<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, s: str) -> int:
        pos = defaultdict(lambda: [inf, -inf])
        for i,c in enumerate(s):
            pos[c] = [min(pos[c][0], i), max(pos[c][1], i)] # min and max char pos in s
        freq = defaultdict(int)
        max_len = 1
        left = 0
        for right in range(len(s)):
            freq[s[right]] += 1
            while len(freq) > 2 or len(freq) == 2 and min(freq.values()) > 1: # if violates statement #1
                freq[s[left]] -= 1
                if freq[s[left]] == 0:
                    del freq[s[left]] # clean up the freq counter.... also could be Counter() :/
                left += 1
            max_char = max(freq.keys(), key=lambda x: freq[x]) # find max freq element
            max_len = max(max_len, freq[max_char] + (pos[max_char][0] < left or right < pos[max_char][1])) # statement #2
        return max_len
16800