<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 that returns the length of the longest substring
        # where each character appears at least k times.
        def longest_substring_util(start, end):
            # Base case: if the substring length is less than k, return 0
            if end - start < k:
                return 0
            
            # Count frequency of each character in the current substring
            count = [0] * 26
            for i in range(start, end):
                count[ord(s[i]) - ord('a')] += 1
            
            # Traverse the substring and split at the points where the
            # character frequency is less than k
            for i in range(start, end):
                if count[ord(s[i]) - ord('a')] < k:
                    # If a character's frequency is less than k, recursively find
                    # the longest substring in the left and right parts
                    left_part = longest_substring_util(start, i)
                    right_part = longest_substring_util(i + 1, end)
                    # Return the maximum of the two parts
                    return max(left_part, right_part)
            
            # If all characters in the substring have frequency at least k,
            # return the length of the substring
            return end - start
        
        # Call the helper function with the entire string
        return longest_substring_util(0, len(s))

# Time Complexity: O(n*26) where n is the length of the string.
# The function iterates through each character and counts frequency in O(n),
# and the worst-case scenario can split the string up to O(26) times.
# Space Complexity: O(26) for the count array, which is O(1) since the size
# of the alphabet is constant.

# Create an instance of the Solution class
solution = Solution()

# Test with the provided examples
print(solution.longestSubstring("aaabb", 3)) # Output: 3
print(solution.longestSubstring("ababbc", 2)) # Output: 5

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