<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 {
public:
    int maxRepOpt1(string text) {
        int n = text.size();
        vector<int> dp(n, 1);
        for (int i = 0; i < n; i++) {
            if (i > 0 && text[i-1] == text[i]) {
                dp[i] = 1 + dp[i-1];
            }
        }
        // for (int n : dp) cout<<n <<" ";
        unordered_map<char,int> cnt;
        for (char c : text) cnt[c]++;
        int res = 1;
        for (int i = 0; i < n; i++) {
            int pre = i - dp[i] - 1;
            if (i - dp[i] - 1 >= 0 && text[i] == text[pre] && dp[i] + dp[pre] + 1 <= cnt[text[i]]) {
                res = max(res, dp[i] + dp[pre] + 1);
            } else {
                res = max(res, dp[i]);
                if (dp[i] < cnt[text[i]]) {
                    res = max(res, dp[i] + 1); 
                }
            }
        }
        return res;
    }
};
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