<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:
        # lets make them into groups by letters
        # if we have adjacent groups of the same letter next to each other we can combine them. We can also add one if there is another of this string somewhere
        # groupsOf count of groups per character type
        groups = dict()
        groupsL = [["", 0]]
        for c in text:
            if c == groupsL[-1][0]:
                groupsL[-1][1] += 1
            else:
                if c not in groups:
                    groups[c] = 0
                groups[c] += 1
                groupsL.append([c, 1])
        if len(groupsL) == 2:
            return groupsL[-1][1]
        maxC = 0
        for i in range(1, len(groupsL)):
            groupSize = groupsL[i][1]
            groupLetter = groupsL[i][0]
            maxC = max(maxC, groupSize)
            if i + 2 < len(groupsL) and groupsL[i + 1][1] == 1 and groupsL[i + 2][0] == groupLetter:
                groupSize += groupsL[i + 2][1]
                if groups[groupLetter] >= 3:
                    groupSize += 1
                maxC = max(maxC, groupSize)
            elif groups[groupLetter] >= 2:
                maxC = max(maxC, groupSize + 1)
        return maxC
17900