<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 count_freq(s):
            freq = [[0] * 26 for _ in range(len(s) + 1)]
            for i in range(len(s)):
                for j in range(26):
                    freq[i+1][j] = freq[i][j]
                freq[i+1][ord(s[i]) - ord('a')] += 1
            return freq

        def get_count(freq, l, r, ch):
            return freq[r+1][ch] - freq[l][ch]

        def longest_substring(freq, s, l, r, k):
            if r - l + 1 < k:
                return 0
            for i in range(l, r + 1):
                if get_count(freq, l, r, ord(s[i]) - ord('a')) < k:
                    return max(longest_substring(freq, s, l, i - 1, k),
                               longest_substring(freq, s, i + 1, r, k))
            return r - l + 1

        if not s:
            return 0
        freq = count_freq(s)
        return longest_substring(freq, s, 0, len(s) - 1, k)
20683