Maximizar la diferencia entre dos subconjuntos de un conjunto con negativos

Dada una cantidad de números enteros de tamaño N. La tarea es separar estos números enteros en dos grupos g1 y g2 de modo que (suma de elementos de g1) – (suma de elementos de g2) sea máxima. Su tarea es imprimir el valor del resultado. Podemos mantener un subconjunto como vacío.
Ejemplos: 
 

Entrada: 3, 7, -4, 10, -11, 2 
Salida: 37 
Explicación: 
g1: 3, 7, 10, 2 
g2: -4, -11 
resultado = ( 3 + 7 + 10 + 2 ) – ( ​​- 4 + -11) = 22 – (-15) = 37
Entrada: 2, 2, -2, -2 
Salida: 8 

La idea es agrupar los números enteros según su valor de signo, es decir, agrupamos los números enteros positivos como g1 y los números enteros negativos como g2. 
Dado que, – ( -g2 ) = +g2 
Por lo tanto, el resultado se convierte en g1 + |g2|. 
 

// CPP program to make two subsets with
// maximum difference.
#include <bits/stdc++.h>
using namespace std;
 
int maxDiff(int arr[], int n)
{
    int sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference
    for (int i = 0; i < n; i++)
         sum = sum + abs(arr[i]);
    
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 7, -4, 10, -11, 2 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maxDiff(arr, n);
    return 0;
}

// Java program to make two subsets with
// maximum difference.
import java.util.*;
 
class solution
{
 
static int maxDiff(int arr[], int n)
{
    int sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference
    for (int i = 0; i < n; i++)
        sum = sum + Math.abs(arr[i]);
     
    return sum;
}
 
// Driver Code
public static void main(String args[])
{
    int []arr = { 3, 7, -4, 10, -11, 2 };
    int n = arr.length;
    System.out.println(maxDiff(arr, n));
}
}

# Python3 program to make two subsets
# with maximum difference.
 
def maxDiff(arr, n) :
 
    sum = 0
 
    # We move all negative elements into
    # one set. So we add negation of negative
    # numbers to maximize difference
    for i in range(n) :
        sum += abs(arr[i])
     
    return sum
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 3, 7, -4, 10, -11, 2 ]
    n = len(arr)
    print(maxDiff(arr, n))
 
# This code is contributed by Ryuga

using System;
 
// C# program to make two subsets with
// maximum difference.
 
public class solution
{
 
public static int maxDiff(int[] arr, int n)
{
    int sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference 
    for (int i = 0; i < n; i++)
    {
        sum = sum + Math.Abs(arr[i]);
    }
 
    return sum;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = new int[] {3, 7, -4, 10, -11, 2};
    int n = arr.Length;
    Console.WriteLine(maxDiff(arr, n));
}
}
 
  // This code is contributed by Shrikant13

<?php
// PHP program to make two subsets
// with maximum difference.
function maxDiff($arr, $n)
{
    $sum = 0;
 
    // We move all negative elements
    // into one set. So we add negation
    // of negative numbers to maximize
    // difference
    for ($i = 0; $i < $n; $i++)
        $sum = $sum + abs($arr[$i]);
     
    return $sum;
}
 
// Driver Code
$arr = array( 3, 7, -4, 10, -11, 2 );
$n = sizeof($arr);
echo maxDiff($arr, $n);
     
// This code is contributed by Sachin.
?>

<script>
// javascript program to make two subsets with
// maximum difference.
function maxDiff(arr , n)
{
    var sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference
    for (i = 0; i < n; i++)
        sum = sum + Math.abs(arr[i]);
     
    return sum;
}
 
// Driver Code
var arr = [ 3, 7, -4, 10, -11, 2 ];
var n = arr.length;
document.write(maxDiff(arr, n));
 
// This code is contributed by Princi Singh
</script>

37

Complejidad de tiempo: O(n)