Longest Nice Substring
<p>A string <code>s</code> is <strong>nice</strong> if, for every letter of the alphabet that <code>s</code> contains, it appears <strong>both</strong> in uppercase and lowercase. For example, <code>&quot;abABB&quot;</code> is nice because <code>&#39;A&#39;</code> and <code>&#39;a&#39;</code> appear, and <code>&#39;B&#39;</code> and <code>&#39;b&#39;</code> appear. However, <code>&quot;abA&quot;</code> is not because <code>&#39;b&#39;</code> appears, but <code>&#39;B&#39;</code> does not.</p>

<p>Given a string <code>s</code>, return <em>the longest <strong>substring</strong> of <code>s</code> that is <strong>nice</strong>. If there are multiple, return the substring of the <strong>earliest</strong> occurrence. If there are none, return an empty string</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;YazaAay&quot;
<strong>Output:</strong> &quot;aAa&quot;
<strong>Explanation: </strong>&quot;aAa&quot; is a nice string because &#39;A/a&#39; is the only letter of the alphabet in s, and both &#39;A&#39; and &#39;a&#39; appear.
&quot;aAa&quot; is the longest nice substring.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;Bb&quot;
<strong>Output:</strong> &quot;Bb&quot;
<strong>Explanation:</strong> &quot;Bb&quot; is a nice string because both &#39;B&#39; and &#39;b&#39; appear. The whole string is a substring.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;c&quot;
<strong>Output:</strong> &quot;&quot;
<strong>Explanation:</strong> There are no nice substrings.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists of uppercase and lowercase English letters.</li>
</ul>
Bit Manipulation; Hash Table; String; Divide and Conquer; Sliding Window
func longestNiceSubstring(s string) string {
	n := len(s)
	k, mx := -1, 0
	for i := 0; i < n; i++ {
		ss := map[byte]bool{}
		for j := i; j < n; j++ {
			ss[s[j]] = true
			ok := true
			for a := range ss {
				b := a ^ 32
				if !(ss[a] && ss[b]) {
					ok = false
					break
				}
			}
			if ok && mx < j-i+1 {
				mx = j - i + 1
				k = i
			}
		}
	}
	if k < 0 {
		return ""
	}
	return s[k : k+mx]
}