# Solving Leetcode Interviews in Seconds with AI: Product of the Last K Numbers


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1352" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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?

	# Explanation
	Here's a breakdown of the approach, complexity, and the Python implementation:

*   **High-Level Approach:**
    *   Maintain a prefix product list. Each element at index `i` in the list stores the product of all numbers seen so far up to index `i` in the stream.
    *   If a `0` is encountered in the stream, reset the prefix product list.  This is because any subsequent product calculation involving this `0` will be `0`. Restart prefix product calculation from the next number.
    *   `getProduct(k)` can be answered in O(1) time by dividing the last prefix product by the prefix product `k` elements before. Handle the edge case where a zero exists within the last `k` elements.

*   **Complexity:**
    *   Runtime Complexity: `add` is O(1), `getProduct` is O(1).
    *   Storage Complexity: O(N), where N is the number of elements added to the stream.

	
	# Code
	```python
	class ProductOfNumbers:

    def __init__(self):
        self.prefix_products = [1]  # Initialize with 1 for the base case of empty stream

    def add(self, num: int) -> None:
        if num == 0:
            self.prefix_products = [1]  # Reset if we encounter 0
        else:
            self.prefix_products.append(self.prefix_products[-1] * num)

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


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