<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>

from collections import Counter, defaultdict 
def f(l, r, k, s):
    if l>r:
        return 0
    c = Counter(list(s[l:r+1]))
    if d[l, r]!=-1:
        return d[l, r]
    ans = 0
    fl = True
    for i in range(l, r+1):
        if c[s[i]]<k:
            fl=False
            d[l, r] = max(ans, f(l, i-1, k, s), f(i+1, r, k, s))
            return d[l, r]
    return r-l+1
    # if fl:
    #     ans = max(ans, r-l+1)
    
    
    # d[l, r] =  ans

    # return d[l, r]

class Solution:
    def longestSubstring(self, s: str, k: int) -> int:
        # c = Counter(list(s))
        global d
        d=defaultdict(lambda:-1)
        return f(0, len(s)-1, k, s)
        # counts = Counter(s)
        # for char, freq in counts.items():
        #     if freq < k:
        #         return max(self.longestSubstring(substring, k) for substring in s.split(char))
        # return len(s)


        
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