<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) {
        vector<int> f(150,0);
        for(int i=0;i<text.size();i++)
        {
            f[text[i]]++;
        }
        int n=text.size();

        int l=1;
        int h=n;
        int ans=0;
        while(l<=h)
        {
            int mid=l+(h-l)/2;
            vector<int>v(150,0);
            int temp=-1;
            for(int k=0;k<n;k++)
            {
                v[text[k]]++;
                if(k+1<mid)
                continue;
                int mx=-1;
                for(int m=95;m<=130;m++)
                {
                    if(v[m]==mid || (v[m]==mid-1 && f[m]>v[m]))
                    {
                        mx=mid;
                    }
                }
                if(mx!=-1)
                {
                    temp=mid;
                    break;
                }
                v[text[k-mid+1]]--;

            }
            if(temp>=ans)
            {
                ans=temp;
                l=mid+1;
            }
            else
            h=mid-1;
        }
        return ans;
    }
};
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