<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:
        if not s or k > len(s):
            return 0

        freq = collections.Counter(s) 

        for i, char in enumerate(s):
            if freq[char] < k: # every character in substring must have frequency >= k
                return max(self.longestSubstring(s[:i], k),  self.longestSubstring(s[i+1:], k))
        
        return len(s)


        """
        Sliding window/ 2 pointers approach
        """
        # number of unique characters available
        max_chars = len(set(s))
        n = len(s)
        ans = 0

        # for all char from 1 to max_chars 
        for available_char in range(1,max_chars+1):

            h = {}
            i = j = 0


            # simple sliding window approach
            while(j < n):

                if(s[j] not in h):
                    h[s[j]] = 0
                h[s[j]] += 1

                # if len of h is less than no of available chars
                if(len(h) < available_char):
                    j += 1

                # if equal then check all have values >=k
                elif(len(h) == available_char):
                    count = 0
                    for x in h.values():
                        if(x >= k):
                            count += 1
                    if(count == available_char):
                        ans = max(ans,j - i + 1)
                    j += 1

                # if greater than remove from starting
                else:
                    while(len(h) != available_char):
                        h[s[i]] -= 1
                        if(h[s[i]] == 0):
                            del h[s[i]]
                        i += 1
                    j += 1

        return ans
54971