1352\. Product of the Last K Numbers

Medium

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`
*   <code>1 <= k <= 4 * 10<sup>4</sup></code>
*   At most <code>4 * 10<sup>4</sup></code> 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.