# Check if given permutation of 1 to N can be counted in clockwise or anticlockwise direction

• Last Updated : 30 May, 2022

Given an integer array arr of size N which contains distinct elements from 1 to N. The task is to check if a position in the array can be found such that all numbers from 1 to N can be counted in a clockwise direction or counterclockwise direction.
Examples:

```Input: arr[] = [2, 3, 4, 5, 1]
Output: YES
Explanation:
1   2
5       3
4
If counting is start at index 4
then all numbers can be counted
from 1 to N in a clockwise order.

Input: arr[] = {1, 2, 3, 5, 4]
Output: NO
Explanation:
There is no any index in array
from which given array can count
1 to N in clockwise order or
counterclockwise order.```

Approach: The above problem can be solved by observation and analysis.

1. An index in the array can be found only if the count of the absolute difference between the consecutive elements greater than 1 is exactly 1, because then only it will be possible to count from 1 to N in clockwise or counterclockwise order.
2. If the count of the absolute difference between the adjacent elements is more than 1 Then it will be impossible to count from 1 to N in clockwise or counterclockwise order.

Below is the basic implementation of the above approach:

## C++

 `// C++ program to check Clockwise or``// counterclockwise order in an array` `#include ``using` `namespace` `std;` `bool` `check_order(vector<``int``> arr)``{``    ``int` `cnt = 0;``    ``for` `(``int` `i = 0; i < arr.size() - 1; i++) {``        ``if` `(``abs``(arr[i + 1] - arr[i]) > 1)``            ``cnt++;``    ``}``    ``// Comparing the first and last``    ``// value of array``    ``if` `(``abs``(arr - arr[arr.size() - 1]) > 1)``        ``cnt++;` `    ``// If the Count is greater``    ``// than 1 then it can't be``    ``// represented in required order``    ``if` `(cnt > 1)``        ``return` `false``;``    ``return` `true``;``}` `// Driver function``int` `main()``{``    ``vector<``int``> arr = { 2, 3, 4, 5, 1 };``    ``if` `(check_order(arr))``        ``cout << ``"YES"``;``    ``else``        ``cout << ``"NO"``;``    ``return` `0;``}`

## Java

 `// Java program to check clockwise or``// counterclockwise order in an array``class` `GFG{` `static` `boolean` `check_order(``int` `[]arr)``{``    ``int` `cnt = ``0``;``    ``for``(``int` `i = ``0``; i < arr.length - ``1``; i++)``    ``{``        ``if` `(Math.abs(arr[i + ``1``] -``                     ``arr[i]) > ``1``)``            ``cnt++;``    ``}``    ` `    ``// Comparing the first and last``    ``// value of array``    ``if` `(Math.abs(arr[``0``] -``                 ``arr[arr.length - ``1``]) > ``1``)``        ``cnt++;` `    ``// If the Count is greater``    ``// than 1 then it can't be``    ``// represented in required order``    ``if` `(cnt > ``1``)``        ``return` `false``;``        ` `    ``return` `true``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `[]arr = { ``2``, ``3``, ``4``, ``5``, ``1` `};``    ` `    ``if` `(check_order(arr))``        ``System.out.print(``"YES"``);``    ``else``        ``System.out.print(``"NO"``);``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program to check clockwise or``# counterclockwise order in an array``def` `check_order(arr):``    ` `    ``cnt ``=` `0``    ``for` `i ``in` `range``(``len``(arr) ``-` `1``):``        ``if` `(``abs``(arr[i ``+` `1``] ``-` `arr[i]) > ``1``):``            ``cnt ``+``=` `1` `    ``# Comparing the first and last``    ``# value of array``    ``if` `(``abs``(arr[``0``] ``-` `arr[``len``(arr) ``-` `1``]) > ``1``):``        ``cnt ``+``=` `1` `    ``# If the count is greater``    ``# than 1 then it can't be``    ``# represented in required order``    ``if` `(cnt > ``1``):``        ``return` `False``        ` `    ``return` `True` `# Driver code``arr ``=` `[ ``2``, ``3``, ``4``, ``5``, ``1` `]` `if` `(check_order(arr)):``    ``print``(``"YES"``)``else``:``    ``print``(``"NO"``)` `# This code is contributed by Vishal Maurya.`

## C#

 `// C# program to check clockwise or``// counterclockwise order in an array``using` `System;` `class` `GFG{` `static` `bool` `check_order(``int` `[]arr)``{``    ``int` `cnt = 0;``    ` `    ``for``(``int` `i = 0; i < arr.Length - 1; i++)``    ``{``        ``if` `(Math.Abs(arr[i + 1] -``                   ``arr[i]) > 1)``            ``cnt++;``    ``}``    ` `    ``// Comparing the first and last``    ``// value of array``    ``if` `(Math.Abs(arr -``                 ``arr[arr.Length - 1]) > 1)``        ``cnt++;` `    ``// If the Count is greater``    ``// than 1 then it can't be``    ``// represented in required order``    ``if` `(cnt > 1)``        ``return` `false``;``        ` `    ``return` `true``;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 2, 3, 4, 5, 1 };``    ` `    ``if` `(check_order(arr))``        ``Console.Write(``"YES"``);``    ``else``        ``Console.Write(``"NO"``);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`YES`

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.

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