<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:

        def helper(s, start, end, k):
            # Base case: if the length of the substring is less than k
            if end - start + 1 < k:
                return 0
            
            # Count the frequency of each character in the current substring
            counts = [0] * 26
            for i in range(start, end + 1):
                counts[ord(s[i]) - ord('a')] += 1
            
            # Look for a character that doesn't meet the frequency requirement
            for i in range(26):
                if counts[i] > 0 and counts[i] < k:
                    split = chr(i + ord('a'))
                    # Split the string by the under-represented character
                    for j in range(start, end + 1):
                        if s[j] == split:
                            # Recursively call the helper for the two split parts
                            return max(helper(s, start, j - 1, k), helper(s, j + 1, end, k))
            
            # If all characters in the substring meet the frequency requirement
            return end - start + 1

        return helper(s, 0, len(s) - 1, k)

        
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