Java If-Else | HackerRank Solution

Hello coders, today we are going to solve Java If-Else HackerRank Solution.

Java If-Else


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:


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

Input Format

A single line containing a positive integer, n.


1 <= n <= 100

Output Format

Print Weird if the number is weird; otherwise, print Not Weird.

Sample input 0


Sample output 0


Sample input 1


Sample output 1

 Not Weird


Sample Case 0: n=3, where n is odd and odd numbers are weird, so we print Weird.

Sample Case 1: n=24, where n>20 and n is even, so it isn’t weird. Thus, we print Not Weird.

Solution – Java If-Else

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

public class Solution {


    private static final Scanner scanner = new Scanner(;

    public static void main(String[] args) {
        int N = scanner.nextInt();

      if (N%2==1){
      else if(N%2==0 && (N>=2 && N<=5)){
          System.out.println("Not Weird");
      else if(N%2==0 && (N>5 && N<=20)){
      else if(N%2==0 && (N>20 && N<=100)){
          System.out.println("Not Weird");


Disclaimer: The above Problem (Java If-Else) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

Leave a Comment

Your email address will not be published. Required fields are marked *