<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 start = 0;
        int end = 0;
        int res = 0;
        map<char,int> m;
        int max_chars_till_now = 0;

        while (end < s.length()) {
            m[s[end]]++;
            max_chars_till_now = max(max_chars_till_now, m[s[end]]);

            // increase start
            while ((end-start+1) - max_chars_till_now > k) {
                m[s[start]]--;
                if (m[s[start]] == 0)
                    m.erase(s[start]);
                start++;
            }

            res = max(res, end-start+1);
            end++;
        }

        return res;
    }
};
/*
Basic idea:
Keep max characters till now
How?
1. keep a variable max_chars_till_now = 0
2. add to map<char,freq>
3. if (end-start+1) - max_chars_till_now   > k
this means that number of chars to swap is greater than k
So, increment start.


*/
10500