Belzabar Software Design Hiring Coding Question | Computer Scientist (Developer position)

Question: Belzabar is 19 years old. On this occasion, we formulated the Belzabar Number. A positive integer is a Belzabar number if it can be represented either as (n * (n + 19)) OR (n * (n — 19)), where n is a prime number.

Write a function named ‘isBelzabarNumber( )’ which accepts a number as input and determines if it is a Belzabar Number (or not). Input to the function will be a number and the output will be Boolean.

For bonus points:

1. Write a function that calculates and prints the count of Belzabar numbers less than or equal to 1 million.
2. Write a function that calculates and prints the count of Belzabar numbers from bonus question #1 above that are prime.

There are additional bonus points for elegance, adequate code comments describing the algorithm(s) used, focus on coding conventions, optimal speed and time complexity and readability.

Explanation: Here we have to find if the given input is belzabar number or not and belzabar number is represented as (n * (n +19)) or (n * (n -19)) where n is a prime number. Let the given number is b then it must be equivalent to (n * (n +19)) or (n * (n -19)).

Belzabar number b = (n * (n +19)) or (n * (n -19))

e.g. b = 66 is a belzabar number.

n = 2 (starting from 2 because 0 and 1 are not prime number)

b = (n * (n + 19)) => (2 * ( 2 +19) => 42 != 66 (n = 2)

b = (n * (n + 19)) => (3* (3 + 19) => 66 == 66 (n = 3)

import java.util.*;
public class GFG {
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int b = 1550;
        GFG gf = new GFG();
        // Printing is number is belzabar
      	System.out.println("Belzabar Hiring | Coding question");
      
	if(gf.isBelzabarNumber(b)){
        	System.out.println(b + " is Belzabar Number");
        } else{
          System.out.println(b + " is not a Belzabar Number");
        }
        sc.close();
    }

    boolean isBelzabarNumber(int b)
    {
        for (int n = 2; n < 1000; n++) {
            // Taking the number n from 2 to 1000 as 1 is
            // already not a prime

            if (isPrime(n)) {
                // If number is prime then we check if it is
                // representable in the form of (n*(n-19))
                // or (n*(n-19))

                if ((b == (n * (n + 19)))
                    || (b == (n * (n - 19)))) {

                    // If the number b is equal to the above
                    // equation then it returns true
                    return true;
                }
            }
        }
        return false;
    }

  // Checking if the number n is prime or not
    boolean isPrime(int n)
    {
        if (n == 2)
            return true; // 2 is already prime number

        // If number n is prime then we return true
        for (int i = 3; i < n / 2; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
}

Note: Dear readers, This code is just for your reference. Kindly try to design your own logic and please don’t copy-paste the code.

Thanks!