Leecode_SQL50_1661. Average Time of Process per Machine
Problem description
Table: Activity
±---------------±--------+
| Column Name | Type |
±---------------±--------+
| machine_id | int |
| process_id | int |
| activity_type | enum |
| timestamp | float |
±---------------±--------+
The table shows the user activities for a factory website.
(machine_id, process_id, activity_type) is the primary key (combination of columns with unique values) of this table.
machine_id is the ID of a machine.
process_id is the ID of a process running on the machine with ID machine_id.
activity_type is an ENUM (category) of type (‘start’, ‘end’).
timestamp is a float representing the current time in seconds.
‘start’ means the machine starts the process at the given timestamp and ‘end’ means the machine ends the process at the given timestamp.
The ‘start’ timestamp will always be before the ‘end’ timestamp for every (machine_id, process_id) pair.
It is guaranteed that each (machine_id, process_id) pair has a ‘start’ and ‘end’ timestamp.
There is a factory website that has several machines each running the same number of processes. Write a solution to find the average time each machine takes to complete a process.
The time to complete a process is the ‘end’ timestamp minus the ‘start’ timestamp. The average time is calculated by the total time to complete every process on the machine divided by the number of processes that were run.
The resulting table should have the machine_id along with the average time as processing_time, which should be rounded to 3 decimal places.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input:
Activity table:
±-----------±-----------±--------------±----------+
| machine_id | process_id | activity_type | timestamp |
±-----------±-----------±--------------±----------+
| 0 | 0 | start | 0.712 |
| 0 | 0 | end | 1.520 |
| 0 | 1 | start | 3.140 |
| 0 | 1 | end | 4.120 |
| 1 | 0 | start | 0.550 |
| 1 | 0 | end | 1.550 |
| 1 | 1 | start | 0.430 |
| 1 | 1 | end | 1.420 |
| 2 | 0 | start | 4.100 |
| 2 | 0 | end | 4.512 |
| 2 | 1 | start | 2.500 |
| 2 | 1 | end | 5.000 |
±-----------±-----------±--------------±----------+
Output:
±-----------±----------------+
| machine_id | processing_time |
±-----------±----------------+
| 0 | 0.894 |
| 1 | 0.995 |
| 2 | 1.456 |
±-----------±----------------+
Explanation:
There are 3 machines running 2 processes each.
Machine 0’s average time is ((1.520 - 0.712) + (4.120 - 3.140)) / 2 = 0.894
Machine 1’s average time is ((1.550 - 0.550) + (1.420 - 0.430)) / 2 = 0.995
Machine 2’s average time is ((4.512 - 4.100) + (5.000 - 2.500)) / 2 = 1.456
My solution
WITH st AS (SELECT machine_id, process_id, timestampFROM ActivityWHERE activity_type = 'start'
),
en AS (SELECT machine_id, process_id, timestampFROM ActivityWHERE activity_type = 'end'
),
b AS (SELECT st.machine_id, st.process_id, (en.timestamp - st.timestamp) AS time_takenFROM st JOIN en ON st.machine_id = en.machine_id AND st.process_id = en.process_id
)SELECT b.machine_id, ROUND(AVG(time_taken), 3) AS processing_time
FROM b
GROUP BY b.machine_id
Other solutions
select a1.machine_id, round(avg(a2.timestamp-a1.timestamp), 3) as processing_time
from Activity a1join Activity a2 on a1.machine_id=a2.machine_id and a1.process_id=a2.process_idand a1.activity_type='start' and a2.activity_type='end'
group by a1.machine_id
Select a.machine_id,ROUND(AVG(b.timestamp-a.timestamp), 3)as processing_time
from Activity a, Activity b
where a.process_id=b.process_id AND a.machine_id=b.machine_idAND a.activity_type="start" AND b.activity_type="end"
group by machine_id;