1357. Apply Discount Every n Orders
Description
There is a supermarket that is frequented by many customers. The products sold at the supermarket are represented as two parallel integer arrays products
and prices
, where the ith
product has an ID of products[i]
and a price of prices[i]
.
When a customer is paying, their bill is represented as two parallel integer arrays product
and amount
, where the jth
product they purchased has an ID of product[j]
, and amount[j]
is how much of the product they bought. Their subtotal is calculated as the sum of each amount[j] * (price of the jth product)
.
The supermarket decided to have a sale. Every nth
customer paying for their groceries will be given a percentage discount. The discount amount is given by discount
, where they will be given discount
percent off their subtotal. More formally, if their subtotal is bill
, then they would actually pay bill * ((100 - discount) / 100)
.
Implement the Cashier
class:
Cashier(int n, int discount, int[] products, int[] prices)
Initializes the object withn
, thediscount
, and theproducts
and theirprices
.double getBill(int[] product, int[] amount)
Returns the final total of the bill with the discount applied (if any). Answers within10-5
of the actual value will be accepted.
Example 1:
Input ["Cashier","getBill","getBill","getBill","getBill","getBill","getBill","getBill"] [[3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]],[[1,2],[1,2]],[[3,7],[10,10]],[[1,2,3,4,5,6,7],[1,1,1,1,1,1,1]],[[4],[10]],[[7,3],[10,10]],[[7,5,3,1,6,4,2],[10,10,10,9,9,9,7]],[[2,3,5],[5,3,2]]] Output [null,500.0,4000.0,800.0,4000.0,4000.0,7350.0,2500.0] Explanation Cashier cashier = new Cashier(3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]); cashier.getBill([1,2],[1,2]); // return 500.0. 1st customer, no discount. // bill = 1 * 100 + 2 * 200 = 500. cashier.getBill([3,7],[10,10]); // return 4000.0. 2nd customer, no discount. // bill = 10 * 300 + 10 * 100 = 4000. cashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]); // return 800.0. 3rd customer, 50% discount. // Original bill = 1600 // Actual bill = 1600 * ((100 - 50) / 100) = 800. cashier.getBill([4],[10]); // return 4000.0. 4th customer, no discount. cashier.getBill([7,3],[10,10]); // return 4000.0. 5th customer, no discount. cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0. 6th customer, 50% discount. // Original bill = 14700, but with // Actual bill = 14700 * ((100 - 50) / 100) = 7350. cashier.getBill([2,3,5],[5,3,2]); // return 2500.0. 6th customer, no discount.
Constraints:
1 <= n <= 104
0 <= discount <= 100
1 <= products.length <= 200
prices.length == products.length
1 <= products[i] <= 200
1 <= prices[i] <= 1000
- The elements in
products
are unique. 1 <= product.length <= products.length
amount.length == product.length
product[j]
exists inproducts
.1 <= amount[j] <= 1000
- The elements of
product
are unique. - At most
1000
calls will be made togetBill
. - Answers within
10-5
of the actual value will be accepted.
Solution
class Cashier:
def __init__(self, n: int, discount: int, products: List[int], prices: List[int]):
self.n = n
self.count = 0
self.discount = discount
self.products = products
self.prices = prices
def getBill(self, P: List[int], A: List[int]) -> float:
self.count += 1
price = 0
for p,a in zip(P,A):
price += self.prices[self.products.index(p)] * a
if self.count%self.n == 0:
price -= (price * (self.discount/100))
return price
# Your Cashier object will be instantiated and called as such:
# obj = Cashier(n, discount, products, prices)
# param_1 = obj.getBill(product,amount)