# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `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 `

Output:

```270
```

Time Complexity: O(N2)

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

• Prefix sum of the elements can be calculated.
• The left array sum is the value at ith position.
• The right array sum is the value at the last position – ith position.
• Now calculate k, the product of sum 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 ` `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 ` `    ``// with arr element ` `    ``prefixArraySum = arr; ` ` `  `    ``// 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 ` `    ``// with arr 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 ` `    ``# with arr 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 ` `    ``// with arr element ` `    ``prefixArraySum = arr; ` ` `  `    ``// 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 `

Output:

```270
```

Time Complexity: O(n)

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