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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Drawing Book
  5. Discussions

Drawing Book

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  • lemming0061
     + 87 comments

    Extremely simple solution in Python

    #!/bin/python

import sys


n = int(raw_input().strip())
p = int(raw_input().strip())

print min(p/2,n/2-p/2)

  • Kanahaiya
     + 10 comments

    Hello friends,

    In this video tutorial, I have explained hackerrank drawing book solution algorithm. hackerrank drawing book problem can be solved by using mathematics. The complexity of drawing book hackerrank solution is O (1)

    If interested to know more about the generic algorithm in details-

    click here for the video explanation of generic algorithm with complexity analysis.

    or you can click on the image too to follow youtube tutorial.

    Here is the working solution:-

    source code :

    static int pageCount(int n, int p) {

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

		return Math.min(targetPageTurnCountFromFront, targetPageTurnCountFromBack);

	}

    Would really appreciate your feedback like, dislike , comment etc. on my video.

    Do not forget to upvote, if you find it useful.

  • Aswath3167
     + 24 comments

    My solution in C...Hope it helps!!!

    #include<stdio.h>
int main()
{
int n=0,p=0,min;
scanf("%d",&n);
scanf("%d",&p);
min=(n/2)-(p/2);
if(min>p/2)
min=p/2;
printf("%d",min);
return 0;
}

  • sakai_stanley
     + 10 comments

    My JavaScript solution:

    function pageCount(n, p) {
    /*
		  n: the number of pages in the book
		  p: the page number to turn to
    */

    const pageTurns = Math.floor(p / 2);
    const totalTurns = Math.floor(n / 2);

    /* Returns the total number of page turns it takes to get
    to a page or how many it requires if starting from the back */

    return Math.min(pageTurns, totalTurns - pageTurns);
}

  • aruncveli
     + 2 comments

    Input for test case #26 has to be on two lines.

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