Java If-Else hackerrank solution

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, n, perform the following conditional actions:

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

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

Code Solution:

//Java If-Else hackerrank 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();
        if(N%2!=0)
        System.out.println("Weird");
        else if((N%2==0) && 2<= N & N<6)
        System.out.println("Not Weird");
        else if((N%2==0) && 6<= N & N<22)
        System.out.println("Weird");
        else 
        System.out.println("Not Weird");
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        scanner.close();
    }
}


//Java If-Else hackerrank solution

Disclaimer: This problem is originally created and published by HackerRank, we only provide solutions to this problem. Hence, doesn’t guarantee the truthfulness of the problem. This is only for information purposes.