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
class Solution {
    private String s;
    private int k;

    public int longestSubstring(String s, int k) {
        this.s = s;
        this.k = k;
        return dfs(0, s.length() - 1);
    }

    private int dfs(int l, int r) {
        int[] cnt = new int[26];
        for (int i = l; i <= r; ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        char split = 0;
        for (int i = 0; i < 26; ++i) {
            if (cnt[i] > 0 && cnt[i] < k) {
                split = (char) (i + 'a');
                break;
            }
        }
        if (split == 0) {
            return r - l + 1;
        }
        int i = l;
        int ans = 0;
        while (i <= r) {
            while (i <= r && s.charAt(i) == split) {
                ++i;
            }
            if (i > r) {
                break;
            }
            int j = i;
            while (j <= r && s.charAt(j) != split) {
                ++j;
            }
            int t = dfs(i, j - 1);
            ans = Math.max(ans, t);
            i = j;
        }
        return ans;
    }
}