<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:
        lk = {}
        prev = text[0]
        length = len(text)
        st = 0
        ct = Counter(text)
        print(ct)
        for i in range(length):
            t = text[i]
            if t != prev:
                if prev in lk:
                    lk[prev].append([st,i])
                else:
                    lk[prev] = [[st,i]]
                st = i
            prev = t
        if prev in lk:
            lk[prev].append([st,length])
        else:
            lk[prev] = [[st,length]]
        ans = 0
        for k,v in lk.items():
            prev = (0,0)
            for st,end in v:
                ans = max(ans,(min(ct[k],(end-st+1))))
                if (st==(prev[1]+1)):
                    ans = max(ans,min(ct[k],end-prev[0]))
                prev = (st,end)
        return ans




18900