Interview guides · Guy Carpenter

Guy Carpenter Software Engineer Interview Questions & Guide 2026

Guy Carpenter software engineer prep on partial failure: a timed-out request has an unknown outcome, not a failed one.

The 60-second brief

As a Software Engineer at Guy Carpenter, you play a pivotal role in developing and maintaining software solutions that enhance the effectiveness of the reinsurance industry. This position is crucial because it directly influences the company's ability to analyze and manage risk, ensuring that clients receive reliable and innovative solutions tailored to their needs. You will work with advanced technologies and methodologies to create systems that not only support internal processes but also improve client-facing offerings.

State every complexity claim with the assumption sitting under it. Hash lookup is O(1) on average and only for a hash that spreads your actual keys; comparison-based sorting cannot beat n log n, though counting or radix sort can when the keys are bounded integers.

Focus areas

Trace a symptom to a mechanism under load
Bound every outbound call with a timeout
Paginate large result sets with keyset cursors

Overview

This role is particularly interesting due to the complexity and scale of the projects you will be involved in. You will contribute to the development of catastrophe models and risk assessment tools, allowing Guy Carpenter to provide unparalleled insights to its clients. Your work will be integral to creating products that are not only functional but also strategically aligned with the company’s goals, impacting how reinsurance is approached globally.

Expect to collaborate with cross-functional teams, leveraging your technical skills to drive projects that are at the forefront of technology and innovation in the industry. This position offers a unique opportunity to blend software engineering with a deep understanding of the reinsurance landscape, making it both a challenging and rewarding career path.

The Guy Carpenter interview process

Guy Carpenter candidates report 4 rounds · ≈ 3-5 weeks. The stages below are what candidates describe, not a published process.

Initial Screening

The title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.

What to demonstrate

Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year

How to prepare

Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter

Technical Assessment

What this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.

What to demonstrate

Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly

How to prepare

Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live

Behavioral Interview

What you say here is written down by each interviewer and compared afterwards, so the unit of evaluation is a claim someone else could check, not a well-told narrative. Two things make a story checkable: detail only a participant would hold, and a clean line around which part was yours. Vague ownership is the usual failure and it is usually accidental, because engineers say we about the team's work and we about their own, so the thing they personally built disappears into the plural. Name the part you wrote, and name who did the rest.

What to demonstrate

Whether your details are ones a participant would hold and an observer would not: the constraint that ruled out the obvious approach, the first attempt that failed, the person who objected and on what grounds
Whether ownership survives a direct question, since a follow-up to we decided is routinely who decided, and an answer that stays plural at that point is read as the work belonging to someone else
Whether the numbers you quote are ones you would say identically to a former colleague with the dashboard open

How to prepare

Go through each story replacing every we with either I or a named role (the on-call engineer, the reviewer, the other team) and check the story still holds together. Wherever it stops making sense you have found a part you cannot actually speak to
Open the artefacts for two of your stories, the pull request, the design doc, the incident notes, and read them for dates and figures you have been rounding in the retelling. Correct your version to match
For each story write the single sentence you would least want repeated to a former teammate, then either make it accurate or take it out

Final Evaluation

Nobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.

What to demonstrate

Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience

How to prepare

Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed

Where candidates lose points

Letting a slow dependency consume unbounded concurrency

The failure that takes a service down is usually not an error but a delay. A dependency answering in thirty seconds instead of fifty milliseconds holds each request's worker or connection six hundred times longer, and since required concurrency is arrival rate times latency, a fleet sized for sixty in-flight requests now needs thirty-six thousand to sustain the same rate - so it queues, and requests whose clients have already abandoned them still occupy resources. Retries make it precisely worse: a policy of three attempts triples the load on a dependency at the exact moment it is least able to serve, which is how one slow dependency becomes an outage of everything sharing that pool. Containment is four specific things - a timeout on every outbound call shorter than the caller's remaining budget, a bounded pool per dependency so one cannot starve the others, backoff with full jitter rather than a fixed delay so retries do not resynchronise, and a circuit that stops sending once the failure rate makes an attempt pointless.

Paginating with LIMIT/OFFSET over a set that changes while the client is reading it

OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.

Issuing one query per row of a result set

Fetch related rows in a single batched query keyed by the ids you already hold, or join them into the original query. A per-row round trip multiplies network latency by the row count, and it looks perfectly fine against the ten rows in your development database.

Naming no test cases at all

State what you would test before being asked: empty input, a single element, all elements equal, the maximum permitted size, and the input that exercises the branch you just wrote. It costs thirty seconds and is much of what separates someone who has shipped code from someone who has only solved puzzles.

Questions & practice

Write a function to sort an array of integers.

Medium · data structures and algorithms

Approach

Choose the data structure from the access pattern, not from familiarity.
State the target complexity and say which constraint rules the naive version out.
Restate the input: its shape, its size, and what is guaranteed about it.

Follow-up

What is the worst case, and how likely is it on real data?
Which test case would catch an off-by-one here?

Can you explain the time complexity of your solution?

Medium · data structures and algorithms

Approach

State the target complexity and say which constraint rules the naive version out.
Restate the input: its shape, its size, and what is guaranteed about it.
Walk one small example through your approach before writing the whole thing.

Follow-up

What is the worst case, and how likely is it on real data?
How does this change if the input no longer fits in memory?

Solve a coding problem live, explaining your thought process as you work through it.

Medium · data structures and algorithms

Approach

Choose the data structure from the access pattern, not from familiarity.
Restate the input: its shape, its size, and what is guaranteed about it.
Walk one small example through your approach before writing the whole thing.

Follow-up

What is the worst case, and how likely is it on real data?
How does this change if the input no longer fits in memory?

Find overlapping job attempts and peak concurrency from lease records

Medium · sweep line · intervals · leases

A day of job_run history yields about 50,000,000 attempt records: (job_run_id, job_type, attempt, started_at, finished_at which is NULL when the worker died, lease_expires_at). Leases expire on a clock, so a job that outran its lease ran twice. Produce (a) every job_run_id whose attempts overlapped in wall-clock time and (b) the peak number of simultaneously running attempts per job_type with the minute it occurred. Target O(n log n). State how you treat a NULL finished_at and what clock skew does to your answer.

Approach

Define the interval before sorting anything: an attempt occupies [started_at, COALESCE(finished_at, lease_expires_at)). finished_at is observed and lease_expires_at is only a promise, so every attempt without a finish contributes an estimate and the whole result is a lower bound on overlap rather than an exact count.
For peak concurrency, sweep: emit 2n endpoints, sort by (timestamp, kind) with ends ordered before starts at equal timestamps, then walk the sequence maintaining a counter per job_type and record each type's maximum with its timestamp. O(n log n) dominated by the sort, O(n) space, or O(1) extra if the sort is external and the walk streams.
For overlap detection, do not compare attempts pairwise. A single global sort by (job_run_id, started_at) gives both the grouping and the order; within a group, keep the maximum end seen so far and report an overlap exactly when the next start is less than that running maximum, which is one linear pass after the sort.
Half-open intervals matter and are easy to get wrong: with closed intervals an attempt ending at the same millisecond another begins reads as concurrency two, and across 50,000,000 records that artefact swamps the real signal.
State the clock caveat: started_at and finished_at are written by different workers, so under skew of a few hundred milliseconds an apparent overlap shorter than that bound is not evidence. Filter reported overlaps by a minimum duration, or prefer timestamps written by whichever component heartbeats the lease.
Scale the sort rather than assuming it fits: the sweep emits two endpoints per attempt, so 50,000,000 records become 100,000,000 endpoints, and at roughly 24 bytes each, an 8-byte timestamp plus a 4-byte job_type plus a kind flag padded to alignment, that is about 2.4 GB of sort keys before any scratch space. Either push the ordering into the database behind an index on (job_type, started_at) or run an external merge sort in chunks; the overlap pass sorts n records rather than 2n, so it is the cheaper of the two.

Worked solution (30 min)

Write the interval derivation with the COALESCE and state in one line which of the two end sources is observed and which is assumed.
Write the concurrency sweep: the endpoint tuples, the sort key including the end-before-start tie-break, and the per-job_type counter.
Hand-trace four attempts of one job, two disjoint and two overlapping by three seconds, and confirm the overlap detector fires exactly once.
Add the skew filter as a minimum overlap duration, state the value you chose, and justify it from how the timestamps are written.

Expected result: One global sort by (job_run_id, started_at) driving a running-maximum-end check for overlaps, plus a sweep over 2n endpoints with ends ordered before starts for per-job_type peak concurrency, both O(n log n), with half-open intervals, NULL finished_at falling back to lease_expires_at, and overlaps shorter than the clock-skew bound excluded.

Verify it yourself

Two attempts touching at exactly the same instant report concurrency 1, not 2.
A job with a single attempt never appears in the overlap output regardless of its lease.
The reported peak timestamp for a job_type lies inside at least one of that type's intervals.

Follow-up

A handler is not idempotent and you have found 400 overlapping jobs. Which of them actually caused damage, and what would you query to find out?
Peak concurrency for one job_type is 4 against a configured cap of 4. Is the cap working, or is the data hiding attempts that never started?
How would you compute both answers incrementally as records arrive rather than in a daily batch?

Denormalise tenant onto revisions and backfill it live

Hard · denormalisation · online migration · backfill

resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.

Approach

Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.
Step three, backfill: batch by primary key rather than by created_at so the cursor is dense and resumable — UPDATE resource_revision rr SET tenant_id = r.tenant_id FROM resource r WHERE r.resource_id = rr.resource_id AND rr.revision_id > $1 AND rr.revision_id <= $1 + 5000 AND rr.tenant_id IS NULL — committing per batch and persisting the cursor. Throttle on replica replay lag and on dead-tuple count, since each batch writes 5,000 new row versions. Run the backfill before the index exists so those updates can stay HOT.
Step four, index then enforce then contract: CREATE INDEX CONCURRENTLY (cannot run inside a transaction block, scans the table twice, waits on open transactions, and leaves an INVALID index to drop concurrently if it fails); ADD CONSTRAINT ... CHECK (tenant_id IS NOT NULL) NOT VALID, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE, after which SET NOT NULL uses the validated check instead of re-scanning on PostgreSQL 12 and later. Only then move the audit reads onto the column and, in a later deploy, delete the join path.

Worked solution (40 min)

Write the five steps as separate scripts and state, for each, the lock mode it acquires and the deploy it pairs with.
On a 20M-row copy, run the ADD COLUMN while a 30-second transaction holds a lock on the table, and record how long unrelated queries queue behind it.
Run the batched backfill at 5,000 rows, kill it mid-run, restart from the persisted cursor, and confirm no row is processed twice and none is skipped.
Build the index concurrently under concurrent write load, then add the CHECK ... NOT VALID, VALIDATE it and SET NOT NULL, timing each.
Compare the audit-feed plan before and after: join-and-filter versus an index seek with no Sort.

Expected result: ADD COLUMN completes in milliseconds once it holds the lock, but with a 30-second transaction open it stalls all traffic on that table for the full 30 seconds unless lock_timeout cuts it off. The backfill resumes exactly at the persisted revision_id. After VALIDATE, SET NOT NULL returns without a table scan. The audit feed drops from scanning the tenant's history to reading about 50 rows.

Verify it yourself

SELECT count(*) FROM resource_revision WHERE tenant_id IS NULL returns 0, re-checked after the dual-writer has been live for a full deploy cycle.
A sampled join confirms rr.tenant_id = r.tenant_id across 100k random revisions; the composite foreign key makes a mismatch unwritable from then on.
pg_stat_activity shows no transaction older than one batch's duration while the backfill runs.
EXPLAIN of the audit feed shows an Index Scan with no Sort node.

Follow-up

The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
How do you verify the backfill actually finished, given rows are still being inserted while it runs?
A resource must now be movable between tenants. What does that do to the composite foreign key and to the revisions already written?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.

Approach

Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.
Answer the two negatives explicitly. owner_user_id alone is not a left prefix of the new index, so it degrades to a full scan of the index at best. Ordered by updated_at, the query still seeks on the (tenant, owner) pair but must sort, because only created_at is ordered within that pair.
Measure both sides with EXPLAIN (ANALYZE, BUFFERS) and compare estimated against actual rows at the lowest node — a 2M-versus-200 misestimate there is usually what chose the plan, and adding an index will not fix a statistics problem.

Follow-up

90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
A dashboard runs this for 40 owners in one page load. What changes about the design?
How do you roll this index out on a table taking 1.2k writes/second, and what does it cost on every insert from then on?

How would you design a scalable architecture for a web application?

Medium · system design and architecture

Approach

State the consistency you need, and where you are willing to be stale.
Choose a partition key and say what query it makes expensive.
Name the read and write paths separately; they rarely have the same bottleneck.

Follow-up

What breaks first when traffic grows ten times?
What would you drop to keep the system up under load?

Explain how you would ensure data integrity in a distributed system.

Medium · system design and architecture

Approach

Name the failure you are designing for, then the recovery path.
Name the read and write paths separately; they rarely have the same bottleneck.
State the consistency you need, and where you are willing to be stale.

Follow-up

What breaks first when traffic grows ten times?
How does this behave when that dependency is down for an hour?

How do you approach debugging a complex software issue?

Medium

Approach

Say what you would check first and why it is the highest-information step.
Work from the requirement backwards to the design.
Clarify what is being asked and what a complete answer contains.

Follow-up

What assumption would you test first?
How would you know your answer was wrong?

Design the bulk write endpoint a migration script retries blindly

Medium · batch api · partial failure · idempotency

A customer's migration script pushes 2 million resources through POST /v1/resources:batch, up to 500 items per call, and retries any call that errors or times out. Within a call some items fail validation, some collide with rows that already exist, and some succeed. Specify the request and response shape, whether a batch is atomic or per-item, how idempotency works for the call and for each item, the status code for a mixed outcome, the size and item-count limits with their error codes, and exactly what the script does after a timeout mid-batch.

Approach

Choose atomicity deliberately and price it. All-or-nothing means one transaction holding locks for the whole batch on a primary already absorbing about 1.2k writes/second, which bounds batch size by lock duration, and it turns one bad row into 499 rejections the script must resend. Per-item partial success is the right default for a migration, and the contract's job is then to make a partial outcome impossible to miss.
Require a client-supplied id on every item and echo it in every result. Deriving a per-item key from the array index breaks the first time the script resends a batch with the failures removed: the indices shift, previously-succeeded items acquire new keys, and they are created a second time.
Key the effects at two levels. The call's Idempotency-Key covers an exact resend of the same bytes; per-item keys of (tenant_id, client_item_id) make a partially-applied batch safe to resend whole. Resending an identical batch must reproduce the same per-item results, not 500 conflicts the script has to interpret.
Answer a mixed outcome with one status plus per-item detail: 200, or 207 borrowed from WebDAV if you prefer it - document whichever you pick - carrying an array of client_item_id, per-item status, and either resource_id or an error code from the same taxonomy the single-item endpoint uses. Reserve 4xx for the request as a whole: unparseable body, too many items, payload over the limit (413). Put a failed count at the top level so that even a script checking only the cheapest thing cannot conclude success while rows were dropped.
Cap the request before doing any of it: item count, total bytes, and concurrent batches per tenant, since one tenant's migration otherwise consumes write capacity everyone shares. Anything that cannot finish inside the request deadline belongs on job_run behind a 202, not in a synchronous call that will time out halfway.
Script behaviour after a timeout: the outcome is unknown and no results were received, so resend the identical batch with the same keys and read the results. Never resend 'only the items I have no result for' - a timeout yields no results at all, and that rule silently means resend everything anyway.

Worked solution (30 min)

Write the request schema with the per-item client id, and the response schema with per-item status and a top-level failed count.
Write the atomicity decision and the sentence of justification that names the lock cost or the resend cost.
Define both key levels and trace a resend of a half-applied batch through them, item by item.
List the whole-request rejections with their status codes and limits.
Write the script's timeout rule and the pacing it should apply between calls.

Expected result: A per-item batch with client-supplied item ids, per-item idempotency plus a call-level key, a mixed-outcome response that a status-only check cannot misread, explicit request caps, and a resend rule that is correct when no results were received.

Verify it yourself

Resending an identical batch twice leaves exactly the same set of resource rows as sending it once.
Resending a batch with the successful items removed does not create duplicates of anything.
A caller that reads only the top-level fields can still tell that some items failed.

Follow-up

At 500 items per call, what is the wall-clock time for 2 million rows, and what pacing do you publish so the migration does not become an incident?
One item in every batch fails with the same code. How does the script discover that without a human reading logs?

Listing latency scales with page size, not with filters

Easy · n+1 · latency · query batching

The tenant listing endpoint reads resource filtered by tenant_id and status, ordered by updated_at DESC, and returns each row plus the owner's display name from app_user and the actor of that resource's latest resource_revision. p99 is 55 ms at 10 rows per page and 1.4 s at 200. Database telemetry shows 401 statements per request, each under 1 ms, and nothing in the slow-query log. Diagnose the cause and give the fix, stating the statement count per request and the p99 you expect afterwards.

Approach

Read the counters before forming a theory. 401 statements for 200 rows is one driver query plus two per row, and sub-millisecond execution with an empty slow-query log rules out a bad plan. The time is round trips, which is why it is invisible in every per-query metric and scales with rows returned rather than with filter selectivity.
Name the two per-row statements from their normalised text: a single-row app_user lookup by user_id, and a resource_revision lookup by resource_id ordered by version DESC LIMIT 1. Confirm by dropping those two response fields and watching the statement count fall to one. That locates the calls in the serialisation layer, not the repository.
Check that the arithmetic accounts for the whole gap. Measure one round trip to the replica in isolation; 400 trips at roughly 3 ms of network plus 0.2 ms of execution is about 1.3 s on top of a 55 ms baseline, which matches. If the multiplication had fallen short, the N+1 would only be part of the story and you would keep looking.
Batch both lookups. Collect owner_user_ids and resource_ids from the driver query, then issue WHERE tenant_id = $1 AND user_id = ANY($2) for the users, and PostgreSQL's SELECT DISTINCT ON (resource_id) ... WHERE resource_id = ANY($2) ORDER BY resource_id, version DESC for the latest revision, which the UNIQUE (resource_id, version) index serves directly. On an engine without DISTINCT ON, use a lateral join or a row_number window. Three statements per request at any page size.
Keep the tenant predicate in the batched query. The per-row version was implicitly scoped because its ids came from tenant-scoped rows; a batched user_id = ANY(...) with no tenant_id is an unscoped read that behaves correctly only as long as the id list is trustworthy.
Re-measure at 10, 50 and 200 rows and confirm the statement count is constant. Latency should now track bytes returned.

Follow-up

The page size is capped at 200 today. What breaks first if it is raised to 2,000, and is it still this bug?
How do you stop the next N+1 from reaching production, given that no individual query is slow and the endpoint's tests pass?
The latest-revision actor is only used to render an avatar. Make the case for denormalising it onto resource, and name the write anomaly that introduces.

Suggested practice checklist

Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.

Day 1 — Coding, one pass at shallow depth

Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
Fix nothing today. The value of the pass is the unfixed record.

Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.

Day 2 — Design, one pass at shallow depth

Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.

Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.

Day 3 — Fundamentals and the practical rounds

Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.

Deliverable: Eight scored short answers and one written reading of unfamiliar code.

Day 4 — The rounds that are about you, and the map

Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.

Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.

Day 5 — First chosen area, to the depth you skipped

Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.

Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.

Day 6 — Second chosen area, where the gap is coverage rather than speed

Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.

Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.

Day 7 — Reassemble the loop

Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
Reduce the week to one page holding only the rules you can state without reading them.

Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.

The failure that was yours

Keep one story where the bad call was yours rather than a dependency's or a manager's. Name the check that would have caught it, whether you added that check afterwards, and whether it has fired since. Answers that route blame outward end the conversation early; answers that end in a guardrail someone still relies on tend to open it up.

Discuss your experience with data analysis and how it relates to software development.

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
Name the disagreement and how you resolved it with evidence.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

What would you do differently if you ran that again?
What did you decide not to do, and why?

What motivates you to perform well in a team environment?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Close with what you would do differently, concretely.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

How did you know your change caused the improvement?
What did you decide not to do, and why?

Unblock an engineer without taking the keyboard

Easy · mentoring · leases · at-least-once

A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.

Approach

Ask before diagnosing, and ask for things answerable from data they already have: the attempt count on the job rows that produced duplicates, the handler's observed duration against its lease expiry, and whether the duplicate rows share a natural key that a unique constraint could have caught.
Teach the shape rather than the answer. A lease cannot distinguish a dead worker from a slow one, so a handler that outruns its lease is running twice by design, and deploys deliver the other half by killing handlers mid-run on every rollout. Both of their candidate theories produce identical duplicate rows, which is why the evidence has to come from timings rather than from argument.
Hand over a checklist they execute: a natural key on every write the handler performs so the second copy collides rather than appends, the record of intent written before any external effect, a lease heartbeat while running, and the metric that shows it working.
Keep ownership with them deliberately. Pair on the first write, then step back; if you finish it yourself you have closed one ticket and left the same person stuck on the next redelivery.
Close on the systemic gap that let two days pass, which is usually a missing dashboard for attempt counts or an undocumented at-least-once contract, and fix that rather than only the bug.

Follow-up

How would you distinguish a genuine double-delivery from a lease expiry using only the data already stored?
Their handler calls an external endpoint before recording that it did. What do you tell them to change first?
What do you do the third time the same person brings you the same class of bug?