Find Latest Group Of Size M - Leetcode Solution

Problem Description:

Given a string with lowercase letters only, if you are allowed to replace no more than ‘k’ letters with any letter, find the length of the longest substring having the same letters after replacement.

Example:

Input: s = "abbcb", k = 2

Output: 4

Explanation: Replace the 'c' and 'b' with 'a' to have a string of "aaaaa". The length of the longest substring with same letters is 4.

Solution:

We can solve this problem using the sliding window technique. We need to maintain a window of size m and keep a count of the maximum frequency of any character in this window. We will then find the length of the window by subtracting the count of the maximum frequency from m. If the length of the window is greater than k, then we need to shrink the window by moving the start of the window till the length of the window becomes less than or equal to k.

Algorithm:

1. Initialize two pointers, start and end, to the beginning of the string.
2. Initialize a variable, maxLength, to store the length of the longest substring with same characters.
3. Initialize a dictionary, freq, to keep track of the frequency of each character in the window.
4. Move the end pointer till it reaches the end of the string.
5. Increment the frequency of the character pointed by the end pointer in the dictionary.
6. Calculate the maximum frequency of any character in the window.
7. If the length of the window, m minus the maximum frequency, is greater than k, then we need to shrink the window.
8. Decrement the frequency of the character pointed by the start pointer from the dictionary.
9. Move the start pointer one step to the right.
10. Update the maxLength with the length of the window.
11. Return the maxLength.

Code:

Here is the Python code to solve this problem:

def findLatestGroup(s: str, m: int) -> int: n = len(s) start = 0 freq = {} maxLength = -1 maxFreq = 0

for end in range(n):
    freq[s[end]] = freq.get(s[end], 0) + 1
    maxFreq = max(maxFreq, freq[s[end]])
    
    if end - start + 1 - maxFreq > m:
        freq[s[start]] -= 1
        start += 1
        
    maxLength = max(maxLength, end - start + 1) if end - start + 1 == maxFreq else maxLength

return maxLength

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