Here is Drawing Book solution in Python, Java, C++, C and Javascript programming - https://programmingoneonone.com/hackerrank-drawing-book-problem-solution.html

Page Turns in Rust: Solving the Book Page Count Problem in O(1)

fn pageCount(n: i32, p: i32) -> i32 {
    if n/2 < p { //from the end
        return n / 2 - p / 2;
    } else { //from beginning
        return p / 2;
    }
}

This function computes the minimum number of page turns needed to reach page p in a book of n pages. The student can start turning pages either from the front (page 1) or from the back (page n).

Key idea:

Number of page turns from the front: from_front = p/2 Because every page turn advances 2 pages (left + right).

Number of page turns from the back: from_back=n/2−p/2 Here n/2 is the total number of “spreads” in the book, and p/2 is how many spreads we’ve passed from the front.

Time complexity: The function performs only a couple of integer operations and a conditional -->O(1) time.

Space complexity: It uses only fixed scalar variables, no extra data structures --> O(1) space.

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)

Simple Python Solution:

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