Number of Trusted Contacts of a Customer — LeetCode 1364 Python Solution
MediumLeetCode PremiumDatabase
- Problem
- #1364
- Reading time
- 4 min
- Source
- leetcode.com
Table schema
SQL
Table: Customers +---------------+---------+ | Column Name | Type | +---------------+---------+ | customer_id | int | | customer_name | varchar | | email | varchar | +---------------+---------+ customer_id is the column of unique values for this table. Each row of this table contains the name and the email of a customer of an online shop.Example
SQL
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| customer_id | int |
| customer_name | varchar |
| email | varchar |
+---------------+---------+
customer_id is the column of unique values for this table.
Each row of this table contains the name and the email of a customer of an online shop.Python solution
Python
import duckdb
import pandas as pd
def solution(customers: pd.DataFrame, contacts: pd.DataFrame, invoices: pd.DataFrame) -> pd.DataFrame:
con = duckdb.connect()
con.register("Customers", customers)
con.register("Contacts", contacts)
con.register("Invoices", invoices)
return con.execute("""SELECT
invoice_id,
t2.customer_name,
price,
COUNT(t3.user_id) AS contacts_cnt,
COUNT(t4.email) AS trusted_contacts_cnt
FROM
Invoices AS t1
LEFT JOIN Customers AS t2 ON t1.user_id = t2.customer_id
LEFT JOIN Contacts AS t3 ON t1.user_id = t3.user_id
LEFT JOIN Customers AS t4 ON t3.contact_email = t4.email
GROUP BY invoice_id
ORDER BY invoice_id;""").df()Complexity
| Measure | Complexity |
|---|---|
| Time | O(n log n) (typical) |
| Space | O(n) auxiliary |
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