LeetCode #1357 — MEDIUM

Apply Discount Every n Orders

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 i^th 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 j^th 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 j^th product).

The supermarket decided to have a sale. Every n^th 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. 1^st customer, no discount.
                                                     // bill = 1 * 100 + 2 * 200 = 500.
cashier.getBill([3,7],[10,10]);                      // return 4000.0. 2^nd 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. 3^rd customer, 50% discount.
                                                     // Original bill = 1600
                                                     // Actual bill = 1600 * ((100 - 50) / 100) = 800.
cashier.getBill([4],[10]);                           // return 4000.0. 4^th customer, no discount.
cashier.getBill([7,3],[10,10]);                      // return 4000.0. 5^th customer, no discount.
cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0. 6^th 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.  7^th 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.

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Design

Example 1

["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]]]

Core Insight

What unlocks the optimal approach

Keep track of the count of the customers.
Check if the count of the customers is divisible by n then apply the discount formula.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1357: Apply Discount Every n Orders
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);
 */

// Accepted solution for LeetCode #1357: Apply Discount Every n Orders
type Cashier struct {
	i        int
	n        int
	discount int
	d        map[int]int
}

func Constructor(n int, discount int, products []int, prices []int) Cashier {
	d := map[int]int{}
	for i, product := range products {
		d[product] = prices[i]
	}
	return Cashier{0, n, discount, d}
}

func (this *Cashier) GetBill(product []int, amount []int) (ans float64) {
	this.i++
	dis := 0
	if this.i%this.n == 0 {
		dis = this.discount
	}
	for j, p := range product {
		x := float64(this.d[p] * amount[j])
		ans += x - (float64(dis)*x)/100.0
	}
	return
}

/**
 * Your Cashier object will be instantiated and called as such:
 * obj := Constructor(n, discount, products, prices);
 * param_1 := obj.GetBill(product,amount);
 */

# Accepted solution for LeetCode #1357: Apply Discount Every n Orders
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)

// Accepted solution for LeetCode #1357: Apply Discount Every n Orders
use std::collections::HashMap;

struct Cashier {
    n: usize,
    index: usize,
    discount: f64,
    inventory: HashMap<i32, f64>,
}

impl Cashier {
    fn new(n: i32, discount: i32, products: Vec<i32>, prices: Vec<i32>) -> Self {
        let n = n as usize;
        let index = 0;
        let discount = (100 - discount) as f64 / 100.0;
        let mut inventory: HashMap<i32, f64> = HashMap::new();
        for (id, price) in products.into_iter().zip(prices.into_iter()) {
            inventory.insert(id, price as f64);
        }
        Cashier {
            n,
            index,
            discount,
            inventory,
        }
    }
    fn get_bill(&mut self, products: Vec<i32>, amount: Vec<i32>) -> f64 {
        let mut res = 0.0;
        for (id, amount) in products.into_iter().zip(amount.into_iter()) {
            res += self.inventory[&id] * amount as f64;
        }
        self.index += 1;
        if self.index == self.n {
            self.index = 0;
            res * self.discount
        } else {
            res
        }
    }
}

#[test]
fn test() {
    use assert_approx_eq::assert_approx_eq;
    let mut cashier = Cashier::new(
        3,
        50,
        vec![1, 2, 3, 4, 5, 6, 7],
        vec![100, 200, 300, 400, 300, 200, 100],
    );
    assert_approx_eq!(cashier.get_bill(vec![1, 2], vec![1, 2]), 500.0);
    assert_approx_eq!(cashier.get_bill(vec![3, 7], vec![10, 10]), 4000.0);
    assert_approx_eq!(
        cashier.get_bill(vec![1, 2, 3, 4, 5, 6, 7], vec![1, 1, 1, 1, 1, 1, 1]),
        800.0
    );
    assert_approx_eq!(cashier.get_bill(vec![4], vec![10]), 4000.0);
    assert_approx_eq!(cashier.get_bill(vec![7, 3], vec![10, 10]), 4000.0);
    assert_approx_eq!(
        cashier.get_bill(vec![7, 5, 3, 1, 6, 4, 2], vec![10, 10, 10, 9, 9, 9, 7]),
        7350.0
    );
    assert_approx_eq!(cashier.get_bill(vec![2, 3, 5], vec![5, 3, 2]), 2500.0);
}

// Accepted solution for LeetCode #1357: Apply Discount Every n Orders
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1357: Apply Discount Every n Orders
// 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);
//  */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.