This Leetcode problem Trips and Users LeetCode Solution is done in SQL.

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Trips and Users LeetCode Solution

Problem Statement

Column Name	Type
id	int
client_id	int
driver_id	int
city_id	int
status	enum
request_at	date

id is the primary key (column with unique values) 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 (category) type of (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’).

Column Name	Type
users_id	int
banned	enum
role	enum

users_id is the primary key (column with unique values) 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’). banned is an ENUM (category) type of (‘Yes’, ‘No’).

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.

Write a solution 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". Round Cancellation Rate to two decimal points.

Return the result table in any order.

The result format is in the following example.

Example 1:
Input:

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

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_driver	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	13	cancelled_by_driver	2013-10-03

Output:

Day	Cancellation Rate
2013-10-01	0.33
2013-10-02	0.00
2013-10-03	0.50

Explanation:
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

Trips and Users LeetCode Solution MySQL

with stats as (
  select 
    Request_at, 
    T.Status <> 'completed' as IsCancelled
  from Trips T 
  join Users C on (Client_Id = C.Users_Id and C.Banned = 'No') 
  join Users D on (Driver_Id = D.Users_Id and D.Banned = 'No') 
  where
    Request_at between '2013-10-01' and '2013-10-03'
)
select 
  Request_at as Day,
  Round(
    cast(sum(IsCancelled) as real) / cast(count(*) as real),
    2
  ) as 'Cancellation Rate'
from stats
group by Request_at
;Code language: JavaScript (javascript)

Trips and Users LeetCode Solution Pandas

import pandas as pd

def trips_and_users(trips: pd.DataFrame, users: pd.DataFrame) -> pd.DataFrame:

    users = users[users.banned == 'No']
    trips = trips[trips.request_at
                     .between('2013-10-01','2013-10-03')
                    ].rename(columns = {'request_at': 'Day'})

    trips['can'] = trips.status.str.startswith('can')

    trips = trips[(trips.client_id.isin(users.users_id)) &
                  (trips.driver_id.isin(users.users_id))
                  ].groupby('Day')['can'].agg(['sum','size']).reset_index()

    trips['Cancellation Rate'] = (trips['sum']/trips['size']).round(2)
    
    return trips[['Day','Cancellation Rate']]Code language: JavaScript (javascript)