Longest Substring with At Least K Repeating Characters
<p>Given a string <code>s</code> and an integer <code>k</code>, return <em>the length of the longest substring of</em> <code>s</code> <em>such that the frequency of each character in this substring is greater than or equal to</em> <code>k</code>.</p>

<p data-pm-slice="1 1 []">if no such substring exists, return 0.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aaabb&quot;, k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest substring is &quot;aaa&quot;, as &#39;a&#39; is repeated 3 times.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ababbc&quot;, k = 2
<strong>Output:</strong> 5
<strong>Explanation:</strong> The longest substring is &quot;ababb&quot;, as &#39;a&#39; is repeated 2 times and &#39;b&#39; is repeated 3 times.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of only lowercase English letters.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>
Hash Table; String; Divide and Conquer; Sliding Window
func longestSubstring(s string, k int) int {
	var dfs func(l, r int) int
	dfs = func(l, r int) int {
		cnt := [26]int{}
		for i := l; i <= r; i++ {
			cnt[s[i]-'a']++
		}
		var split byte
		for i, v := range cnt {
			if v > 0 && v < k {
				split = byte(i + 'a')
				break
			}
		}
		if split == 0 {
			return r - l + 1
		}
		i := l
		ans := 0
		for i <= r {
			for i <= r && s[i] == split {
				i++
			}
			if i > r {
				break
			}
			j := i
			for j <= r && s[j] != split {
				j++
			}
			t := dfs(i, j-1)
			ans = max(ans, t)
			i = j
		}
		return ans
	}
	return dfs(0, len(s)-1)
}