โจทย์
LINQ's Join matches items from two sequences by a shared key, much like a database join. Here, customers are matched with their orders by customer id.
Customer i has id customer_ids[i] and name customer_names[i]. Each order is [customer_id, amount].
Join the two: for every customer with at least one order, add up the amounts of their orders. Customers without orders are left out, and orders whose customer id is unknown are ignored.
Implement join_customers_orders(customer_ids, customer_names, orders).
Input
customer_ids, customer_names: The customers (ids are distinct).
orders: [customer_id, amount] pairs.
Output
Return "<name>:<total>" for each customer with orders, sorted by total from highest to lowest, and by name (ordinal order) when totals are equal.
เงื่อนไข
1 ≤ number of customers ≤ 10,000
0 ≤ number of orders ≤ 100,000; 1 ≤ amount ≤ 10^6
ตัวอย่าง
อินพุต:
orders = [
[1, 50],
[2, 30],
[1, 20]
]
customer_ids = [1, 2, 3]
customer_names = ["Ada", "Lin", "Bo"]
ผลลัพธ์ที่คาดหวัง:
expected = ["Ada:70", "Lin:30"]อินพุต:
std::vector<std::vector<long long>> orders = {
{1, 50},
{2, 30},
{1, 20}
};
std::vector<long long> customer_ids = {1, 2, 3};
std::vector<std::string> customer_names = {"Ada", "Lin", "Bo"};
ผลลัพธ์ที่คาดหวัง:
std::vector<std::string> expected = {"Ada:70", "Lin:30"};อินพุต:
long[][] orders = new long[][] {
new long[] {1, 50},
new long[] {2, 30},
new long[] {1, 20}
};
long[] customer_ids = new long[] {1, 2, 3};
string[] customer_names = new string[] {"Ada", "Lin", "Bo"};
ผลลัพธ์ที่คาดหวัง:
string[] expected = new string[] {"Ada:70", "Lin:30"};คำอธิบาย
Ada has two orders (50 + 20 = 70) and Lin one (30); Bo has none and is left out: ["Ada:70", "Lin:30"].
อินพุต:
orders = [
[8, 40],
[7, 40],
[9, 99]
]
customer_ids = [7, 8]
customer_names = ["Zoe", "Max"]
ผลลัพธ์ที่คาดหวัง:
expected = ["Max:40", "Zoe:40"]อินพุต:
std::vector<std::vector<long long>> orders = {
{8, 40},
{7, 40},
{9, 99}
};
std::vector<long long> customer_ids = {7, 8};
std::vector<std::string> customer_names = {"Zoe", "Max"};
ผลลัพธ์ที่คาดหวัง:
std::vector<std::string> expected = {"Max:40", "Zoe:40"};อินพุต:
long[][] orders = new long[][] {
new long[] {8, 40},
new long[] {7, 40},
new long[] {9, 99}
};
long[] customer_ids = new long[] {7, 8};
string[] customer_names = new string[] {"Zoe", "Max"};
ผลลัพธ์ที่คาดหวัง:
string[] expected = new string[] {"Max:40", "Zoe:40"};คำอธิบาย
Both total 40, so they are ordered by name; the order for unknown id 9 is ignored: ["Max:40", "Zoe:40"].
แนวทางการแก้
ลองทำเองก่อน แล้วค่อยเปิดคำใบ้ทีละข้อ
คำใบ้ 1
First add up the order amounts per customer id.
คำใบ้ 2
Then go through the customers and keep those that have a total.
คำใบ้ 3
Sort by total descending, then by name in ordinal order.
ความซับซ้อนที่คาดหวัง
O(n log n) time, O(n) extra space
แหล่งอ้างอิง
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