Java End Of File

 Hackerrank Java End Of File Problem By Java Tech Solutionz

The challenge here is to read  lines of input until you reach EOF, then number and print all  lines of content.

Hint: Java's Scanner.hasNext() method is helpful for this problem.

Input Format

Read some unknown  lines of input from stdin(System.in) until you reach EOF; each line of input contains a non-empty String.

Output Format

For each line, print the line number, followed by a single space, and then the line content received as input.

Sample Input

Hello world
I am a file
Read me until end-of-file.

Sample Output

1 Hello world
2 I am a file
3 Read me until end-of-file.

Solution:

public class Solution {
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner scanner = new Scanner(System.in);
        int i=0;
        while(scanner.hasNextLine()){
            System.out.println(++i + " "+scanner.nextLine());
        }
    }
}

Java static initializer block

 Hackerrank Java static initializer block By Java Tech Solutionz

Static initialization blocks are executed when the class is loaded, and you can initialize static variables in those blocks.

It's time to test your knowledge of Static initialization blocks. You can read about it here.

You are given a class Solution with a main method. Complete the given code so that it outputs the area of a parallelogram with breadth  and height . You should read the variables from the standard input.

If  or  , the output should be "java.lang.Exception: Breadth and height must be positive" without quotes.

Input Format

There are two lines of input. The first line contains : the breadth of the parallelogram. The next line contains : the height of the parallelogram.

Constraints

Output Format

If both values are greater than zero, then the main method must output the area of the parallelogram. Otherwise, print "java.lang.Exception: Breadth and height must be positive" without quotes.

Sample input 1

1
3

Sample output 1

3

Sample input 2

-1
2

Sample output 2

java.lang.Exception: Breadth and height must be positive

Solution:

static int B;
static int H;
static boolean flag = false;
static{
    Scanner scanner = new Scanner(System.in);
     B = scanner.nextInt();
     H = scanner.nextInt();
    
    if(B>0 && H>0){
        flag = true;
    }else{
        System.out.println( "java.lang.Exception: Breadth and height must be positive");
    }
}

Note:

Static Initialization Blocks
A static initialization block is a normal block of code enclosed in braces, { }, and preceded by the static keyword. Here is an example:
static {
    // whatever code is needed for initialization goes here
}
A class can have any number of static initialization blocks, and they can appear anywhere in the class body. The runtime system guarantees that static initialization blocks are called in the order that they appear in the source code.
There is an alternative to static blocks — you can write a private static method:
class Whatever {
    public static varType myVar = initializeClassVariable();
        
    private static varType initializeClassVariable() {

        // initialization code goes here
    }
}
The advantage of private static methods is that they can be reused later if you need to reinitialize the class variable

Java If Else Problem

 Java If Else Problem By Java Tech Solutionz

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:

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:

public static void main(String[] args) {

        int n = scanner.nextInt();

        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        scanner.close();

        

        if((n%2==1) || (n%2==0 && (n<=20 && n>=6))){

            System.out.println("Weird");

        }else if((n%2==0) && ((n<=5 && n>=2) || (n>20))){

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

        }

        

    }