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

Drawing Book

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2108 Discussions

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  • nice00guy69
     + 0 comments

    There are two types of people: Type 1:

    #return min((p//2), ((n//2)-(p//2)))
'
Type 2:
'
listi = []
    for i in range(0, n+1, 2):
        listi.append((i, i+1))
    for i in range(len(listi)):
        if p in listi[i]:
            front = i
    back = len(listi) - front - 1
    return min(front, back)

  • 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;
      }