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# Minimum absolute difference of adjacent elements in a Circular Array

Given a circular array arr[] of length N, the task is to find the minimum absolute difference between any adjacent pair. If there are many optimum solutions, output any of them.

Examples:

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

Input: arr[] = {10, 20, 30, 40}
Output: 10
Explanation: |10 – 20| = 10 is the minimum, 20 30 or 30 40 could be the answer too.

Approach: Below is the idea to solve the problem

Traverse from second element to last an check the difference of every adjacent pair and store the minimum value. The edge case is to check difference between last element and first element.

Follow the steps below to implement the idea:

• Create a variable res to store the minimum difference between any adjacent pair.
• Run a for loop of i from 1 to N-1 and for each iteration.
• Calculate the absolute difference of Arr[i] and Arr[i-1].
• Update the value of res if the value of | Arr[i] – Arr[i-1] | is smaller than res.
• Return res as the required answer.

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 code``int` `main()``{``    ``int` `a[] = { 10, 12, 13, 15, 10 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``  ` `      ``//Function Call``    ``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

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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.

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