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Leetcode1352-Product Of The Last K Numbers

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

Implement the class ProductOfNumbers that supports two methods:

  1. add(int num)

Adds the number num to the back of the current list of numbers.

  1. 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.
At any time, the product of any contiguous sequence of numbers will fit into a single 32-bit integer without overflowing.

Example

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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:

  • There will be at most 40000 operations considering both add and getProduct.
  • 0 <= num <= 100
  • 1 <= k <= 40000

Solution

  • Time Complexity:
  • Space Complexity: 
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class ProductOfNumbers {

    private List<Integer> prefixProduction;
    public ProductOfNumbers() {
        prefixProduction = new ArrayList<>();
        prefixProduction.add(1);
    }
    
    public void add(int num) {
        if (num > 0) {
            prefixProduction.add(prefixProduction.get(prefixProduction.size() - 1) * num);
        }
        else {
            prefixProduction = new ArrayList<>();
            prefixProduction.add(1);
        }
    }
    
    public int getProduct(int k) {
        int len = prefixProduction.size();
        return k >= len ? 0 : prefixProduction.get(len - 1) / prefixProduction.get(len - 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);
 */