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