<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)
    {
        // At level h : count of characters that have frequency = h in our sliding window
        // loc maps our char to its level
        // height is maximum height of our tower right now
        vector<int> counts(2);
        vector<int> loc(26);
        counts[0] = 25;
        counts[1] = 1;
        loc[s[0]-'A'] = 1;

        int i = 0, j = 0;
        int height = 1;
        int ans = 1;
        for (j = 1; j < s.length(); j++) {
            char c = s[j];

            counts[loc[c-'A']]--;
            if (loc[c-'A']+1 > height) {
                height++;
                counts.push_back(0);
            }
            loc[c-'A']++;
            counts[loc[c-'A']]++;

            // We get the total number of char in our window, and count the number of char
            // which are other than the most frequent one
            int total = j - i + 1;
            int excess = total - height;
            while (excess > k) {
                char d = s[i];
                counts[loc[d-'A']]--;
                loc[d-'A']--;
                counts[loc[d-'A']]++;
                if (counts[height] == 0)
                    height--;
                i++;
                total = j - i + 1;
                excess = total - height;
            }
            ans = max(ans, total);
        }
        return ans;
    }
};
9000