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

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2107 Discussions

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  • NaveenG_EVOL
     + 0 comments

    Simple Python Solution:

    def pageCount(n, p):
    return min( p, n-p +(n%2!=1) )//2

  • gracegovjobs
     + 0 comments
    def pageCount(n, p):
    papers=(n//2)+1
    paper=(p//2)+1
    if (papers-paper)<paper:
        return (papers-paper)
    else:
        return paper-1

  • Nafeh
     + 0 comments

    The problem sounds tough at first but if you think about it, it's a simple division problem. Here's an easy C# solution -

    	public static int pageCount(int n, int p)
    {
        // When iterating from 1 take the celing value
        double pageTurnNeededFromStart = Math.Ceiling((double)(p-1)/2);
        
        // when iterating from end - take the ceiling value if the end page is even, otherwise take floor value
        double pageTurnNeededFromEnd = n%2==0? Math.Ceiling((double)(n-p)/2) : Math.Floor((double)(n-p)/2);
        
        return (int)Math.Min(pageTurnNeededFromStart, pageTurnNeededFromEnd);
    }

  • vonsats
     + 1 comment
      int pz = n / 2;
      int p2 = p / 2;
      int high = pz - p2;
      int low  = p2 - 0;

      if(low < high)
      {
        return low;

      }
      else
      {
        return high;
      }

  • brian54
     + 0 comments

    The AI bros doing L33tCode are heavy now. These 1 liners are ridiculous and help no person. But below hopefully is the logic illustrated by some of the ChatGPT coders.

    public static int pageCount(int n, int p)
    {
        if (p > n)
            return -1; // early exit, can't be done
        
        
        var evenPages = n % 2 == 0;
        
        // p == begin of book, end of book, or left page end of book if odd pages
        if (p == 1 || n == p || (!evenPages && p == n - 1))
            return 0;
        else if (evenPages && p == n - 1)
            return 1;
        
        var fromLeftTurns = (int)Math.Floor((decimal)p / 2);
        var fromRightTurns = (int)Math.Floor((decimal)(n - p) / 2);
        
        return fromLeftTurns <= fromRightTurns ? fromLeftTurns : fromRightTurns;
    }