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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Angry Professor
  5. Discussions

Angry Professor

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1978 Discussions

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  • sam_a_d
     + 0 comments

    `

    def angryProfessor(k, a):

        return "NO" if len([i for i in a if i <=0]) >= k else "YES"

`

  • alban_tyrex
     + 0 comments

    Here are my c++ solutions for this problem, you can watch the explanation here : https://youtu.be/MKqtPYhaNrs Solution 1 O(N)

    string angryProfessor(int k, vector<int> a) {
    int attendees = 0;
    for(int el: a) if(el <= 0) attendees++;
    return attendees >= k ? "NO":"YES";
}

    Solution 2 O(nLog(n)) because of the sorting, it's not the best option here, but it's still interesting to know because it can be useful in case you received a sorted array in a similar problem.

    string angryProfessor(int k, vector<int> a) {
    sort(a.begin(), a.end());
    return a[k-1] <= 0 ? "NO":"YES";
}

  • vlad_suslov
     + 0 comments

    Problem wording is confusing, seems misleading

  • highsoft_soi
     + 0 comments

    Perl:

    sub angryProfessor {
    my ($k, $a) = @_;
    
    return ($k > scalar(my @t = grep { $_ <= 0 } @$a)) ? "YES" : "NO";
}

  • somaditya
     + 0 comments
    public static String angryProfessor(int k, List<Integer> a) {
        int count = 0;
        
        for (int arvT : a) {
            if (arvT <= 0) count++;
        }

        return count < k ? "YES" : "NO";
    }