<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:
bool valid(const map<char,int>& m,int& value) {
    for (const auto& elem : m) {
        if (elem.second < value) {
            return false;
        }
    }
    return true;
}
    int longestSubstring(string s, int k) {
        int ans=0;
        if(k>s.length())return 0;
        if(k==1)return s.length();
       for(int i=0;i<=s.length()-k;i++)
       {
        char key=s[i];
        map<char,int>m;
        m[key]=1;
        int c=1;
        int j=i+1;
        while(j<s.length() &&s[j]!='\0')
        {
            key=s[j];
            j++;
        if(m.find(key)==m.end())
        {
            m[key]=1;
            c++;
        }
        else
        {
            m[key]++;
            c++;
            if(valid(m,k))
            {
             ans=max(ans,c);
            }
        }
        }
       }
       return ans;
    }
};
18994