<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 ans = 0;
    void solve(string &s, int &k, int begin, int end, vector<int> &record){
        // priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> small_mp;
        unordered_map<int, int> small_mp;
        int index = end;
        for(int i = begin; i < end; i++){
            if(record[s[i] - 'a'] < k){
                solve(s, k, i + 1, end, record);
                index = i;
                break;
            }
            small_mp[s[i] - 'a']++;
        }
        for(int i = begin; i < index; i++){
            if(small_mp[s[i] - 'a'] < k){
                solve(s, k, begin, i , record);
                solve(s, k, i + 1, index, record);
                return;
            }
        }

        ans = max(ans, index - begin);
    }
    int longestSubstring(string s, int k) {
        int len = s.size();
        vector<int> mp(26, 0);
        for(int i = 0; i < len; i++){
            mp[s[i] - 'a']++;
        }
        solve(s, k, 0, len, mp);
        return ans;
    }
};
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