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
["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.

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.

Python3JavaC++Go

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()
# obj.add(num)
# param_2 = obj.getProduct(k)

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

    public ProductOfNumbers() {
        s.add(1);
    }

    public void add(int num) {
        if (num == 0) {
            s.clear();
            s.add(1);
            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();
 * obj.add(num);
 * 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();
 * obj->add(num);
 * int 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();
 * obj.Add(num);
 * param_2 := obj.GetProduct(k);
 */

1352. Product of the Last K Numbers

Description

Solutions

Solution 1: Prefix Product

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