# 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[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: 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[0]);     cout << minSum(arr, n);        return 0; }

## 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

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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