My math skill sucks pants, java brute force approach for fun

public static int pageCount(int n, int p) {
// Write your code here
List<List<Integer>> book = new ArrayList();
    System.out.println(String.format("n=%d, p=%d ", n, p));
int numOfPages = n + 1; // +1 is page zero which will always present
int l = (int) Math.ceil (numOfPages/2.0);

System.out.println("l=" + l);

//create the books
for (int i = 0; i < l; i++) {
    List<Integer> pages = new ArrayList<>();
    int number = i * 2;
    pages.add(number);
    pages.add(number+1);
    book.add(new ArrayList<>(pages));
}

//7 pages
//should be 4 paper
//7 / 2
System.out.println("book: " + book);    

int ftbFlip = 0;
for (int i = 0; i < book.size(); i++) {
    List<Integer> pages = book.get(i);
    System.out.println(String.format("iftb=%d pages=%s", i, pages.toString()));

    if (pages.contains(p)) {
        ftbFlip = i;
        break;
    }
}
System.out.println("ftbFlip: " + ftbFlip);

int btfFlip = 0;
for (int i = book.size() - 1; i > 0; i--) {
    List<Integer> pages = book.get(i);
    System.out.println("pages: " + pages);
    if (pages.contains(p)) {
        break;
    } else {
        btfFlip++;
    }
}
System.out.println("btfFlip: " + btfFlip);

int result = Math.min(ftbFlip, btfFlip);
return result;
}

Scala

def pageCount(n: Int, p: Int): Int = {
     //Sample book: 1 - 2 3 - 4 5 - 6 7 - 8 9 - 10 11
        val lastPage = if(n % 2 == 0) n+1 else n
        val firstPage = 0
        
        val fromFirst: Int = p/2
        val fromLast: Int = (lastPage - p)/2
        
        Array(fromFirst, fromLast).min
    }

short c++ solution. the main idea is that we can calculate how many pages to turn to get number 'x' by turning 'x//2' pages (divide with rounding down e.g. 5//2 = 2). It's because on every display of 2 pages there are 2 consecutive pages e.g. (2,3), (3,4) etc. All we gotta do is just see if we can get from 0->p by turning p//2 pages or from n->p which is n//2-p//2.

int pageCount(int n, int p) {
    int sza = n/2;
    int pp = p/2;
    return min(pp, sza-pp);
}