2026 July 25 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.

1876. Substrings of Size Three with Distinct Characters Easy 76.9%

3956. Maximum Sum of M Non-Overlapping Subarrays I Hard 26.2%

3957. Maximum Sum of M Non-Overlapping Subarrays II Hard 15.5%

Here is the meeting url:

https://us05web.zoom.us/j/82553858456?pwd=fnK5Vb0CcfFpv0lnouILjeEs2z2Khc.1 

Full details:

Topic: Leet Code meeting
Time: Mar 28, 2026 10:00 AM Pacific Time (US and Canada)
        Every week on Sat, 110 occurrence(s)
Please download and import the following iCalendar (.ics) files to your calendar system.
Weekly: https://us05web.zoom.us/meeting/tZYocuqgqzMuH9IuHUphHIiZ_4T4IndJLToX/ics?icsToken=DJFc88pL6-me_TneLwAALAAAAA70ShjYgLH6RfEnqUm5LK03v2OgtbXujwlKXRMVgIVWCY9aONPJMCycMlkhyPZ3XSrMVVgXDSYlGF_r7DAwMDAwMQ&meetingMasterEventId=MGJmt5lJTd6mPygcS0hiAQ
Join Zoom Meeting
https://us05web.zoom.us/j/82553858456?pwd=fnK5Vb0CcfFpv0lnouILjeEs2z2Khc.1

Meeting ID: 825 5385 8456
Passcode: 182596

[Anuj Patnaik]

class Solution:

def countGoodSubstrings(self, s: str) -> int:

if len(s) < 3:

return 0

if len(s) == 3:

if s[0] != s[1] and s[1] != s[2] and s[0] != s[2]:

return 1

else:

return 0

count = 0

for i in range(len(s) - 2):

if s[i] != s[i+1] and s[i+1] != s[i+2] and s[i] != s[i+2]:

count += 1

return count

[Anuj Patnaik]

class Solution:

def countGoodSubstrings(self, s: str) -> int:

if len(s) < 3:

return 0

count = 0

for i in range(len(s) - 2):

if s[i] != s[i+1] and s[i+1] != s[i+2] and s[i] != s[i+2]:

count += 1

return count

[Carlos Green]

/**

* @param {string} s

* @return {number}

*/

var countGoodSubstrings = function(s) {

let count = 0

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

const slice = new Set(s.slice(i,i+3))

if (slice.size === 3) {

count++

}

}

return count

};

[Jagrut]

1876. Substrings of Size Three with Distinct Characters 

class Solution:

def countGoodSubstrings(self, s: str) -> int:

res = 0

k = 3

window_counts = {}

left = 0

for right, in_char in enumerate(s):

if in_char not in window_counts:

window_counts[in_char] = 0

window_counts[in_char] += 1

if right - left + 1 > k:

out_char = s[left]

window_counts[out_char] -= 1

if window_counts[out_char] == 0:

del window_counts[out_char]

left += 1

if right - left + 1 == k:

if len(window_counts) == k:

res += 1

return res

[Allen S.]

// isGood() abstracted
func countGoodSubstrings(s string) int {
    count := 0
    for i := 0; i <= len(s) - 3; i++ {
        if isGood(s[i:i+3]) {
            count++
        }
    }
    return count
}

func isGood(s string) bool {
    return s[0] != s[1] && s[1] != s[2] && s[0] != s[2]

[Jagrut]

3956. Maximum Sum of M Non-Overlapping Subarrays I 

class Solution:

def maximumSum(self, nums: List[int], m: int, l: int, r: int) -> int:

n = len(nums)

NEG_INF = float('-inf')

# pf[i] = sum of nums[0..i], inclusive

pf = [0] * n

pf[0] = nums[0]

for i in range(1, n):

pf[i] = pf[i - 1] + nums[i]

def prefix_sum(idx):

# sum of nums[0..idx-1] -- i.e. everything strictly before idx

if idx == 0:

return 0

return pf[idx - 1]

dp = [[NEG_INF] * (n + 1) for _ in range(m + 1)]

for i in range(n + 1):

dp[0][i] = 0

for j in range(1, m + 1):

window = deque() # holds (idx, g_value), g decreasing front-to-back

for i in range(n + 1):

candidate_idx = i - l

if candidate_idx >= 0 and dp[j - 1][candidate_idx] != NEG_INF:

g_value = dp[j - 1][candidate_idx] - prefix_sum(candidate_idx)

while window and window[-1][1] <= g_value:

window.pop()

window.append((candidate_idx, g_value))

while window and window[0][0] < i - r:

window.popleft()

best = dp[j][i - 1] if i >= 1 else NEG_INF

if window:

candidate_value = prefix_sum(i) + window[0][1]

if candidate_value > best:

best = candidate_value

dp[j][i] = best

return max(dp[j][n] for j in range(1, m + 1))

[Gowrima Jayaramu]

class Solution:

def countGoodSubstrings(self, s: str) -> int:

count = 0

i, j = 0, 0

substr = ""

while j < len(s):

if s[j] not in substr:

substr += s[j]

if len(substr) == 3:

count += 1

substr = ""

i += 1

j = i

continue

j += 1

else:

substr = ""

i += 1

j = i

return count

'''

Time complexity: O(n)

Space complexity: O(1)

'''