LeetCode 1352. Product of the Last K Numbers Solution in Java, C++, Python & More | Explanation + Code

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.

 

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

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 + 1th 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.

PythonJavaC++GoTypeScriptJavaScript

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)(code-box)

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); */(code-box)

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); */(code-box)

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); */(code-box)

class ProductOfNumbers { s = [1]; add(num: number): void { if (num === 0) { this.s = [1]; } else { const i = this.s.length; this.s[i] = this.s[i - 1] * num; } } getProduct(k: number): number { const i = this.s.length; if (k > i - 1) return 0; return this.s[i - 1] / this.s[i - k - 1]; } }(code-box)

class ProductOfNumbers { s = [1]; add(num) { if (num === 0) { this.s = [1]; } else { const i = this.s.length; this.s[i] = this.s[i - 1] * num; } } getProduct(k) { const i = this.s.length; if (k > i - 1) return 0; return this.s[i - 1] / this.s[i - k - 1]; } }(code-box)