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 * 10⁴</code>

• At most <code>4 * 10⁴</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.