<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:
    result = 0
    def divide(self, s, left, right, k):
        if abs(right-left) < k:
            return
        hashmap = {}
        index_map = {}
        for i in range(left, right):
            hashmap[s[i]] = hashmap.get(s[i], 0) + 1
            if s[i] not in index_map:
                index_map[s[i]] = i
        min_index = float('inf')
        for key in hashmap:
            if hashmap[key] < k:
                min_index = min(min_index, index_map[key])
        if min_index == float('inf'):
            self.result = max(self.result, right-left)
        else:
            self.divide(s, left, min_index, k)
            self.divide(s, min_index+1, right, k)
            
    def longestSubstring(self, s, k):
        self.result = 0
        self.divide(s, 0, len(s), k)
        return self.result
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