# HackerRank Angry Professor Solution

Hello Programmers, In this post, you will know how to solve the HackerRank Angry Professor Solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## HackerRank Angry Professor Solution

A Discrete Mathematics professor has a class of students. Frustrated with their lack of discipline, the professor decides to cancel class if fewer than some number of students are present when class starts. Arrival times go from on time (arrivalTime <= 0) to arrived late (arrivalTime > 0).

Given the arrival time of each student and a threshhold number of attendees, determine if the class is cancelled.

Example

n = 5
k = 3
a = [-2, -1, 0, 1, 2]
The first 3 students arrived on. The last 2 were late. The threshold is 3 students, so class will go on. Return `YES`.

Note: Non-positive arrival times (a[i] <= 0) indicate the student arrived early or on time; positive arrival times (a[i] > 0) indicate the student arrived a[i] minutes late.

Function Description

Complete the angryProfessor function in the editor below. It must return `YES` if class is cancelled, or `NO` otherwise.

angryProfessor has the following parameter(s):

• int k: the threshold number of students
• int a[n]: the arrival times of the n students

Returns

• string: either `YES` or `NO`

Input Format

The first line of input contains t, the number of test cases.

Each test case consists of two lines.

The first line has two space-separated integers, n and k, the number of students (size of a) and the cancellation threshold.
The second line contains n space-separated integers (a, a, . . . ,a[n]) that describe the arrival times for each student.

Constraints

• 1 <= t <= 10
• 1 <= n <= 1000
• 1 <= k <= n
• -100 <= a[i] <= 100, where i belongs to [1, . . . ,n]

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. The professor wants at least 3 students in attendance, but only 2 have arrived on time (-3 and -1) so the class is cancelled.

For the second test case,k = 2 . The professor wants at least 2 students in attendance, and there are 2 who arrived on time (0 and -1). The class is not cancelled.

## HackerRank Angry Professor Solution

### Angry Professor Solution in C

```#include<stdio.h>
int main()
{
int count, i, K, N, T, time;
scanf("%d", &T);
while (T--)
{
scanf("%d %d", &N, &K);
count = 0;
for (i = 0; i < N; i++)
{
scanf("%d", &time);
if (time <= 0)
count++;
}
puts((count < K) ? "YES" : "NO");
}
return 0;
}```

### Angry Professor Solution in Cpp

```#include <bits/stdc++.h>
using namespace std;
int main()
{
int a; cin >> a;
for (int g=0; g<a; g++)
{
int b,c; cin >> b >> c;
int num=0;
for (int g=0; g<b; g++)
{
int d; cin >> d;
if (d<=0) num++;
}
if (num>=c)
{
cout << "NO" << '\n';
}
else cout << "YES" << '\n';
}
return 0;
}```

### Angry Professor Solution in Java

```import java.io.BufferedReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class AngryProf {
public static void main(String args[] ) throws Exception {
PrintWriter w = new PrintWriter(System.out);

while(t-- > 0) {
int n = ip(st1.nextToken());
int k = ip(st1.nextToken());

int a[] = new int[n];
for(int i=0;i<n;i++)
a[i] = ip(st2.nextToken());

int count = 0;
for(int i=0;i<n;i++)
if(a[i]<=0)	count++;
w.println(count < k ? "YES" : "NO");
}

w.close();

}

public static int ip(String s){
return Integer.parseInt(s);
}

}```

### Angry Professor Solution in Python

```for _ in xrange(input()):
n, m = map(int, raw_input().split())
A = map(int, raw_input().split())
for x in A:
if x <= 0:
m -= 1
if m <= 0:
print "NO"
else:
print "YES"```

### Angry Professor Solution using JavaScript

```function processData(input) {
input = input.split('\n');
var length = input.length;
var N, K;
var buf,j;
for(var i = 1; i < length; i += 2){
count = 0;
buf = input[i].split(" ");
N = Number(buf);
K = Number(buf);
buf = input[i+1].split(" ");
for(j = 0; j < N; ++j){
if(Number(buf[j] ) <= 0)
++count;
}
if(count >= K)
console.log("NO");
else
console.log("YES");
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});```

### Angry Professor Solution in Scala

```object Solution {
def main(args: Array[String]) {
for (_ <- 1 to t) {
val studentsInTime = ts.count(_ <= 0)
if (k <= studentsInTime)
println("NO")
else
println("YES")
}
}
}```

### Angry Professor Solution in Pascal

```var
id,t,n,k,i,sum,m:longint;
begin
for id:=1 to t do begin
sum:=0;
for i:=1 to n do begin
if m<=0 then inc(sum);
end;
if sum<k then writeln('YES') else writeln('NO');
end;
end.```

Disclaimer: This problem (Angry Professor) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.