<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:
        length = len(set(s))
        res = 0
        
        
        for x in range(1,length + 1):
            countValid = 0
            l = 0
            dic = defaultdict(int)#这三个要放在里面 每次都重置
            for i in range(len(s)):
                dic[s[i]] += 1
                if dic[s[i]] == k:
                    countValid += 1
                while   len(dic) > x:
                    dic[s[l]] -= 1
                    if dic[s[l]] == k-1:
                        countValid -=1
                    if dic[s[l]] == 0:
                        del dic[s[l]]
                    l += 1
                if countValid == len(dic) ==x:
                    res = max(res,i-l+1)
        return res
16463