Array element with minimum sum of absolute differences

Given an array arr[] of N integers, the task is to find an element x from the array such that |arr[0] – x| + |arr[1] – x| + |arr[2] – x| + … + |arr[n – 1] – x| is minimized, then print the minimized sum.

Examples:

Input: arr[] = {1, 3, 9, 3, 6}
Output: 11
The optimal solution is to choose x = 3, which produces the sum
|1 – 3| + |3 – 3| + |9 – 3| + |3 – 3| + |6 – 3| = 2 + 0 + 6 + 0 + 3 = 11

Input: arr[] = {1, 2, 3, 4}
Output: 4



A simple solution is to iterate through every element and check if it gives optimal solution or not. Time Complexity of this solution is O(n*n)

An Efficient Approach: is to always pick x as the median of the array. If n is even and there are two medians then both the medians are optimal choices. The time complexity for the approach is O(n * log(n)) because the array will have to be sorted in order to find the median. Calculate and print the minimized sum when x is found (median of the array).

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimized sum
int minSum(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
  
    // Median of the array
    int x = arr[n / 2];
  
    int sum = 0;
  
    // Calculate the minimized sum
    for (int i = 0; i < n; i++)
        sum += abs(arr[i] - x);
  
    // Return the required sum
    return sum;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 3, 9, 3, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minSum(arr, n);
  
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to return the minimized sum 
def minSum(arr, n) :
      
    # Sort the array 
    arr.sort(); 
  
    # Median of the array 
    x = arr[n // 2]; 
  
    sum = 0
  
    # Calculate the minimized sum 
    for i in range(n) :
        sum += abs(arr[i] - x); 
  
    # Return the required sum 
    return sum
  
# Driver code 
if __name__ == "__main__"
      
    arr = [ 1, 3, 9, 3, 6 ]; 
    n = len(arr)
    print(minSum(arr, n));
  
# This code is contributed by Ryuga

chevron_right


Output:

11

The time complexity of the above solution is O(n Log n). We can further optimize it to work in O(n) using linear time algortihm to find k-th largest element.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Ryuga