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Checkout Facade

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

배우게 될 내용

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

문제

번역본 보기

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.

제약 조건

1 ≤ n ≤ 1,000

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

예시

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

예상 출력:
expected = 10
입력:
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};

예상 출력:
long long expected = 10;
입력:
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};

예상 출력:
long expected = 10;

설명

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

예시 2
입력:
cart = [
    [1, 1],
    [1, 2]
]
stock = [5, 2]
prices = [3, 20]

예상 출력:
expected = -1
입력:
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};

예상 출력:
long long expected = -1;
입력:
long[][] cart = new long[][] {
    new long[] {1, 1},
    new long[] {1, 2}
};
long[] stock = new long[] {5, 2};
long[] prices = new long[] {3, 20};

예상 출력:
long expected = -1;

설명

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

접근 방법

먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.

힌트 1

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

힌트 2

Compare those totals with the stock before computing any price.

힌트 3

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

예상 복잡도

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

참고 자료

이 문제는 CodeUtility에 맞게 수정되었습니다. 위의 설명과 제한 조건을 따르세요.

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