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Checkout Facade

Hide stock checks and price totals behind one simple checkout call.

Das lernst du

  • Hide several subsystems behind one call
  • Check stock before charging
  • Combine repeated cart lines per item

Problem

Übersetzung anzeigen

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:

  1. Check the inventory: if the total quantity of any item in the cart is greater than its stock, the order fails.
  2. 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.

Einschränkungen

1 ≤ n ≤ 1,000

0 ≤ stock[i], prices[i] ≤ 100,000; 0 ≤ cart lines ≤ 1,000; 1 ≤ quantity ≤ 1,000

Beispiele

Beispiel 1
Eingabe:
cart = [
    [0, 2],
    [2, 4]
]
stock = [5, 2, 10]
prices = [3, 20, 1]

Erwartete Ausgabe:
expected = 10
Eingabe:
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};

Erwartete Ausgabe:
long long expected = 10;
Eingabe:
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};

Erwartete Ausgabe:
long expected = 10;

Erläuterung

Both items are in stock. 2 × 3 + 4 × 1 = 10.

Beispiel 2
Eingabe:
cart = [
    [1, 1],
    [1, 2]
]
stock = [5, 2]
prices = [3, 20]

Erwartete Ausgabe:
expected = -1
Eingabe:
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};

Erwartete Ausgabe:
long long expected = -1;
Eingabe:
long[][] cart = new long[][] {
    new long[] {1, 1},
    new long[] {1, 2}
};
long[] stock = new long[] {5, 2};
long[] prices = new long[] {3, 20};

Erwartete Ausgabe:
long expected = -1;

Erläuterung

Item 1 appears twice for a total of 3 units, but only 2 are in stock, so the order fails: -1.

So gehst du vor

Versuch es zuerst allein und öffne dann einen Hinweis nach dem anderen.

Hinweis 1

Add up the quantity of each item over all cart lines first.

Hinweis 2

Compare those totals with the stock before computing any price.

Hinweis 3

If any item is short, return -1 straight away.

Erwartete Komplexität

O(n + cart lines) time, O(n) extra space

Referenzen

Diese Aufgabe wurde für CodeUtility angepasst. Beachte die oben genannten Vorgaben und Einschränkungen.

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