Sum of absolute differences of all pairs in a given array - GeeksforGeeks

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

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).

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].

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)

// 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;

}

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