<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:
        group = []
        dic = {}
        i = 0

        while i < len(text):
            char = text[i]
            count = 0
            while i < len(text) and text[i] == char:
                count += 1
                i += 1
            group.append((char, count))
            dic[char] = dic.get(char, 0) + count

        res = 0

        for i in range(len(group)):
            (k, count) = group[i]
            res = max(res, min(count + 1, dic[k]))
        for i in range(1, len(group) - 1):
            if group[i - 1][0] == group[i + 1][0] and group[i][1] == 1:
                count = min(group[i - 1][1] + group[i + 1][1] + 1, dic[group[i - 1][0]])
                res = max(res, count)
        return res
        
17800