In this HackerRank Day 25: Running Time and Complexity problem solution we have given a number, n, determine and print whether it is Prime or Not prime.

## Problem solution in Python programming.

```import math
n = int(input())
for i in range(n):
isPrime = True
num = int(input())
if num <= 1:
isPrime = False
else:
for n in range(2, math.ceil(math.sqrt(num))):
if num%n == 0:
isPrime = False
break
if(isPrime):
print( 'Prime' )
else:
print( 'Not prime' )```

### Problem solution in Java programming.

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
/* Enter your code here. */
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();

for(int idx = 0; idx < n; ++idx)
{
System.out.println(isPrime(scan.nextInt()) ? "Prime" : "Not prime");
}
}

public static boolean isPrime(int value)
{
if(value<= 1) return false;
if(value == 2) return true;

//Check values up to the square root.
int sqrt = (int)Math.ceil(Math.sqrt(value));
for(int idx = 2; idx <= sqrt; ++idx)
{
if(value % idx == 0)
{
return false;
}
}

return true;
}
}```

### Problem solution in C++ programming.

```#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

bool p(int a){
for(int i=2; i*i<=a; ++i)
if(a%i == 0) return false;
return true && a != 1;
}

int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int t;
cin >> t;
while(t--){
int a;
cin >> a;
if(p(a)) cout << "Prime" << endl;
else cout << "Not prime" << endl;
}
return 0;
}```

### Problem solution in C programming.

```#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int T;
scanf("%d\n", &T);
for (int i = 0; i < T; i++) {
int num;
scanf("%d", &num);
int k;
int square_root = (int)sqrt(num);
if (num > 1) {
for (k = 2; k < square_root; k++) {
if ((num % k && num != k) == 0) {
printf("Not prime\n");
break;
}
}
if (k == square_root || num == k) {
printf("Prime\n");
}
} else {
printf("Not prime\n");
}
}
return 0;
}```

### Problem solution in Java programming.

```function processData(input) {
input = input.split('\n').slice(1);
main_loop: for(var n of input){
if(n==1){
console.log('Not prime');
continue;
}
if(n==2){
console.log('Prime');
continue;
}
for(var i=2; i<= Math.ceil(Math.sqrt(n));i++){
if(n%i==0){
console.log('Not prime');
continue main_loop;
}
}
console.log('Prime');
}
}

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});

process.stdin.on("end", function () {
processData(_input);
});```