Suma mínima de diferencias con un elemento en una array

Dada una array, necesitamos encontrar la suma de los elementos de una array después de cambiar el elemento, ya que arr[i] se convertirá en abs(arr[i]-x) donde x es un elemento de array.

Ejemplos: 

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

La idea se basa en el hecho de que el elemento medio causaría una suma mínima de diferencias. Cuando hay un número par de elementos, podemos tomar cualquiera de los dos elementos del medio. Podemos verificar este hecho tomando algunos ejemplos. 

A continuación se muestra la implementación de la idea anterior:  

// 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

// 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
}

# 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# 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
}

<?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
?>

<script>
 
// Javascript program to find minimum sum of absolute
// differences with an array element.
 
// Function to find min sum after operation
function absSumDidd(a)
{
     
    // Sort the array
    a.sort((a, b) => a - b);
     
    // Pick middle value
    var midValue = a[a.length / 2];
 
    // Sum of absolute differences.
    var sum = 0;
    for(var i = 0; i < a.length; i++)
    {
        sum = sum + Math.abs(a[i] - midValue);
    }
    return sum;      
}
 
// Driver Code
var arr = [ 5, 11, 14, 10, 17, 15 ];
document.write(absSumDidd(arr)); 
 
// This code is contributed by shikhasingrajput
 
</script>

20

Complejidad de tiempo: O(n Log n) 
Podemos optimizar aún más la solución anterior a O(n) usando un algoritmo de tiempo lineal para encontrar la mediana.