How do you solve Trips and Users on LeetCode?

Trips and Users (LeetCode #262) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to join tables effectively is crucial.

What is the brute force solution for Trips and Users?

The brute force approach for Trips and Users is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each trip individually to see if both the client and driver are unbanned. This method is straightforward but inefficient, as it requires multiple queries and checks for each trip. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Trips and Users?

Trips and Users uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Trips and Users?

Always clarify the requirements and edge cases. Discuss your thought process out loud to show your understanding. Practice writing SQL queries to become more comfortable.

What are common mistakes when solving Trips and Users?

Not filtering out banned users correctly. Forgetting to round the cancellation rate.

#262

Trips and Users

Hard

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages SQL's aggregation functions and joins to efficiently calculate the cancellation rates in a single query without needing multiple checks for each trip.

⚙️

Algorithm

4 steps

1Step 1: Join the Trips table with the Users table for both clients and drivers.
2Step 2: Filter for unbanned users and the specified date range.
3Step 3: Use aggregation functions to count total trips and canceled trips in one go.
4Step 4: Calculate the cancellation rate and round it to two decimal points.

solution.py6 lines

1SELECT DATE(request_at) AS day, ROUND(SUM(CASE WHEN status IN ('cancelled_by_driver', 'cancelled_by_client') THEN 1 ELSE 0 END) / COUNT(*), 2) AS cancellation_rate
2FROM Trips
3JOIN Users AS clients ON Trips.client_id = clients.users_id AND clients.banned = 'No'
4JOIN Users AS drivers ON Trips.driver_id = drivers.users_id AND drivers.banned = 'No'
5WHERE DATE(request_at) BETWEEN '2013-10-01' AND '2013-10-03'
6GROUP BY day;

ℹ

Complexity note: This complexity is due to the single pass through the Trips and Users tables, allowing us to aggregate results efficiently.

1Understanding how to join tables effectively is crucial.
2Recognizing the importance of filtering out banned users early in the process can simplify calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.