<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 cc[26] = {0};
        char mc = s[0];
        cc[s[0]-'A']=1;
        int diversity = 0;
        int dInxSet=0, dInxGet=0, len = s.size();
        int dQueue[len];
        int i=0,j=1;
        int ans = 0;
        char prev = mc;
        while(j<len) {
           cc[s[j]-'A']++;
           if(prev!=s[j]) dQueue[dInxSet++] = j;
            if(s[j]!=mc) {
                diversity++;
                while(diversity>k) {
                    //cout<<j<<" "<<diversity<<" "<<ans<<endl;
                    int cur = j-i+1+k-diversity;
                    if(cur>ans) {
                        ans=cur;
                        cout<<ans<<":"<<s.substr(i,ans)<<endl;
                    }
                    cc[s[i]-'A'] -= (dQueue[dInxGet]-i);
                    diversity+=cc[s[i]-'A'];
                    //cout<<diversity<<" "<<i<<endl;
                    i=dQueue[dInxGet++];
                    //if(dInxGet>dInxSet) check
                    mc = s[i];
                    diversity-=cc[mc-'A'];
                    //cout<<diversity<<" "<<i<<endl;
                    //if(diversity<0) diversity=0;
                }
            }
            prev = s[j++];
        }
        //cout<<ans<<"-"<<diversity<<"-"<<i<<endl;
        while(dInxGet<dInxSet) {
            //cout<<j<<" "<<i<<" "<<diversity<<" "<<ans<<endl;
            int cur = j-i+k-diversity;
            if(cur>len) cur = len;
            if(cur>ans) {
                ans=cur;
                cout<<ans<<":"<<s.substr(i,ans)<<endl;
            }
            cc[s[i]-'A'] -= (dQueue[dInxGet]-i);
            diversity+=cc[s[i]-'A'];
            //cout<<diversity<<" "<<i<<endl;
            i=dQueue[dInxGet++];
            //if(dInxGet>dInxSet) check
            mc = s[i];
            diversity-=cc[mc-'A'];
            //cout<<diversity<<" "<<i<<endl;
            //if(diversity<0) diversity=0;
        }
        int curr = j-i+k-diversity;
        if(curr>len) curr = len;
        if(ans<curr) ans=curr;
        return ans;
    }
};

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