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

• Difficulty Level : Basic
• Last Updated : 28 Sep, 2022

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
// alet 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(N2)
Auxiliary Space: O(1)

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