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Join Customers and Orders

Join customers with their orders like a LINQ Join and report each customer's total spend.

Das lernst du

  • Match two collections on a shared key
  • Sum values per group
  • Sort by a value and break ties by name

Problem

Übersetzung anzeigen

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.

Einschränkungen

1 ≤ number of customers ≤ 10,000

0 ≤ number of orders ≤ 100,000; 1 ≤ amount ≤ 10^6

Beispiele

Beispiel 1
Eingabe:
orders = [
    [1, 50],
    [2, 30],
    [1, 20]
]
customer_ids = [1, 2, 3]
customer_names = ["Ada", "Lin", "Bo"]

Erwartete Ausgabe:
expected = ["Ada:70", "Lin:30"]
Eingabe:
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"};

Erwartete Ausgabe:
std::vector<std::string> expected = {"Ada:70", "Lin:30"};
Eingabe:
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"};

Erwartete Ausgabe:
string[] expected = new string[] {"Ada:70", "Lin:30"};

Erläuterung

Ada has two orders (50 + 20 = 70) and Lin one (30); Bo has none and is left out: ["Ada:70", "Lin:30"].

Beispiel 2
Eingabe:
orders = [
    [8, 40],
    [7, 40],
    [9, 99]
]
customer_ids = [7, 8]
customer_names = ["Zoe", "Max"]

Erwartete Ausgabe:
expected = ["Max:40", "Zoe:40"]
Eingabe:
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"};

Erwartete Ausgabe:
std::vector<std::string> expected = {"Max:40", "Zoe:40"};
Eingabe:
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"};

Erwartete Ausgabe:
string[] expected = new string[] {"Max:40", "Zoe:40"};

Erläuterung

Both total 40, so they are ordered by name; the order for unknown id 9 is ignored: ["Max:40", "Zoe:40"].

So gehst du vor

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

Hinweis 1

First add up the order amounts per customer id.

Hinweis 2

Then go through the customers and keep those that have a total.

Hinweis 3

Sort by total descending, then by name in ordinal order.

Erwartete Komplexität

O(n log n) time, O(n) extra space

Referenzen

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

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