Sum of absolute differences of all pairs in a given array
Last Updated :
23 Apr, 2023
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++
#include<bits/stdc++.h>
using namespace std;
int sumPairs(int arr[],int n)
{
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;
}
int main()
{
int arr[] = {1, 8, 9, 15, 16};
int n = sizeof(arr)/sizeof(arr[0]);
cout << sumPairs(arr, n);
return 0;
}
|
Java
class GFG {
static int sumPairs(int arr[], int n)
{
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;
}
public static void main(String[] args)
{
int arr[] = { 1, 8, 9, 15, 16 };
int n = arr.length;
System.out.println(sumPairs(arr, n));
}
}
|
Python3
def sumPairs(arr, n):
sum = 0;
for i in range(n):
for j in range(i + 1, n):
sum += abs(arr[i]-arr[j]);
return sum;
arr = [1, 8, 9, 15, 16];
n = len(arr);
print(sumPairs(arr, n));
|
C#
using System;
class GFG {
static int sumPairs(int[] arr, int n)
{
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;
}
public static void Main(string[] args)
{
int[] arr = { 1, 8, 9, 15, 16 };
int n = arr.Length;
Console.WriteLine(sumPairs(arr, n));
}
}
|
Javascript
function sumPairs(arr, n)
{
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;
}
let arr = [1, 8, 9, 15, 16];
let n = arr.length;
console.log(sumPairs(arr, n));
|
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++
#include<bits/stdc++.h>
using namespace std;
int sumPairs(int arr[],int n)
{
int sum = 0;
for (int i=n-1; i>=0; i--)
sum += i*arr[i] - (n-1-i)*arr[i];
return sum;
}
int main()
{
int arr[] = {1, 8, 9, 15, 16};
int n = sizeof(arr)/sizeof(arr[0]);
cout << sumPairs(arr, n);
return 0;
}
|
C
#include <stdio.h>
int sumPairs(int arr[], int n)
{
int sum = 0;
for (int i = n - 1; i >= 0; i--)
sum += i * arr[i] - (n - 1 - i) * arr[i];
return sum;
}
int main()
{
int arr[] = { 1, 8, 9, 15, 16 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d", sumPairs(arr, n));
return 0;
}
|
Java
class GFG {
static int sumPairs(int arr[], int n)
{
int sum = 0;
for (int i = n - 1; i >= 0; i--)
sum += i * arr[i] - (n - 1 - i) * arr[i];
return sum;
}
public static void main(String arg[])
{
int arr[] = { 1, 8, 9, 15, 16 };
int n = arr.length;
System.out.print(sumPairs(arr, n));
}
}
|
Python3
def sumPairs(arr, n):
sum = 0
for i in range(n - 1, -1, -1):
sum += i*arr[i] - (n-1-i) * arr[i]
return sum
arr = [1, 8, 9, 15, 16]
n = len(arr)
print(sumPairs(arr, n))
|
C#
using System;
class GFG {
static int sumPairs(int []arr, int n)
{
int sum = 0;
for (int i = n - 1; i >= 0; i--)
sum += i * arr[i] - (n - 1 - i)
* arr[i];
return sum;
}
public static void Main()
{
int []arr = { 1, 8, 9, 15, 16 };
int n = arr.Length;
Console.Write(sumPairs(arr, n));
}
}
|
PHP
<?php
function sumPairs($arr,$n)
{
$sum = 0;
for ($i=$n-1; $i>=0; $i--)
$sum = $sum + $i*$arr[$i] - ($n-1-$i)*$arr[$i];
return $sum;
}
$arr = array(1, 8, 9, 15, 16);
$n = sizeof($arr)/sizeof($arr[0]);
echo sumPairs($arr, $n);
?>
|
Javascript
<script>
function sumPairs( arr, n)
{
let sum = 0;
for (let i=n-1; i>=0; i--)
sum += i*arr[i] - (n-1-i)*arr[i];
return sum;
}
let arr = [ 1, 8, 9, 15, 16 ];
let n = arr.length;
document.write(sumPairs(arr, n));
</script>
|
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++
#include<bits/stdc++.h>
using namespace std;
int sumPairs(int arr[], int n)
{
sort(arr, arr+n);
int sum = 0;
for(int i = 0; i < n; i++){
sum += i*arr[i] - (n-1-i)*arr[i];
}
return sum;
}
int main()
{
int arr[] = {16, 8, 9, 1, 15};
int n = sizeof(arr)/sizeof(arr[0]);
cout<<sumPairs(arr, n);
return 0;
}
|
Java
import java.util.Arrays;
public class Main {
static int sumPairs(int[] arr, int n) {
Arrays.sort(arr);
int sum = 0;
for(int i = 0; i < n; i++) {
sum += i*arr[i] - (n-1-i)*arr[i];
}
return sum;
}
public static void main(String[] args) {
int[] arr = {16, 8, 9, 1, 15};
int n = arr.length;
System.out.println(sumPairs(arr, n));
}
}
|
Python3
def sumPairs(arr, n):
arr.sort()
sum = 0
for i in range(n):
sum += i*arr[i] - (n-1-i)*arr[i]
return sum
arr = [16, 8, 9, 1, 15]
n = len(arr)
print(sumPairs(arr, n))
|
C#
using System;
class Program {
static int SumPairs(int[] arr, int n)
{
Array.Sort(arr);
int sum = 0;
for(int i = 0; i < n; i++){
sum += i*arr[i] - (n-1-i)*arr[i];
}
return sum;
}
static void Main(string[] args) {
int[] arr = {16, 8, 9, 1, 15};
int n = arr.Length;
Console.WriteLine(SumPairs(arr, n));
}
}
|
Javascript
function sumPairs(arr, n) {
arr.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < n; i++) {
sum += i*arr[i] - (n-1-i)*arr[i];
}
return sum;
}
let arr = [16, 8, 9, 1, 15];
let n = arr.length;
console.log(sumPairs(arr, n));
|
Time Complexity: O(n*log(n))
Auxiliary space: O(1)
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