<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();
      int ans = 0;
      vector<int> arr[100100];
      vector<int> fre(28,0);

      for(int i=0;i<n;i++)
      {
          fre[s[i]-'a']++;
          arr[i] = fre;
      }

      for(int i=0;i<n;i++)
      {
        for(int j=i;j<n;j++)
        {
             if(i==0)
             {
                int tt = 0;
               for(int k=0;k<26;k++)
               {
                   if(arr[j][k]!=0 && arr[j][k] < K)
                   {
                       tt = 1;
                       break;
                   } 
               } 
               if(tt==0)
               ans = max(ans,j+1); 
             }
             else
             {
               int tt = 0;
               for(int k=0;k<26;k++)
               {
                   if((arr[j][k] - arr[i-1][k])!=0 && (arr[j][k] - arr[i-1][k] < K))
                   {
                       tt = 1;
                       break;
                   }
               } 
               if(tt==0)
               ans = max(ans,j-i+1); 
             }
        }
      }  
         

       return ans;
    }
};
13643