Maximize sum of consecutive differences in a circular array

Given an array of n elements. Consider array as circular array i.e element after an is a1. The task is to find maximum sum of the difference between consecutive elements with rearrangement of array element allowed i.e after rearrangement of element find |a1 – a2| + |a2 – a3| + …… + |an – 1 – an| + |an – a1|.

Examples:

Input : arr[] = { 4, 2, 1, 8 }
Output : 18
Rearrange given array as : { 1, 8, 2, 4 }
Sum of difference between consecutive element
= |1 - 8| + |8 - 2| + |2 - 4| + |4 - 1|
= 7 + 6 + 2 + 3
= 18.

Input : arr[] = { 10, 12, 15 }
Output : 10



The idea is to use Greedy Approach and try to bring elements having greater difference closer.
Consider the sorted permutation of the given array a1, a1, a2,…., an – 1, an such that a1 < a2 < a3…. < an – 1 < an.
Now, to obtain the answer having maximum sum of difference between consecutive element, arrange element in following manner:
a1, an, a2, an-1,…., an/2, a(n/2) + 1
We can observe that the arrangement produces the optimal answer, as all a1, a2, a3,….., a(n/2)-1, an/2 are subtracted twice while a(n/2)+1, a(n/2)+2, a(n/2)+3,….., an – 1, an are added twice.

C++

// C++ program to maximize the sum of difference
// between consecutive elements in circular array
#include <bits/stdc++.h>
using namespace std;

// Return the maximum Sum of difference between
// consecutive elements.
int maxSum(int arr[], int n)
{
    int sum = 0;

    // Sorting the array.
    sort(arr, arr + n);

    // Subtracting a1, a2, a3,....., a(n/2)-1, an/2
    // twice and adding a(n/2)+1, a(n/2)+2, a(n/2)+3,.
    // ...., an - 1, an twice.
    for (int i = 0; i < n/2; i++)
    {
        sum -= (2 * arr[i]);
        sum += (2 * arr[n - i - 1]);
    }

    return sum;
}

// Driver Program
int main()
{
    int arr[] = { 4, 2, 1, 8 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maxSum(arr, n) << endl;
    return 0;
}

Java

// Java program to maximize the sum of difference
// between consecutive elements in circular array
import java.io.*;
import java.util.Arrays;
 
class MaxSum
{
    // Return the maximum Sum of difference between
    // consecutive elements.
    static int maxSum(int arr[], int n)
    {
        int sum = 0;
 
        // Sorting the array.
        Arrays.sort(arr);
     
        // Subtracting a1, a2, a3,....., a(n/2)-1, 
        // an/2 twice and adding a(n/2)+1, a(n/2)+2, 
        // a(n/2)+3,....., an - 1, an twice.
        for (int i = 0; i < n/2; i++)
        {
            sum -= (2 * arr[i]);
            sum += (2 * arr[n - i - 1]);
        }
     
        return sum;
    }

    // Driver Program
    public static void main (String[] args)
    {
        int arr[] = { 4, 2, 1, 8 };
        int n = arr.length;
        System.out.println(maxSum(arr, n));
    }
}
/*This code is contributed by Prakriti Gupta*/

Python3

# Python3 program to maximize the sum of difference
# between consecutive elements in circular array

# Return the maximum Sum of difference 
# between consecutive elements
def maxSum(arr, n):
    sum = 0

    # Sorting the array
    arr.sort()

    # Subtracting a1, a2, a3,....., a(n/2)-1, an/2
    # twice and adding a(n/2)+1, a(n/2)+2, a(n/2)+3,.
    # ...., an - 1, an twice.
    for i in range(0, int(n / 2)) :
        sum -= (2 * arr[i])
        sum += (2 * arr[n - i - 1])

    return sum


# Driver Program
arr = [4, 2, 1, 8]
n = len(arr)
print (maxSum(arr, n))

# This code is contributed by Shreyanshi Arun.

C#

// C# program to maximize the sum of difference
// between consecutive elements in circular array
using System;

class MaxSum {
    
    // Return the maximum Sum of difference 
    // between consecutive elements.
    static int maxSum(int[] arr, int n)
    {
        int sum = 0;

        // Sorting the array.
        Array.Sort(arr);

        // Subtracting a1, a2, a3, ....., a(n/2)-1,
        // an/2 twice and adding a(n/2)+1, a(n/2)+2,
        // a(n/2)+3, ....., an - 1, an twice.
        for (int i = 0; i < n / 2; i++) {
            sum -= (2 * arr[i]);
            sum += (2 * arr[n - i - 1]);
        }

        return sum;
    }

    // Driver Program
    public static void Main()
    {
        int[] arr = { 4, 2, 1, 8 };
        int n = arr.Length;
        Console.WriteLine(maxSum(arr, n));
    }
}

//This Code is contributed by vt_m.


Output :

18

Time Complexity: O(nlogn).
Auxiliary Space : O(1)

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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