Product of the Last K Numbers

Design an algorithm that accepts a stream of integers and retrieves the product of the last `k` integers of the stream. Implement the `ProductOfNumbers` class: - `ProductOfNumbers()` Initializes the object with an empty stream. - `void add(int num)` Appends the integer `num` to the stream. - `int getProduct(int k)` Returns the product of the last `k` numbers in the current list. You can assume that always the current list has at least `k` numbers. The test cases are generated so that, at any time, the product of any contiguous sequence of numbers will fit into a single 32-bit integer without overflowing.   **Example:** **Input** ["ProductOfNumbers","add","add","add","add","add","getProduct","getProduct","getProduct","add","getProduct"] [[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]] **Output** [null,null,null,null,null,null,20,40,0,null,32] **Explanation** ProductOfNumbers productOfNumbers = new ProductOfNumbers(); productOfNumbers.add(3); // [3] productOfNumbers.add(0); // [3,0] productOfNumbers.add(2); // [3,0,2] productOfNumbers.add(5); // [3,0,2,5] productOfNumbers.add(4); // [3,0,2,5,4] productOfNumbers.getProduct(2); // return 20. The product of the last 2 numbers is 5 * 4 = 20 productOfNumbers.getProduct(3); // return 40. The product of the last 3 numbers is 2 * 5 * 4 = 40 productOfNumbers.getProduct(4); // return 0. The product of the last 4 numbers is 0 * 2 * 5 * 4 = 0 productOfNumbers.add(8); // [3,0,2,5,4,8] productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers is 4 * 8 = 32 ```   **Constraints:** - `0 <= num <= 100` - `1 <= k <= 4 * 104` - At most `4 * 104` calls will be made to `add` and `getProduct`. - The product of the stream at any point in time will fit in a **32-bit** integer.   **Follow-up: **Can you implement **both** `GetProduct` and `Add` to work in `O(1)` time complexity instead of `O(k)` time complexity?