Leetcode #1364: Number of Trusted Contacts of a Customer
In this guide, we solve Leetcode #1364 Number of Trusted Contacts of a Customer in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+---------------+---------+
| 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
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
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.