<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:
        # If the length of the string is less than k, return 0
        if len(s) < k:
            return 0
        
        # Count the frequency of each character in the string
        char_freq = {}
        for char in s:
            if char not in char_freq:
                char_freq[char] = 1
            else:
                char_freq[char] += 1
        
        # Find the index of the first character with a frequency less than k
        for i, char in enumerate(s):
            if char_freq[char] < k:
                # Split the string into two parts and recursively find the longest substring in each part
                left = self.longestSubstring(s[:i], k)
                right = self.longestSubstring(s[i+1:], k)
                # Return the maximum length of the two substrings
                return max(left, right)
        
        # If all characters have a frequency greater than or equal to k, return the length of the string
        return len(s)
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