問題
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.
制約
1 ≤ number of customers ≤ 10,000
0 ≤ number of orders ≤ 100,000; 1 ≤ amount ≤ 10^6
例
入力：
orders = [
[1, 50],
[2, 30],
[1, 20]
]
customer_ids = [1, 2, 3]
customer_names = ["Ada", "Lin", "Bo"]
期待される出力：
expected = ["Ada:70", "Lin:30"]入力：
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"};
期待される出力：
std::vector<std::string> expected = {"Ada:70", "Lin:30"};入力：
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"};
期待される出力：
string[] expected = new string[] {"Ada:70", "Lin:30"};解説
Ada has two orders (50 + 20 = 70) and Lin one (30); Bo has none and is left out: ["Ada:70", "Lin:30"].
入力：
orders = [
[8, 40],
[7, 40],
[9, 99]
]
customer_ids = [7, 8]
customer_names = ["Zoe", "Max"]
期待される出力：
expected = ["Max:40", "Zoe:40"]入力：
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"};
期待される出力：
std::vector<std::string> expected = {"Max:40", "Zoe:40"};入力：
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"};
期待される出力：
string[] expected = new string[] {"Max:40", "Zoe:40"};解説
Both total 40, so they are ordered by name; the order for unknown id 9 is ignored: ["Max:40", "Zoe:40"].
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
First add up the order amounts per customer id.
ヒント 2
Then go through the customers and keep those that have a total.
ヒント 3
Sort by total descending, then by name in ordinal order.
目安の計算量
O(n log n) time, O(n) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。