<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 longestSubstring(string s, int k) {
        int countMap[26];
        int maxUnique = getMaxUniqueLetters(s);
        int result = 0;
        for (int currUnique = 1; currUnique <= maxUnique; currUnique++) {
            // reset countMap
            memset(countMap, 0, sizeof(countMap));
            int windowStart = 0, windowEnd = 0,
             idx = 0, unique = 0, countAtLeastK = 0;
            while (windowEnd < s.size()) {
                // expand the sliding window
                if (unique <= currUnique) {
                    idx = s[windowEnd] - 'a';
                    if (countMap[idx] == 0) unique++;
                    countMap[idx]++;
                    if (countMap[idx] == k) countAtLeastK++;
                    windowEnd++;
                }
                // shrink the sliding window
                else {
                    idx = s[windowStart] - 'a';
                    if (countMap[idx] == k) countAtLeastK--;
                    countMap[idx]--;
                    if (countMap[idx] == 0) unique--;
                    windowStart++;
                }
                if (unique == currUnique && unique == countAtLeastK)
                    result = max(windowEnd - windowStart, result);
            }
        }

        return result;
    }

    // get the maximum number of unique letters in the string s
    int getMaxUniqueLetters(string s) {
        bool map[26] = {0};
        int maxUnique = 0;
        for (int i = 0; i < s.length(); i++) {
            if (!map[s[i] - 'a']) {
                maxUnique++;
                map[s[i] - 'a'] = true;
            }
        }
        return maxUnique;
    }
};

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