# 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 – 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: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

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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