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).
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 absolutre 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; } |
Java
// Java program to find sum of absolutre // 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 Anant Agarwal. |
Python3
# Python3 program to find sum of # absolutre 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 absolutre // 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 absolutre 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); ?> |
Output:
74
Time Complexity : O(n)
Auxiliary space: O(1)e
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)
This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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