Maximum sum of pairwise product in an array with negative allowed

Given an array of n elements. Find maximum sum of pairwise multiplications. Sum can be larger so take mod with 10^9+7. If there are odd elements, then we can add any one element (without forming a pair) to the sum.

Examples:

Input : arr[] = {-1, 4, 5, -7, -4, 9, 0}
Output : 77
So to get the maximum sum, the arrangement will 
be {-7, -4}, {-1, 0}, {9, 5} and {4}.
So the answer is (-7*(-4))+((-1)*0)+(9*5)+(4) ={77}.

Input : arr[] = {8, 7, 9}
Output : 79
Answer is (9*8) +(7) = 79.

1- Sort the given array.
2- First, multiply the negative numbers pairwise from the starting and add to the total_sum.
3- Second, multiply the positive numbers pairwise from the last and to the total_sum.
4- Check if negative and positive both counts are odd, then add the product of last pair
i.e. last negative and positive left.
5- Or if any of the one counts is odd, then add that element left.
6- Return sum.

C++



filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for above implementation
#include <bits/stdc++.h>
#define Mod 1000000007
using namespace std;
  
// Function to find the maximum sum
long long int findSum(int arr[], int n)
{
    long long int sum = 0;
  
    // Sort the array first
    sort(arr, arr + n);
  
    // First multiply negative numbers pairwise
    // and sum up from starting as to get maximum 
    // sum. 
    int i = 0;
    while (i < n && arr[i] < 0) {
        if (i != n - 1 && arr[i + 1] <= 0) {
            sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;
            i += 2;
        }
        else
            break;
    }
  
    // Second multiply positive numbers pairwise
    // and summed up from the last as to get maximum 
    // sum.
    int j = n - 1;
    while (j >= 0 && arr[j] > 0) {
        if (j != 0 && arr[j - 1] > 0) {
            sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;
            j -= 2;
        }
        else
            break;
    }
  
    // To handle case if positive and negative
    // numbers both are odd in counts.
    if (j > i)
        sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;
  
    // If one of them occurs odd times
    else if (i == j)
        sum = (sum + arr[i]) % Mod;
  
    return sum;
}
  
// Drivers code
int main()
{
    int arr[] = { -1, 9, 4, 5, -4, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSum(arr, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for above implementation
import java.io.*;
import java.util.*;
  
class GFG {
  
static int Mod = 1000000007;
  
// Function to find the maximum sum
static long findSum(int arr[], int n) {
    long sum = 0;
  
    // Sort the array first
    Arrays.sort(arr);
  
    // First multiply negative numbers 
    // pairwise and sum up from starting 
    // as to get maximum sum.
    int i = 0;
    while (i < n && arr[i] < 0) {
    if (i != n - 1 && arr[i + 1] <= 0) {
        sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;
        i += 2;
    
    else
        break;
    }
  
    // Second multiply positive numbers 
    // pairwise and summed up from the 
    // last as to get maximum sum.
    int j = n - 1;
    while (j >= 0 && arr[j] > 0) {
    if (j != 0 && arr[j - 1] > 0) {
        sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;
        j -= 2;
    } else
        break;
    }
  
    // To handle case if positive and negative
    // numbers both are odd in counts.
    if (j > i)
    sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;
  
    // If one of them occurs odd times
    else if (i == j)
    sum = (sum + arr[i]) % Mod;
  
    return sum;
}
  
// Drivers code
public static void main(String args[]) {
    int arr[] = {-1, 9, 4, 5, -4, 7};
    int n = arr.length;
    System.out.println(findSum(arr, n));
}
}
  
/*This code is contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 code for above implementation
Mod= 1000000007
  
# Function to find the maximum sum
def findSum(arr, n):
    sum = 0
      
    # Sort the array first
    arr.sort()
      
    # First multiply negative numbers 
    # pairwise and sum up from starting
    # as to get maximum sum.
    i = 0
    while i < n and arr[i] < 0:
        if i != n - 1 and arr[i + 1] <= 0:
            sum = (sum + (arr[i] * arr[i + 1]) 
                                 % Mod) % Mod
            i += 2
        else:
            break
          
    # Second multiply positive numbers
    # pairwise and summed up from the 
    # last as to get maximum sum.
    j = n - 1
    while j >= 0 and arr[j] > 0:
        if j != 0 and arr[j - 1] > 0:
            sum = (sum + (arr[j] * arr[j - 1])
                                 % Mod) % Mod
            j -= 2
        else:
            break
          
    # To handle case if positive 
    # and negative numbers both
    # are odd in counts.
    if j > i:
        sum = (sum + (arr[i] * arr[j]) % Mod) 
                                       % Mod
          
    # If one of them occurs odd times
    elif i == j:
        sum = (sum + arr[i]) % Mod
      
    return sum
  
# Driver code
arr = [ -1, 9, 4, 5, -4, 7 ]
n = len(arr) 
print(findSum(arr, n))
  
# This code is contributed by "Sharad_Bhardwaj".

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for above implementation
using System;
  
class GFG {
  
    static int Mod = 1000000007;
  
    // Function to find the maximum sum
    static long findSum(int[] arr, int n)
    {
        long sum = 0;
  
        // Sort the array first
        Array.Sort(arr);
  
        // First multiply negative numbers
        // pairwise and sum up from starting
        // as to get maximum sum.
        int i = 0;
        while (i < n && arr[i] < 0) {
            if (i != n - 1 && arr[i + 1] <= 0) {
                sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;
                i += 2;
            }
            else
                break;
        }
  
        // Second multiply positive numbers
        // pairwise and summed up from the
        // last as to get maximum sum.
        int j = n - 1;
        while (j >= 0 && arr[j] > 0) {
            if (j != 0 && arr[j - 1] > 0) {
                sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;
                j -= 2;
            }
            else
                break;
        }
  
        // To handle case if positive and negative
        // numbers both are odd in counts.
        if (j > i)
            sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;
  
        // If one of them occurs odd times
        else if (i == j)
            sum = (sum + arr[i]) % Mod;
  
        return sum;
    }
  
    // Drivers code
    public static void Main()
    {
        int[] arr = { -1, 9, 4, 5, -4, 7 };
        int n = arr.Length;
        Console.WriteLine(findSum(arr, n));
    }
}
  
/*This code is contributed by vt_m.*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program for above implementation
  
$Mod = 1000000007;
  
// Function to find the maximum sum
function findSum(&$arr, $n)
{
    global $Mod;
    $sum = 0;
  
    // Sort the array first
    sort($arr);
  
    // First multiply negative numbers 
    // pairwise and sum up from starting 
    // as to get maximum sum. 
    $i = 0;
    while ($i < $n && $arr[$i] < 0) 
    {
        if ($i != $n - 1 && $arr[$i + 1] <= 0) 
        {
            $sum = ($sum + ($arr[$i] *
                    $arr[$i + 1]) % $Mod) % $Mod;
            $i += 2;
        }
        else
            break;
    }
  
    // Second multiply positive numbers pairwise
    // and summed up from the last as to get 
    // maximum sum.
    $j = $n - 1;
    while ($j >= 0 && $arr[$j] > 0)
    {
        if ($j != 0 && $arr[$j - 1] > 0)
        {
            $sum = ($sum + ($arr[$j] * 
                    $arr[$j - 1]) % $Mod) % $Mod;
            $j -= 2;
        }
        else
            break;
    }
  
    // To handle case if positive and negative
    // numbers both are odd in counts.
    if ($j > $i)
        $sum = ($sum + ($arr[$i] * 
                $arr[$j]) % $Mod) % $Mod;
  
    // If one of them occurs odd times
    else if ($i == $j)
        $sum = ($sum + $arr[$i]) % Mod;
  
    return $sum;
}
  
// Driver code
$arr = array (-1, 9, 4, 5, -4, 7 );
$n = sizeof($arr);
echo findSum($arr, $n);
  
// This code is contributed by ita_c
?>

chevron_right



Output:

87


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : chitranayal



Article Tags :
Practice Tags :


3


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.