Sum of absolute differences of all pairs in a given array

Given a sorted array of distinct elements, the task is to find the summation of absolute differences of all pairs in the given array.

Examples: 

Input : arr[] = {1, 2, 3, 4}
Output: 10
Sum of |2-1| + |3-1| + |4-1| +
       |3-2| + |4-2| + |4-3| = 10

Input : arr[] = {1, 8, 9, 15, 16}
Output: 74

Input : arr[] = {1, 2, 3, 4, 5, 7, 9, 11, 14}
Output: 188

A simple solution for this problem is to one by one look for each pair take their difference and sum up them together. The time complexity for this approach is O(n²).

C++

// C++ program to find sum of absolute differences
// in all pairs in a sorted array of distinct numbers
#include<bits/stdc++.h>
using namespace std;

// Function to calculate sum of absolute difference
// of all pairs in array
// arr[] --> array of elements
int sumPairs(int arr[],int n)
{
    // final result
    int sum = 0;
    for(int i=0;i<n;i++)
    {
        for(int j=i+1;j<n;j++)
        {
            sum+=abs(arr[i]-arr[j]);
        }
    }
    return sum;
}

// Driver program to run the case
int main()
{
    int arr[] = {1, 8, 9, 15, 16};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << sumPairs(arr, n);
    return 0;
}

// This code is contributed by Pushpesh Raj.

Java

// Java program to find sum of absolute differences
// in all pairs in a sorted array of distinct numbers
class GFG {
    // Function to calculate sum of absolute difference
    // of all pairs in array
    // arr[] --> array of elements
    static int sumPairs(int arr[], int n)
    {
        // final result
        int sum = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                sum += Math.abs(arr[i] - arr[j]);
            }
        }
        return sum;
    }

    // Driver program to run the case
    public static void main(String[] args)
    {
        int arr[] = { 1, 8, 9, 15, 16 };
        int n = arr.length;
        System.out.println(sumPairs(arr, n));
    }
}

// This code is contributed by karandeep1234.

Python3

#  Python3 program to find sum of absolute differences
# in all pairs in a sorted array of distinct numbers

# Function to calculate sum of absolute difference
# of all pairs in array
# arr[] --> array of elements
def sumPairs(arr, n):

    # final result
    sum = 0;
    for i in range(n):
        for j in range(i + 1, n):

            sum += abs(arr[i]-arr[j]);

    return sum;

# Driver program to run the case
arr = [1, 8, 9, 15, 16];
n = len(arr);
print(sumPairs(arr, n));

# This code is contributed by phasing17

C#

// C# program to find sum of absolute differences
// in all pairs in a sorted array of distinct numbers
using System;

class GFG {
    // Function to calculate sum of absolute difference
    // of all pairs in array
    // arr[] --> array of elements
    static int sumPairs(int[] arr, int n)
    {
        // final result
        int sum = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                sum += Math.Abs(arr[i] - arr[j]);
            }
        }
        return sum;
    }

    // Driver program to run the case
    public static void Main(string[] args)
    {
        int[] arr = { 1, 8, 9, 15, 16 };
        int n = arr.Length;
        Console.WriteLine(sumPairs(arr, n));
    }
}

// This code is contributed by phasing17.

Javascript

//  JS program to find sum of absolute differences
// in all pairs in a sorted array of distinct numbers

// Function to calculate sum of absolute difference
// of all pairs in array
// arr[] --> array of elements
function sumPairs(arr, n)
{
    // final result
    let sum = 0;
    for(var i=0;i<n;i++)
    {
        for(var j=i+1;j<n;j++)
        {
            sum+= Math.abs(arr[i]-arr[j]);
        }
    }
    return sum;
}

// Driver program to run the case
let  arr = [1, 8, 9, 15, 16];
let n = arr.length;
 console.log(sumPairs(arr, n));

// This code is contributed by phasing17

Output

74

The space complexity of this program is O(1), because it does not use any extra space that depends on the input size. The only space used is for the input array, which has a fixed size of n, so the space complexity is constant.

An efficient solution for this problem needs a simple observation. Since array is sorted and elements are distinct when we take sum of absolute difference of pairs each element in the i’th position is added ‘i’ times and subtracted ‘n-1-i’ times. 

For example in {1,2,3,4} element at index 2 is arr[2] = 3 so all pairs having 3 as one element will be (1,3), (2,3) and (3,4), now when we take summation of absolute difference of pairs, then for all pairs in which 3 is present as one element summation will be = (3-1)+(3-2)+(4-3). We can see that 3 is added i = 2 times and subtracted n-1-i = (4-1-2) = 1 times. 

The generalized expression for each element will be sum = sum + (i*a[i]) – (n-1-i)*a[i]. 

C++

// C++ program to find sum of absolute differences
// in all pairs in a sorted array of distinct numbers
#include<bits/stdc++.h>
using namespace std;

// Function to calculate sum of absolute difference
// of all pairs in array
// arr[]  --> array of elements
int sumPairs(int arr[],int n)
{
    // final result
    int sum = 0;
    for (int i=n-1; i>=0; i--)
        sum += i*arr[i] - (n-1-i)*arr[i];
    return sum;
}

// Driver program to run the case
int main()
{
    int arr[] = {1, 8, 9, 15, 16};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << sumPairs(arr, n);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

C

// C program to find sum of absolute differences
// in all pairs in a sorted array of distinct numbers
#include <stdio.h>

// Function to calculate sum of absolute difference
// of all pairs in array
// arr[]  --> array of elements
int sumPairs(int arr[], int n)
{
    // final result
    int sum = 0;
    for (int i = n - 1; i >= 0; i--)
        sum += i * arr[i] - (n - 1 - i) * arr[i];
    return sum;
}

// Driver program to run the case
int main()
{
    int arr[] = { 1, 8, 9, 15, 16 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d", sumPairs(arr, n));
    return 0;
}

// This code is contributed by Sania Kumari Gupta

Java

// Java program to find sum of absolute differences in all
// pairs in a sorted array of distinct numbers
class GFG {

    // Function to calculate sum of absolute
    // difference of all pairs in array
    // arr[] --> array of elements
    static int sumPairs(int arr[], int n)
    {
        // final result
        int sum = 0;
        for (int i = n - 1; i >= 0; i--)
            sum += i * arr[i] - (n - 1 - i) * arr[i];
        return sum;
    }

    // Driver program
    public static void main(String arg[])
    {
        int arr[] = { 1, 8, 9, 15, 16 };
        int n = arr.length;
        System.out.print(sumPairs(arr, n));
    }
}

// This code is contributed by Sania Kumari Gupta

Python3

# Python3 program to find sum of
# absolute differences in all pairs
# in a sorted array of distinct numbers

# Function to calculate sum of absolute
# difference of all pairs in array
# arr[] --> array of elements
def sumPairs(arr, n):

    # final result
    sum = 0
    for i in range(n - 1, -1, -1):
        sum += i*arr[i] - (n-1-i) * arr[i]
    return sum

# Driver program
arr = [1, 8, 9, 15, 16]
n = len(arr)
print(sumPairs(arr, n))

# This code is contributed by Anant Agarwal.

C#

// C# program to find sum of absolute
// differences in all pairs in a sorted
// array of distinct numbers
using System;
class GFG {

    // Function to calculate sum of absolute
    // difference of all pairs in array
    // arr[] --> array of elements
    static int sumPairs(int []arr, int n)
    {

        // final result
        int sum = 0;
        for (int i = n - 1; i >= 0; i--)
            sum += i * arr[i] - (n - 1 - i)
                                  * arr[i];

        return sum;
    }

    // Driver program
    public static void Main()
    {
        int []arr = { 1, 8, 9, 15, 16 };
        int n = arr.Length;

        Console.Write(sumPairs(arr, n));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php

// PHP program to find sum of absolute differences
// in all pairs in a sorted array of distinct numbers

// Function to calculate sum of absolute difference
// of all pairs in array
// arr[] --> array of elements
function sumPairs($arr,$n)
{
    // final result
    $sum = 0;
    for ($i=$n-1; $i>=0; $i--)
        $sum =  $sum + $i*$arr[$i] - ($n-1-$i)*$arr[$i];
    return $sum;
}

// Driver program to run the case
$arr = array(1, 8, 9, 15, 16);
    $n = sizeof($arr)/sizeof($arr[0]);
    echo sumPairs($arr, $n);
?>

Javascript

<script>

// JavaScript program to find
// sum of absolute differences
// in all pairs in a sorted array
// of distinct numbers

// Function to calculate sum of absolute difference
// of all pairs in array
// arr[]  --> array of elements
function sumPairs( arr, n)
{
    // final result
    let sum = 0;
    for (let i=n-1; i>=0; i--)
        sum += i*arr[i] - (n-1-i)*arr[i];
    return sum;
}

// Driver program to run the case
let arr = [ 1, 8, 9, 15, 16 ];
let n = arr.length;
document.write(sumPairs(arr, n));

</script>

Output

74

Time Complexity: O(n) 
Auxiliary space: O(1)

What if array is not sorted?
The efficient solution is also better for the cases where array is not sorted. We can sort the array first in O(n Log n) time and then find the required value in O(n). So overall time complexity is O(n Log n) which is still better than O(n²)

Below is the code for above approach.

C++

// C++ program for above approach

#include<bits/stdc++.h>
using namespace std;

// Function to calculate sum of absolute
// difference of all pairs in array
int sumPairs(int arr[], int n)
{
    // sorting the array
    sort(arr, arr+n);

    // Initialising the variable
    // to store the sum
    int sum = 0;
    // Iterating through each element
    // and adding it i times and
    // subtracting it (n-1-i) times
    for(int i = 0; i < n; i++){
        sum += i*arr[i] - (n-1-i)*arr[i];
    }

    return sum;
}

// Driver program
int main()
{
    int arr[] = {16, 8, 9, 1, 15};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout<<sumPairs(arr, n);

    return 0;
}

Java

// Java program for above approach
import java.util.Arrays;

public class Main {
    // Function to calculate sum of absolute
    // difference of all pairs in array
    static int sumPairs(int[] arr, int n) {
        // sorting the array
        Arrays.sort(arr);

        // Initialising the variable
        // to store the sum
        int sum = 0;
        // Iterating through each element
        // and adding it i times and
        // subtracting it (n-1-i) times
        for(int i = 0; i < n; i++) {
            sum += i*arr[i] - (n-1-i)*arr[i];
        }

        return sum;
    }

    // Driver program
    public static void main(String[] args) {
        int[] arr = {16, 8, 9, 1, 15};
        int n = arr.length;
        System.out.println(sumPairs(arr, n));
    }
}
// This code is contributed by bhardwajji

Python3

# Python program for above approach

# Function to calculate sum of absolute
# difference of all pairs in array
def sumPairs(arr, n):

    # sorting the array
    arr.sort()

    # Initialising the variable
    # to store the sum
    sum = 0
    # Iterating through each element
    # and adding it i times and
    # subtracting it (n-1-i) times
    for i in range(n):
        sum += i*arr[i] - (n-1-i)*arr[i]

    return sum

# Driver program
arr = [16, 8, 9, 1, 15]
n = len(arr)
print(sumPairs(arr, n))

# This code is contributed by Aman Kumar.

C#

using System;

class Program {

  // Function to calculate sum of absolute
  // difference of all pairs in array
  static int SumPairs(int[] arr, int n)
  {
    // sorting the array
    Array.Sort(arr);

    // Initialising the variable
    // to store the sum
    int sum = 0;

    // Iterating through each element
    // and adding it i times and
    // subtracting it (n-1-i) times
    for(int i = 0; i < n; i++){
      sum += i*arr[i] - (n-1-i)*arr[i];
    }

    return sum;
  }

  // Driver program
  static void Main(string[] args) {
    int[] arr = {16, 8, 9, 1, 15};
    int n = arr.Length;
    Console.WriteLine(SumPairs(arr, n));
  }
}

// This code is contributed by divyansh2212

Javascript

// JavaScript program for above approach

// Function to calculate sum of absolute
// difference of all pairs in array
function sumPairs(arr, n) {

    // sorting the array
    arr.sort((a, b) => a - b);

    // Initialising the variable
    // to store the sum
    let sum = 0;

    // Iterating through each element
    // and adding it i times and
    // subtracting it (n-1-i) times
    for (let i = 0; i < n; i++) {
        sum += i*arr[i] - (n-1-i)*arr[i];
    }

return sum;
}

// Driver code
let arr = [16, 8, 9, 1, 15];
let n = arr.length;
console.log(sumPairs(arr, n));


// This code is contributed by phasing17

Output

74

Time Complexity: O(n*log(n))
Auxiliary space: O(1)

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.