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] β x| + |arr[1] β x| + |arr[2] β x| + β¦ + |arr[n β 1] β 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 β 3| + |3 β 3| + |9 β 3| + |3 β 3| + |6 β 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).
Algorithm:
Below is the implementation of the approach:
C++
// C++ code for the approach #include<bits/stdc++.h> using namespace std; // Function to find the minimum sum of absolute differences int findMinSum( int arr[], int n) { // Initialize the minimum sum and minimum element int minSum = INT_MAX, minElement = -1; // Traverse through all elements of the array for ( int i = 0; i < n; i++) { int sum = 0; // Calculate the sum of absolute differences // of each element with the current element for ( int j = 0; j < n; j++) { sum += abs (arr[i] - arr[j]); } // Update the minimum sum and minimum element if (sum < minSum) { minSum = sum; minElement = arr[i]; } } // Return the minimum sum return minSum; } // Driver code int main() { int arr[] = { 1, 3, 9, 3, 6 }; int n = sizeof (arr)/ sizeof (arr[0]); // Find the minimum sum of absolute differences int minSum = findMinSum(arr, n); // Print the minimum sum cout << minSum << endl; return 0; } |
Java
// Java code for the approach import java.util.*; public class GFG { // Function to find the minimum sum of absolute differences public static int findMinSum( int [] arr, int n) { // Initialize the minimum sum and minimum element int minSum = Integer.MAX_VALUE, minElement = - 1 ; // Traverse through all elements of the array for ( int i = 0 ; i < n; i++) { int sum = 0 ; // Calculate the sum of absolute differences // of each element with the current element for ( int j = 0 ; j < n; j++) { sum += Math.abs(arr[i] - arr[j]); } // Update the minimum sum and minimum element if (sum < minSum) { minSum = sum; minElement = arr[i]; } } // Return the minimum sum return minSum; } // Driver code public static void main(String[] args) { int [] arr = { 1 , 3 , 9 , 3 , 6 }; int n = arr.length; // Find the minimum sum of absolute differences int minSum = findMinSum(arr, n); // Print the minimum sum System.out.println(minSum); } } |
Python3
# Python3 code for the approach import sys # Function to find the minimum sum of absolute differences def findMinSum(arr, n): # Initialize the minimum sum and minimum element minSum = sys.maxsize minElement = - 1 # Traverse through all elements of the array for i in range (n): sum = 0 # Calculate the sum of absolute differences # of each element with the current element for j in range (n): sum + = abs (arr[i] - arr[j]) # Update the minimum sum and minimum element if ( sum < minSum): minSum = sum minElement = arr[i] # Return the minimum sum return minSum # Driver code if __name__ = = '__main__' : arr = [ 1 , 3 , 9 , 3 , 6 ] n = len (arr) # Find the minimum sum of absolute differences minSum = findMinSum(arr, n) # Print the minimum sum print (minSum) |
C#
using System; class Program { // Function to find the minimum sum of absolute differences static int FindMinSum( int [] arr, int n) { // Initialize the minimum sum int minSum = int .MaxValue; // Traverse through all elements of the array for ( int i = 0; i < n; i++) { int sum = 0; // Calculate the sum of absolute differences // of each element with the current element for ( int j = 0; j < n; j++) { sum += Math.Abs(arr[i] - arr[j]); } // Update the minimum sum if (sum < minSum) { minSum = sum; } } // Return the minimum sum return minSum; } // Driver code static void Main() { int [] arr = { 1, 3, 9, 3, 6 }; int n = arr.Length; // Find the minimum sum of absolute differences int minSum = FindMinSum(arr, n); // Print the minimum sum Console.WriteLine(minSum); } } |
Javascript
// Function to find the minimum sum of absolute differences function findMinSum(arr) { // Initialize the minimum sum and minimum element let minSum = Infinity; let minElement = -1; // Traverse through all elements of the array for (let i = 0; i < arr.length; i++) { let sum = 0; // Calculate the sum of absolute differences // of each element with the current element for (let j = 0; j < arr.length; j++) { sum += Math.abs(arr[i] - arr[j]); } // Update the minimum sum and minimum element if (sum < minSum) { minSum = sum; minElement = arr[i]; } } // Return the minimum sum return minSum; } // Driver code const arr = [1, 3, 9, 3, 6]; const minSum = findMinSum(arr); // Print the minimum sum console.log(minSum); // This code is contributed by shivamgupta0987654321 |
11
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 |
Javascript
<script> //Javascript implementation of the approach // Function to return the minimized sum function minSum(arr, n) { // Sort the array arr.sort(); // Median of the array let x = arr[(Math.floor(n / 2))]; let sum = 0; // Calculate the minimized sum for (let i = 0; i < n; i++) sum += Math.abs(arr[i] - x); // Return the required sum return sum; } // Driver code let arr = [ 1, 3, 9, 3, 6 ]; let n = arr.length; document.write(minSum(arr, n)); // This code is contributed by Mayank Tyagi </script> |
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.