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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 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.

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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class Cashier:
    def __init__(self, n: int, discount: int, products: List[int], prices: List[int]):
        self.i = 0
        self.n = n
        self.discount = discount
        self.d = {product: price for product, price in zip(products, prices)}

    def getBill(self, product: List[int], amount: List[int]) -> float:
        self.i += 1
        discount = self.discount if self.i % self.n == 0 else 0
        ans = 0
        for p, a in zip(product, amount):
            x = self.d[p] * a
            ans += x - (discount * x) / 100
        return ans


# Your Cashier object will be instantiated and called as such:
# obj = Cashier(n, discount, products, prices)
# param_1 = obj.getBill(product,amount)
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class Cashier {
    private int i;
    private int n;
    private int discount;
    private Map<Integer, Integer> d = new HashMap<>();

    public Cashier(int n, int discount, int[] products, int[] prices) {
        this.n = n;
        this.discount = discount;
        for (int j = 0; j < products.length; ++j) {
            d.put(products[j], prices[j]);
        }
    }

    public double getBill(int[] product, int[] amount) {
        int dis = (++i) % n == 0 ? discount : 0;
        double ans = 0;
        for (int j = 0; j < product.length; ++j) {
            int p = product[j], a = amount[j];
            int x = d.get(p) * a;
            ans += x - (dis * x) / 100.0;
        }
        return ans;
    }
}

/**
 * Your Cashier object will be instantiated and called as such:
 * Cashier obj = new Cashier(n, discount, products, prices);
 * double param_1 = obj.getBill(product,amount);
 */
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class Cashier {
public:
    Cashier(int n, int discount, vector<int>& products, vector<int>& prices) {
        this->n = n;
        this->discount = discount;
        for (int j