Maximum Product Subarray

To find the maximum product subarray in an array, we can use dynamic programming. Let’s break down the problem into smaller subproblems and build a recursive function to solve it.
Let’s define the function maxProduct(nums, start, end) that takes an array of integers nums, and two indices start and end as input. This function will return the maximum product subarray ending at index end with the given range [start, end].
Here’s the recursive function to find the maximum product subarray:

def maxProduct(nums, start, end):
if start == end:
return nums[start]
max_product = float('-inf')
current_product = 1
for i in range(start, end + 1):
current_product *= nums[i]
max_product = max(max_product, current_product)
return max_product

The base case is when start is equal to end, which means we have only one element in the subarray. In this case, we simply return that element as the maximum product.
For all other cases, we initialize max_product and current_product to negative infinity. We then iterate over the range [start, end] and update current_product by multiplying each element with the previous product. We also keep track of the maximum product encountered so far and update it whenever we find a larger product.
Finally, we return the maximum product found.
To find the maximum product subarray in the entire array, we can call the function with start = 0 and end = len(nums) - 1.
Let’s see an example to understand the solution better. Consider the array nums = [-2, 3, -4]. We can find the maximum product subarray by calling maxProduct(nums, 0, 2). The function will return -8, which is the maximum product subarray in this case. The subarray is [-2, 3, -4].
Now let’s solve a more complex example to demonstrate the power of this solution. Consider the array nums = [10, -2, -10]. We can find the maximum product subarray by calling maxProduct(nums, 0, 2). The function will return -20, which is the maximum product subarray in this case. The subarray is [-2, -10].
It’s important to note that if there are multiple maximum product subarrays with different lengths, the function will return the maximum product from any of those subarrays.
The time complexity of this solution is O(n), where n is the length of the input array. This is because we iterate over each element in the array once. The space complexity is O(1) since we only use a constant amount of additional space to store variables during the recursive calls.