The description says: "The first 3 students arrived on. The last 2 were late. The threshold is 3 students, so class will go on. Return YES." So yes when the class is will hold.

But the solution requires the oppsite, so yes when cancelled.

Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-angry-professor-problem-solution.html

Why this is wrong answer althrough the algo is same.

string angryProfessor(int k, vector a) { // int attendees = 0; // for(int el: a) if(el <= 0) attendees++; // return attendees >= k ? "NO":"YES"; int atee = 0; for(int i = 0; i < a.size(); i++){ if(a[i] <= 0){ atee += 1;

    }
} 
if(atee >= k){
    return "YES";
}
else return "NO";

}

Why this is wrong answer althrough the algo is same.

string angryProfessor(int k, vector a) { // int attendees = 0; // for(int el: a) if(el <= 0) attendees++; // return attendees >= k ? "NO":"YES"; int atee = 0; for(int i = 0; i < a.size(); i++){ if(a[i] <= 0){ atee += 1;

    }
} 
if(atee >= k){
    return "YES";
}
else return "NO";

}

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";
}