Apply Discount Every n Orders - Practice Coding Problems

Apply Discount Every n Orders - Problem

🛒 Supermarket Cashier System

You're implementing a smart cashier system for a busy supermarket! The supermarket has decided to run an exciting promotion: every n-th customer gets a discount on their purchase.

Here's how it works:

Each product has a unique ID and a price
When customers check out, they bring multiple products with different quantities
Every n-th customer (3rd, 6th, 9th, etc.) gets a percentage discount on their total bill
Other customers pay the full amount

Your task: Design a Cashier class that tracks customer orders and applies discounts automatically!

Example: If n=3 and discount=50%, then the 3rd, 6th, 9th customers get 50% off their bills.

The class needs two methods:

Cashier(n, discount, products[], prices[]) - Initialize the system
getBill(product[], amount[]) - Calculate final bill for current customer

Input & Output

example_1.py — Basic Usage

$ Input: Cashier cashier = new Cashier(3, 50, [1,2,3,4,5,6,7], [100,200,300,400,500,600,700]); cashier.getBill([1,2], [1,2]); // Customer 1 cashier.getBill([3,7], [10,10]); // Customer 2 cashier.getBill([1,2,3,4,5,6,7], [1,1,1,1,1,1,1]); // Customer 3

› Output: 500.0 10000.0 1400.0

💡 Note: Customer 1: 1×100 + 2×200 = 500 (no discount). Customer 2: 10×300 + 10×700 = 10000 (no discount). Customer 3: 100+200+300+400+500+600+700 = 2800, but gets 50% discount = 1400.

example_2.py — Small Discount

$ Input: Cashier cashier = new Cashier(2, 10, [1,2], [5,3]); cashier.getBill([1], [2]); // Customer 1 cashier.getBill([2], [3]); // Customer 2

› Output: 10.0 8.1

💡 Note: Customer 1: 2×5 = 10 (no discount). Customer 2: 3×3 = 9, gets 10% discount = 9×0.9 = 8.1.

example_3.py — Multiple Discount Customers

$ Input: Cashier cashier = new Cashier(3, 25, [1,2,3], [10,20,30]); cashier.getBill([1], [1]); // Customer 1 cashier.getBill([2], [1]); // Customer 2 cashier.getBill([3], [1]); // Customer 3 (discount) cashier.getBill([1,2], [1,1]); // Customer 4 cashier.getBill([2,3], [2,1]); // Customer 5 cashier.getBill([1,2,3], [1,1,1]); // Customer 6 (discount)

› Output: 10.0 20.0 22.5 30.0 70.0 45.0

💡 Note: Customers 3 and 6 get 25% discount. Customer 3: 30×0.75=22.5. Customer 6: (10+20+30)×0.75=45.0.

Visualization

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Understanding the Visualization

System Setup

Store creates price lookup table and sets promotion rules (every nth customer gets discount)

Fast Checkout

Customer arrives with cart - system instantly looks up each item's price using hash map

Smart Discount

System checks customer count and automatically applies discount if they're the lucky nth customer

Key Takeaway

🎯 Key Insight: Hash maps transform O(m) searches into O(1) lookups, making the cashier system blazingly fast even with large product catalogs!

Time & Space Complexity

Time Complexity

⏱️

O(k)

k is number of items in customer's cart - each price lookup is O(1) with hash map

✓ Linear Growth

Space Complexity

O(m)

m is number of products in catalog - we store all product-price pairs in hash map

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁴
0 ≤ discount ≤ 100
1 ≤ products.length ≤ 200
1 ≤ products[i] ≤ 200
1 ≤ prices[i] ≤ 1000
products contains unique values
1 ≤ product.length ≤ products.length
product[i] exists in products
1 ≤ amount[i] ≤ 1000
At most 300 calls will be made to getBill

Asked in

a Amazon 45 G Google 32 ⊞ Microsoft 28 f Meta 22

The optimal solution uses a hash map to store product-price mappings for O(1) lookups and a simple counter with modulo to track discount eligibility. This achieves O(k) time per checkout where k is the cart size, compared to O(m×k) for the brute force approach.

Common Approaches

Approach	Time	Space	Notes
✓ Hash Map for Price Lookup (Optimal)	O(k)	O(m)	Use hash map for O(1) price lookups and simple counter for discounts
Linear Search for Prices	O(m × k)	O(1)	Search through products array for each item's price

Hash Map for Price Lookup (Optimal) — Algorithm Steps

In constructor, create hash map from product IDs to prices
Initialize customer counter to 0
For each getBill call, increment customer counter
For each product in cart, lookup price in O(1) time using hash map
Calculate subtotal by summing (price × quantity) for all items
Check if customer number is divisible by n for discount
Apply discount percentage if applicable and return final amount

Visualization

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Step-by-Step Walkthrough

Build hash map

During initialization, map all product IDs to prices

Instant lookup

For each cart item, get price in O(1) time

Apply discount

Check customer number and apply discount if needed

Code -

solution.c — C

#include <stdlib.h>

#define HASH_SIZE 1009

typedef struct HashNode {
    int productId;
    int price;
    struct HashNode* next;
} HashNode;

typedef struct {
    int n, discount, customerCount;
    HashNode* priceMap[HASH_SIZE];
} Cashier;

int hash(int key) {
    return key % HASH_SIZE;
}

void insertPrice(Cashier* obj, int productId, int price) {
    int index = hash(productId);
    HashNode* node = (HashNode*)malloc(sizeof(HashNode));
    node->productId = productId;
    node->price = price;
    node->next = obj->priceMap[index];
    obj->priceMap[index] = node;
}

int getPrice(Cashier* obj, int productId) {
    int index = hash(productId);
    HashNode* node = obj->priceMap[index];
    while (node) {
        if (node->productId == productId) {
            return node->price;
        }
        node = node->next;
    }
    return 0;
}

Cashier* cashierCreate(int n, int discount, int* products, int productsSize, int* prices, int pricesSize) {
    Cashier* obj = (Cashier*)malloc(sizeof(Cashier));
    obj->n = n;
    obj->discount = discount;
    obj->customerCount = 0;
    
    // Initialize hash map
    for (int i = 0; i < HASH_SIZE; i++) {
        obj->priceMap[i] = NULL;
    }
    
    // Create hash map for O(1) price lookups
    for (int i = 0; i < productsSize; i++) {
        insertPrice(obj, products[i], prices[i]);
    }
    
    return obj;
}

double cashierGetBill(Cashier* obj, int* product, int productSize, int* amount, int amountSize) {
    obj->customerCount++;
    double subtotal = 0.0;
    
    // Calculate subtotal using O(1) hash map lookups
    for (int i = 0; i < productSize; i++) {
        int productId = product[i];
        int quantity = amount[i];
        int price = getPrice(obj, productId);
        subtotal += price * quantity;
    }
    
    // Apply discount if this is nth customer
    if (obj->customerCount % obj->n == 0) {
        subtotal = subtotal * (100 - obj->discount) / 100.0;
    }
    
    return subtotal;
}

void cashierFree(Cashier* obj) {
    for (int i = 0; i < HASH_SIZE; i++) {
        HashNode* node = obj->priceMap[i];
        while (node) {
            HashNode* temp = node;
            node = node->next;
            free(temp);
        }
    }
    free(obj);
}

Time & Space Complexity

Time Complexity

⏱️

O(k)

k is number of items in customer's cart - each price lookup is O(1) with hash map

✓ Linear Growth

Space Complexity

O(m)

m is number of products in catalog - we store all product-price pairs in hash map

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁴
0 ≤ discount ≤ 100
1 ≤ products.length ≤ 200
1 ≤ products[i] ≤ 200
1 ≤ prices[i] ≤ 1000
products contains unique values
1 ≤ product.length ≤ products.length
product[i] exists in products
1 ≤ amount[i] ≤ 1000
At most 300 calls will be made to getBill

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🛒 Supermarket Cashier System

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Hash Map for Price Lookup (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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