<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.size(), lastIndex = -1, ans = 0;
        
        while(lastIndex < n-1){
            int i = lastIndex+1;
            int currLastIndex = lastIndex;

            unordered_map<int,int> count;
            while(i<n){
                count[s[i]-'a']++;
                bool canBe = true;
                
                for(auto x : count){
                    if(x.second < k){
                        canBe = false;
                        break;
                    }
                }

                if(canBe){
                    ans = max(ans, i-lastIndex);
                    currLastIndex = i;
                }

                i++;

            }

            if(currLastIndex==lastIndex){
                lastIndex++;
            }
            else lastIndex = currLastIndex;
        }

        return ans;

    }
};
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