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. 7th 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.
Solutions
Solution 1: Hash Table + Simulation
Use a hash table $d$ to store the product ID and price, then traverse the product ID and quantity, calculate the total price, and then calculate the price after discount based on the discount.
The time complexity of initialization is $O(n)$, where $n$ is the number of products. The time complexity of the getBill
function is $O(m)$, where $m$ is the number of products purchased. The space complexity is $O(n)$.
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