Maximum sum of absolute difference of any permutation

• Difficulty Level : Easy
• Last Updated : 24 May, 2021

Given an array, we need to find the maximum sum of absolute difference of any permutation of the given array.
Examples:

Input : { 1, 2, 4, 8 }
Output : 18
Explanation : For the given array there are
several sequence possible
like : {2, 1, 4, 8}
{4, 2, 1, 8} and some more.
Now, the absolute difference of an array sequence will be
like for this array sequence {1, 2, 4, 8}, the absolute
difference sum is
= |1-2| + |2-4| + |4-8| + |8-1|
= 14
For the given array, we get the maximum value for
the sequence {1, 8, 2, 4}
= |1-8| + |8-2| + |2-4| + |4-1|
= 18

To solve this problem, we have to think greedily that how can we maximize the difference value of the elements so that we can have a maximum sum. This is possible only if we calculate the difference between some very high values and some very low values like (highest – smallest). This is the idea which we have to use to solve this problem. Let us see the above example, we will have maximum difference possible for sequence {1, 8, 2, 4} because in this sequence we will get some high difference values, ( |1-8| = 7, |8-2| = 6 .. ). Here, by placing 8(highest element) in place of 1 and 2 we get two high difference values. Similarly, for the other values, we will place next highest values in between other, as we have only one left i.e 4 which is placed at last.
Algorithm: To get the maximum sum, we should have a sequence in which small and large elements comes alternate. This is done to get maximum difference.
For the implementation of the above algorithm ->
1. We will sort the array.
2. Calculate the final sequence by taking one smallest element and largest element from the sorted array and make one vector array of this final sequence.
3. Finally, calculate the sum of absolute difference between the elements of the array.
Below is the implementation of above idea :

C++

 // CPP implementation of// above algorithm#include using namespace std; int MaxSumDifference(int a[], int n){    // final sequence stored in the vector    vector finalSequence;     // sort the original array    // so that we can retrieve    // the large elements from    // the end of array elements    sort(a, a + n);     // In this loop first we will insert    // one smallest element not entered    // till that time in final sequence    // and then enter a highest element    // (not entered till that time) in    // final sequence so that we    // have large difference value. This    // process is repeated till all array    // has completely entered in sequence.    // Here, we have loop till n/2 because    // we are inserting two elements at a    // time in loop.    for (int i = 0; i < n / 2; ++i) {        finalSequence.push_back(a[i]);        finalSequence.push_back(a[n - i - 1]);    }     // If there are odd elements, push the    // middle element at the end.    if (n % 2 != 0)        finalSequence.push_back(a[n/2]);     // variable to store the    // maximum sum of absolute    // difference    int MaximumSum = 0;     // In this loop absolute difference    // of elements for the final sequence    // is calculated.    for (int i = 0; i < n - 1; ++i) {        MaximumSum = MaximumSum + abs(finalSequence[i] -                                  finalSequence[i + 1]);    }     // absolute difference of last element    // and 1st element    MaximumSum = MaximumSum + abs(finalSequence[n - 1] -                                      finalSequence);     // return the value    return MaximumSum;} // Driver functionint main(){    int a[] = { 1, 2, 4, 8 };    int n = sizeof(a) / sizeof(a);     cout << MaxSumDifference(a, n) << endl;}

Java

 // Java implementation of// above algorithmimport java.io.*;import java.util.*; public class GFG {         static int MaxSumDifference(Integer []a, int n)    {                 // final sequence stored in the vector        List finalSequence =                        new ArrayList();             // sort the original array        // so that we can retrieve        // the large elements from        // the end of array elements        Arrays.sort(a);             // In this loop first we will insert        // one smallest element not entered        // till that time in final sequence        // and then enter a highest element        // (not entered till that time) in        // final sequence so that we        // have large difference value. This        // process is repeated till all array        // has completely entered in sequence.        // Here, we have loop till n/2 because        // we are inserting two elements at a        // time in loop.        for (int i = 0; i < n / 2; ++i) {            finalSequence.add(a[i]);            finalSequence.add(a[n - i - 1]);        }         // If there are odd elements, push the        // middle element at the end.        if (n % 2 != 0)            finalSequence.add(a[n/2]);         // variable to store the        // maximum sum of absolute        // difference        int MaximumSum = 0;             // In this loop absolute difference        // of elements for the final sequence        // is calculated.        for (int i = 0; i < n - 1; ++i) {            MaximumSum = MaximumSum +                  Math.abs(finalSequence.get(i)                   - finalSequence.get(i + 1));        }             // absolute difference of last element        // and 1st element        MaximumSum = MaximumSum +              Math.abs(finalSequence.get(n - 1)                       - finalSequence.get(0));             // return the value        return MaximumSum;    }         // Driver Code    public static void main(String args[])    {        Integer []a = { 1, 2, 4, 8 };        int n = a.length;             System.out.print(MaxSumDifference(a, n));    }} // This code is contributed by// Manish Shaw (manishshaw1)

Python3

 import numpy as npclass GFG:         def MaxSumDifference(a,n):        # sort the original array        # so that we can retrieve        # the large elements from        # the end of array elements        np.sort(a);             # In this loop first we will        # insert one smallest element        # not entered till that time        # in final sequence and then        # enter a highest element(not        # entered till that time) in        # final sequence so that we        # have large difference value.        # This process is repeated till        # all array has completely        # entered in sequence. Here,        # we have loop till n/2 because        # we are inserting two elements        # at a time in loop.        j = 0        finalSequence = [0 for x in range(n)]        for i in range(0, int(n / 2)):            finalSequence[j] = a[i]            finalSequence[j + 1] = a[n - i - 1]            j = j + 2         # If there are odd elements, push the        # middle element at the end.        if (n % 2 != 0):           finalSequence[n-1] = a[n//2 + 1]         # variable to store the        # maximum sum of absolute        # difference        MaximumSum = 0             # In this loop absolute        # difference of elements        # for the final sequence        # is calculated.        for i in range(0, n - 1):            MaximumSum = (MaximumSum +                          abs(finalSequence[i] -                              finalSequence[i + 1]))             # absolute difference of last        # element and 1st element        MaximumSum = (MaximumSum +                      abs(finalSequence[n - 1] -                         finalSequence));             # return the value        print (MaximumSum)     # Driver Codea = [ 1, 2, 4, 8 ]n = len(a)GFG.MaxSumDifference(a, n);     # This code is contributed# by Prateek Bajaj

C#

 // C# implementation of// above algorithmusing System;using System.Collections.Generic;class GFG {         static int MaxSumDifference(int []a, int n)    {                 // final sequence stored in the vector        List finalSequence = new List();             // sort the original array        // so that we can retrieve        // the large elements from        // the end of array elements        Array.Sort(a);             // In this loop first we will insert        // one smallest element not entered        // till that time in final sequence        // and then enter a highest element        // (not entered till that time) in        // final sequence so that we        // have large difference value. This        // process is repeated till all array        // has completely entered in sequence.        // Here, we have loop till n/2 because        // we are inserting two elements at a        // time in loop.        for (int i = 0; i < n / 2; ++i) {            finalSequence.Add(a[i]);            finalSequence.Add(a[n - i - 1]);        }             // If there are odd elements, push the        // middle element at the end.        if (n % 2 != 0)            finalSequence.Add(a[n/2]);         // variable to store the        // maximum sum of absolute        // difference        int MaximumSum = 0;             // In this loop absolute difference        // of elements for the final sequence        // is calculated.        for (int i = 0; i < n - 1; ++i) {            MaximumSum = MaximumSum + Math.Abs(finalSequence[i] -                                    finalSequence[i + 1]);        }             // absolute difference of last element        // and 1st element        MaximumSum = MaximumSum + Math.Abs(finalSequence[n - 1] -                                        finalSequence);             // return the value        return MaximumSum;    }         // Driver Code    public static void Main()    {        int []a = { 1, 2, 4, 8 };        int n = a.Length;             Console.WriteLine(MaxSumDifference(a, n));    }} // This code is contributed by// Manish Shaw (manishshaw1)



Javascript



Output :

18

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