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";
}

Java 8:

public static String angryProfessor(int k, List<Integer> a) {
    a.removeIf(n -> (n > 0));
    if (a.size() >= k) {
        return "NO";
    }
    return "YES";
}

C# Solution

return k > a.Where(x => x <= 0).Count() ? "YES": "NO";

Each student's arrival time is given as either negative (on time or early) or positive (late). The professor sets a threshold (k) for the minimum number of on-time arrivals needed to avoid canceling class. If the number of on-time arrivals is less than k, the class is canceled; otherwise, it continues. You need to count how many students arrive on time and compare that count with k to determine if the class is canceled or not for each test case. If the count of on-time arrivals is less than k, return "YES" (class canceled); otherwise, return "NO" (class not canceled). This type of calculation can be done through any kind of software on desktop or PC.

JS/Javascript:-

function angryProfessor(k, a) {
    let onTime = 0;
    for (let i = 0;i<a.length;i++) {
        if (a[i] <= 0) onTime++;
        if (onTime >= k) return 'NO';
    }
    return 'YES';
}