<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:
        total = [0 for _ in range(26)]
        less = []
        less_value =[-1]
        large = 0
        for i in range(len(s)):
            total[ord(s[i])-97]+=1
        for i in range(26):
            if total[i] < k and total[i]!=0:
                less.append(i)
        for i in range(len(s)):
            if ord(s[i])-97 in less:
                less_value.append(i)
        less_value.append(len(s))
        if len(less_value)==2:
            return len(s)
        for i in range(len(less_value)-1):
            if less_value[i+1] - less_value[i] > max(2,large) :
              large = max(large,self.longestSubstring(s[less_value[i]+1:less_value[i+1]],k))
        return large
16991