SQL LeetCode: 262. Trips and Users

Pallavi Mirajkar Dantkale
3 min readApr 4, 2021

262. Trips and Users

Hard

Table: Trips

+-------------+----------+
| Column Name | Type |
+-------------+----------+
| Id | int |
| Client_Id | int |
| Driver_Id | int |
| City_Id | int |
| Status | enum |
| Request_at | date |
+-------------+----------+
Id is the primary key for this table.
The table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are foreign keys to the Users_Id at the Users table.
Status is an ENUM type of (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’).

Table: Users

+-------------+----------+
| Column Name | Type |
+-------------+----------+
| Users_Id | int |
| Banned | enum |
| Role | enum |
+-------------+----------+
Users_Id is the primary key for this table.
The table holds all users. Each user has a unique Users_Id, and Role is an ENUM type of (‘client’, ‘driver’, ‘partner’).
Status is an ENUM type of (‘Yes’, ‘No’).

Write a SQL query to find the cancellation rate of requests with unbanned users (both client and driver must not be banned) each day between "2013-10-01" and "2013-10-03".

The cancellation rate is computed by dividing the number of canceled (by client or driver) requests with unbanned users by the total number of requests with unbanned users on that day.

Return the result table in any order. Round Cancellation Rate to two decimal points.

The query result format is in the following example:

Trips table:
+----+-----------+-----------+---------+---------------------+------------+
| Id | Client_Id | Driver_Id | City_Id | Status | Request_at |
+----+-----------+-----------+---------+---------------------+------------+
| 1 | 1 | 10 | 1 | completed | 2013-10-01 |
| 2 | 2 | 11 | 1 | cancelled_by_driver | 2013-10-01 |
| 3 | 3 | 12 | 6 | completed | 2013-10-01 |
| 4 | 4 | 13 | 6 | cancelled_by_client | 2013-10-01 |
| 5 | 1 | 10 | 1 | completed | 2013-10-02 |
| 6 | 2 | 11 | 6 | completed | 2013-10-02 |
| 7 | 3 | 12 | 6 | completed | 2013-10-02 |
| 8 | 2 | 12 | 12 | completed | 2013-10-03 |
| 9 | 3 | 10 | 12 | completed | 2013-10-03 |
| 10 | 4 | 13 | 12 | cancelled_by_driver | 2013-10-03 |
+----+-----------+-----------+---------+---------------------+------------+
Users table:
+----------+--------+--------+
| Users_Id | Banned | Role |
+----------+--------+--------+
| 1 | No | client |
| 2 | Yes | client |
| 3 | No | client |
| 4 | No | client |
| 10 | No | driver |
| 11 | No | driver |
| 12 | No | driver |
| 13 | No | driver |
+----------+--------+--------+
Result table:
+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+
On 2013-10-01:
- There were 4 requests in total, 2 of which were canceled.
- However, the request with Id=2 was made by a banned client (User_Id=2), so it is ignored in the calculation.
- Hence there are 3 unbanned requests in total, 1 of which was canceled.
- The Cancellation Rate is (1 / 3) = 0.33
On 2013-10-02:
- There were 3 requests in total, 0 of which were canceled.
- The request with Id=6 was made by a banned client, so it is ignored.
- Hence there are 2 unbanned requests in total, 0 of which were canceled.
- The Cancellation Rate is (0 / 2) = 0.00
On 2013-10-03:
- There were 3 requests in total, 1 of which was canceled.
- The request with Id=8 was made by a banned client, so it is ignored.
- Hence there are 2 unbanned request in total, 1 of which were canceled.
- The Cancellation Rate is (1 / 2) = 0.50

SELECT t.Request_at AS Day, Round(COUNT(CASE WHEN t.Status LIKE “cancelled%” THEN 1 END)/COUNT(*),2) AS ‘Cancellation Rate’
FROM Trips t
JOIN Users u ON t.Client_Id=u.Users_Id
JOIN Users s ON t.Driver_Id=s.Users_Id
WHERE u.Banned = ‘NO’and s.Banned =’NO’
GROUP BY 1
HAVING Day BETWEEN ‘2013–10–01’ AND ‘2013–10–03’
ORDER BY 1 ASC

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Pallavi Mirajkar Dantkale

QA Engineer / Data Analyst — Highly committed to Quality Assurance and data analysis, advocate for quality and add the right value to the organization.