LeetCode #1661 — EASY

Average Time of Process per Machine

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers":{"Activity":["machine_id","process_id","activity_type","timestamp"]},"rows":{"Activity":[[0,0,"start",0.712],[0,0,"end",1.52],[0,1,"start",3.14],[0,1,"end",4.12],[1,0,"start",0.55],[1,0,"end",1.55],[1,1,"start",0.43],[1,1,"end",1.42],[2,0,"start",4.1],[2,0,"end",4.512],[2,1,"start",2.5],[2,1,"end",5]]}}

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1661: Average Time of Process per Machine
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #1661: Average Time of Process per Machine
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1661: Average Time of Process per Machine
// # Write your MySQL query statement below
// SELECT
//     machine_id,
//     ROUND(
//         AVG(
//             CASE
//                 WHEN activity_type = 'start' THEN -timestamp
//                 ELSE timestamp
//             END
//         ) * 2,
//         3
//     ) AS processing_time
// FROM Activity
// GROUP BY 1;
// "#
// }

// Accepted solution for LeetCode #1661: Average Time of Process per Machine
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1661: Average Time of Process per Machine
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1661: Average Time of Process per Machine
// # Write your MySQL query statement below
// SELECT
//     machine_id,
//     ROUND(
//         AVG(
//             CASE
//                 WHEN activity_type = 'start' THEN -timestamp
//                 ELSE timestamp
//             END
//         ) * 2,
//         3
//     ) AS processing_time
// FROM Activity
// GROUP BY 1;
// "#
// }

# Accepted solution for LeetCode #1661: Average Time of Process per Machine
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #1661: Average Time of Process per Machine
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #1661: Average Time of Process per Machine
# # Write your MySQL query statement below
# SELECT
#     machine_id,
#     ROUND(
#         AVG(
#             CASE
#                 WHEN activity_type = 'start' THEN -timestamp
#                 ELSE timestamp
#             END
#         ) * 2,
#         3
#     ) AS processing_time
# FROM Activity
# GROUP BY 1;
# "#
# }

// Accepted solution for LeetCode #1661: Average Time of Process per Machine
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #1661: Average Time of Process per Machine
# Write your MySQL query statement below
SELECT
    machine_id,
    ROUND(
        AVG(
            CASE
                WHEN activity_type = 'start' THEN -timestamp
                ELSE timestamp
            END
        ) * 2,
        3
    ) AS processing_time
FROM Activity
GROUP BY 1;
"#
}

// Accepted solution for LeetCode #1661: Average Time of Process per Machine
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1661: Average Time of Process per Machine
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1661: Average Time of Process per Machine
// # Write your MySQL query statement below
// SELECT
//     machine_id,
//     ROUND(
//         AVG(
//             CASE
//                 WHEN activity_type = 'start' THEN -timestamp
//                 ELSE timestamp
//             END
//         ) * 2,
//         3
//     ) AS processing_time
// FROM Activity
// GROUP BY 1;
// "#
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.