<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:
    // cut the string at char num < k
    // and keep doing the same for the cutted sub string
    // until find the substring that can't be cut, return the substring length
    // or when substring len < k, return -1;
    int longestSubstring(string s, int k) {
        
        int len = s.length();

        if (len < k) return 0;
        vector<int> map(26, 0);
        int left = 0;
        int right = 0;
        int valid_cnt = 0;
        for (int i = 0; i < len; i++)
        {
            map[s[i] - 'a']++;
        }

        for (int j = 0; j < len; j++)
        {
            if (map[s[j] - 'a'] < k)
            {
                //if (j - 1 >= 0)
                if (j >= k)// dont need to check smaller j
                {
                    int tmpleft = longestSubstring(s.substr(0, j), k);
                    left = max(tmpleft, left);
                }

                //if (j + 1 < len)
                if (len - j > k) // dont need to check larger j
                {
                    int tmpright = longestSubstring(s.substr(j + 1, len - j - 1), k);
                    right = max(tmpright, right);
                }

                return max(right, left); // only need to cut once in this layer
                
            }
            else
            {
                valid_cnt++;
            }
        }

        return valid_cnt == len ? len: max(left, right);
    }
};
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