<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> alphabet(26,0);
         int ans = 1;
         for(char &ch : text)
            alphabet[ch-'a']++;
          int n = text.size(), val = 1;
             vector<int> v(n);
             v[0] = 1;
        for(int i = 1; i < n; i++)
        {
            if(text[i] == text[i-1])
                v[i] = v[i-1]+1;
            else
                v[i] = 1;
        } 
        for(int i = n-1; i > 0; i--)
        {
            if(text[i] == text[i-1])
                val++;
            else
            {
                if(i == 1 || text[i] != text[i-2])
                {
                    val += 1;
                    val = min(val, alphabet[text[i]-'a']);
                    ans = max(val, ans);
                    val = 1;
                }
                else
                {
                    val += v[i-2]+1;
                    val = min(val, alphabet[text[i]-'a']);
                    ans = max(val, ans);
                    val = 1;
                } 
            }
            ans = max(min(v[i]+1, alphabet[text[i]-'a']), ans);
        }
        return max(ans,val);
    }
};
8600