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 = 11Input: arr[] = {1, 2, 3, 4}
Output: 4
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 <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 |
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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