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 <bits/stdc++.h>` `using` `namespace` `std;` `void` `minAdjDifference(` `int` `arr[], ` `int` `n)` `{` ` ` `if` `(n < 2)` ` ` `return` `;` ` ` `// Checking normal adjacent elements` ` ` `int` `res = ` `abs` `(arr[1] - arr[0]);` ` ` `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[0]));` ` ` `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[0]);` ` ` `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[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]));` ` ` `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

`<?php` `// PHP program to find maximum` `// difference between adjacent` `// elements in a circular array.` `function` `minAdjDifference(` `$arr` `, ` `$n` `)` `{` ` ` `if` `(` `$n` `< 2)` ` ` `return` `;` ` ` ` ` `// Checking normal` ` ` `// adjacent elements` ` ` `$res` `= ` `abs` `(` `$arr` `[1] - ` `$arr` `[0]);` ` ` `for` `(` `$i` `= 2; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `$res` `= min(` `$res` `,` ` ` `abs` `(` `$arr` `[` `$i` `] -` ` ` `$arr` `[` `$i` `- 1]));` ` ` ` ` `// Checking circular link` ` ` `$res` `= min(` `$res` `, ` `abs` `(` `$arr` `[` `$n` `- 1] -` ` ` `$arr` `[0]));` ` ` ` ` `echo` `"Min Difference = "` `, ` `$res` `;` `}` `// Driver Code` `$a` `= ` `array` `(10, 12, 13, 15, 10);` `$n` `= ` `count` `(` `$a` `);` `minAdjDifference(` `$a` `, ` `$n` `);` `//This code is contributed by anuj_67` `//and improved by Anuj Sharma.` `?>` |

## Javascript

`<script>` `// Javascript program to find maximum difference` `// between adjacent elements in a circular array.` `function` `minAdjDifference( arr, n)` `{` ` ` `if` `(n < 2)` ` ` `return` `;` ` ` `// Checking normal adjacent elements` ` ` `let res = Math.abs(arr[1] - arr[0]);` ` ` `for` `(let 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]));` ` ` `document.write(` `"Min Difference = "` `+ res);` `}` `// driver code` ` ` `let a = [ 10, 12, 13, 15, 10 ];` ` ` `let n = a.length;` ` ` `minAdjDifference(a, n);` `</script>` |

**Output: **

Min Difference = 0

**Time complexity:** O(n)

https://youtu.be/G

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