# Java If-Else HackerRank Solution

#### ByBrokenprogrammers

Dec 4, 2022

Hello Programmers, In this post, you will know how to solve the Java If-Else HackerRank Solution. This problem is a part of the HackerRank Java Programming Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## 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:

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.

1<=n<=100

#### Output Format

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

Sample Input 0

```3
```

Sample Output 0

```Weird
```

#### Example 1 :-

Sample Input 1

```24
```

Sample Output1

```Not Weird
```

#### Explanation

Sample Case 0: n=3n is odd and odd numbers are weird, so we print `Weird`.Sample Case 1:n=24 n>24 and n is even, so it isn’t weird. Thus, we print `Not Weird`.

### Java If-Else HackerRank Solutions

```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&&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();
}
}```

Disclaimer: The above Problem (Java If-Else) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.