Problem URL: Product of the last k numbers

We will calculate the prefix product while adding number to the queue. Then while getting the product, we will divive the last item with the last kth item to get the product of last k elements.

class ProductOfNumbers:
    def __init__(self):
        self.array = [1]

    def add(self, num: int) -> None:
        if num == 0:
            self.array = [1]
        else:
            self.array.append(self.array[-1] * num)

    def getProduct(self, k: int) -> int:
        if k >= len(self.array): 
            return 0
        return self.array[-1] // self.array[-k-1]

# Your ProductOfNumbers object will be instantiated and called as such:
# obj = ProductOfNumbers()
# obj.add(num)
# param_2 = obj.getProduct(k)

Time Complexity: O(1) for each operation
Space Complexity: O(n)