<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(object):
    def longestSubstring(self, s, k):
        i,j=0,0
        hashMap=defaultdict(int)
    #storing all chars in map with their counts
        for i in range(len(s)):
            hashMap[s[i]]+=1
        #now since we know that charcters having lesser frequncy in map than k given,we need to escape them,so we dividing problems in subproblems
        #d&C approach logn
        for index,letter in enumerate(s):
            if hashMap[letter]<k:
                leftString=self.longestSubstring(s[:index],k)
                rightString=self.longestSubstring(s[index+1:],k)
                longest= max (leftString,rightString)
                return longest
        return len(s)
     
      
        
53388