2026 May 2 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.

1221. Split a String in Balanced Strings Easy 87.4%

Carry over problems from last week:

3853. Merge Close Characters Medium 54.6%

3841. Palindromic Path Queries in a Tree Hard 35.5%

We will have a different Zoom meeting that Bhupathi is providing this week:

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.
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Join Zoom Meeting
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Meeting ID: 825 5385 8456
Passcode: 182596

[Bhupathi Kakarlapudi]

class Solution:

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

balance:int = 0

num_of_balance_strings:int = 0

for ch in s:

if ch == 'L':

balance += 1

else:

balance -= 1

if balance == 0:

num_of_balance_strings += 1

return num_of_balance_strings

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[Allen S.]

func balancedStringSplit(s string) int {

balance := 0

count := 0

for _, v := range s {

if v == 'L' {

balance++

} else {

balance--

}

if balance == 0 {

count++

}

}

return count

}

[Allen S.]

func mergeCharacters(s string, k int) string {

latestIndex := map[byte]int{}

keyOffset := map[byte]int{}

res := []byte{}

for i := range s {

// close ?

if j, ok := latestIndex[s[i]]; ok && i - j - keyOffset[s[i]] <= k { // yes

for key := range keyOffset {

keyOffset[key]++

}

} else { // not close

res = append(res, s[i])

latestIndex[s[i]] = i

keyOffset[s[i]] = 0

}

}

return string(res)

}

[Anuj Patnaik]

class Solution:

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

balanced_str = 0

ct = 0

for i in s:

if i == 'R':

balanced_str -= 1

else:

balanced_str += 1

if balanced_str == 0:

ct += 1

return ct

[Bhupathi Kakarlapudi]

class Solution:

def mergeCharacters(self, s: str, k: int) -> str:

arr = list(s)

while True:

merged = False

n = len(arr)

for i in range(n):

for j in range(i + 1, min(i+1+k,n)):

if arr[i] == arr[j]:

arr.pop(j) # right merges into left

merged = True

break

if merged:

break

if not merged:

break

return "".join(arr)

To view this discussion visit https://groups.google.com/d/msgid/leetcode-meetup/2d26cec9-7d20-4892-a3a2-2bd6f342c555n%40googlegroups.com.

[Rudy W]

class Solution:

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

# intuition: consider char 'L' equal to integer -1 value

# and char 'R' -> integer +1

# since s is a balanced string, that means the number of 'L' chars is equal to the number of 'R' chars

# that also means the total value of string s is 0

# here create a hash map of char 'L' and char 'R', assign both to -1 and 1 respectively

hash = {'L': -1, 'R': 1}

max = 0

current_total = 0

for c in s:

current_total += hash[c]

# when the current total is equal to 0

# that is when the substring has an equal quantity of 'L' and 'R' chars

if current_total == 0:

max += 1

return max

class Solution:

def mergeCharacters(self, s: str, k: int) -> str:

# intuition: store the up to date/current index in the hash map

# need to keep track of number of deletion/merge

# hash map will store the char index

hash = defaultdict(int)

new_str = ''

# keep track deletion count so that the char index is up to date to the new_str index

del_count = 0

for i, char in enumerate(s):

# print(f'new_str={new_str}, del_count={del_count}')

if char not in hash:

hash[char] = i - del_count

new_str += char

elif char in hash:

if i - hash[char] - del_count > k:

hash[char] = i - del_count

new_str += char

else:

# remove char -> don't write to new_str

del_count += 1

return new_str

To view this discussion visit https://groups.google.com/d/msgid/leetcode-meetup/CA%2BohYXhwq%3Dfp61Tm6CrsAbbz0_WU1hfiYRjJQimHtSKb%3D2vVSg%40mail.gmail.com.