Minimum number of page turns to get to a desired page

11 Aug, 2021
Given a book of N pages, the task is to calculate the minimum number of page turns to get to a give desired page K. We can either start turning pages from the front side of the book (i.e from page 1) or from the back side of the book (i.e page number N). Each page has two sides, front and back, except the first page, which has only back side and the last page which may only have back side depending on the number of pages of the book. Examples :  
Input : N = 6 and K = 2.
Output : 1.
From front, (1) -> (2, 3), page turned = 1.
From back, (6) -> (4, 5) -> (2,3), page turned = 2.
So, Minimum number of page turned = 1.

Input : N = 5 and K = 4.
Output : 1.
From front, (1) -> (2, 3) -> (4,5), page turned = 2.
From back, (4, 5) page turned = 1. From back, it is 2nd page, since 4 is on other side of page 5 and page 5 is the first one from back
So, Minimum number of page turned = 1.
 
The idea is to calculate distance of the desired page from the front and from the back of the book, minimum of this is the required answer. Now, Consider there is page 0, which is front of the first page. And if N is even, consider there is page N+1, which is back of the last page, so total number of pages are N+1. To calculate the distance, 1. If K is even, front distance = (K – 0)/2 and back distance = (N – 1 – K)/2. 2. If K is odd, front distance = (K – 1)/2 and back distance = (N – K)/2.  
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find minimum number of page// turns to reach a page#include<bits/stdc++.h>using namespace std; int minTurn(int n, int k){    // Considering back of last page.    if (n%2 == 0)        n++;     // Calculating Distance from front and    // back of the book and return the min    return min((k + 1)/2, (n - k + 1)/2);} // Driven Programint main(){    int n = 6, k = 2;    cout << minTurn(n,k) << endl;    return 0;} // This code is modified by naveenkonda
// Java program to find minimum// number of page turns to// reach a pageimport java.io.*; public class GFG{ // Function to calculate// minimum number of page// turns requiredstatic int minTurn(int n, int k){         // Considering back of last page.    if (n % 2 == 0)        n++;     // Calculating Distance from front and    // back of the book and return the min    Math.min((k + 1) / 2, (n - k + 1) / 2);}     // Driver Code    static public void main (String[] args)    {        int n = 6, k = 2;        System.out.println(minTurn(n, k));    }} // This code is contributed by vt_m.
# Python3 program to find minimum number# of page turns to reach a pagedef minTurn(n, k):         # Considering back of last page.    if (n % 2 == 0):        n += 1     // Calculating Distance from front and    // back of the book and return the min    return min((k + 1) / 2, (n - k + 1) / 2) # Driver Codeif __name__ == '__main__':    n = 6    k = 2    print(int(minTurn(n, k)))     # This code is contributed by# Surendra_Gangwar
// C# program to find minimum// number of page turns to// reach a pageusing System; public class GFG{ // Function to calculate// minimum number of page// turns requiredstatic int minTurn(int n, int k){         // Considering back of last page.    if (n % 2 == 0)        n++;                            // Calculating Distance from front and    // back of the book and return the min    return Math.Min((k + 1) / 2,                    (n - k + 1) / 2);}     // Driver Code    static public void Main (String[] args)    {        int n = 6, k = 2;        Console.WriteLine(minTurn(n, k));    }} // This code is contributed by vt_m.
<?php// PHP program to find minimum number// of page turns to reach a page function minTurn($n, $k){    // Considering back of last page.    if ($n % 2 == 0)        $n++;     // Calculating Distance from front and    // back of the book and return the min    return min(($k + 1) / 2,               ($n - $k + 1) / 2);} // Driver Code$n = 6; $k = 2;echo minTurn($n, $k) ; // This code is contributed by nitin mittal.?>
<script> // Javascript program to find minimum// number of page turns to// reach a page // Function to calculate// minimum number of page// turns requiredfunction minTurn(n, k){           // Considering back of last page.    if (n % 2 == 0)        n++;       // Calculating Distance from front and    // back of the book and return the min    let x = Math.min((k + 1) / 2, (n - k + 1) / 2);    return Math.floor(x);}      // Driver code            let n = 6, k = 2;        document.write(minTurn(n, k));                   </script>
Output :  
1
Time Complexity : O(1)Auxiliary Space: O(1) This article is contributed by Anuj Chauhan(anuj0503). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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