<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:
        # Helper function to find the longest valid substring recursively
        def longest_substring_helper(s, k):
            if len(s) < k:
                return 0  # Base case: no valid substring if length is less than k

            # Count frequency of each character in the string
            char_count = {}
            for char in s:
                char_count[char] = char_count.get(char, 0) + 1
            
            # Iterate through the string and split if a character occurs less than k times
            for i, char in enumerate(s):
                if char_count[char] < k:
                    # Split the string at the current character and check both parts
                    left_part = longest_substring_helper(s[:i], k)
                    right_part = longest_substring_helper(s[i + 1:], k)
                    return max(left_part, right_part)

            # If all characters in the string appear at least k times, the whole string is valid
            return len(s)

        # Call the recursive function on the entire string
        return longest_substring_helper(s, k)

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