<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) {

        unordered_map<char, int > freq;
        int m = INT_MIN;
        unordered_set<char> s;
         
         const int n = text.size();
        for (int i = 0 ; i < n; i++)
        {
              freq[text[i]] += 1;
              s.insert(text[i]);
        }

        for(auto c: s)
        {
             
             int start = 0 ; 
             int k = 1;
             int z = 0;


             for(int i = 0 ; i < n ; i++)
             {
                if(text[i] != c && k)
                {
                    k =0;
                }
                else if( text[i] != c && !k)
                {
                       while(!k)
                       {
                          if(text[start] != c)
                          {
                             k =1;
                          }
                          start += 1;
                       }
                       k = 0;
                }
                m = max(m, min(freq[c] , i - start + 1));
             }
        }
        return m;
    }
};
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