Swap For Longest Repeated Character Substring
<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>
Hash Table; String; Sliding Window
function maxRepOpt1(text: string): number {
    const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
    const cnt: number[] = new Array(26).fill(0);
    for (const c of text) {
        cnt[idx(c)]++;
    }
    let ans = 0;
    let i = 0;
    const n = text.length;
    while (i < n) {
        let j = i;
        while (j < n && text[j] === text[i]) {
            ++j;
        }
        const l = j - i;
        let k = j + 1;
        while (k < n && text[k] === text[i]) {
            ++k;
        }
        const r = k - j - 1;
        ans = Math.max(ans, Math.min(cnt[idx(text[i])], l + r + 1));
        i = j;
    }
    return ans;
}
