LeetCode 238 - Product of Array Except Self - 2022.12.31 solution

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        """ How to solve it: The normal way to do this would be to multiply all the numbers in the input into
        a single product, then go through the input again and divide by each number to get the product for that
        index.  Since we can't use the division operation, we need to figure out how to only use the multiplication
        operator.  So we need two additional arrays, one for the product to the left of the given index, the
        other for the product to the right of the given index.
        To do it in O(1) extra space complexity, we can use the result list to store the left-products and then
        iterate from the right again with a single variable to store the right product.
        """
        res = []

        # Fill out the left-products
        for i in range(len(nums)):
            if not res:
                res.append(1)
            else:
                res.append(nums[i-1] * res[i-1])

        right_product = 1
        for i in range(len(nums)-1, -1, -1):
            if i == len(nums) - 1:
                continue
            else:
                right_product *= nums[i+1]
                res[i] = res[i] * right_product
        return res