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:

Wikipedia if-else flow chart

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])?");


        scanner.close();

        if(N%2==0){

            if(N<=5 && N>=2){

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

            }

            if(N<=20 && N>=6){

                System.out.println("Weird");

            }

            if(N>20){

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

            }

        }

        else{

            System.out.println("Weird");

        }

    }

}



🙏THANKS FOR VISIT🙏


Java If-Else || Hackerrank Solution Java If-Else || Hackerrank Solution Reviewed by CodexRitik on November 06, 2020 Rating: 5

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