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1352. Product of the Last K Numbers

Difficulty Topics

Description

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.

Solution

product-of-the-last-k-numbers.py
class ProductOfNumbers:

    def __init__(self):
        self.A = [1]


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

    def getProduct(self, k: int) -> int:
        n = len(self.A)
        if k >= n : return 0

        return self.A[-1] // self.A[-k-1]


# Your ProductOfNumbers object will be instantiated and called as such:
# obj = ProductOfNumbers()
# obj.add(num)
# param_2 = obj.getProduct(k)