Leetcode #1635: Hopper Company Queries I
In this guide, we solve Leetcode #1635 Hopper Company Queries I 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: Drivers +-------------+---------+ | Column Name | Type | +-------------+---------+ | driver_id | int | | join_date | date | +-------------+---------+ driver_id is the primary key (column with unique values) for this table. Each row of this table contains the driver's ID and the date they joined the Hopper company.
Quick Facts
- Difficulty: Hard
- 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 |
+-------------+---------+
| driver_id | int |
| join_date | date |
+-------------+---------+
driver_id is the primary key (column with unique values) for this table.
Each row of this table contains the driver's ID and the date they joined the Hopper company.
Python Solution
import duckdb
import pandas as pd
def solution(drivers: pd.DataFrame, rides: pd.DataFrame, accepted_rides: pd.DataFrame) -> pd.DataFrame:
con = duckdb.connect()
con.register("Drivers", drivers)
con.register("Rides", rides)
con.register("AcceptedRides", accepted_rides)
return con.execute("""WITH RECURSIVE
Months AS (
SELECT
1 AS month
UNION ALL
SELECT
month + 1
FROM Months
WHERE month < 12
),
Ride AS (
SELECT MONTH(requested_at) AS month, COUNT(1) AS cnt
FROM
Rides AS r
JOIN AcceptedRides AS a ON r.ride_id = a.ride_id AND YEAR(requested_at) = 2020
GROUP BY month
)
SELECT
m.month,
COUNT(driver_id) AS active_drivers,
IFNULL(r.cnt, 0) AS accepted_rides
FROM
Months AS m
LEFT JOIN Drivers AS d
ON (m.month >= MONTH(d.join_date) AND YEAR(d.join_date) = 2020) OR YEAR(d.join_date) < 2020
LEFT JOIN Ride AS r ON m.month = r.month
GROUP BY month;""").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.