<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 hlpr(low,high):
            if high-low+1<k:
                return 0

            mp={}
            for i in range(low,high+1):
                if s[i] in mp:
                    mp[s[i]]+=1
                else:
                    mp[s[i]]=1
            
            #mid=low+(high-low)//2
            for mid in range(low,high+1):
                if mp[s[mid]]<k:
                    lflen=hlpr(low,mid-1)
                    rtlen=hlpr(mid+1,high)

                    return max(lflen,rtlen)

            return high-low+1
            
        return hlpr(0,len(s)-1)

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