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

Drawing Book

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2101 Discussions

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  • melvy948
     + 0 comments

    I was overthinking this problem at first, but once I broke it down it made sense—it’s really just about finding the minimum turns from either side of the book. My Python version looks like this:

    def pageCount(n, p): return min(p // 2, (n // 2) - (p // 2))

    Clean and efficient—sometimes the simplest logic is the most powerful! this delta key helps alot

  • greivin_lopez
     + 0 comments
    # Drawing Book 📖
def page_count(n, p):
    s = p // 2
    e = n // 2 - p // 2
    return min(s, e)

  • briandeandean460
     + 0 comments

    I initially overcomplicated it by trying to simulate the entire book, but then I realized it's just about comparing the number of page turns from the front and back. Here's a simple Python solution: python

    def pageCount(n, p): return min(p // 2, (n // 2 - p // 2))

    Interestingly, this concept is also applied in healthcare systems. For instance, revenue cycle management services help streamline processes by efficiently handling tasks, much like minimizing unnecessary steps in a problem-solving approach. You can learn more about this here: https://swiftmds.com/services/healthcare-revenue-cycle-management-services/.

    Hope this helps!

  • kbigdelysh
     + 0 comments

    Isn't it should be 'back' instead of 'front' in the following sentence from the problem? "...The last page may only be printed on the front, given the length of the book."

    I don't understand that sentence

  • kepliu
     + 1 comment
     public static int pageCount(int n, int p) {
        int totalFaces = 1 + (n-1)/2 + (n-1) % 2; // a face is the left the right page
        int forwardTurns = p/2;
        int reverseTurns = (totalFaces-1) - forwardTurns; //totalFaces-1 is the index of the last face; the first face's index is 0

        return Math.min(forwardTurns, reverseTurns);
}