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1352-product-of-the-last-k-numbers.py
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1352-product-of-the-last-k-numbers.py
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"""
Problem Link: https://leetcode.com/problems/product-of-the-last-k-numbers/
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.
2. 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:
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
"""
class ProductOfNumbers:
def __init__(self):
self.product = [1]
def add(self, num: int) -> None:
if num == 0:
self.product = [1]
else:
self.product.append(self.product[-1]*num)
def getProduct(self, k: int) -> int:
if k >= len(self.product):
return 0
return self.product[-1] // self.product[-k-1]
# Your ProductOfNumbers object will be instantiated and called as such:
# obj = ProductOfNumbers()
# obj.add(num)
# param_2 = obj.getProduct(k)