<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 {
    bool good(unordered_map<char, int> &w, unordered_map<char, int> &m) {
        if(w.size() == 1) return true;
        if(w.size() > 2) return false;
        char a = '#', b;
        for(auto p : w) {
            a == '#' ? a = p.first : b = p.first;
        }
        if(w[a] == 1 && m.find(b) != m.end() || w[b] == 1 && (m.find(a) != m.end())) return true;
        return false;
    }
public:
    int maxRepOpt1(string text) {
        unordered_map<char, int> m, w;
        for(char c : text) m[c]++;

        int n = text.size(), i = 0, ans = 0;
        for(int j=0; j<n; j++) {
            w[text[j]]++;
            m[text[j]]--;
            if(!m[text[j]]) m.erase(text[j]);
            
            while(!good(w, m)) {
                w[text[i]]--;
                m[text[i]]++;
                if(!w[text[i]]) w.erase(text[i]);
                i++;
            }
            ans = max(ans, j - i + 1);
        }
        return ans;
    }
};
10400