1357. Apply Discount Every n Orders LeetCode Solution
In this guide, you will get 1357. Apply Discount Every n Orders LeetCode Solution with the best time and space complexity. The solution to Apply Discount Every n Orders problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Apply Discount Every n Orders
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 with n, the discount, and the products and their prices.
double getBill(int[] product, int[] amount) Returns the final total of the bill with the discount applied (if any). Answers within 10-5 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 in products.
1 <= amount[j] <= 1000
The elements of product are unique.
At most 1000 calls will be made to getBill.
Answers within 10-5 of the actual value will be accepted.
Complexity Analysis
Time Complexity: O(|\texttt{product})
Space Complexity: O(|\texttt{products}|)
1357. Apply Discount Every n Orders LeetCode Solution in C++
class Cashier {
public:
Cashier(int n, int discount, vector<int>& products, vector<int>& prices)
: n(n), discount(discount) {
for (int i = 0; i < products.size(); ++i)
productToPrice[products[i]] = prices[i];
}
double getBill(vector<int> product, vector<int> amount) {
++count;
int total = 0;
for (int i = 0; i < product.size(); ++i)
total += productToPrice[product[i]] * amount[i];
return count % n == 0 ? total * (1 - discount / 100.0) : total;
}
private:
const int n;
const int discount;
unordered_map<int, int> productToPrice;
int count = 0;
}; /* code provided by PROGIEZ */
1357. Apply Discount Every n Orders LeetCode Solution in Java
class Cashier {
public Cashier(int n, int discount, int[] products, int[] prices) {
this.n = n;
this.discount = discount;
for (int i = 0; i < products.length; ++i)
productToPrice.put(products[i], prices[i]);
}
public double getBill(int[] product, int[] amount) {
++count;
int total = 0;
for (int i = 0; i < product.length; ++i)
total += productToPrice.get(product[i]) * amount[i];
return count % n == 0 ? total * (1 - discount / 100.0) : total;
}
private int n;
private int discount;
private Map<Integer, Integer> productToPrice = new HashMap<>();
private int count = 0;
} // code provided by PROGIEZ
1357. Apply Discount Every n Orders LeetCode Solution in Python
class Cashier:
def __init__(
self,
n: int,
discount: int,
products: list[int],
prices: list[int],
):
self.n = n
self.discount = discount
self.productToPrice = dict(zip(products, prices))
self.count = 0
def getBill(self, product: list[int], amount: list[int]) -> float:
self.count += 1
total = sum(self.productToPrice[p] * amount[i]
for i, p in enumerate(product))
if self.count % self.n == 0:
return total * (1 - self.discount / 100)
return total # code by PROGIEZ
