# Minimum number of page turns to get to a desired page

• Difficulty Level : Basic
• Last Updated : 08 Feb, 2023

Given a book of N pages, the task is to calculate the minimum number of page turns to get to a given 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++

 `// C++ program to find minimum number of page``// turns to reach a page``#include``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 Program``int` `main()``{``    ``int` `n = 6, k = 2;``    ``cout << minTurn(n,k) << endl;``    ``return` `0;``}` `// This code is modified by naveenkonda`

## Java

 `// Java program to find minimum``// number of page turns to``// reach a page``import` `java.io.*;` `public` `class` `GFG``{` `// Function to calculate``// minimum number of page``// turns required``static` `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

 `# Python3 program to find minimum number``# of page turns to reach a page``def` `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 Code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `6``    ``k ``=` `2``    ``print``(``int``(minTurn(n, k)))``    ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to find minimum``// number of page turns to``// reach a page``using` `System;` `public` `class` `GFG``{` `// Function to calculate``// minimum number of page``// turns required``static` `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

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## Javascript

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Output :

`1`

Time Complexity : O(1)
Auxiliary Space: O(1)
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