Maximize the Product of Sum by converting Array elements into given two types

Given an array of positive integer arr[] of length N and an integer Z, (Z > arr[i] for all 0 ? i ? N – 1). Each integer of the array can be converted into the following two types:

• Keep it unchanged
• Change it to Z – arr[i].

The task is to maximize the product of the sum of these two types of elements.

Note: There should be present at least one element of each type. 

Examples:

Input: N = 5, arr[] = {500, 100, 400, 560, 876}, Z = 1000
Output: 290400
Explanation: arr[] = {500, 100, 400, 560, 876}
Convert elements present at indices 0, 3 and 4 to first type =  (500, 560, 876)
Convert elements present at indices 1 and 2 to second type 
= (Z-arr[1],  Z-arr[2]) = (1000 – 100, 1000 – 400) = (900, 600)
Sum of all first type elements = 500+560+876 = 1936
Sum of all second type elements = 900 + 600 = 1500  
Product of each type sum = 1936*1500 = 290400. 
Which is maximum possible for this case.

Input: N = 4, arr[] = {1, 4, 6, 3}, Z = 7
Output: 100
Explanation: Change the 1st and last element to 2nd type, i.e., 
{7-1, 7-3} = {6, 4}. The sum is (6 + 4) = 10.
Keep the 2nd and third element as it is. Their sum = (4 + 6) = 10 .
Product is 10*10 = 100. This is the maximum product possible.

Approach: The problem can be solved using Sorting based on the following idea:

The idea is to sort the arr[] in decreasing order, Calculate product of all possible combinations of the types. Obtain maximum product among all combinations. 

Illustration:

Input: N = 4, arr[] = {1, 4, 6, 3}, Z = 7

After sorting arr[] in decreasing order = {6, 4, 3, 1}

Now we have 3 possible combinations for choosing all elements as first or second type:

• 1 of first type, 3 of second type 
• 2 of first type, 2 of second type 
• 3 of first type, 1 of second type 

Let’s see the product and sum at each combination for decreasing ordered arr[]:

Choosing first element as first type and next 3 elements as second type:

• Sum of first type elements = 6
• Sum of second type elements = ((7 – 4)+(7 – 3)+(7 – 1))= 13
• Product of first and second = 6 * 13 = 78

Choosing first two elements as first type and last 2 elements as second type:

• Sum of first type elements = 6 + 4 = 10
• Sum of second type elements = (7 – 3)+(7 – 1))= 10
• Product of first and second types = 10 * 10 = 100

Choosing first three elements as first type and last element as second type:

• Sum of first type elements = 6 + 4 + 3 = 13
• Sum of second type elements = (7 – 1)) = 6
• Product of first and second types = 13 * 6 = 78

As we can clearly see that 2nd combination has maximum value of product.Therefore, output for this case is : 

Maximum Product: 100

Follow the steps to solve the problem:

• Sort the input array arr[].
• Traverse from the end of the array to calculate the product for all possible combinations:
  • Consider all the element till index i as first type, and the suffix elements as second type.
  • Calculate the product of this combination.
  • Update the maximum product accordingly.
• Print the maximum product obtained.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to obtain maximum product
// along with sum of X and Y
long Max_Product(int n, vector<long> arr, long Z)
{
    // Variable to store maximum product
    long product = INT_MIN;

    // Sorting arr[]
    sort(arr.begin(), arr.end());

    // Variable to Hold sum of first type
    long sum1 = 0;

    // Variable to hold Maximum value
    // of first type sum
    long X = INT_MIN;

    // Variable to hold Maximum value
    // of second type sum
    long Y = INT_MAX;

    // Loop for iterating on sorted arr[]
    // from right to left for decreasing order
    for (int i = n - 1; i > 0; i--)
    {
        sum1 += arr[i];
        long sum2 = 0;
        for (int j = i - 1; j >= 0; j--)
        {
            sum2 = sum2 + (Z - arr[j]);
        }
        if (sum1 * sum2 > product)
        {
            product = sum1 * sum2;
            X = sum1;
            Y = sum2;
        }
    }

    return (product);
}
// Driver code
int main()
{
    vector<long> arr = {500, 100, 400, 560, 876};
    int N = arr.size();
    long Z = 1000;

    // Function call
    cout << (Max_Product(N, arr, Z));
}
// This code is contributed by Potta Lokesh

Java

// Java code to implement the approach

import java.util.*;

class GFG {

    // Driver code
    public static void main(String[] args)
    {
        long[] arr = { 500, 100, 400, 560, 876 };
        int N = arr.length;
        long Z = 1000;

        // Function call
        System.out.println(Max_Product(N, arr, Z));
    }

    // Function to obtain maximum product
    // along with sum of X and Y
    static long Max_Product(int n, long[] arr, long Z)
    {
        // Variable to store maximum product
        long product = Long.MIN_VALUE;

        // Sorting arr[]
        Arrays.sort(arr);

        // Variable to Hold sum of first type
        long sum1 = 0;

        // Variable to hold Maximum value
        // of first type sum
        long X = Integer.MIN_VALUE;

        // Variable to hold Maximum value
        // of second type sum
        long Y = Integer.MAX_VALUE;

        // Loop for iterating on sorted arr[]
        // from right to left for decreasing order
        for (int i = n - 1; i > 0; i--) {
            sum1 += arr[i];
            long sum2 = 0;
            for (int j = i - 1; j >= 0; j--) {
                sum2 = sum2 + (Z - arr[j]);
            }
            if (sum1 * sum2 > product) {
                product = sum1 * sum2;
                X = sum1;
                Y = sum2;
            }
        }

        return (product);
    }
}

Python3

# Python code to implement the approach

# Function to obtain maximum product
# along with sum of X and Y
def Max_Product(n, arr, Z):

    # Variable to store maximum product
    product = -10000000000000000000000

    # Sorting arr[]
    arr.sort()

    # Variable to Hold sum of first type
    sum1 = 0

    # Variable to hold Maximum value
    # of first type sum
    X = -10000000000000000000000

    # Variable to hold Maximum value
    # of second type sum
    Y = 100000000000000000000000

    # Loop for iterating on sorted arr[]
    # from right to left for decreasing order
    for i in range(n-1, 0, -1):
        sum1 += arr[i]
        sum2 = 0
        for j in range(i-1, -1, -1):
            sum2 = sum2 + (Z - arr[j])
            if (sum1 * sum2 > product):
                product = sum1 * sum2
                X = sum1
                Y = sum2

    return product

# Driver code
if __name__ == "__main__":
    arr = [500, 100, 400, 560, 876]
    N = len(arr)
    Z = 1000

    # Function call
    print(Max_Product(N, arr, Z))

# This code is contributed by Rohit Pradhan

C#

// C# code to implement the approach

using System;

public class GFG{

  static public void Main (){

    // Code
    long[] arr = { 500, 100, 400, 560, 876 };
    int N = arr.Length;
    long Z = 1000;

    // Function call
    Console.WriteLine(Max_Product(N, arr, Z));
  }

  // Function to obtain maximum product
  // along with sum of X and Y
  static long Max_Product(int n, long[] arr, long Z)
  {
    #pragma warning disable 219
      // Variable to store maximum product
      long product = Int64.MinValue;

    // Sorting arr[]
    Array.Sort(arr);

    // Variable to Hold sum of first type
    long sum1 = 0;

    // Variable to hold Maximum value
    // of first type sum
    long X = Int32.MinValue;

    // Variable to hold Maximum value
    // of second type sum
    long Y = Int32.MaxValue;

    // Loop for iterating on sorted arr[]
    // from right to left for decreasing order
    for (int i = n - 1; i > 0; i--) {
      sum1 += arr[i];
      long sum2 = 0;
      for (int j = i - 1; j >= 0; j--) {
        sum2 = sum2 + (Z - arr[j]);
      }
      if (sum1 * sum2 > product) {
        product = sum1 * sum2;
        X = sum1;
        Y = sum2;
      }
    }

    return product;
  }
}

// This code is contributed by lokeshmvs21.

Javascript

<script>
// JS code to implement the approach

    // Function to obtain maximum product
    // along with sum of X and Y
    function Max_Product(n, arr, Z)
    {
        // Variable to store maximum product
        let product = Number.MIN_VALUE;

        // Sorting arr[]
        arr.sort();

        // Variable to Hold sum of first type
        let sum1 = 0;

        // Variable to hold Maximum value
        // of first type sum
        let X =Number.MIN_VALUE;

        // Variable to hold Maximum value
        // of second type sum
        let Y = Number.MAX_VALUE;

        // Loop for iterating on sorted arr[]
        // from right to left for decreasing order
        for (let i = n - 1; i > 0; i--) {
            sum1 += arr[i];
            let sum2 = 0;
            for (let j = i - 1; j >= 0; j--) {
                sum2 = sum2 + (Z - arr[j]);
            }
            if (sum1 * sum2 > product) {
                product = sum1 * sum2;
                X = sum1;
                Y = sum2;
            }
        }

        return (product);
    }

// Driver code

        let arr = [ 500, 100, 400, 560, 876 ];
        let N = arr.length;
        let Z = 1000;

        // Function call
        document.write(Max_Product(N, arr, Z));

// This code is contributed by sanjoy_62.
</script>

Output

2904000

Time Complexity: O(N*N+NlogN)
Auxiliary Space: O(1)