Angry Professor - Hacker Rank Solution

A Discrete Mathematics professor has a class of N$N$ students. Frustrated with their lack of discipline, he decides to cancel class if fewer than K$K$ students are present when class starts.
Given the arrival time of each student, determine if the class is canceled.

Input Format

The first line of input contains T$T$, the number of test cases.
Each test case consists of two lines. The first line has two space-separated integers, N$N$ (students in the class) and K$K$ (the cancelation threshold).
The second line contains N$N$ space-separated integers (a1,a2,…,aN$a_1,a_2,\dots ,a_N$) describing the arrival times for each student.
Note: Non-positive arrival times (ai≤0$a_i\le 0$) indicate the student arrived early or on time; positive arrival times (ai>0$a_i>0$) indicate the student arrived ai$a_i$ minutes late.

Output Format

For each test case, print the word YES if the class is canceled or NO if it is not.
Constraints

• 1≤T≤10$1\le T\le 10$
• 1≤N≤1000$1\le N\le 1000$
• 1≤K≤N$1\le K\le N$
• −100≤ai≤100,where i∈[1,N]$-100\le a_i\le 100,wherei\in [1,N]$

Note
If a student arrives exactly on time (ai=0)$(a_i=0)$, the student is considered to have entered before the class started.

Sample Input

2
4 3
-1 -3 4 2
4 2
0 -1 2 1

Sample Output

YES
NO

Explanation

For the first test case, K=3$K=3$. The professor wants at least 3$3$ students in attendance, but only 2$2$ have arrived on time (−3$-3$ and −1$-1$). Thus, the class is canceled.
For the second test case, K=2$K=2$. The professor wants at least 2$2$ students in attendance, and there are 2$2$ who have arrived on time (0$0$ and −1$-1$). Thus, the class is not canceled.

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  Angry Professor -  Hacker Rank Solution

 

#include <math.h>

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <assert.h>

#include <limits.h>

#include <stdbool.h>

int main(){

    int t;

    scanf("%d",&t);

    for(int a0 = 0; a0 < t; a0++){

        int n;

        int k;

        int count=0;

        scanf("%d %d",&n,&k);

        int a[n];

        for(int a_i = 0; a_i < n; a_i++){

           scanf("%d",&a[a_i]);

        }

       

           for(int i=0;i<n;i++)

        {

             if(a[i]<=0) count++;

           }

          if(count>=k)

              printf("NO \n");

          else

              printf("YES \n");

    }

    return 0;

}

Angry Professor -  Hacker Rank Solution