Maximum product of sum of two contiguous subarrays of an array

Given an array arr[] of N positive integers, the task is to split the array into two contiguous subarrays such that the product of the sum of two contiguous subarrays is maximum. 

Examples:

Input: arr[] = {4, 10, 1, 7, 2, 9} 
Output: 270 
All possible partitions and their product of sum are: 
{4} and {10, 1, 7, 2, 9} -> product of sum = 116 
{4, 10} and {1, 7, 2, 9} -> product of sum = 266 
{4, 10, 1} and {7, 2, 9} -> product of sum = 270 
{4, 10, 1, 7} and {2, 9} -> product of sum = 242 
{4, 10, 1, 7, 2} and {9} -> product of sum = 216

Input: arr[] = {4, 10, 11, 10, 4} 
Output: 350 

Naive approach: A simple approach is to consider all the possible partitions for the subarrays one by one and calculate the maximum product of the sum of the subarrays.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the maximum
// product of sum for any partition
int maxProdSum(int arr[], int n)
{
    int leftArraySum = 0, maxProduct = 0;

    // Traversing the array
    for (int i = 0; i < n; i++) {

        // Compute left array sum
        leftArraySum += arr[i];

        // Compute right array sum
        int rightArraySum = 0;
        for (int j = i + 1; j < n; j++) {
            rightArraySum += arr[j];
        }

        // Multiplying left and right subarray sum
        int k = leftArraySum * rightArraySum;

        // Checking for the maximum product
        // of sum of left and right subarray
        if (k > maxProduct) {
            maxProduct = k;
        }
    }

    // Printing the maximum product
    return maxProduct;
}

// Driver code
int main()
{
    int arr[] = { 4, 10, 1, 7, 2, 9 };
    int n = sizeof(arr) / sizeof(int);

    cout << maxProdSum(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the maximum
// product of sum for any partition
static int maxProdSum(int arr[], int n)
{
    int leftArraySum = 0, maxProduct = 0;

    // Traversing the array
    for (int i = 0; i < n; i++)
    {

        // Compute left array sum
        leftArraySum += arr[i];

        // Compute right array sum
        int rightArraySum = 0;
        for (int j = i + 1; j < n; j++)
        {
            rightArraySum += arr[j];
        }

        // Multiplying left and right subarray sum
        int k = leftArraySum * rightArraySum;

        // Checking for the maximum product
        // of sum of left and right subarray
        if (k > maxProduct)
        {
            maxProduct = k;
        }
    }

    // Printing the maximum product
    return maxProduct;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 10, 1, 7, 2, 9 };
    int n = arr.length;

    System.out.print(maxProdSum(arr, n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Function to return the maximum
# product of sum for any partition
def maxProdSum(arr, n):
    leftArraySum = 0;
    maxProduct = 0;

    # Traversing the array
    for i in range(n):

        # Compute left array sum
        leftArraySum += arr[i];

        # Compute right array sum
        rightArraySum = 0;
        for j in range(i + 1, n):
            rightArraySum += arr[j];

        # Multiplying left and right subarray sum
        k = leftArraySum * rightArraySum;

        # Checking for the maximum product
        # of sum of left and right subarray
        if (k > maxProduct):
            maxProduct = k;

    # Printing the maximum product
    return maxProduct;

# Driver code
if __name__ == '__main__':
    arr = [ 4, 10, 1, 7, 2, 9 ];
    n = len(arr);

    print(maxProdSum(arr, n));

# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the maximum
// product of sum for any partition
static int maxProdSum(int []arr, int n)
{
    int leftArraySum = 0, maxProduct = 0;

    // Traversing the array
    for (int i = 0; i < n; i++)
    {

        // Compute left array sum
        leftArraySum += arr[i];

        // Compute right array sum
        int rightArraySum = 0;
        for (int j = i + 1; j < n; j++)
        {
            rightArraySum += arr[j];
        }

        // Multiplying left and right subarray sum
        int k = leftArraySum * rightArraySum;

        // Checking for the maximum product
        // of sum of left and right subarray
        if (k > maxProduct)
        {
            maxProduct = k;
        }
    }

    // Printing the maximum product
    return maxProduct;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 10, 1, 7, 2, 9 };
    int n = arr.Length;

    Console.Write(maxProdSum(arr, n));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript implementation of the approach

// Function to return the maximum
// product of sum for any partition
function maxProdSum(arr, n) {
    let leftArraySum = 0, maxProduct = 0;

    // Traversing the array
    for (let i = 0; i < n; i++) {

        // Compute left array sum
        leftArraySum += arr[i];

        // Compute right array sum
        let rightArraySum = 0;
        for (let j = i + 1; j < n; j++) {
            rightArraySum += arr[j];
        }

        // Multiplying left and right subarray sum
        let k = leftArraySum * rightArraySum;

        // Checking for the maximum product
        // of sum of left and right subarray
        if (k > maxProduct) {
            maxProduct = k;
        }
    }

    // Printing the maximum product
    return maxProduct;
}

// Driver code

let arr = [4, 10, 1, 7, 2, 9];
let n = arr.length;

document.write(maxProdSum(arr, n));
</script>

Output:

270

Time Complexity: O(N²)

Auxiliary Space: O(1)

Efficient approach: A better approach is to use the concept prefix array sum which helps in calculating the sum of both the contiguous subarrays.  

• The prefix sum of the elements can be calculated.
• The left array sum is the value at i^th position.
• The right array sum is the value at the last position – i^th position.
• Now calculate k, the product of some of these left and right subarrays.
• Find and print the maximum of the product of these arrays.

Below is the implementation of the above approach: 

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the maximum
// product of sum for any partition
int maxProdSum(int arr[], int n)
{
    int prefixArraySum[n], maxProduct = 0;

    // Initialise prefixArraySum[0]
    // with arr[0] element
    prefixArraySum[0] = arr[0];

    // Traverse array elements
    // to compute prefix array sum
    for (int i = 1; i < n; i++) {
        prefixArraySum[i] = prefixArraySum[i - 1]
                            + arr[i];
    }

    for (int i = 0; i < n - 1; i++) {
        // Compute left and right array sum
        int leftArraySum = prefixArraySum[i];
        int rightArraySum = prefixArraySum[n - 1]
                            - prefixArraySum[i];

        // Multiplying left and right subarray sum
        int k = leftArraySum * rightArraySum;

        // Checking for maximum product of
        // the sum of left and right subarray
        if (k > maxProduct) {
            maxProduct = k;
        }
    }

    // Printing the maximum value
    return maxProduct;
}

// Driver code
int main()
{
    int arr[] = { 4, 10, 1, 7, 2, 9 };
    int n = sizeof(arr) / sizeof(int);

    cout << maxProdSum(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the maximum
// product of sum for any partition
static int maxProdSum(int arr[], int n)
{
    int []prefixArraySum = new int[n];
    int maxProduct = 0;

    // Initialise prefixArraySum[0]
    // with arr[0] element
    prefixArraySum[0] = arr[0];

    // Traverse array elements
    // to compute prefix array sum
    for (int i = 1; i < n; i++)
    {
        prefixArraySum[i] = prefixArraySum[i - 1]
                            + arr[i];
    }

    for (int i = 0; i < n - 1; i++)
    {
        // Compute left and right array sum
        int leftArraySum = prefixArraySum[i];
        int rightArraySum = prefixArraySum[n - 1]
                            - prefixArraySum[i];

        // Multiplying left and right subarray sum
        int k = leftArraySum * rightArraySum;

        // Checking for maximum product of
        // the sum of left and right subarray
        if (k > maxProduct)
        {
            maxProduct = k;
        }
    }

    // Printing the maximum value
    return maxProduct;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 10, 1, 7, 2, 9 };
    int n = arr.length;

    System.out.print(maxProdSum(arr, n));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python implementation of the approach

# Function to return the maximum
# product of sum for any partition
def maxProdSum(arr, n):

    prefixArraySum = [0] * n
    maxProduct = 0

    # Initialise prefixArraySum[0]
    # with arr[0] element
    prefixArraySum[0] = arr[0]

    # Traverse array elements
    # to compute prefix array sum
    for i in range(1, n):
        prefixArraySum[i] = prefixArraySum[i - 1] + arr[i]

    for i in range(n - 1):

        # Compute left and right array sum
        leftArraySum = prefixArraySum[i]
        rightArraySum = prefixArraySum[n - 1] - \
                        prefixArraySum[i]

        # Multiplying left and right subarray sum
        k = leftArraySum * rightArraySum

        # Checking for maximum product of
        # the sum of left and right subarray
        if (k > maxProduct):
            maxProduct = k

    # Printing the maximum value
    return maxProduct

# Driver code
arr = [4, 10, 1, 7, 2, 9]
n = len(arr)
print(maxProdSum(arr, n))

# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the maximum
// product of sum for any partition
static int maxProdSum(int []arr, int n)
{
    int []prefixArraySum = new int[n];
    int maxProduct = 0;

    // Initialise prefixArraySum[0]
    // with arr[0] element
    prefixArraySum[0] = arr[0];

    // Traverse array elements
    // to compute prefix array sum
    for (int i = 1; i < n; i++)
    {
        prefixArraySum[i] = prefixArraySum[i - 1]
                            + arr[i];
    }

    for (int i = 0; i < n - 1; i++)
    {
        // Compute left and right array sum
        int leftArraySum = prefixArraySum[i];
        int rightArraySum = prefixArraySum[n - 1]
                            - prefixArraySum[i];

        // Multiplying left and right subarray sum
        int k = leftArraySum * rightArraySum;

        // Checking for maximum product of
        // the sum of left and right subarray
        if (k > maxProduct)
        {
            maxProduct = k;
        }
    }

    // Printing the maximum value
    return maxProduct;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 10, 1, 7, 2, 9 };
    int n = arr.Length;

    Console.Write(maxProdSum(arr, n));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript implementation of the approach

// Function to return the maximum
// product of sum for any partition
function maxProdSum(arr, n)
{
    let prefixArraySum = [], maxProduct = 0;

    // Initialise prefixArraySum[0]
    // with arr[0] element
    prefixArraySum[0] = arr[0];

    // Traverse array elements
    // to compute prefix array sum
    for (let i = 1; i < n; i++) {
        prefixArraySum[i] = prefixArraySum[i - 1]
                            + arr[i];
    }

    for (let i = 0; i < n - 1; i++) {
        // Compute left and right array sum
        let leftArraySum = prefixArraySum[i];
        let rightArraySum = prefixArraySum[n - 1]
                            - prefixArraySum[i];

        // Multiplying left and right subarray sum
        let k = leftArraySum * rightArraySum;

        // Checking for maximum product of
        // the sum of left and right subarray
        if (k > maxProduct) {
            maxProduct = k;
        }
    }

    // Printing the maximum value
    return maxProduct;
}

// Driver code
let arr = [ 4, 10, 1, 7, 2, 9 ];
let n = arr.length;

document.write(maxProdSum(arr, n));

// This code is contributed by Samim Hossain Mondal.
</script>

Output:

270

Time Complexity: O(n)

Auxiliary Space: O(n)