<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) {

        if(k==1){
            return s.length();
        }

        int ans = 0;

       
        for(int i=0; i<s.length(); i++){
            string temp = "";
            int charFreq[26]={0};
            for(int j=i; j<s.length(); j++){
                temp.push_back(s[j]);
                charFreq[s[j]-'a']++;
                if(temp.length()>=k){
                    int flag = 1;
                    for(int K=0; K<26; K++){
                        if(charFreq[K] != 0 &&charFreq[K] < k){
                            flag = 0;
                        }
                    }
                    if(flag==1){
                       int tempAns = temp.length();
                       ans = max(tempAns,ans);
                    }
                }
            }
            
        }

        return ans;
        
    }
};
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