1352. Product of the Last K Numbers

Problem

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 * 10^4

• At most 4 * 10^4 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.

Solution (Java)

class ProductOfNumbers {
    List<Integer> prod;
    int p = 1;
    public ProductOfNumbers() {
        prod = new ArrayList();
        prod.add(1);
    }

    public void add(int num) {
        if (num == 0) {
            p = 1;
            prod = new ArrayList();
            prod.add(1);
        } else {
            p *= num;
            prod.add(p);
        }
    }

    public int getProduct(int k) {
        if (k >= prod.size()) {
            return 0;
        }

        int last = prod.get(prod.size() - 1);
        int prev = prod.get(prod.size() - 1 - k);
        return last / prev;
    }
}

/**
 * Your ProductOfNumbers object will be instantiated and called as such:
 * ProductOfNumbers obj = new ProductOfNumbers();
 * obj.add(num);
 * int param_2 = obj.getProduct(k);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).