<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:
    def characterReplacement(self, s: str, k: int) -> int:
        count = {}
        res = 0
        maxfreq = 0
        l = 0
        for r in range(len(s)):
            count[s[r]] = 1+count.get(s[r],0)
            maxfreq = max(maxfreq,count[s[r]] )
            
            while (r-l+1) - maxfreq > k:
                count[s[l]] -= 1
                l+=1
            res = max(res, r-l+1)
        return res
16300