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
This approach involves using a data structure like a HashMap to store the start and end timestamps for each process on each machine. For each machine and process, compute the time taken using these timestamps. Then, calculate the average processing time for all machines.
This Python solution uses a dictionary to collect timestamps by process and machine. It then calculates process times and averages them by machine.
JavaScript
Time Complexity: O(n), where n is the number of rows in the input. We iterate once through the list to collect timestamps.
Space Complexity: O(m), where m is the number of unique machine and process combinations.
In this approach, we first sort the data by machine_id and process_id, ensuring that each start and end appears consecutively. We then do a single linear scan to compute the time taken for each process and maintain a cumulative sum for each machine, eventually computing the average processing time.
This Java implementation sorts the activity data and employs a dual-index scan to compute and accumulate the process time for each machine. The average times are then calculated and displayed.
C++
Time Complexity: O(n log n) due to sorting the list of activities.
Space Complexity: O(m), for storing accumulated processing times and counts by machine.
| Approach | Complexity |
|---|---|
| Using HashMap for Storing Start and End Times | Time Complexity: O(n), where n is the number of rows in the input. We iterate once through the list to collect timestamps. |
| Sorting and Linear Scan Method | Time Complexity: O(n log n) due to sorting the list of activities. |
Average Time of Process per Machine | Leetcode 1661 | Crack SQL Interviews in 50 Qs #mysql • Learn With Chirag • 16,697 views views
Watch 9 more video solutions →Practice Average Time of Process per Machine with our built-in code editor and test cases.
Practice on FleetCode