<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:
        result = 0
        count = dict()
        last = dict()
        run = []
        for i in range(26):
            run.append([0])
        for i, c in enumerate(s):
            count[c] = count.get(c, 0) + 1
            last[c] = i

            for j in range(26):
                run[j].append(run[j][-1] + int(c == chr(j + 97)))

            right = -1
            toremove = []
            for j in range(26):
                u = chr(j+97)
                if count.get(u, 0) > 0 and count.get(u, 0) < k:
                    toremove.append(u)
            while len(toremove) > 0:
                for u in toremove:
                    right = max(right, last[u])
                toremove.clear()
                for j in range(26):
                    u = chr(j+97)
                    if count.get(u, 0) >= k and last[u] > right:
                        print(u, count[u], last[u], run[j][right+1])
                        if count[u] - run[j][right+1] < k:
                            toremove.append(u)
            result = max(result, i-right)

        return result

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