1477. Product Of The Last K Numbers¶

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1477. Product of the Last K Numbers

Medium

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.

Follow-up: Can you implement both GetProduct and Add to work in O(1) time complexity instead of O(k) time complexity?

Solution¶

class ProductOfNumbers {
    static ArrayList<Integer> res; 
    public ProductOfNumbers() {
        res = new ArrayList<>();
    }
    public void add(int num) {
        res.add(num);
    }
    public int getProduct(int k) {
        int sum = 1;
        int len = res.size() - 1;
        while (k > 0) {
            sum = sum * res.get(len--);
            k--;
        }
        return sum;
    }
}

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

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here