Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 14

Description

Product of the Last K Numbers

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 {
  vector<long> result;
  set<int> zeroPos;
public:
  ProductOfNumbers() {
    result.push_back(1);
  }
  
  void add(int num) {
    if(!num) {
      num = 1;
      zeroPos.insert(result.size());
      result.push_back(1);
    } else {
      result.push_back(result.back() * num);
    }
  }
  
  int getProduct(int k) {
    if(zeroPos.lower_bound(result.size() - k) != zeroPos.end()) return 0;
    return result.back() / *(result.rbegin() + k);
  }
};

// Accepted
// 33/33 cases passed (15 ms)
// Your runtime beats 85.88 % of cpp submissions
// Your memory usage beats 5.54 % of cpp submissions (79.5 MB)