# 1352. Product of the Last K Numbers

## Description

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
[[],[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.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.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.

## Solutions

Solution 1: Prefix Product

We initialize an array $s$, where $s[i]$ represents the product of the first $i$ numbers.

When calling add(num), we judge whether num is $0$. If it is, we set $s$ to [1]. Otherwise, we multiply the last element of $s$ by num and add the result to the end of $s$.

When calling getProduct(k), we now judge whether the length of $s$ is less than or equal to $k$. If it is, we return $0$. Otherwise, we return the last element of $s$ divided by the $k + 1$th element from the end of $s$. That is, $s[-1] / s[-k - 1]$.

The time complexity is $O(1)$, and the space complexity is $O(n)$. Where $n$ is the number of times add is called.

• class ProductOfNumbers {
private List<Integer> s = new ArrayList<>();

public ProductOfNumbers() {
}

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

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

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

• class ProductOfNumbers {
public:
ProductOfNumbers() {
s.push_back(1);
}

void add(int num) {
if (num == 0) {
s.clear();
s.push_back(1);
return;
}
s.push_back(s.back() * num);
}

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

private:
vector<int> s;
};

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

• class ProductOfNumbers:
def __init__(self):
self.s = [1]

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

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

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


• type ProductOfNumbers struct {
s []int
}

func Constructor() ProductOfNumbers {
return ProductOfNumbers{[]int{1} }
}

func (this *ProductOfNumbers) Add(num int) {
if num == 0 {
this.s = []int{1}
return
}
this.s = append(this.s, this.s[len(this.s)-1]*num)
}

func (this *ProductOfNumbers) GetProduct(k int) int {
n := len(this.s)
if n <= k {
return 0
}
return this.s[len(this.s)-1] / this.s[len(this.s)-k-1]
}

/**
* Your ProductOfNumbers object will be instantiated and called as such:
* obj := Constructor();
* param_2 := obj.GetProduct(k);
*/