# Minimum absolute difference of adjacent elements in a circular array

• Difficulty Level : Basic
• Last Updated : 08 Jun, 2022

Given n integers, which form a circle. Find the minimal absolute value of any adjacent pair. If there are many optimum solutions, output any of them.
Note: they are in circle
Examples:

```Input : arr[] = {10, 12, 13, 15, 10}
Output : 0
Explanation: |10 - 10| = 0 which is the
minimum possible.

Input : arr[] = {10, 20, 30, 40}
Output : 10
Explanation: |10 - 20| = 10 which is the
minimum, 2 3 or 3 4 can be the answers also.  ```

Initially consider the minimum value to be of first and second elements. Traverse from second element to last. Check the difference of every adjacent pair and store the minimum value. When last element is reached, check its difference with first element.
Below is the implementation of the above approach.

## C++

 `// C++ program to find maximum difference``// between adjacent elements in a circular array.``#include ``using` `namespace` `std;` `void` `minAdjDifference(``int` `arr[], ``int` `n)``{``    ``if` `(n < 2)``        ``return``;` `    ``// Checking normal adjacent elements``    ``int` `res = ``abs``(arr - arr);``    ``for` `(``int` `i = 2; i < n; i++)``        ``res = min(res, ``abs``(arr[i] - arr[i - 1]));` `    ``// Checking circular link``    ``res = min(res, ``abs``(arr[n - 1] - arr));` `    ``cout << ``"Min Difference = "` `<< res;``}` `// driver program to check the above function``int` `main()``{``    ``int` `a[] = { 10, 12, 13, 15, 10 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``minAdjDifference(a, n);``    ``return` `0;``}`

## Java

 `// Java program to find maximum difference``// between adjacent elements in a circular``// array.``class` `GFG {` `    ``static` `void` `minAdjDifference(``int` `arr[], ``int` `n)``    ``{``        ``if` `(n < ``2``)``            ``return``;` `        ``// Checking normal adjacent elements``        ``int` `res = Math.abs(arr[``1``] - arr[``0``]);``        ``for` `(``int` `i = ``2``; i < n; i++)``            ``res = Math.min(res, Math.abs(arr[i] - arr[i - ``1``]));` `        ``// Checking circular link``        ``res = Math.min(res, Math.abs(arr[n - ``1``] - arr[``0``]));` `        ``System.out.print(``"Min Difference = "` `+ res);``    ``}` `    ``// driver code``    ``public` `static` `void` `main(String arg[])``    ``{` `        ``int` `a[] = { ``10``, ``12``, ``13``, ``15``, ``10` `};``        ``int` `n = a.length;` `        ``minAdjDifference(a, n);``    ``}``}` `// This code is contributed by Anant Agarwal``// and improved by Anuj Sharma.`

## Python3

 `# Python3 program to find maximum``# difference between adjacent``# elements in a circular array.` `def` `minAdjDifference(arr, n):` `    ``if` `(n < ``2``): ``return` `    ``# Checking normal adjacent elements``    ``res ``=` `abs``(arr[``1``] ``-` `arr[``0``])``    ` `    ``for` `i ``in` `range``(``2``, n):``        ``res ``=` `min``(res, ``abs``(arr[i] ``-` `arr[i ``-` `1``]))` `    ``# Checking circular link``    ``res ``=` `min``(res, ``abs``(arr[n ``-` `1``] ``-` `arr[``0``]))` `    ``print``(``"Min Difference = "``, res)` `# Driver Code``a ``=` `[``10``, ``12``, ``13``, ``15``, ``10``]``n ``=` `len``(a)``minAdjDifference(a, n)` `# This code is contributed by Anant Agarwal``# and improved by Anuj Sharma.`

## C#

 `// C# program to find maximum difference``// between adjacent elements in a circular array.``using` `System;` `class` `GFG {` `    ``static` `void` `minAdjDifference(``int``[] arr, ``int` `n)``    ``{``        ``if` `(n < 2)``            ``return``;` `        ``// Checking normal adjacent elements``        ``int` `res = Math.Abs(arr - arr);``        ``for` `(``int` `i = 2; i < n; i++)``            ``res = Math.Min(res, Math.Abs(arr[i] - arr[i - 1]));` `        ``// Checking circular link``        ``res = Math.Min(res, Math.Abs(arr[n - 1] - arr));` `        ``Console.Write(``"Min Difference = "` `+ res);``    ``}` `    ``// driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] a = { 10, 12, 13, 15, 10 };``        ``int` `n = a.Length;``        ``minAdjDifference(a, n);``    ``}``}` `// This code is contributed by Anant Agarwal``// and improved by Anuj Sharma.`

## PHP

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## Javascript

 ``

Output:

`Min Difference =  0`

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.

https://youtu.be/G

-vouVriEvY
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