<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:
        def longestSubstringR(start, end):
            # char -> times in appears in the string
            freqs = defaultdict(int)

            # char -> first location
            locations = defaultdict(list)

            for i in range(start, end):
                c = s[i]
                freqs[c] += 1
                locations[c].append(i)

            # divide and conquer: we find the first invalid char (one that appears less than k times) and
            # check the other two substrings
            for c, freq in freqs.items():
                if freq < k:
                    maxLength = 0
                    for i in range(len(locations[c]) + 1):
                        currStart = locations[c][i - 1] + 1 if i > 0 else start
                        currEnd = locations[c][i] if i < len(locations[c]) else end
                        maxLength = max(maxLength, longestSubstringR(currStart, currEnd))
                    return maxLength

            # no char is invalid -> max length is the length of the current string
            return end - start

        return longestSubstringR(0, len(s))


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