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).
Algorithm:
Below is the implementation of the approach:
// 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 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 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)
|
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);
}
} |
// 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++ 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 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 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# 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 |
<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.