<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 longestSubstring(string s, int k) {
        if (s.length() == 0 or s.length() < k) {
            return 0;
        }   
        if (!k) {
            return s.length();
        }
        map<char, int> m;
        for (int i = 0; i < s.length(); i++) {
            m[s[i]]++;
        }
        int idx = 0;
        while (idx < s.length() and m[s[idx]] >= k) {
            idx++;
        }
        if (idx == s.length()) {
            return s.length();
        }
        // idx is the first char with freq < k
        string left_string = s.substr(0, idx);

        string right_string = s.substr(idx + 1, s.length() - idx - 1);
        int left = longestSubstring(left_string, k);
        int right = longestSubstring(right_string, k);
        return max(left, right);
    }
};
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