1352. Product of the Last K Numbers

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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:

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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 * 10^4
  • At most 4 * 10^4 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?

Hints/Notes

Solution

Language: C++

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class ProductOfNumbers {
public:
    vector<int> preSum;

    ProductOfNumbers() {
        preSum.push_back(1);
    }

    void add(int num) {
        if (num) {
            preSum.push_back(preSum.back() * num);
        } else {
            preSum = {1};
        }
    }

    int getProduct(int k) {
        int n = preSum.size();
        if (k >= n) {
            return 0;
        }
        return preSum.back() / preSum[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);
 */