<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:
        d = Counter(s)

        @cache
        def rec(i, j):
            if i > j:
                return 0
            if i == len(s) - 1 or j == 0:
                return 1 if k == 1 else 0
            if sum(1 if v >= k else 0 for v in d.values()) == sum(1 if v > 0 else 0 for v in d.values()):
                return j - i + 1
            if sum(1 if v >= k else 0 for v in d.values()) == 0:
                return 0
            d[s[j]] -= 1
            m = rec(i, j - 1)
            d[s[j]] += 1
            d[s[i]] -= 1
            m = max(m, rec(i + 1, j))
            d[s[i]] += 1
            return m

        return rec(0, len(s) - 1)
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