<p>You are given a string <code>s</code> and an integer <code>k</code>. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most <code>k</code> times.</p>

<p>Return <em>the length of the longest substring containing the same letter you can get after performing the above operations</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ABAB&quot;, k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> Replace the two &#39;A&#39;s with two &#39;B&#39;s or vice versa.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;AABABBA&quot;, k = 1
<strong>Output:</strong> 4
<strong>Explanation:</strong> Replace the one &#39;A&#39; in the middle with &#39;B&#39; and form &quot;AABBBBA&quot;.
The substring &quot;BBBB&quot; has the longest repeating letters, which is 4.
There may exists other ways to achieve this answer too.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of only uppercase English letters.</li>
	<li><code>0 &lt;= k &lt;= s.length</code></li>
</ul>

class Solution {
public:
    int characterReplacement(string s, int k) {
        int n = s.size();
        int j = 0;
        int i = 0;
        int max_len = 0;
        int max_freq = 0;
        vector<int>occurence(26,0);
        while(j<n){
             occurence[s[j]-'A']++;
             max_freq = max(max_freq , occurence[s[j]-'A']);
            while((j-i+1-max_freq)>k){
                occurence[s[i]-'A']--;
                max_freq = 0;
                for(auto &it:occurence){
                    if(it>max_freq){
                        max_freq = it;
                    }
                }
                i++;
            }
            max_len = max(max_len,j-i+1);
            j++;
        }
        return max_len;
    }
};
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