Java If-Else|||| hackerrank || Java language || programming_info

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

Source: Wikipedia

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

• If  is odd, print Weird
• If  is even and in the inclusive range of  to , print Not Weird
• If  is even and in the inclusive range of  to , print Weird
• If  is even and greater than , print Not Weird

Complete the stub code provided in your editor to print whether or not  is weird.

Input Format

A single line containing a positive integer, .

Constraints

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

Sample Case 1: 
 and  is even, so it isn't weird. Thus, we print Not Weird.

Now here's the solution in Java:-

code snippet:
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==1)
        {
            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();
    }
}

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