14_Product_of_the_Last_K_Numbers.cpp

// 1352. 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?

class ProductOfNumbers
{
public:
    vector<int> list;
    int prod = 1;

    ProductOfNumbers() {}

    void add(int num)
    {
        if (num == 0)
        {
            list.clear();
            prod = 1;
        }
        else
        {
            prod *= num;
            list.push_back(prod);
        }
    }

    int getProduct(int k)
    {
        if (list.size() < k)
            return 0;
        if (list.size() == k)
            return list.back();
        return list.back() / list[list.size() - k - 1];
    }
};

/*
This code implements a class that maintains a running product of numbers in a stream. 
Instead of storing the actual numbers, it stores the running product at each position.
When a number is added:
- If it's 0, the list is cleared since any future product involving 0 will be 0
- Otherwise, multiply it with previous product and store
For getting product of last k numbers:
- If list size < k, means there was a 0, so return 0
- If list size = k, return the last product directly
- Otherwise, divide the last product by the product at (size-k-1) position to get last k numbers' product
This gives O(1) time complexity for both operations.
*/