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++

// 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;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to return the minimized sum
static int minSum(int arr[], int n)
{
    // Sort the array
    Arrays.sort(arr);

    // Median of the array
    int x = arr[(int)n / 2];

    int sum = 0;

    // Calculate the minimized sum
    for (int i = 0; i < n; i++)
        sum += Math.abs(arr[i] - x);

    // Return the required sum
    return sum;
}

// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 3, 9, 3, 6 };
    int n = arr.length;
    System.out.println(minSum(arr, n));
}
}

// This code is contribute by
// Surendra_Gangwar

Python3

# 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

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the minimized sum
static int minSum(int []arr, int n)
{
    // Sort the array
    Array.Sort(arr);

    // Median of the array
    int x = arr[(int)(n / 2)];

    int sum = 0;

    // Calculate the minimized sum
    for (int i = 0; i < n; i++)
        sum += Math.Abs(arr[i] - x);

    // Return the required sum
    return sum;
}

// Driver code
static void Main()
{
    int []arr = { 1, 3, 9, 3, 6 };
    int n = arr.Length;
    Console.WriteLine(minSum(arr, n));
}
}

// This code is contributed by mits

Javascript

<script>

//Javascript implementation of the approach


// Function to return the minimized sum
function minSum(arr, n)
{
    // Sort the array
    arr.sort();

    // Median of the array
    let x = arr[Math.floor(n / 2)];

    let sum = 0;

    // Calculate the minimized sum
    for (let i = 0; i < n; i++)
        sum += Math.abs(arr[i] - x);

    // Return the required sum
    return sum;
}

// Driver code
    let arr = [ 1, 3, 9, 3, 6 ];
    let n = arr.length;
    document.write(minSum(arr, n));

// This code is contributed by Mayank Tyagi

</script>

Output

11

Time Complexity: O(n log n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required.

We can further optimize it to work in O(n) using linear time algorithm to find k-th largest element.