Problème
The facade pattern gives a simple entry point to a complicated subsystem. A shop's checkout call hides two subsystems: the inventory and the price list.
Items are numbered 0 to n − 1. stock[i] is how many units of item i are available and prices[i] is its price. The cart is a list of [item, quantity] lines; the same item may appear on several lines.
checkout(stock, prices, cart) must:
- Check the inventory: if the total quantity of any item in the cart is greater than its stock, the order fails.
- Otherwise return the total price of the cart.
Implement checkout_facade(stock, prices, cart).
Input
stock, prices: n values each.
cart: [item, quantity] lines.
Output
Return the total price, or -1 if any item does not have enough stock.
Contraintes
1 ≤ n ≤ 1,000
0 ≤ stock[i], prices[i] ≤ 100,000; 0 ≤ cart lines ≤ 1,000; 1 ≤ quantity ≤ 1,000
Exemples
Entrée :
cart = [
[0, 2],
[2, 4]
]
stock = [5, 2, 10]
prices = [3, 20, 1]
Sortie attendue :
expected = 10Entrée :
std::vector<std::vector<long long>> cart = {
{0, 2},
{2, 4}
};
std::vector<long long> stock = {5, 2, 10};
std::vector<long long> prices = {3, 20, 1};
Sortie attendue :
long long expected = 10;Entrée :
long[][] cart = new long[][] {
new long[] {0, 2},
new long[] {2, 4}
};
long[] stock = new long[] {5, 2, 10};
long[] prices = new long[] {3, 20, 1};
Sortie attendue :
long expected = 10;Explication
Both items are in stock. 2 × 3 + 4 × 1 = 10.
Entrée :
cart = [
[1, 1],
[1, 2]
]
stock = [5, 2]
prices = [3, 20]
Sortie attendue :
expected = -1Entrée :
std::vector<std::vector<long long>> cart = {
{1, 1},
{1, 2}
};
std::vector<long long> stock = {5, 2};
std::vector<long long> prices = {3, 20};
Sortie attendue :
long long expected = -1;Entrée :
long[][] cart = new long[][] {
new long[] {1, 1},
new long[] {1, 2}
};
long[] stock = new long[] {5, 2};
long[] prices = new long[] {3, 20};
Sortie attendue :
long expected = -1;Explication
Item 1 appears twice for a total of 3 units, but only 2 are in stock, so the order fails: -1.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
Add up the quantity of each item over all cart lines first.
Indice 2
Compare those totals with the stock before computing any price.
Indice 3
If any item is short, return -1 straight away.
Complexité attendue
O(n + cart lines) time, O(n) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.