<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:
        res = 0
        max_chars = len(set(s))
        for available_char in range(1, max_chars+1):
            freq = defaultdict(int)
            l, r = 0, 0
            for r in range(len(s)):
                
                freq[s[r]] += 1

                if len(freq) < available_char:
                    continue
                
                elif len(freq) == available_char:
                    count = 0
                    for i in freq:
                        if freq[i] >= k:
                            count += 1
                    if count == available_char:
                        res = max(res, r - l + 1)
                else:
                    while len(freq) > available_char:
                        if freq[s[l]] > 1:
                            freq[s[l]] -= 1
                        else:
                            del freq[s[l]]
                        l += 1
        return res


             
16463