problemmediumalgorithmsleetcode-1352leetcode 1352leetcode1352

Product of the Last K Numbers

MediumUpdated: Aug 2, 2025
Practice on:
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Video Solutions:

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.

Examples

Example 1:

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 * 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

Method 1 - Using Prefix Product

To efficiently solve this, we use prefix products (cumulative product array):

  1. Maintain a prefix list where each element at index i holds the product of all the numbers added so far up to the i-th index.
  2. Adding a number (add(num)):
    • If num != 0, append prefix[-1] * num to prefix.
    • If num == 0, reset the prefix list to [1], since any product involving zero is 0.
  3. Querying (getProduct(k)):
    • If we want the product of the last k numbers, calculate it as prefix[-1] / prefix[-k-1] (or directly divide products to exclude earlier terms).
    • If k exceeds the size of the prefix, return 0.

Video explanation

Here is the video explaining this method in detail. Please check it out:

<div class="youtube-embed"><iframe src="https://www.youtube.com/embed/RYC1iRntyio" frameborder="0" allowfullscreen></iframe></div>

Code

Java
class ProductOfNumbers {
    private List<Integer> prefix;

    public ProductOfNumbers() {
        init();
    }

	private void init() {
		prefix = new ArrayList<>();
        prefix.add(1);
	}

    public void add(int num) {
        if (num == 0) {
            init();
        } else {
            prefix.add(prefix.get(prefix.size() - 1) * num);
        }
    }

    public int getProduct(int k) {
		int n = prefix.size();
        if (k >= n) {
            return 0;
        }
        return prefix.get(n - 1) / prefix.get(n - k - 1);
    }
}
Python
class ProductOfNumbers:
    def __init__(self) -> None:
        self.prefix: list[int] = [1]

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

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

Complexity

  • ⏰ Time complexity:
    • Adding add(num): O(1)
    • Querying getProduct(k): O(1)
  • 🧺 Space complexity: O(n) for storing the prefix product array, where n is the number of add calls made.

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