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Array element with minimum sum of absolute differences

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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:

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)


C#




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

11


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.



Last Updated : 23 Nov, 2023
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