1352. 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?

Approach 01:

#include <vector>

class ProductOfNumbers {
public:
    ProductOfNumbers() { prefixProducts = {1}; }

    void add(int num) {
        if (num == 0)
            prefixProducts = {1}; 
        else
            prefixProducts.push_back(prefixProducts.back() * num);
    }

    int getProduct(int k) {
        if (k >= prefixProducts.size())
            return 0;
        return prefixProducts.back() / prefixProducts[prefixProducts.size() - k - 1];
    }

private:
    std::vector<int> prefixProducts; 
};

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

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

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

Time Complexity:

• add(num):

  Appending an element to the prefix product vector takes \( O(1) \).

• getProduct(k):

  Accessing elements and performing division takes \( O(1) \).

• Overall Time Complexity:

  \( O(1) \) for both operations.

Space Complexity:

• Prefix Product Storage:

  Stores all prefix products, leading to \( O(N) \) space usage in the worst case.

• Auxiliary Variables:

  A few integer variables are used, contributing \( O(1) \) extra space.

• Overall Space Complexity:

  \( O(N) \).