1352. Product of the Last K Numbers LeetCode Solution

Last updated: February 23, 2025

In this guide, you will get 1352. Product of the Last K Numbers LeetCode Solution with the best time and space complexity. The solution to Product of the Last K Numbers problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Product of the Last K Numbers solution in C++
Product of the Last K Numbers solution in Java
Product of the Last K Numbers solution in Python
Additional Resources

1352. Product of the Last K Numbers LeetCode Solution image

Problem Statement of 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?

Example not found

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?

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(|\texttt{add()}|)

1352. Product of the Last K Numbers LeetCode Solution in C++

class ProductOfNumbers {
 public:
  ProductOfNumbers() : prefix{1} {}

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

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

 private:
  vector<int> prefix;
};
/* code provided by PROGIEZ */

1352. Product of the Last K Numbers LeetCode Solution in Java

class ProductOfNumbers {
  public ProductOfNumbers() {
    prefix = new ArrayList<>(List.of(1));
  }

  public void add(int num) {
    if (num == 0)
      prefix = new ArrayList<>(List.of(1));
    else
      prefix.add(prefix.get(prefix.size() - 1) * num);
  }

  public int getProduct(int k) {
    return k >= prefix.size() ? 0
                              : prefix.get(prefix.size() - 1) / prefix.get(prefix.size() - k - 1);
  }

  private List<Integer> prefix = new ArrayList<>();
}
// code provided by PROGIEZ

1352. Product of the Last K Numbers LeetCode Solution in Python

class ProductOfNumbers:
  def __init__(self):
    self.prefix = [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:
    return 0 if k >= len(self.prefix) else self.prefix[-1] // self.prefix[len(self.prefix) - k - 1]
 # code by PROGIEZ

Additional Resources

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