<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:
    def maxRepOpt1(self, text: str) -> int:
        freq = Counter(text)
        groups = []
        curr = ''
        currlen = 0
        text += ' '
        for i,a in enumerate(text):
            if a == curr:
                currlen+=1
            else:
                if curr:
                    groups.append((curr,currlen))
                curr = a
                currlen = 1
        N = len(groups)
        res = max([b for a,b in groups])
        for i,(a,l) in enumerate(groups):
            if l!=freq[a]:
                res = max(res,l+1)
            if i<N-2 and a == groups[i+2][0] and 1 == groups[i+1][1]:
                if l+groups[i+2][1]<freq[a]:
                    res = max(res,l+groups[i+2][1]+1)
                else:
                    res = max(res,l+groups[i+2][1])
        return res
18000