<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:
    bool check(string& s,int len,int k)
    {
        vector<int> freq(26,0);

        for(int i=0;i<len;i++)
        {
            freq[s[i]-'A']++;
        }
        int maxi=*max_element(freq.begin(),freq.end());
        int rem=len-maxi;
        if(rem<=k) return true;
        int n=s.length();
        for(int i=len;i<n;i++)
        {
            char prev=s[i-len];
            freq[prev-'A']--;
            char curr=s[i];
            freq[curr-'A']++;
            maxi=*max_element(freq.begin(),freq.end());
            rem=len-maxi;
            if(rem<=k) return true;
        }
        return false;

    }
    int characterReplacement(string s, int k) {
        int n=s.length();
        int start=1,end=n;

        int ans=-1;

        while(start<=end)
        {
            int mid=start+(end-start)/2;
            if(check(s,mid,k))
            {
                ans=mid;
                start=mid+1;
            }
            else end=mid-1;
        }
        return ans;
    }
};
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