2025 January 24 Problems

[daryl...@gmail.com]

Feel free to work on any of the problems you want; we'll have people present their solutions at 11:30.

Please post your solutions to this thread so others can use this as a reference.

3707. Equal Score Substrings Easy 56.7%

3654. Minimum Sum After Divisible Sum Deletions Medium 45.7%

3510. Minimum Pair Removal to Sort Array II Hard 15.3%

We're going to use the same Google meet link as the last few times: https://meet.google.com/xnn-kifj-jnx?pli=1

[Anuj Patnaik]

class Solution:

def scoreBalance(self, s: str) -> bool:

ind_char = list(s)

for i in range(len(ind_char)):

ind_char[i] = ord(ind_char[i]) - ord("a") + 1

total_sum = sum(ind_char)

l = 0

for j in range(len(ind_char)):

l += ind_char[j]

r = total_sum - l

if r == l:

return True

return False

[saurabh s]

class Solution:

def scoreBalance(self, s: str) -> bool:

total_sum = sum(ord(c)-ord('a')+1 for c in s)

if total_sum % 2 != 0:

return False

target = total_sum // 2

current_sum = 0

for c in s:

current_sum += ord(c) - ord('a') + 1

if current_sum == target:

return True

return False

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[Allen S.]

func scoreBalance(s string) bool {

/*

two passes solution

*/

const offset = 'a' - 1

var totalScore rune // rune type is an alias for int32

for _, v := range s {

totalScore += v - offset

}

var prefixScore rune

for _, v := range s {

prefixScore += v - offset

if prefixScore * 2 == totalScore {

return true

}

}

return false

}

On Saturday, January 24, 2026 at 9:57:47 AM UTC-8 daryl...@gmail.com wrote:

[Bhupathi Kakarlapudi]

class Solution:

def scoreBalance(self, s: str) -> bool:

score_s = [ord(s[i])-ord('a') + 1 for i in range(len(s))]

lidx:int = 1

ridx:int = 1

while (ridx < len(score_s)):

# left:list = score_s[:lidx]

# right:list = score_s[ridx:]

if (sum(score_s[:lidx]) == sum(score_s[ridx:])):

return True

else:

lidx += 1

ridx += 1

return False

--

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[vinay]

3707. Equal Score Substrings Easy 56.7%

using prefix sum here. 

time o(n), space o(1)

class Solution {

    public boolean scoreBalance(String s) {

        int totalScore = 0;

        int score = 0;

       

        for(int i = 0; i < s.length(); i++){

            totalScore +=  s.charAt(i) - 'a' +1;

        }

        for(int i = 0; i < s.length(); i++){

            score += s.charAt(i) - 'a' +1;

            if(score == totalScore - score){

                return true;

            }

        }

        return false;

    }

}

On Sat, Jan 24, 2026 at 10:48 AM Allen S. <allen....@gmail.com> wrote:

--

[Wang Lijun]

class Solution:

def scoreBalance(self, s: str) -> bool:

total = 0

for c in s:

total += ord(c) - ord('a') + 1

if total % 2 != 0:

return False

target = total // 2

total = 0

for c in s:

total += ord(c) - ord('a') + 1

if total == target:

return True

return False

[Jagrut Sharma]

3707. Equal Score Substrings Easy 56.7%

class Solution:

def scoreBalance(self, s: str) -> bool:

letter_values = {chr(i): i - ord('a') + 1 for i in range(ord('a'), ord('z') + 1)}

sum = 0

target_sum = 0

for i in range(len(s)):

sum += letter_values[s[i]]

if sum % 2 == 1:

return False

target_sum = sum / 2

sum = 0

for i in range(len(s)):

sum += letter_values[s[i]]

if sum == target_sum:

return True

return False

3654. Minimum Sum After Divisible Sum Deletions Medium 45.7%

class Solution:

def minArraySum(self, nums: List[int], k: int) -> int:

dp = [float('inf')] * k

res = 0

dp[0] = 0

for n in nums:

res += n # prefix sum

r = res % k

res = min(res, dp[r]) # jump back to smaller sum with same remainder

dp[r] = res

return res

To view this discussion visit https://groups.google.com/d/msgid/leetcode-meetup/b123e0d3-2219-4c13-856b-6071f6b01e7dn%40googlegroups.com.

--

Jagrut

[Rag Mur]

3707. Equal Score Substrings Easy 56.7%

class Solution:

def scoreBalance(self, s: str) -> bool:

'''

l = 100

this is a rolling sum

bin search wont make muchdifference since you anyway need to look at all the letters

'''

OFFSET = 96

atoi = lambda l: ord(l) - OFFSET

totalScore = sum([atoi(l) for l in s])

if totalScore % 2 > 0:

return False

half = totalScore // 2

rolling_sum = 0

for l in s:

rolling_sum += atoi(l)

if rolling_sum == half:

return True

elif rolling_sum > half:

return False

return False

3654. Minimum Sum After Divisible Sum Deletions Medium 45.7%

class Solution:

def minArraySum(self, nums: List[int], k: int) -> int:

'''

the problem can be simpified to find all the disjoint subarrays that are divisible by k and turn them to 0.

this is based on prefix sum theory

prefix_sum[i] means sum(nums[:i+1])

a subarray [i:j] is divisible by k is

(prefix_sum[j] - prefix_sum[i-1]) % k == 0

this means prefix_sum[i-1]) % k == prefix_sum[j]) % k

'''

inf = float('inf')

output = 0

dp = [0] + [inf]*k

for n in nums:

output += n

output = dp[output%k] = min(dp[output%k], output)

return output

-

Raghu

On Saturday, January 24, 2026 at 9:57:47 AM UTC-8 daryl...@gmail.com wrote: