<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();
        if(n<k) return 0;
        int freq[26];
        memset(freq, 0, sizeof(freq));

        for(int i=0;i<n;i++){
            freq[s[i]-'a'] +=1;
        }  

for(int i=0;i<26;i++){
            cout<<freq[i]<< " ";
        }  
        int maxLen =0;

        for(int end=0;end<n;end++){
            if(freq[s[end]-'a']<k){
                string str1 = s.substr(0, end);
                string str2 = s.substr(end+1, n);
                return max(longestSubstring(str1, k), longestSubstring(str2, k));
            }
        }
        return n;
    }
};
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