<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,vector<int>> map;
        int cnt = 0;
        int cnt1 = 0; //for maintaining gap between non-consecutive
        int res = 0;
        int mx = 0;
        
        for(int i=0; i<text.size(); i++)
            map[text[i]].push_back(i);
        
        vector<int> temp;
        for(auto m : map)
        {
            temp = m.second;
            cnt = 1;
            cnt1 = 0;
            mx = 0;
            for(int i=1; i<temp.size(); i++)
            {
                if(temp[i]==temp[i-1]+1)
                    cnt++;
                else
                {
                    if(temp[i]==temp[i-1]+2)
                        cnt1 = cnt;
                    else
                        cnt1 = 0;
                    cnt = 1;
                }
                mx = max(mx,cnt+cnt1);
            }
            mx = mx + ((temp.size()>mx)?1:0);
            res = max(res,mx);
        }
        return res;
            
    }
};
11000