LeetCode 1352 Product Of The Last K Numbers - Medium

1352. Product Of The Last K Numbers — Medium

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Problem

1352. Product of the Last K Numbers -- Medium

Implement the class ProductOfNumbers that supports two methods:

1. add(int num)

Adds the number num to the back of the current list of numbers. 2. 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. 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:

There will be at most 40000 operations considering both add and getProduct. 0 <= num <= 100 1 <= k <= 40000

Solution

# Prefix Product:
class ProductOfNumbers:
    
    def __init__(self):
        self.stack = [1]
 
    def add(self, num: int) -> None:
        if num == 0:
            self.stack = [1]
        else:
            self.stack.append(self.stack[-1] * num)
 
    def getProduct(self, k: int) -> int:
        if k >= len(self.stack):
            return 0
        return self.stack[-1] // self.stack[-k - 1]
 
 
# Your ProductOfNumbers object will be instantiated and called as such:
# obj = ProductOfNumbers()
# obj.add(num)
# param_2 = obj.getProduct(k)