<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 n = s.length();
        if(k==1){
            return n; 
        }
        vector<vector<int>> ocurr(26);
        for(int i=0; i<n; i++){
            ocurr[s[i]-'a'].push_back(i);
        }
        int ans =0; 
        for(int i=0; i<n; i++){
            map<char,int> hm; 
            set<pair<int,char>> st; 
            for(int j=i; j<n; j++){
                // s[j]-'a'
                int ind = s[j]-'a';
                int freq = ocurr[ind].end() - lower_bound(ocurr[ind].begin(),ocurr[ind].end(),i);
                if(freq<k){
                    break; 
                }
                else{
                    hm[s[j]]++;
                    if(hm[s[j]]==1){
                        st.insert({1,s[j]});
                    } 
                    else{
                        st.erase({hm[s[j]]-1,s[j]});
                        st.insert({hm[s[j]],s[j]});
                    }
                }
               if(st.begin()->first>=k){
                 ans = max(ans,j-i+1);
               }
            }
        }
        return ans; 
    }
};
/*
we can brute force to find the
if substring starting from i we want to know there is atleast k instances of every char from;  
*/
33264