Java If-Else-HackerRank Solution

In this challenge,  test your knowledge of using if-else conditional statements to automate decision-making processes. An if-else statement has the following logical flow:

Task

Given an integer,n, perform the following conditional actions:

• If  n is odd, print Weird
• If n is even and in the inclusive range of 2 to 5, print Not Weird
• If n is even and in the inclusive range of 6 to 20, print Weird
• If n is even and greater than 20, print Not Weird

Complete the stub code provided in your editor to print whether n is weird.

Input Format

A single line containing a positive integer,n,

Constraints

1<=n<=100
Output Format
Print Weird if the number is weird; otherwise, print Not Weird.

Sample Input 0

3

Sample Output 0

Weird

Sample Input 1

24

Sample Output 1

Not Weird

Explanation

Sample Case 0: n =3

n is odd and odd numbers are weird, so we print Weird.

Sample Case 1: n = 24

n> 20 and n is even, so it isn't weird. Thus, we print Not Weird.

Solution:

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) {
        int N = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
        if(N%2!=0)
        {
            System.out.println("Weird");
        }
        else if(N%2 == 0){
            if(N>=2&&N<=5){
                System.out.println("Not Weird");
            }
            else if(N>=6&&N<=20){
                System.out.println("Weird");
            }
            else if(N>20) {
                System.out.println("Not Weird");
            }
        }

        scanner.close();
    }
}