<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 partition the string based on characters that appear fewer than k times
        def part(l, r):
            # Frequency count of characters in the substring
            char_count = {}
            for i in range(l, r):
                char_count[s[i]] = char_count.get(s[i], 0) + 1

            # Find the first character that appears less than k times
            for mid in range(l, r):
                if char_count[s[mid]] < k:
                    # Divide and conquer: check both parts and return the max length
                    left_part = part(l, mid)
                    right_part = part(mid + 1, r)
                    return max(left_part, right_part)
            
            # If all characters appear at least k times, return the length of the current substring
            return r - l

        # Call the partition helper function on the entire string
        return part(0, len(s))

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