<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.size();
        // unordered_set<char>st;
        vector<int>dict(26,0);
        int ans=0;
        int maxunique=0;
        for(int i=0;i<s.size();i++){
            if(dict[s[i]-'a']==0)maxunique++;
            dict[s[i]-'a']++;
        }
        
        for(int curunique=1;curunique<=maxunique;curunique++){
            int l=0;
            int cntk=0;
            unordered_map<char,int>mp;
            for(int r=0;r<n;r++){
                mp[s[r]]++;
                if(mp[s[r]]==k)cntk++;
                while(mp.size()>curunique){
                    mp[s[l]]--;
                    if(mp[s[l]]==k-1)cntk--;
                    if(mp[s[l]]==0)mp.erase(s[l]);
                    l++;
                }
                 if(curunique==cntk){
                    cout<<r<<' '<<l<<' '<<curunique<<endl;
                    ans=max(ans,r-l+1);
                }
            }
        }
        return ans;
    }
};
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