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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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.



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