Formatted question description: https://leetcode.ca/all/1352.html

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

Related Topics:
Array, Design

Solution 1. Prefix Product

// OJ: https://leetcode.com/problems/product-of-the-last-k-numbers/

// Time: O(1) for all functions
// Space: O(N)
class ProductOfNumbers {
    vector<int> v{1};
public:
    ProductOfNumbers() {}
    
    void add(int num) {
        if (num) v.push_back(v.back() * num);
        else v = {1};
    }
    
    int getProduct(int k) {
        return k < v.size() ? v.back() / v[v.size() - k - 1] : 0;
    }
};

All Problems

All Solutions