main.go

// Source: https://leetcode.com/problems/product-of-the-last-k-numbers
// Title: Product of the Last K Numbers
// Difficulty: Medium
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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 * 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?
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

// Use prefix products
// When add zero, reset the struct
type ProductOfNumbers struct {
	prefixs []int // prefix product
	total   int   // total product
}

func Constructor() ProductOfNumbers {
	return ProductOfNumbers{
		total: 1,
	}
}

func (this *ProductOfNumbers) Add(num int) {
	if num == 0 {
		this.total = 1
		this.prefixs = nil
	} else {
		this.prefixs = append(this.prefixs, this.total)
		this.total *= num
	}
}

func (this *ProductOfNumbers) GetProduct(k int) int {
	if k > len(this.prefixs) {
		return 0
	}
	return this.total / this.prefixs[len(this.prefixs)-k]
}