<p>You are given a string <code>text</code>. You can swap two of the characters in the <code>text</code>.</p>

<p>Return <em>the length of the longest substring with repeated characters</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;ababa&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can swap the first &#39;b&#39; with the last &#39;a&#39;, or the last &#39;b&#39; with the first &#39;a&#39;. Then, the longest repeated character substring is &quot;aaa&quot; with length 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;aaabaaa&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> Swap &#39;b&#39; with the last &#39;a&#39; (or the first &#39;a&#39;), and we get longest repeated character substring &quot;aaaaaa&quot; with length 6.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;aaaaa&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> No need to swap, longest repeated character substring is &quot;aaaaa&quot; with length is 5.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= text.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>text</code> consist of lowercase English characters only.</li>
</ul>

class Solution {
public:
    int maxRepOpt1(string text) {
        
        map<char,int> freq;
        for(auto c : text) freq[c]++;

        int n = text.size();
        int index = 0;
        int maxlen = INT_MIN;

        while(index<n){
            
            int samecharindex = index;

            while(samecharindex<n && text[samecharindex]==text[index]){
                samecharindex++;
            }

            int samecharlen = samecharindex - index;

            int nextcharindex = samecharindex + 1;

            while(nextcharindex<n && text[nextcharindex]==text[index]){
                nextcharindex++;
            }
            int nextcharlen = nextcharindex - samecharindex - 1;

            maxlen = max( maxlen, min( freq[text[index]], nextcharlen+samecharlen+1 ) );

            index = samecharindex;
        }
        return maxlen;
    }
};
9300