Java If-Else – HackerRank Solution

Java If-Else – HackerRank Solution

Problem :

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 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 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.

Solution :

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

public class Solution {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();            
        String ans="";

        if(n>=1 && n<=100){
            if(n%2==1){
                ans = "Weird";
            }
            else{
                if(n>=2 && n<=5 ){
                   ans= "Not Weird";
                }
                else if (n>=6 && n<=20){
                    ans = "Weird";
                }
                else if(n>20) {
                    ans = "Not Weird";
                }
            }
        }
        else{
            ans="Invalid no! Enter a number between 1 and 100.";
        }

        System.out.println(ans);

    }
}

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