<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:
    def longestSubstring(self, s: str, k: int) -> int:
        """
        Input: s = "aaabb", k = 3
        Explanation: The longest substring is "aaa", as 'a' is repeated 3 times.
        s = "ababbc", k = 2
        5

        - counting the frequency of each character
        - knowing when to start and when to end the substring
            - how do you know whether you will see a character more than k times
        """
        return self.longest_substring_helper(s, k)

    def longest_substring_helper(self, s, k):
        freqs = {}
        for char in s:
            freqs[char] = freqs.get(char, 0) + 1
        length = 0
        for i in range(len(s)):
            if freqs[s[i]] < k:
                mid = i
                l = self.longest_substring_helper(s[0:mid], k)
                r = self.longest_substring_helper(s[mid+1: len(s)], k)
                return max(l, r)
            else:
                length+=1
        return length
46531