LeetCode in Kotlin

1352. Product of the Last K Numbers

Medium

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.

Solution

class ProductOfNumbers {
    private var ints = ArrayList<Int>()

    fun add(num: Int) {
        if (num == 0) ints.clear() else ints.add(if (ints.isEmpty()) num else num * ints[ints.size - 1])
    }

    fun getProduct(k: Int): Int {
        val n = ints.size
        if (k > n) return 0
        return if (k == n) ints[n - 1] else ints[n - 1] / ints[n - 1 - k]
    }
}
/*
 * Your ProductOfNumbers object will be instantiated and called as such:
 * var obj = ProductOfNumbers()
 * obj.add(num)
 * var param_2 = obj.getProduct(k)
 */

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