Java If-Else Hackkerrank Solution in Java

Problem-3:-

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.

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  && N>=2 && N<=5){

            System.out.println("Not Weird");

        }

        else if(N%2 == 0  && N>=6 && N<=20){

            System.out.println("Weird");

        }

        else if(N%2 == 0  && N>20 && N<=100){

            System.out.println("Not Weird");

        }

        else{

            System.out.println("Invalid Number");

        }

        scanner.close();

    }

}

Hackerrank Completion Proof:-

Hope, it Helps!!