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An on-call engineer is staring at fifty thousand log lines and cannot see the pattern for the noise, because every line carries a different job id, retry counter or key. They want the lines grouped by template — the message with its numbers masked — so the loudest kind of message is at the top.
Write group_by_template(lines). It takes lines, a list of raw log-line strings in the order they were written, and returns a list of dicts {"template": str, "count": int, "first_seen": int}, one per distinct template.
Function to write
group_by_template(lines: list[str]) -> list[dict]One dict per template with its count and first index, ordered by count descending then first_seen ascending.
How to approach it
Turn each line into its template first — mask digit runs, collapse whitespace — and use that string as the dict key. Everything else is counting.
Sample cases
+ 2 held back until you submit
retry counters are one template
Three retry lines differ only in their numbers, so they are one template seen 3 times.
Input
Argument 1
[
'retry 3 of 5',
'job 17 started',
'retry 4 of 5',
'job 18 started',
'retry 1 of 5',
'disk full'
]Returns
| template | count | first_seen |
|---|---|---|
| retry <n> of <n> | 3 | 0 |
| job <n> started | 2 | 1 |
| disk full | 1 | 5 |
digits glued to a word
A digit run inside a word is still masked: user42 and user7 both become user<n>.
Input
Argument 1
[
'user42 logged in',
'user7 logged in',
'user42 logged out'
]Returns
| template | count | first_seen |
|---|---|---|
| user<n> logged in | 2 | 0 |
| user<n> logged out | 1 | 2 |
spacing differs, the template does not
A double space, padding and a tab all collapse to one space, and 30s becomes <n>s.
Input
Argument 1
[
'cache miss for key 9',
' cache miss for key 10 ',
'cache miss for key 11',
'timeout after 30s'
]Returns
| template | count | first_seen |
|---|---|---|
| cache miss for key <n> | 3 | 0 |
| timeout after <n>s | 1 | 3 |
a tie on count goes to the first seen
Both templates have 2 lines; worker <n> idle appeared first, so it comes first, not alphabetically.
Input
Argument 1
[
'worker 2 idle',
'flush 10 rows',
'flush 12 rows',
'worker 5 idle'
]Returns
| template | count | first_seen |
|---|---|---|
| worker <n> idle | 2 | 0 |
| flush <n> rows | 2 | 1 |
no lines
An empty log returns an empty list rather than None.
Input
Argument 1
[] (empty list)
Returns
[] (empty list)
Constraints
<n>, then every run of any whitespace collapsed to a single space, then stripped. "retry 3 of 5" becomes "retry <n> of <n>"."user42 logged in" becomes "user<n> logged in", and "30s" becomes "<n>s".count is how many lines have that template; first_seen is the index in lines of the first line with that template.count descending, then by first_seen ascending — never alphabetically."" and is grouped like any other. An empty lines list returns [].Worked example
Take ["retry 3 of 5", "job 17 started", "retry 4 of 5", "job 18 started", "retry 1 of 5", "disk full"]. Compute each line's template and use it as a dict key: index 0 gives retry <n> of <n> (new: first_seen 0, count 1), index 1 gives job <n> started (new: first_seen 1), index 2 is retry <n> of <n> again (count 2), index 3 bumps job <n> started to 2, index 4 bumps retry to 3, and index 5 disk full is new at 5. Sorted by count then first_seen: retry 3 at 0, job 2 at 1, disk full 1 at 5.
The trap: grouping by the raw line. "retry 3 of 5" and "retry 4 of 5" are different strings, so every line becomes its own group of 1 and the report hides exactly the repetition the engineer was looking for.
What this tests
Hashing by a normalised key: the pattern behind LeetCode 'Group Anagrams', where the key is a canonical form of the record rather than the record itself. The target is O(total characters) — one pass computing keys into a dict — against the O(n²) habit of comparing each line with every group. A data engineer meets it for real clustering logs and alerts, and deduplicating near-identical records. The shape is reported in data engineering loops: 'Group Anagrams' is tagged for data engineers at Google, Oracle, Goldman Sachs and Visa on Exponent.
group_by_template(lines: list[str]) -> list[dict]Submit for review to find out what your query gets right, what it gets wrong, and how it compares with the best working query for this exercise.
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