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:

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: Prefix Sum

• Time complexity: O(1)
• Space complexity: O(add calls)

 

JavaScript

const ProductOfNumbers = function () {
  this.prefixProduct = [1];
};

/**
 * @param {number} num
 * @return {void}
 */
ProductOfNumbers.prototype.add = function (num) {
  if (num === 0) {
    this.prefixProduct = [1];
    return;
  }
  const n = this.prefixProduct.length;
  const prevProduct = this.prefixProduct[n - 1];

  this.prefixProduct.push(prevProduct * num);
};

/**
 * @param {number} k
 * @return {number}
 */
ProductOfNumbers.prototype.getProduct = function (k) {
  const n = this.prefixProduct.length;

  if (k >= n) return 0;

  return this.prefixProduct[n - 1] / this.prefixProduct[n - k - 1];
};

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