<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) {
        int j=0,cnt=0,flag=0,cnt1=0,ans=1,k=0,n=text.size();
        for(int i=0;i<26;i++)
        {
           // cout<<i<<endl;
            char c=('a'+i);
           k=0,j=0;
            vector<pair<int,int>>v;
            while(k<n)
            {
                j=k;
                if(text[j]==c)
                {
                    while(j<n && (text[j]==c))
                    {
                        j++;
                       

                }
                v.push_back({k,j});
                k=j;
                }
                else
                {

                    j++;
                    k=j;
                }
            }
            int nn=v.size();
            for(int i=0;i<nn;i++)
            {

                ans=max(ans,(v[i].second-v[i].first + ((i+1<nn || i-1>=0)?1:0)));
                if(i+1<nn)
                {
                    if(v[i+1].first-v[i].second<=1)
                    {
                        int a=v[i+1].second-v[i+1].first + v[i].second-v[i].first;
                        if(i+2<nn || i-1>=0)
                        a+=1;
                        ans=max(ans,a);
                    }
                }
            }
        }
        return ans;
        
    }
};
9500