Minimum sum of differences with an element in an array

Given an array, we need to find the sum of elements of an array after changing the element as arr[i] will become abs(arr[i]-x) where x is an array element.

Examples:

Input  : {2, 5, 1, 7, 4}
Output : 9
We get minimum sum when we choose
x = 4. The minimum sum is 
abs(2-4) + abs(5-4) + abs(1-4) + (7-4)
abs(4-4) = 9

Input  : {5, 11, 14, 10, 17, 15}
Output : 20
We can either choose x = 14 or x = 11


The idea is based on fact that middle element would cause minimum sum of differences. When there are even number of elements, we can take any of the middle two elements. We can verify this fact by taking few examples.

Below is the implementation of above idea:

C++

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// C++ program to find minimum sum of absolute
// differences with an array element.
#include<bits/stdc++.h>
using namespace std;
    // function to find min sum after operation
    int absSumDidd(int a[],int n)
    {
        // Sort the array
        sort(a,a+n);
          
        // Pick middle value
        int midValue = a[(int)(n / 2)];
  
        // Sum of absolute differences.
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum = sum + abs(a[i] - midValue);
        }
          
        return sum;     
    }
  
    // Driver Code
    int main()
    {
        int arr[] = { 5, 11, 14, 10, 17, 15 };
        int n=sizeof(arr)/sizeof(arr[0]);
        cout<< absSumDidd(arr,n);     
    }
    // Contributed by mits

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Java

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// Java program to find minimum sum of absolute
// differences with an array element.
import java.lang.*;
import java.util.*;
  
public class GFG {
  
    // function to find min sum after operation
    static int absSumDidd(int a[])
    {
        // Sort the array
        Arrays.sort(a);
          
        // Pick middle value
        int midValue = a[a.length / 2];
  
        // Sum of absolute differences.
        int sum = 0;
        for (int i = 0; i < a.length; i++) {
            sum = sum + Math.abs(a[i] - midValue);
        }
         
        return sum;       
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 5, 11, 14, 10, 17, 15 };
        System.out.print(absSumDidd(arr));        
    }
    // Contributed by Saurav Jain
}

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Python3

# Python3 program to find minimum sum of
# absolute differences with an array element.

# function to find min sum after operation
def absSumDidd(a, n):

# Sort the array
a.sort()

# Pick middle value
midValue = a[(int)(n // 2)]

# Sum of absolute differences.
sum = 0
for i in range(n):
sum = sum + abs(a[i] – midValue)
return sum

# Driver Code
arr = [5, 11, 14, 10, 17, 15]
n = len(arr)
print(absSumDidd(arr, n))

# This code is contributed
# by sahishelangia

C#

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// C# program to find minimum sum of absolute 
// differences with an array element. 
using System;
  
  
 class GFG { 
    
    // function to find min sum after operation 
    static int absSumDidd(int []a) 
    
        // Sort the array 
        Array.Sort(a); 
            
        // Pick middle value 
        int midValue = a[a.Length / 2]; 
    
        // Sum of absolute differences. 
        int sum = 0; 
        for (int i = 0; i < a.Length; i++) { 
            sum = sum + Math.Abs(a[i] - midValue); 
        
           
        return sum;        
    
    
    // Driver Code 
    public static void Main() 
    
        int []arr = { 5, 11, 14, 10, 17, 15 }; 
        Console.Write(absSumDidd(arr));         
    
    // Contributed by Subhadeep

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PHP

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<?php
// PHP program to find minimum 
// sum of absolute differences
// with an array element.
  
// function to find min sum
// after operation
function absSumDidd($a, $n)
{
    // Sort the array
    sort($a);
      
    // Pick middle value
    $midValue = $a[($n / 2)];
  
    // Sum of absolute differences.
    $sum = 0;
    for ( $i = 0; $i < $n; $i++)
    {
        $sum = $sum + abs($a[$i] - 
               $midValue);
    }
      
    return $sum;     
}
  
// Driver Code
$arr = array(5, 11, 14, 10, 17, 15 );
$n = count($arr);
echo absSumDidd($arr, $n); 
  
// This code is contributed
// by anuj_67
?>

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Output:

20

Time Complexity: O(n Log n)

We can further optimize above solution to O(n) by using linear time algorithm for median finding.



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