# Leetcode 424 - Longest Repeating Character Replacement

### Understanding the problem

Given a string s that consists of only uppercase English letters, you can perform at most k operations on that string.

In one operation, you can choose any character of the string and change it to any other uppercase English character.

Find the length of the longest substring that contains all repeating letters you can get after performing the above operations.

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### Plan your solution

One approach to solve this problem is to use a sliding window and a count of the most frequent letter in the current window. We can use a hashmap to keep track of the frequency of each letter in the current window.

We start by initializing a left and right pointer, and a count of the most frequent letter. We then move the right pointer to the right and update the hashmap with the new letter. If the new letter is the most frequent letter, we increment the count. If the new letter is not the most frequent letter, we decrement the count.

If the number of operations we have performed, represented by the size of the window minus the count of the most frequent letter, is greater than k, we move the left pointer to the right and update the hashmap accordingly.

We update the maximum length of the substring every time we move the right pointer.

### Implement your solution

```
def characterReplacement(s: str, k: int) -> int:
left = 0
count = {}
max_count = 0
max_length = 0
for right in range(len(s)):
count[s[right]] = count.get(s[right], 0) + 1 #increment the count of current character
max_count = max(max_count, count[s[right]]) #update the max count of any char seen so far
if right - left + 1 - max_count > k:
count[s[left]] -= 1 #decrement the count of char at left pointer
left += 1 #move left pointer to right
max_length = max(max_length, right - left + 1) #update the max length of substring
return max_length
```

### Test your solution

Let’s test our solution with the example inputs:

```
s = "ABAB"
k = 2
print(characterReplacement(s, k))
```

Output: 4

As we can see, the output is correct and matches the expected output.

In summary, we have successfully implemented a solution to Leetcode Problem 424 by using a sliding window approach and keeping track of the frequency of each letter in the current window. We also kept track of the most frequent letter and the number of operations performed. We moved the left pointer and updated the hashmap accordingly if the number of operations performed was greater than k. Finally, we updated the maximum length of the substring every time we moved the right pointer. This approach has a time complexity of O(n) and a space complexity of O(1), where n is the size of the input string.