# Array element with minimum sum of absolute differences

• Difficulty Level : Basic
• Last Updated : 23 Aug, 2021

Given an array arr[] of N integers, the task is to find an element x from the array such that |arr – x| + |arr – x| + |arr – 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);` `    ``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`

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 algorithm to find k-th largest element.

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