<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 {
public:
        int divideAndConquer(const string& s, int start, int end, int k) {
        if (end - start < k) return 0; // If the length is less than k, it's not possible

        // Frequency map for the current substring
        unordered_map<char, int> freq;
        for (int i = start; i < end; i++) {
            freq[s[i]]++;
        }

        // Iterate over the substring and divide at invalid characters
        for (int i = start; i < end; i++) {
            if (freq[s[i]] < k) {
                // Split and solve for both halves
                int leftPart = divideAndConquer(s, start, i, k);
                int rightPart = divideAndConquer(s, i + 1, end, k);
                return max(leftPart, rightPart);
            }
        }

        // If all characters have frequency >= k, return the length of the substring
        return end - start;
    }
    int longestSubstring(string s, int k) {
        return divideAndConquer(s, 0, s.size(), k);
    }
};
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