समस्या
The decorator pattern adds behaviour by wrapping an object in layers. Each layer here changes the price of the layer inside it.
Start from base_price and apply the decorators in order:
discount p: reduce the price by p percent.tax p: increase the price by p percent.shipping a: add a fixed amount a.coupon a: subtract a, but never go below 0.
Implement price_decorators(base_price, decorators).
Input
base_price: The starting price.
decorators: The layers, in the order they wrap the price. Amounts may have decimals.
Output
Return the final price. Answers within 1e-6 are accepted.
सीमाएँ
0 ≤ base_price ≤ 1,000,000
0 ≤ number of decorators ≤ 100; 0 ≤ p ≤ 100, 0 ≤ a ≤ 100,000
उदाहरण
इनपुट:
base_price = 100
decorators = ["discount 10", "tax 8", "shipping 5"]
अपेक्षित आउटपुट:
expected = 102.2इनपुट:
long long base_price = 100;
std::vector<std::string> decorators = {"discount 10", "tax 8", "shipping 5"};
अपेक्षित आउटपुट:
double expected = 102.2;इनपुट:
long base_price = 100;
string[] decorators = new string[] {"discount 10", "tax 8", "shipping 5"};
अपेक्षित आउटपुट:
double expected = 102.2;व्याख्या
100 − 10% = 90, plus 8% tax = 97.2, plus 5 shipping = 102.2.
इनपुट:
base_price = 20
decorators = ["coupon 25", "shipping 4"]
अपेक्षित आउटपुट:
expected = 4.0इनपुट:
long long base_price = 20;
std::vector<std::string> decorators = {"coupon 25", "shipping 4"};
अपेक्षित आउटपुट:
double expected = 4.0;इनपुट:
long base_price = 20;
string[] decorators = new string[] {"coupon 25", "shipping 4"};
अपेक्षित आउटपुट:
double expected = 4.0;व्याख्या
The coupon would take the price below zero, so it stops at 0; shipping then makes it 4.
कैसे हल करें
पहले खुद कोशिश करें, फिर एक-एक करके संकेत खोलें।
संकेत 1
Apply the layers strictly in the given order.
संकेत 2
Percentages are taken from the current price, not the base price.
संकेत 3
A coupon can never push the price below 0.
अपेक्षित जटिलता
O(number of decorators) time, O(1) extra space
संदर्भ
यह चुनौती CodeUtility के लिए अनुकूलित की गई है। ऊपर दिए गए स्टेटमेंट और सीमाओं का पालन करें।