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1352. Product of the Last K Numbers

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Description

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

Solutions

Python3

class ProductOfNumbers:

    def __init__(self):
        self.pre_product = []

    def add(self, num: int) -> None:
        if num == 0:
            self.pre_product = []
            return
        if not self.pre_product:
            self.pre_product.append(1)
        self.pre_product.append(num * self.pre_product[-1])

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


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

Java

class ProductOfNumbers {
    private List<Integer> preProduct;

    public ProductOfNumbers() {
        preProduct = new ArrayList<>();
    }

    public void add(int num) {
        if (num == 0) {
            preProduct.clear();
            return;
        }
        if (preProduct.isEmpty()) {
            preProduct.add(1);
        }
        preProduct.add(num * preProduct.get(preProduct.size() - 1));
    }

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

/**
 * Your ProductOfNumbers object will be instantiated and called as such:
 * ProductOfNumbers obj = new ProductOfNumbers();
 * obj.add(num);
 * int param_2 = obj.getProduct(k);
 */

...