Maximum product subarray in C++

In this article, we will discuss how to find the maximum product subarray in C++.

Find the largest product of the subarray of positive and negative integers in the given array. O(n) is the predicted time complexity, and the only usable extra space is O(1).

Examples:

Input: arr[] = {4, -5, -6, 0, 2}
Output:   120  // The subarray is {4, -5, -6}

Input: arr[] = {-5, -10, -2, 0, 60}
Output:   0  // The subarray is {0}

Input: arr[] = {-10, -3, 0, -5, -4}
Output:   12  // The subarray is {-3, -4}

Naive Solution:

The goal is to search over all adjacent subarrays, determine the product of each subarray, and then return the highest product from these findings.

Program:

Let's take an example to demonstrate how to find the maximum product subarray in C++:

// C++ program to find Maximum Product Subarray 
#include <bits/stdc++.h> 
using namespace std; 
int maxSubarrayProduct(int arr[], int n) 
{ 
int result = arr[0]; 
for (int i = 0; i < n; i++) 
{ 
int mul = arr[i]; 
for (int j = i + 1; j < n; j++) 
{ 
result = max(result, mul); 
mul *= arr[j]; 
} 
// updating the result for (n-1)th index. 
result = max(result, mul); 
} 
return result; 
} 
int main() 
{ 
int arr[] = {4, -5, -6, 0, 2, -8, -2 }; 
int n = sizeof(arr) / sizeof(arr[0]); 
cout << "Maximum Sub array product is "<< maxSubarrayProduct(arr, n); 
return 0; 
} 

Output:

Complexity:

Time Complexity: O(N2)

Auxiliary Space: O(1)

Efficient Solution:

Program:

#include <bits/stdc++.h> 
using namespace std; 
int maxSubarrayProduct(int arr[], int n) 
{ 
int max_ending_here = 1; 
int min_ending_here = 1; 
int max_so_far = 0; 
int flag = 0; 
for (int i = 0; i < n; i++) 
{ 
if (arr[i] > 0) 
{ 
max_ending_here = max_ending_here * arr[i]; 
min_ending_here = min(min_ending_here * arr[i], 1); 
flag = 1; 
} 
else if (arr[i] == 0) { 
max_ending_here = 1; 
min_ending_here = 1; 
} 
else { 
int temp = max_ending_here; 
max_ending_here = max(min_ending_here * arr[i], 1); 
min_ending_here = temp * arr[i]; 
} 
if (max_so_far < max_ending_here) 
max_so_far = max_ending_here; 
} 
if (flag == 0 && max_so_far == 0) 
return 0; 
return max_so_far; 
} 
int main() 
{ 
int arr[] = {4, -5, -6, 0, 2, -8, -2 }; 
int n = sizeof(arr) / sizeof(arr[0]); 
cout << "Maximum Sub array product is "<< maxSubarrayProduct(arr, n); 
return 0; 
} 

Output:

Complexity:

Time Complexity: O(n)

Auxiliary Space: O(1)

Benefits of Maximum product subarray:

There are several benefits of the Maximum Product Subarray in C++. Some main advantages of the Maximum Product Subarray in C++ are as follows:

Efficiency: The Maximum Product Subarray problem can be solved efficiently using C++, a high-performance language. You can employ dynamic programming or two-pointer algorithms to obtain the ideal temporal complexity.

Flexibility: The Maximum Product Subarray problem can be implemented using a variety of C++ data structures and utilities. You can encode the input data using arrays, vectors, or other data structures to compute the maximum product efficiently.

Standard Template Library (STL): The Maximum Product Subarray problem can be more easily implemented due to the Standard Template Library (STL) in C++. The Standard Template Library (STL) provides a variety of helpful data structures and algorithms. For instance, you can make your code simpler by using the std::vector and std::max methods.

Performance Optimization: C++ allows you to optimize your code for speed. Your code can execute more quickly by utilizing inline assembly and compiler optimizations, which is crucial for solving significant instances of the problem.

Portability: C++ is a widely used and portable language, making it simple to create programs that work across different systems. Software that needs to operate in a variety of contexts and platforms must be portable.

Maintainability: The Maximum Product Subarray problem can be solved by designing clear, maintainable code due to C++'s capabilities for object-oriented programming. Your code can be contained in classes and structures, which will make it simpler to comprehend and update.

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