Array element with minimum sum of absolute differences

Last Updated : 23 Nov, 2023

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).

Algorithm:

Below is the implementation of the approach:

C++

// C++ code for the approach

#include<bits/stdc++.h>
using namespace std;

// Function to find the minimum sum of absolute differences
int findMinSum(int arr[], int n) {

    // Initialize the minimum sum and minimum element
    int minSum = INT_MAX, minElement = -1;

    // Traverse through all elements of the array
    for (int i = 0; i < n; i++) {
        int sum = 0;

        // Calculate the sum of absolute differences
        // of each element with the current element
        for (int j = 0; j < n; j++) {
            sum += abs(arr[i] - arr[j]);
        }

        // Update the minimum sum and minimum element
        if (sum < minSum) {
            minSum = sum;
            minElement = arr[i];
        }
    }

    // Return the minimum sum
    return minSum;
}

// Driver code
int main() {
    int arr[] = { 1, 3, 9, 3, 6 };
    int n = sizeof(arr)/sizeof(arr[0]);

    // Find the minimum sum of absolute differences
    int minSum = findMinSum(arr, n);

    // Print the minimum sum
    cout << minSum << endl;

    return 0;
}

Java

// Java code for the approach

import java.util.*;

public class GFG {
    // Function to find the minimum sum of absolute differences
    public static int findMinSum(int[] arr, int n) {
        // Initialize the minimum sum and minimum element
        int minSum = Integer.MAX_VALUE, minElement = -1;
        
        // Traverse through all elements of the array
        for (int i = 0; i < n; i++) {
            int sum = 0;
            
            // Calculate the sum of absolute differences
            // of each element with the current element
            for (int j = 0; j < n; j++) {
                sum += Math.abs(arr[i] - arr[j]);
            }
            
            // Update the minimum sum and minimum element
            if (sum < minSum) {
                minSum = sum;
                minElement = arr[i];
            }
        }
        
        // Return the minimum sum
        return minSum;
    }
    
    // Driver code
    public static void main(String[] args) {
        int[] arr = {1, 3, 9, 3, 6};
        int n = arr.length;
        
        // Find the minimum sum of absolute differences
        int minSum = findMinSum(arr, n);
        
        // Print the minimum sum
        System.out.println(minSum);
    }
}

Python3

# Python3 code for the approach

import sys

# Function to find the minimum sum of absolute differences


def findMinSum(arr, n):

    # Initialize the minimum sum and minimum element
    minSum = sys.maxsize
    minElement = -1

    # Traverse through all elements of the array
    for i in range(n):
        sum = 0

        # Calculate the sum of absolute differences
        # of each element with the current element
        for j in range(n):
            sum += abs(arr[i] - arr[j])

        # Update the minimum sum and minimum element
        if (sum < minSum):
            minSum = sum
            minElement = arr[i]

    # Return the minimum sum
    return minSum


# Driver code
if __name__ == '__main__':
    arr = [1, 3, 9, 3, 6]
    n = len(arr)

    # Find the minimum sum of absolute differences
    minSum = findMinSum(arr, n)

    # Print the minimum sum
    print(minSum)

using System;

class Program
{
    // Function to find the minimum sum of absolute differences
    static int FindMinSum(int[] arr, int n)
    {
        // Initialize the minimum sum
        int minSum = int.MaxValue;

        // Traverse through all elements of the array
        for (int i = 0; i < n; i++)
        {
            int sum = 0;

            // Calculate the sum of absolute differences
            // of each element with the current element
            for (int j = 0; j < n; j++)
            {
                sum += Math.Abs(arr[i] - arr[j]);
            }

            // Update the minimum sum
            if (sum < minSum)
            {
                minSum = sum;
            }
        }

        // Return the minimum sum
        return minSum;
    }

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

        // Find the minimum sum of absolute differences
        int minSum = FindMinSum(arr, n);

        // Print the minimum sum
        Console.WriteLine(minSum);
    }
}

JavaScript

// Function to find the minimum sum of absolute differences
function findMinSum(arr) {
    // Initialize the minimum sum and minimum element
    let minSum = Infinity;
    let minElement = -1;

    // Traverse through all elements of the array
    for (let i = 0; i < arr.length; i++) {
        let sum = 0;

        // Calculate the sum of absolute differences
        // of each element with the current element
        for (let j = 0; j < arr.length; j++) {
            sum += Math.abs(arr[i] - arr[j]);
        }

        // Update the minimum sum and minimum element
        if (sum < minSum) {
            minSum = sum;
            minElement = arr[i];
        }
    }

    // Return the minimum sum
    return minSum;
}

// Driver code
const arr = [1, 3, 9, 3, 6];
const minSum = findMinSum(arr);

// Print the minimum sum
console.log(minSum);

// This code is contributed by shivamgupta0987654321

Output

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# 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

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.

Print distinct absolute differences of all possible pairs from a given array

vashu_1997

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Array element with minimum sum of absolute differences

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