<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 max(int &a,int b)
    {
        return a>b?a:b;
    }
    int longestSubstring(string s, int k) {
        unordered_map<char,int> mp;
        int maxi=0;
        for(int i=0;i<s.size();i++)
        {
            mp.clear();
            string tmp="";
            for(int j=i;j<s.size();j++)
            {
                tmp+=s[j];
                mp[s[j]]++;
                auto it=mp.begin();
                for(;it!=mp.end();it++)
                {
                    if(it->second<k) break;
                }
                if(it==mp.end()) maxi=max(maxi,tmp.length());
            }
        }
        return maxi;
    }
};
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