Maximum product of sum of two contiguous subarrays of an array
Last Updated :
23 Nov, 2021
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 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 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# 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
<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>
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 ith position.
- The right array sum is the value at the last position - ith 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 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
# 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# 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
<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>
Time Complexity: O(n)
Auxiliary Space: O(n)
Similar Reads
DSA Tutorial - Learn Data Structures and Algorithms
DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. Data structures manage how data is stored and accessed, while algorithms focus on
7 min read
Quick Sort
QuickSort is a sorting algorithm based on the Divide and Conquer that picks an element as a pivot and partitions the given array around the picked pivot by placing the pivot in its correct position in the sorted array. It works on the principle of divide and conquer, breaking down the problem into s
12 min read
Merge Sort - Data Structure and Algorithms Tutorials
Merge sort is a popular sorting algorithm known for its efficiency and stability. It follows the divide-and-conquer approach. It works by recursively dividing the input array into two halves, recursively sorting the two halves and finally merging them back together to obtain the sorted array. Merge
14 min read
Breadth First Search or BFS for a Graph
Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list conta
15+ min read
Bubble Sort Algorithm
Bubble Sort is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in the wrong order. This algorithm is not suitable for large data sets as its average and worst-case time complexity are quite high.We sort the array using multiple passes. After the fir
8 min read
Binary Search Algorithm - Iterative and Recursive Implementation
Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Binary Search AlgorithmConditions to apply Binary Searc
15 min read
Insertion Sort Algorithm
Insertion sort is a simple sorting algorithm that works by iteratively inserting each element of an unsorted list into its correct position in a sorted portion of the list. It is like sorting playing cards in your hands. You split the cards into two groups: the sorted cards and the unsorted cards. T
9 min read
Data Structures Tutorial
Data structures are the fundamental building blocks of computer programming. They define how data is organized, stored, and manipulated within a program. Understanding data structures is very important for developing efficient and effective algorithms. What is Data Structure?A data structure is a st
2 min read
Dijkstra's Algorithm to find Shortest Paths from a Source to all
Given a weighted undirected graph represented as an edge list and a source vertex src, find the shortest path distances from the source vertex to all other vertices in the graph. The graph contains V vertices, numbered from 0 to V - 1.Note: The given graph does not contain any negative edge. Example
12 min read
Selection Sort
Selection Sort is a comparison-based sorting algorithm. It sorts an array by repeatedly selecting the smallest (or largest) element from the unsorted portion and swapping it with the first unsorted element. This process continues until the entire array is sorted.First we find the smallest element an
8 min read