<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:
        repeat = [1]
        count = Counter(s)
        
        for i in range(1, len(s)):
            if s[i]== s[i-1]:
                repeat.append(repeat[i-1]+1)
            else:
                repeat.append(1)
        right = len(s)-1
        currMax = 1
        while right >= 0:
            firstRight = right - (repeat[right] - 1)
            lastLeft = firstRight - 2
            leftLen = 0
            thisSum = repeat[right]
            if lastLeft>=0 and s[lastLeft] == s[right]:
                leftLen = repeat[lastLeft]
                thisSum = leftLen+thisSum
            if thisSum < count[s[right]]:
                thisSum +=1
            currMax = max(currMax,thisSum)
            print(lastLeft, firstRight, currMax)
            right = firstRight-1
        return currMax
                
17000