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Maximize sum of consecutive differences in a circular array

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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
Recommended Practice

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. 

Note:  a(n/2)+1 This term is considered only for even n because for odd n, it is added once and subtracted once and hence cancels out. 

Implementation:

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.


Javascript




<script>
 
// JavaScript program for the above approach
 
    // Return the maximum Sum of difference between
    // consecutive elements.
    function maxSum(arr, n)
    {
        let 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 (let i = 0; i < n/2; i++)
        {
            sum -= (2 * arr[i]);
            sum += (2 * arr[n - i - 1]);
        }
      
        return sum;
    }
 
// Driver Code
    let arr = [ 4, 2, 1, 8 ];
    let n = arr.length;
    document.write(maxSum(arr, n));
 
// This code is contributed by chinmoy1997pal.
</script>


Output

18

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



Last Updated : 08 Jul, 2022
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