<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 s) {
        int n=s.size();
    vector<int>index[26];
    for(int i=0;i<n;i++)
    {
        index[s[i]-'a'].push_back(i);
    }
    int ans=1;
    for(int i=0;i<26;i++)
    {
        int curans=0;
        int curlen=1;
        int prevlen=0;
        for(int j=1;j<index[i].size();j++)
        {
            if((index[i][j])==(index[i][j-1]+1))
            {
                curlen++;
            }
            else
            {
                prevlen=0;
                if(index[i][j]==(index[i][j-1]+2))
                {
                    prevlen=curlen;
                }
                curlen=1;
            }
            curans=max(curans,prevlen+curlen);
        }
        if(curans<(index[i].size()))
        {
            curans++;
        }
        ans=max(ans,curans);
    }
    return ans;
    }
};
9100