Drawing Book in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct]

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Algorithm Solution in Hackerrank | Java Hackerrank Solutions

Hello Programmers/Coders, Today we are going to share solutions of Programming problems of HackerRank, Algorithm Solutions of Problem Solving Section in Java. At Each Problem with Successful submission with all Test Cases Passed, you will get an score or marks. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for Drawing Book in Java-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.

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Introduction To Algorithm

The word Algorithm means “a process or set of rules to be followed in calculations or other problem-solving operations”. Therefore Algorithm refers to a set of rules/instructions that step-by-step define how a work is to be executed upon in order to get the expected results. 

Advantages of Algorithms:

  • It is easy to understand.
  • Algorithm is a step-wise representation of a solution to a given problem.
  • In Algorithm the problem is broken down into smaller pieces or steps hence, it is easier for the programmer to convert it into an actual program.

Link for the Problem – Drawing Book – Hacker Rank Solution

Drawing Book – Hacker Rank Solution

Problem:

A teacher asks the class to open their books to a page number. A student can either start turning pages from the front of the book or from the back of the book. They always turn pages one at a time. When they open the book, page  is always on the right side:

image

When they flip page , they see pages  and . Each page except the last page will always be printed on both sides. The last page may only be printed on the front, given the length of the book. If the book is  pages long, and a student wants to turn to page , what is the minimum number of pages to turn? They can start at the beginning or the end of the book.

Given  and , find and print the minimum number of pages that must be turned in order to arrive at page .

Example

Untitled Diagram(4).png

Using the diagram above, if the student wants to get to page , they open the book to page , flip  page and they are on the correct page. If they open the book to the last page, page , they turn  page and are at the correct page. Return .

Function Description

Complete the pageCount function in the editor below.

pageCount has the following parameter(s):

  • int n: the number of pages in the book
  • int p: the page number to turn to

Returns

  • int: the minimum number of pages to turn

Input Format

The first line contains an integer , the number of pages in the book.
The second line contains an integer, , the page to turn to.

Constraints

Sample Input 0

6
2

Sample Output 0

1

Explanation 0

If the student starts turning from page , they only need to turn  page:

Untitled Diagram(6).png

If a student starts turning from page , they need to turn  pages:

Untitled Diagram(3).png

Return the minimum value, .

Sample Input 1

5
4

Sample Output 1

0

Explanation 1

If the student starts turning from page , they need to turn  pages:

Untitled Diagram(4).png

If they start turning from page , they do not need to turn any pages:

Untitled Diagram(5).png

Return the minimum value, 0.

Drawing Book – Hacker Rank Solution
import java.util.Scanner;

/**
 * @author Techno-RJ
 *
 */
public class DrawingBook {

	static int pageCount(int n, int p) {

		int totalPageTurnCountFromFront = n / 2;
		int targetPageTurnCountFromFront = p / 2;
		int targetPageTurnCountFromBack = totalPageTurnCountFromFront - targetPageTurnCountFromFront;

		return Math.min(targetPageTurnCountFromFront, targetPageTurnCountFromBack);

	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int p = in.nextInt();
		System.out.println(pageCount(n, p));
		in.close();
	}
}

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