum % 2 == 1:
return []
first = 2
sum_ = 0
result = []
while sum_ < finalSum:
sum_ += first
result.append(first)
first += 2
if sum_ > finalSum:
e1 = result.pop()
e2 = result.pop()
result.append(finalSum - (sum_ - (e1 + e2)))
return result | maximum-split-of-positive-even-integers | [Python] O(n) Solution | tejeshreddy111 | 0 | 15 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,224 |
https://leetcode.com/problems/maximum-split-of-positive-even-integers/discuss/1783773/Python3-oror-Explanation-With-Examples | class Solution:
def maximumEvenSplit(self, f: int) -> List[int]:
if f%2!=0: # if odd then return
return []
count= (-1 + sqrt(1 + 4 * f)) // 2 # find x
ans=[]
i=2
while len(ans)!=count:
ans.append(i) # add even integers in list
i=i+2
if sum(ans)==f: # check if sum is euqal to finalSum
return ans
ans[-1]=ans[-1]+(f-sum(ans)) # else add difference in the last element
return ans | maximum-split-of-positive-even-integers | Python3 || Explanation With Examples | rushi_javiya | 0 | 21 | maximum split of positive even integers | 2,178 | 0.591 | Medium | 30,225 |
https://leetcode.com/problems/count-good-triplets-in-an-array/discuss/1783361/Python-3-SortedList-solution-O(NlogN) | class Solution:
def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
hashmap2 = {}
for i in range(n):
hashmap2[nums2[i]] = i
indices = []
for num in nums1:
indices.append(hashmap2[num])
from sortedcontainers import SortedList
left, right = SortedList(), SortedList()
leftCount, rightCount = [], []
for i in range(n):
leftCount.append(left.bisect_left(indices[i]))
left.add(indices[i])
for i in range(n - 1, -1, -1):
rightCount.append(len(right) - right.bisect_right(indices[i]))
right.add(indices[i])
count = 0
for i in range(n):
count += leftCount[i] * rightCount[n - 1 - i]
return count | count-good-triplets-in-an-array | Python 3 SortedList solution, O(NlogN) | xil899 | 13 | 554 | count good triplets in an array | 2,179 | 0.371 | Hard | 30,226 |
https://leetcode.com/problems/count-good-triplets-in-an-array/discuss/1784262/Python-Simple-O(NlogN)-solution-using-bisect | class Solution:
def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
res = 0
m2 = [0] * n
q = []
# Build index map of nums2
for i in range(n):
m2[nums2[i]] = i
for p1 in range(n):
p2 = m2[nums1[p1]] # Position of nums1[p1] in nums2
idx = bisect.bisect(q, p2) # Position smaller than this one so far
q.insert(idx, p2)
before = idx
after = n-1 - p1 - p2 + before # Based on number of unique values before and after are the same
res += before * after
return res | count-good-triplets-in-an-array | [Python] Simple O(NlogN) solution using bisect | slbteam08 | 2 | 94 | count good triplets in an array | 2,179 | 0.371 | Hard | 30,227 |
https://leetcode.com/problems/count-good-triplets-in-an-array/discuss/1783658/Share-My-Code | class Solution:
def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
N = len(nums1)
BIT = [0] * (N+1)
nums = [0] * N
num1 = [0] * N
num2 = [0] * N
for i, n in enumerate(nums1):
num1[n]=i
for i, n in enumerate(nums2):
num2[n]=i
for i, n in enumerate(num1):
nums[n] = num2[i]
def lowbit(i):
return i&(-i)
def add(i):
i+=1
while i <= N:
BIT[i]+=1
i+=lowbit(i)
def query(i):
i += 1
re = 0
while i:
re += BIT[i]
i -= lowbit(i)
return re
smaller = []
bigger = []
for i, n in enumerate(nums):
smaller.append(query(n))
bigger.append(N-1-i-n+smaller[-1])
add(n)
return sum(i*j for i, j in zip(smaller, bigger))
``` | count-good-triplets-in-an-array | Share My Code | Jeff871025 | 0 | 54 | count good triplets in an array | 2,179 | 0.371 | Hard | 30,228 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1785049/lessPython3greater-O(1)-Discrete-Formula-100-faster-1-LINE | class Solution:
def countEven(self, num: int) -> int:
return num // 2 if sum([int(k) for k in str(num)]) % 2 == 0 else (num - 1) // 2 | count-integers-with-even-digit-sum | <Python3> O(1) - Discrete Formula - 100% faster - 1 LINE | drknzz | 9 | 845 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,229 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1785599/Python-Faster-than-100-(Easy-solution) | class Solution:
def countEven(self, num: int) -> int:
count = 0
for i in range(num+1):
sum_of_digits = sum(int(digit) for digit in str(i))
if sum_of_digits % 2 == 0 and sum_of_digits != 0:
count += 1
return count | count-integers-with-even-digit-sum | [Python] Faster than 100% (Easy solution) | yashitanamdeo | 3 | 121 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,230 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1784841/Python-3-(30ms)-or-Simple-Maths-Formula-or-Even-Odd-Solution | class Solution:
def countEven(self, num: int) -> int:
if num%2!=0:
return (num//2)
s=0
t=num
while t:
s=s+(t%10)
t=t//10
if s%2==0:
return num//2
else:
return (num//2)-1 | count-integers-with-even-digit-sum | Python 3 (30ms) | Simple Maths Formula | Even Odd Solution | MrShobhit | 3 | 217 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,231 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1784774/Python3-brute-force | class Solution:
def countEven(self, num: int) -> int:
ans = 0
for x in range(1, num+1):
sm = sum(map(int, str(x)))
if not sm&1: ans += 1
return ans | count-integers-with-even-digit-sum | [Python3] brute-force | ye15 | 3 | 211 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,232 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1877275/Python-easy-solution-full-explanation-oror-99.23-success-rate | class Solution:
def countEven(self, num: int) -> int:
if sum(map(int,str(num))) % 2 == 0:
return num//2
return (num-1)//2 | count-integers-with-even-digit-sum | ✅Python easy solution full explanation || 99.23% success rate | Dev_Kesarwani | 2 | 158 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,233 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1793315/Python-Solution | class Solution:
def countEven(self, num: int) -> int:
t = num
s = 0
while num > 0:
s = s + num%10
num = num//10
return t//2 if s%2 == 0 else (t-1)//2 | count-integers-with-even-digit-sum | Python Solution | ayush_kushwaha | 1 | 55 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,234 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2837383/Python-solution | class Solution:
def countEven(self, num: int) -> int:
nums = [sum(list(map(int,list(str(i))))) for i in range(1,num+1) if (i < 10 and i % 2 == 0) or i >= 10]
nums = [i for i in nums if i % 2 == 0]
return len(nums) | count-integers-with-even-digit-sum | Python solution | lornedot | 0 | 1 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,235 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2836626/Simple-Python-sol | class Solution:
def digitSum(self,n):
s = 0
while n:
s += n%10
n = n//10
return s
def countEven(self, num: int) -> int:
ans = 0
for i in range(2,num+1):
if self.digitSum(i) % 2 == 0:
ans += 1
return ans | count-integers-with-even-digit-sum | Simple Python sol | aruj900 | 0 | 1 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,236 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2778177/python-solution | class Solution:
def countEven(self, num: int) -> int:
count=0
su=0
for i in range(2,num+1):
su=0
s=list(str(i))
for i in s:
su=su+int(i)
if su%2==0:
count+=1
return count | count-integers-with-even-digit-sum | python solution | sindhu_300 | 0 | 3 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,237 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2695120/Python-1-liner-with-explanation | class Solution:
def countEven(self, num: int) -> int:
return int(num/2) if sum(list(map(int, str(num)))) % 2 == 0 else int((num-1)/2) | count-integers-with-even-digit-sum | Python 1 liner with explanation | code_snow | 0 | 7 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,238 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2645231/Python-Simple-and-Easy-Solution | class Solution:
def countEven(self, n: int) -> int:
c = 0
for i in range(2,n+1):
i = str(i) ; l = [int(j) for j in i]
if sum(l) % 2 == 0: c += 1
return c | count-integers-with-even-digit-sum | Python Simple and Easy Solution | SouravSingh49 | 0 | 19 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,239 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2645181/Python-easy-solution-78ms-only | class Solution:
def countEven(self, num: int) -> int:
l=[]
for i in range(1,num+1):
l.append(list(str(i)))
s=0
c=0
for i in l:
for j in i:
s+=int(j)
if s%2==0:
c+=1
s=0
return c | count-integers-with-even-digit-sum | Python easy solution 78ms only | abhint1 | 0 | 3 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,240 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2616026/Python-Beats-98-without-checking-all-numbers-below | class Solution:
def countEven(self, num: int) -> int:
digits = [int(x) for x in list(str(num))]
if sum(digits) % 2 == 0:
return num//2
else:
return (num-1)//2 | count-integers-with-even-digit-sum | [Python] Beats 98%, without checking all numbers below | aaron2k12 | 0 | 20 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,241 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2547704/Python-Simple-Python-Solution-or-Faster-than-89.99 | class Solution:
def countEven(self, num: int) -> int:
def DigitSum(number):
total = 0
while number > 0:
remainder = number % 10
number = number // 10
total = total + remainder
return total
result = 0
for current_num in range(1,num + 1):
if DigitSum(current_num) % 2 == 0:
result = result + 1
return result | count-integers-with-even-digit-sum | [ Python ] ✅✅ Simple Python Solution | Faster than 89.99% 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 0 | 34 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,242 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2536621/Easy-approach-or-Python3 | class Solution:
def countEven(self, n: int) -> int:
c=0
for i in range(1,n+1):
evens=0
odds=0
'''even number even no's. & even number of odd no's. gievs resultant= even no.
similarly, odd even no.& odd odd no.=odd
odd even no. & even odd no.=even
even even no.& odd odd no.=odd'''
for j in str(i):
if int(j)%2==0:
evens+=1
else:
odds+=1
if (evens%2==0 and odds%2==0) or (evens%2!=0 and odds%2==0):
c=c+1
return c | count-integers-with-even-digit-sum | Easy approach | Python3 | keertika27 | 0 | 31 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,243 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2509748/Python-or-Easy-to-read-solution | class Solution:
def countEven(self, num: int) -> int:
counter = 0
for i in range(1, num+1):
res = list(str(i))
res = list(map(int, res))
res = sum(res)
if res % 2 == 0:
if res <= num:
counter += 1
else:
break
return counter | count-integers-with-even-digit-sum | Python | Easy to read solution | Wartem | 0 | 236 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,244 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2466539/Python-for-beginners-naive-approach-optimized-approach | class Solution:
def countEven(self, num: int) -> int:
#Optimized Approach
#Runtime: 57ms
return num//2 if sum([int(i) for i in str(num)])%2==0 else (num-1)//2
"""
#Naive Approach
#Runtime: 81ms
count=0
lis=[str(i) for i in range(1,num+1)]
for i in lis:
check=0
for j in i:
check+=int(j)
if(check%2==0):
count+=1
return count
""" | count-integers-with-even-digit-sum | Python for beginners naive approach, optimized approach | mehtay037 | 0 | 32 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,245 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2407525/Python-Soln | class Solution:
def countEven(self, num: int) -> int:
cnt = 0
for i in range(1, num+1):
if self.sumOfDigits(i)%2==0 and self.sumOfDigits(i) <= num:
# print(i)
cnt += 1
return cnt
def sumOfDigits(self, n):
digitSum = 0
while n:
digitSum += n%10
n = n //10
print(digitSum)
return digitSum | count-integers-with-even-digit-sum | Python Soln | logeshsrinivasans | 0 | 40 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,246 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2380621/Python3-straightforward | class Solution:
def countEven(self, num: int) -> int:
c=0
for i in range(1,num+1):
if sum(map(int,list(str(i))))%2==0:
c+=1
return c | count-integers-with-even-digit-sum | [Python3] straightforward | sunakshi132 | 0 | 35 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,247 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2157385/Count-Integers-With-Even-Digit-Sum-for-Python | class Solution:
def countEven(self, num: int) -> int:
count = 0
for i in range(1, num + 1):
oddI = i
product = 0
while i >= 1:
product += (i % 10)
i //= 10
if oddI < 10 and product % 2 == 0:
count += 1
if product % 2 == 0 and oddI >= 10:
count += 1
return count | count-integers-with-even-digit-sum | Count Integers With Even Digit Sum for Python | YangJenHao | 0 | 26 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,248 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2106162/basic-solution-with-helper-function | class Solution:
def countEven(self, num: int) -> int:
def sum_digits(n):
return sum([int(v) for v in str(n)])
for n in range(1, num + 1):
if not sum_digits(n) % 2:
res += 1
return res | count-integers-with-even-digit-sum | basic solution with helper function | andrewnerdimo | 0 | 36 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,249 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2095645/Python-Solution-with-a-simple-approach | class Solution:
def countEven(self, num: int) -> int:
count = 0
for i in range(1, num+1): # Range: [1, 1000]
check = self.sum_digits(i) # Call this fuction to get the sum of digits
if check%2 == 0:
count += 1
return count
def sum_digits(self, n): # Eg: n = 19
sum_ = 0
while n > 0:
digit = n % 10 # digit = 9 : Step - 1
sum_ += digit # sum_ = 9 : Step - 2
n = n//10 # n = 1, Again Step 1 and 2 as 1 > 0 then the sum_ will be 10
return sum_ | count-integers-with-even-digit-sum | Python Solution with a simple approach | yash921 | 0 | 67 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,250 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2036696/Python-Clean-and-Simple! | class Solution:
def countEven(self, num):
return sum(self.isEven(self.calcDigitSum(n)) for n in range(1,num+1))
def calcDigitSum(self, num):
return sum(int(d) for d in str(num))
def isEven(self, num):
return num % 2 == 0 | count-integers-with-even-digit-sum | Python - Clean and Simple! | domthedeveloper | 0 | 92 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,251 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2036696/Python-Clean-and-Simple! | class Solution:
def countEven(self, num):
return num//2 if self.isEven(self.calcDigitSum(num)) else (num-1)//2
def calcDigitSum(self, num):
return sum(int(d) for d in str(num))
def isEven(self, num):
return num % 2 == 0 | count-integers-with-even-digit-sum | Python - Clean and Simple! | domthedeveloper | 0 | 92 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,252 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/2034306/Python-oneliner | class Solution:
def countEven(self, num: int) -> int:
return len([x for x in range(1,num+1) if sum(map(int,list(str(x))))%2 == 0]) | count-integers-with-even-digit-sum | Python oneliner | StikS32 | 0 | 71 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,253 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1950200/Easy | class Solution:
def countEven(self, num: int) -> int:
c = 0
for i in range(1, num+1):
if sum([int(d) for d in str(i)]) % 2 == 0:
c += 1
return c | count-integers-with-even-digit-sum | Easy | Dario_Adamovic | 0 | 34 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,254 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1944128/Python-dollarolution-(98-Faster) | class Solution:
def countEven(self, num: int) -> int:
count = 0
for i in range(2,num+1):
k = i
s = 0
while k != 0:
s += k%10
k //= 10
if s % 2 == 0:
count += 1
return count | count-integers-with-even-digit-sum | Python $olution (98% Faster) | AakRay | 0 | 85 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,255 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1926825/Python-Simple-Math-Solution-O(Length(Num)) | class Solution:
def countEven(self, num: int) -> int:
s, c = str(num), 0
for i in s:
c += int(i)
return num // 2 if c % 2 == 0 else (num - 1) // 2 | count-integers-with-even-digit-sum | Python Simple Math Solution O(Length(Num)) | Hejita | 0 | 56 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,256 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1854962/Python-light-solution-space-less-than-95 | class Solution:
def countEven(self, num: int) -> int:
count = 0
for i in range(1, num+1):
temp = list(str(i))
digits = [int(x) for x in temp]
if sum(digits) % 2 == 0:
count += 1
return count | count-integers-with-even-digit-sum | Python light solution, space less than 95% | alishak1999 | 0 | 45 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,257 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1840824/python-3-oror-O(log-n) | class Solution:
def countEven(self, num: int) -> int:
digitSum = 0
n = num
while n:
digitSum += n % 10
n //= 10
return (num - 1) // 2 if digitSum % 2 else num // 2 | count-integers-with-even-digit-sum | python 3 || O(log n) | dereky4 | 0 | 70 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,258 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1799915/Python3-accepted-solution | class Solution:
def countEven(self, num: int) -> int:
count=0
for i in range(1,num+1):
if(sum([int(j) for j in str(i)])%2==0):
count+=1
return count | count-integers-with-even-digit-sum | Python3 accepted solution | sreeleetcode19 | 0 | 53 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,259 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1796430/Python-Brute-force-using-str-less-greater-int | class Solution:
def countEven(self, num: int) -> int:
count = 0
for n in range(1, num + 1):
if n < 10:
if n % 2 == 0:
count += 1
else:
ns = str(n)
x = 0
for s in ns:
x += int(s)
if x % 2 == 0:
count += 1
return count | count-integers-with-even-digit-sum | [Python] Brute force using str <-> int | casshsu | 0 | 33 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,260 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1794191/Python-Iterative-Solutions | class Solution:
def countEven(self, num: int) -> int:
def getDidgitSum(num):
digitSum = 0
while(num > 0):
digitSum += num % 10
num //= 10
return digitSum
total = 0
for i in range(1, num+1):
digitSum = getDidgitSum(i)
if digitSum % 2 == 0:
total += 1
return total | count-integers-with-even-digit-sum | Python Iterative Solutions | rushirg | 0 | 47 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,261 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1794191/Python-Iterative-Solutions | class Solution:
def countEven(self, num: int) -> int:
def getDidgitSum(num):
digitSum = 0
for digit in str(num):
digitSum += int(digit)
return digitSum
total = 0
for i in range(1, num+1):
digitSum = getDidgitSum(i)
if digitSum % 2 == 0:
total += 1
return total | count-integers-with-even-digit-sum | Python Iterative Solutions | rushirg | 0 | 47 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,262 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1793947/1-Line-Python-Solution-oror-97-Faster-(33ms)-oror-Memory-less-than-60 | class Solution:
def countEven(self, num: int) -> int:
return num//2 if sum([int(d) for d in str(num)])%2==0 else (num-1)//2 | count-integers-with-even-digit-sum | 1-Line Python Solution || 97% Faster (33ms) || Memory less than 60% | Taha-C | 0 | 59 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,263 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1791391/Memorization-trick-39-ms-96-speed | class Solution:
memo = {1: 0}
keys = [1]
def countEven(self, num: int) -> int:
if num not in Solution.memo:
idx = bisect_right(Solution.keys, num)
Solution.memo[num] = (Solution.memo[Solution.keys[idx - 1]] +
sum(not sum(int(d) for d in str(n)) % 2
for n in range(Solution.keys[idx - 1] + 1,
num + 1)))
Solution.keys.insert(idx, num)
return Solution.memo[num] | count-integers-with-even-digit-sum | Memorization trick, 39 ms 96% speed | EvgenySH | 0 | 45 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,264 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1791006/Intutive-easy-Python-solution | class Solution:
def countEven(self, num: int) -> int:
def digitSum(num):
sum_of_digits = 0
while num:
num, remainder = divmod(num, 10) #num = num/10 & remainder = num%10
sum_of_digits += remainder
return sum_of_digits
return num//2 if digitSum(num)%2 == 0 else (num-1)//2 | count-integers-with-even-digit-sum | Intutive easy Python solution | MaverickEyedea | 0 | 28 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,265 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1787580/Short-fast-python-solution | class Solution:
def countEven(self, num: int) -> int:
def match(v):
return sum(map(int, str(v))) % 2 == 0
return sum(map(match, range(1, num+1))) | count-integers-with-even-digit-sum | Short, fast python solution | emwalker | 0 | 27 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,266 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1785031/python3-solution-brute-force | class Solution:
def countEven(self, num: int) -> int:
count=0
for i in range(1,num+1):
s=str(i)
mylist=[int(val) for val in s]
if sum(mylist)%2==0:
count+=1
return count | count-integers-with-even-digit-sum | python3 solution brute force | Karna61814 | 0 | 17 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,267 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1784965/Python-brute-force | class Solution:
def countEven(self, num: int) -> int:
return sum(sum(map(int, str(n))) % 2 == 0 for n in range(2, num + 1)) | count-integers-with-even-digit-sum | Python, brute force | blue_sky5 | 0 | 28 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,268 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1784833/Python-Solution | class Solution:
def countEven(self, num: int) -> int:
count = 0
for i in range(2,num+1):
a = list(str(i))
sum1 = 0
for item in a:
sum1 += int(item)
if (sum1%2==0):
count+=1
return count | count-integers-with-even-digit-sum | Python Solution | anjaligupta0621 | 0 | 31 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,269 |
https://leetcode.com/problems/count-integers-with-even-digit-sum/discuss/1785205/Python3-%2B-Js-solution | class Solution:
def countEven(self, num: int) -> int:
count = 0
for i in range(1, num + 1):
summetion = 0
for digit in str(i):
summetion += int(digit)
if summetion % 2 == 0 :
count += 1
return count | count-integers-with-even-digit-sum | Python3 + Js solution | shakilbabu | -1 | 37 | count integers with even digit sum | 2,180 | 0.645 | Easy | 30,270 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1784873/Python-3-or-Dummy-Node-O(N)-Time-Solution | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
d=ListNode(0)
t=0
r=ListNode(0,d)
while head:
if head.val!=0:
t+=head.val
else:
print(t)
if t!=0:
d.next=ListNode(t)
d=d.next
t=0
head=head.next
return r.next.next | merge-nodes-in-between-zeros | Python 3 | Dummy Node O(N) Time Solution | MrShobhit | 6 | 576 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,271 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1784780/Python3-simulation | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
dummy = node = ListNode()
chunk = head
while chunk:
chunk = chunk.next
sm = 0
while chunk and chunk.val:
sm += chunk.val
chunk = chunk.next
if sm: node.next = node = ListNode(sm)
return dummy.next | merge-nodes-in-between-zeros | [Python3] simulation | ye15 | 6 | 417 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,272 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1784818/Simple-Python-Solution-in-O(n)-Time | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
current = head
ans = ListNode()
dummy = ans
while current is not None and current.next is not None:
if current.val == 0:
count = 0
current = current.next
while current.val != 0 and current is not None:
count += current.val
current = current.next
dummy.next = ListNode(count)
dummy = dummy.next
return ans.next | merge-nodes-in-between-zeros | Simple Python Solution in O(n) Time | anCoderr | 4 | 391 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,273 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/2787661/Easy-Python-solution-in-O(N)-TC | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
temp=head
sm=0
temp=temp.next
curr=head
while temp:
if temp.val==0:
curr=curr.next
curr.val=sm
sm=0
else:
sm+=temp.val
temp=temp.next
curr.next=None
return head.next | merge-nodes-in-between-zeros | Easy Python solution in O(N) TC | shubham_1307 | 2 | 95 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,274 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1904688/python3-oror-easy-solution-oror-O(n)O(1) | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
res, cur = head, head.next
while cur.next:
if cur.val:
res.val += cur.val
else:
res.next = res = cur
cur = cur.next
res.next = None
return head | merge-nodes-in-between-zeros | python3 || easy solution || O(n)/O(1) | dereky4 | 1 | 104 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,275 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1787614/Python-two-pointers | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
ans = prev = ListNode(-1)
ans.next = curr = head
while curr.next:
if curr.val == 0:
prev.next = curr
prev = curr
else:
prev.val += curr.val
prev.next = curr.next
curr = curr.next
prev.next = None
return ans.next | merge-nodes-in-between-zeros | Python, two pointers | emwalker | 1 | 43 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,276 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/2820924/Python-or-Beats-99.58-or-O(N)-Time-Complexity-or-O(1)-Space-Complexity | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
sumi=0
dummy = curr = head
curr = curr.next
while curr:
if curr.val!=0:
sumi+=curr.val
else:
dummy=dummy.next
dummy.val=sumi
sumi=0
curr=curr.next
dummy.next=None
return head.next | merge-nodes-in-between-zeros | Python | Beats 99.58% | O(N) Time Complexity | O(1) Space Complexity | liontech_123 | 0 | 2 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,277 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/2692527/Python3-Simple-Solution | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
arr = []
s = -1
while head is not None:
if head.val == 0:
if s != -1: arr.append(s)
s = 0
s += head.val
head = head.next
H = ListNode(arr[0])
prev = H
for num in arr[1:]:
tmpNode = ListNode(num)
prev.next = tmpNode
prev = tmpNode
return H | merge-nodes-in-between-zeros | Python3 Simple Solution | mediocre-coder | 0 | 2 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,278 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/2526510/Python-Simple-one-pass-Iterative-solution | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
p1 = p2 = head
while p2 and p2.next:
p2 = p2.next
p1.val += p2.val
if p2.val == 0:
if p2.next:
p1.next = p2
p1 = p2
else:
p1.next = None
return head | merge-nodes-in-between-zeros | [Python] Simple one-pass Iterative solution | kavansoni | 0 | 36 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,279 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/2237455/Python3-SImple-solution | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev = head
curr = head.next
sumNode = 0
while curr:
if curr.val == 0:
# We have reached a 0 node
# Create a new node with sum of all nodes between zeros
new_node = ListNode(sumNode)
prev.next = new_node
# Skip the 0 node
new_node.next = curr.next
# this is needed to link the new sum node (new_node) to prev node (old new_node)
prev = new_node
sumNode = 0
else:
sumNode += curr.val
curr = curr.next
return head.next | merge-nodes-in-between-zeros | [Python3] SImple solution | Gp05 | 0 | 7 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,280 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1962701/Merge-Nodes-in-Between-Zeros-O(n) | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
newList = ListNode(0)
pointer = newList
while head and head.next:
if head.val == 0:
head = head.next
nodesum = 0
while head.val > 0:
nodesum += head.val
head = head.next
pointer.next = ListNode(nodesum)
pointer =pointer.next
return newList.next | merge-nodes-in-between-zeros | Merge Nodes in Between Zeros O(n) | Edithturn | 0 | 35 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,281 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1901147/Python3-simple-solution | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
temp = head
x1 = False
x2 = False
c = 0
head1 = ListNode()
temp1 = head1
while temp:
if temp.val == 0:
if x1:
if x2:
temp1.next = ListNode(c)
temp1 = temp1.next
else:
temp1.val = c
x2 = True
c = 0
else:
x1 = True
else:
c += temp.val
temp = temp.next
return head1 | merge-nodes-in-between-zeros | Python3 simple solution | EklavyaJoshi | 0 | 15 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,282 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1896237/Python-Iterative-2-Approaches-oror-Time-O(N)-Space-O(1) | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
"""
time: O(n)
space: O(1)
"""
if not head:
return head
dummy_node = second_head = ListNode(-1)
curr, prev = head.next, head
while curr:
sum_ = 0
while (prev.val == 0) and (curr.val != 0):
sum_ += curr.val
curr = curr.next
new_node = ListNode(sum_)
dummy_node.next = new_node
dummy_node = dummy_node.next
prev, curr = curr, curr.next
return second_head.next | merge-nodes-in-between-zeros | Python Iterative - 2 Approaches || Time - O(N), Space - O(1) | ChidinmaKO | 0 | 35 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,283 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1896237/Python-Iterative-2-Approaches-oror-Time-O(N)-Space-O(1) | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
"""
time: O(n)
space: O(1)
"""
if not head:
return head
curr, prev, sum_ = head.next, head, 0
while curr:
if curr.val == 0:
prev.next = ListNode(sum_)
prev = prev.next
sum_ = 0
else:
sum_ += curr.val
curr = curr.next
return head.next | merge-nodes-in-between-zeros | Python Iterative - 2 Approaches || Time - O(N), Space - O(1) | ChidinmaKO | 0 | 35 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,284 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1895286/Python3-Solution-with-using-two-pointers | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
dummy = ListNode()
cur = dummy
prev_sum = 0
while head:
if head.val == 0:
cur.val = prev_sum
# skip last zero element
if head.next:
cur.next = ListNode()
cur = cur.next
prev_sum = 0
else:
prev_sum += head.val
head = head.next
return dummy.next | merge-nodes-in-between-zeros | [Python3] Solution with using two pointers | maosipov11 | 0 | 21 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,285 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1877950/Easy-to-understand-python-submission-100-working. | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
ans=ListNode()
temp=ans
s=0
head=head.next
while head:
if head.val!=0:
s+=head.val
else:
temp.val=s
s=0
if head.next!=None:
temp.next=ListNode()
temp=temp.next
head=head.next
return ans | merge-nodes-in-between-zeros | Easy to understand python submission 100% working. | tkdhimanshusingh | 0 | 32 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,286 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1799554/Python-Easy-creating-new-chain | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
root=None
curr=None
while head :
if head.val==0 :
head=head.next
continue
x=0
while head and head.val :
x+=head.val
head=head.next
if not x : continue
if root is None :
root=ListNode(x)
curr=root
else :
curr.next=ListNode(x)
curr=curr.next
return root | merge-nodes-in-between-zeros | Python Easy creating new chain | P3rf3ct0 | 0 | 76 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,287 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1791087/Relink-in-place-one-pass.-85-speed | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
node = head
while node.next:
if node.next.val != 0:
node.val += node.next.val
node.next = node.next.next
elif node.next.next:
node = node.next
else:
node.next = None
return head | merge-nodes-in-between-zeros | Relink in place, one pass. 85% speed | EvgenySH | 0 | 35 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,288 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1785303/Simple-solution-O(n)-(js%2Bpython3) | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
cur = head.next
dummy = ListNode(-1)
newHead = dummy
summetion = 0
while cur :
if cur.val == 0 :
newHead.next = ListNode(summetion)
newHead = newHead.next
summetion = 0
else :
summetion += cur.val
cur = cur.next
return dummy.next | merge-nodes-in-between-zeros | Simple solution - O(n) (js+python3) | shakilbabu | 0 | 36 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,289 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1785274/Python3-O(N)-Recursive-and-Iterative-Solutions | class Solution:
def getMerged(self, head):
if not head.next:
return None, 0
nextList, _sum = self.getMerged(head.next)
if head.val:
return nextList, _sum + head.val
else:
head.next = nextList
head.val = _sum
return head, 0
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
return self.getMerged(head)[0] | merge-nodes-in-between-zeros | [Python3] O(N) - Recursive & Iterative Solutions | dwschrute | 0 | 31 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,290 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1785274/Python3-O(N)-Recursive-and-Iterative-Solutions | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow, fast = head, head.next
accum = 0
while fast:
if not fast.val:
slow.val = accum
if fast.next:
slow = slow.next
else:
slow.next = None
accum = 0
else:
accum += fast.val
fast = fast.next
return head | merge-nodes-in-between-zeros | [Python3] O(N) - Recursive & Iterative Solutions | dwschrute | 0 | 31 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,291 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1785081/Python3-simple-intuitive-solution-explained | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
nodes = []
while head:
nodes.append(head.val)
head = head.next
result = ListNode()
temp = []
cumsum = 0
for i in nodes[1:]:
if i != 0:
cumsum += i
else:
temp.append(cumsum)
cumsum = 0
head = result
for i in temp:
head.next = ListNode(i)
head = head.next
return result.next | merge-nodes-in-between-zeros | Python3 simple intuitive solution explained | s_mugisha | 0 | 23 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,292 |
https://leetcode.com/problems/merge-nodes-in-between-zeros/discuss/1785029/python3-solution-easy-and-clean | class Solution:
def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
temp=head.next
s=0
newHead=dummy=ListNode(-1)
while temp:
if temp.val==0:
newHead.next=ListNode(s)
newHead=newHead.next
s=0
else:
s+=temp.val
temp=temp.next
return dummy.next | merge-nodes-in-between-zeros | python3 solution easy and clean | Karna61814 | 0 | 22 | merge nodes in between zeros | 2,181 | 0.867 | Medium | 30,293 |
https://leetcode.com/problems/construct-string-with-repeat-limit/discuss/1784789/Python3-priority-queue | class Solution:
def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
pq = [(-ord(k), v) for k, v in Counter(s).items()]
heapify(pq)
ans = []
while pq:
k, v = heappop(pq)
if ans and ans[-1] == k:
if not pq: break
kk, vv = heappop(pq)
ans.append(kk)
if vv-1: heappush(pq, (kk, vv-1))
heappush(pq, (k, v))
else:
m = min(v, repeatLimit)
ans.extend([k]*m)
if v-m: heappush(pq, (k, v-m))
return "".join(chr(-x) for x in ans) | construct-string-with-repeat-limit | [Python3] priority queue | ye15 | 23 | 1,100 | construct string with repeat limit | 2,182 | 0.519 | Medium | 30,294 |
https://leetcode.com/problems/construct-string-with-repeat-limit/discuss/1784794/Greedy-Python-Solution-or-100-Runtime | class Solution:
def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
table = Counter(s)
char_set = ['0', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's',
't', 'u', 'v', 'w', 'x', 'y', 'z']
sorted_table = []
for i in range(26,-1,-1):
if char_set[i] in table:
sorted_table.append((char_set[i],table[char_set[i]]))
result = ""
n = len(sorted_table)
for i in range(n):
char, curr_freq = sorted_table[i] # The lexicographically larger character and its frequency
index_to_take_from = i + 1 # We take from this index the next lexicographically larger character(TEMP) if the repeatLimit for A is exceeded
while curr_freq > repeatLimit: # Limit exceeded
result += char*repeatLimit # Add repeatLimit number of character A's
curr_freq -= repeatLimit # Decrease frequency of character A
# Now we search for the next lexicographically superior character that can be used once
while index_to_take_from < n: # Till we run out of characters
ch_avail, freq_avail = sorted_table[index_to_take_from]
if freq_avail == 0: # If its freq is 0 that means that it was previously used. This is done as we are not removing the character from table if its freq becomes 0.
index_to_take_from += 1 # Check for next lexicographically superior character
else:
result += ch_avail # If found then add that character
sorted_table[index_to_take_from] = (ch_avail,freq_avail-1) # Update the new characters frequency
break
else:
break # We cannot find any lexicographically superior character
else:
result += char*curr_freq # If the freq is in limits then just add them
return result | construct-string-with-repeat-limit | Greedy Python Solution | 100% Runtime | anCoderr | 4 | 425 | construct string with repeat limit | 2,182 | 0.519 | Medium | 30,295 |
https://leetcode.com/problems/construct-string-with-repeat-limit/discuss/1973741/python-3-oror-O(n)O(1)-oror-without-priority-queue | class Solution:
def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
count = collections.Counter(s)
chrs = list(map(list, sorted(count.items(), reverse=True)))
res = []
first, second = 0, 1
n = len(chrs)
while second < n:
if chrs[first][1] <= repeatLimit:
res.append(chrs[first][0] * chrs[first][1])
first += 1
while chrs[first][1] == 0:
first += 1
if first >= second:
second = first + 1
else:
res.append(chrs[first][0] * repeatLimit + chrs[second][0])
chrs[first][1] -= repeatLimit
chrs[second][1] -= 1
if chrs[second][1] == 0:
second += 1
res.append(chrs[first][0] * min(repeatLimit, chrs[first][1]))
return ''.join(res) | construct-string-with-repeat-limit | python 3 || O(n)/O(1) || without priority queue | dereky4 | 1 | 52 | construct string with repeat limit | 2,182 | 0.519 | Medium | 30,296 |
https://leetcode.com/problems/construct-string-with-repeat-limit/discuss/1953156/8-Lines-Python-Solution-oror-80-Faster-oror-Memory-less-than-80 | class Solution:
def repeatLimitedString(self, s: str, l: int) -> str:
C=[list(x) for x in sorted(Counter(s).items(),reverse=True)] ; ans=''
if len(C)==1: return C[0][0]*l
while len(C)>1:
ans+=C[0][0]*min(C[0][1],l)
C[0][1]-=min(C[0][1],l)
if C[0][1]>0: ans+=C[1][0] ; C[1][1]-=1
C=[c for c in C if c[1]>0]
return ans+C[0][0]*min(C[0][1],l) | construct-string-with-repeat-limit | 8-Lines Python Solution || 80% Faster || Memory less than 80% | Taha-C | 0 | 44 | construct string with repeat limit | 2,182 | 0.519 | Medium | 30,297 |
https://leetcode.com/problems/construct-string-with-repeat-limit/discuss/1787058/Python-O(n)-Time-Solution | class Solution:
def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
hmap = OrderedDict(Counter(s))
hmap = dict(sorted(hmap.items(), key = lambda x: x[0], reverse=True))
result = ""
while len(hmap) > 0:
keys = list(hmap.keys())
first = keys[0]
if hmap[first] > repeatLimit:
result += first * repeatLimit
hmap[first] -= repeatLimit
if len(hmap) == 1:
return result
result += keys[1]
hmap[keys[1]] -= 1
if hmap[keys[1]] == 0:
del hmap[keys[1]]
else:
result += first * hmap[first]
del hmap[first]
return result | construct-string-with-repeat-limit | [Python] O(n) Time Solution | tejeshreddy111 | 0 | 16 | construct string with repeat limit | 2,182 | 0.519 | Medium | 30,298 |
https://leetcode.com/problems/construct-string-with-repeat-limit/discuss/1786630/Python-3-Heap-and-wait-character | class Solution:
def repeatLimitedString(self, s: str, r: int) -> str:
cnt = Counter(s)
q = [-(ord(x) - ord('a')) for x in cnt]
heapq.heapify(q)
wait = ''
ans = ''
while q:
cur = chr(ord('a') - heappop(q))
if wait:
ans += cur
cnt[cur] -= 1
if cnt[cur]:
heappush(q, -(ord(cur) - ord('a')))
heappush(q, -(ord(wait) - ord('a')))
wait = ""
else:
if cnt[cur] <= r:
ans += cur * cnt[cur]
else:
ans += cur * r
cnt[cur] -= r
wait = cur
return ans | construct-string-with-repeat-limit | [Python 3] Heap and wait character | chestnut890123 | 0 | 28 | construct string with repeat limit | 2,182 | 0.519 | Medium | 30,299 |
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