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Minimum absolute difference of adjacent elements in a Circular Array

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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 <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 code
int main()
{
    int a[] = { 10, 12, 13, 15, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    
      //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[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), 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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Last Updated : 18 Sep, 2023
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