<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>

class Solution {
public:
    int solve(string s, int k, int start, int end) {
        if (start == end) return 0;
        vector<int> v(26, 0);
        for(int i = start; i < end; i++) v[s[i] - 'a']++;
        bool check = false;
        for(int a: v) {
            if (a > 0 && a < k) {
                check = true;
                break;
            }
        }
        if (!check) return end - start;
        int ans = 0, t = start;
        for(int i = start; i < end; i++) {
            if (v[s[i] - 'a'] < k) {
                ans = max(ans, solve(s, k, t, i));
                t = i + 1;
            }
        }
        ans = max(ans, solve(s, k, t, end));
        return ans;
    }
    int longestSubstring(string s, int k) {
        return solve(s, k, 0, (int)s.size());
    }
};
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