# Array element with minimum sum of absolute differences

• Difficulty Level : Basic
• Last Updated : 29 Nov, 2022

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).

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 ``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

 ``

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.

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