Minimum absolute difference of adjacent elements in a circular array

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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Output:

Min Difference =  0

Time complexity: O(n)

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Improved By : vt_m, Anuj_Sharma