<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:
        l, r = 0, 0
        chars = {}
        maxChars = 0
        res = 0
        while r < len(s):
            if maxChars + k >= len(s[l:r]):
                if res < len(s[l:r]):
                    res = len(s[l:r])
                chars[s[r]] = chars.get(s[r], 0) + 1
                if maxChars < chars.get(s[r], 0):
                    maxChars = chars[s[r]]
                r += 1
            else:
                chars[s[l]] = chars.get(s[l], 0) - 1
                l += 1
        if maxChars + k >= len(s[l:r]):
            return len(s[l:r])
        return res
16100