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Adcock Refrigeration and Air Conditioning Software Engineer Interview Questions & Guide 2026

Adcock Refrigeration and Air Conditioning software engineer prep on operations: decide what you would need to see to know a release is going wrong.

The 60-second brief

The role of a Software Engineer at Adcock Refrigeration and Air Conditioning is uniquely positioned at the intersection of advanced mechanical systems and modern digital infrastructure. As a leader in the HVAC and refrigeration sector, Adcock Refrigeration and Air Conditioning relies on its engineering talent to maintain, optimize, and innovate the complex climate control systems that serve critical commercial and industrial clients.

Allocate preparation against your weakest link rather than your favourite topic. A strong algorithm habit usually comes with weak out-loud explanation of tradeoffs, and years of shipping usually come with rusty from-scratch implementation under a clock.

Focus areas

  • Trace a symptom to a mechanism under load
  • Choose indexes from the query's access path
  • Bound every outbound call with a timeout

Overview

Your work directly impacts the reliability and efficiency of high-stakes environments, ranging from retail cold-storage facilities to complex office air conditioning networks. You will be tasked with solving technical challenges that ensure these systems operate within strict performance parameters. This role is not just about writing code; it is about applying engineering principles to real-world hardware, keeping the company's technical solutions robust, scalable, and responsive to its clients' evolving needs.

The Adcock Refrigeration and Air Conditioning interview process

Adcock Refrigeration and Air Conditioning candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Initial Screening

Half of this call is the part candidates treat as small talk: start date, notice period, work authorisation and its timing, location and time zone, on-call, and the number. Those are what kill offers late, after several engineers have each spent a day. Surfacing a hard constraint now costs you nothing and occasionally buys you something, since a loop compressed to fit a competing deadline can usually only be arranged if it is asked for early. The common failure is deflecting the compensation question twice, then discovering at offer stage that the band never reached your number.

What to demonstrate

  • Whether your hard constraints are compatible with the role before a loop gets booked: earliest start, notice period, what authorisation you hold and when it needs action, days on site, willingness to carry a pager
  • Whether you give a compensation range with something behind it, such as current total compensation or a competing timeline, rather than leaving the band untested
  • Whether your stated timeline is real, since a competing deadline raised now is something scheduling can sometimes work around and the same deadline raised at offer stage usually is not

How to prepare

  • Write each constraint down in one line before the call and state them as facts rather than negotiating them live under a question you were not expecting
  • Set your range from two or three current data points for that level and location, and name the structure you are quoting in, so the number is comparable to the one they are holding
  • If another process is running, say where it stands and by when, and ask directly whether this loop can be scheduled inside that window

Technical Interviews

Input bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.

What to demonstrate

  • Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
  • Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
  • Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
  • Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply

How to prepare

  • For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
  • For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
  • Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last

Where candidates lose points

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.

Shipping a migration and the code that depends on it as a single change

During any rolling deploy, and for as long as a rollback remains possible, old and new code execute against the same schema at the same time. A migration that drops or renames a column breaks every instance that has not restarted yet, and code that requires a column the migration has not applied breaks every instance that restarted early. The discipline is expand then contract: add the new column nullable, write both shapes, backfill in batches, move reads across once the backfill is verified, and only then stop writing the old shape and drop it - four deploys, usually spread over days. It feels disproportionate until the first rollback, at which point it is the only reason the previous version still runs.

Assuming the input fits in memory

Ask how large the input is in bytes before committing to an in-memory algorithm; beyond that point the options are a single streaming pass, an external sort with bounded buffers, or a sketch that trades exactness for constant memory. An algorithm that assumes random access to the whole input is a different algorithm from one that sees each element once.

Optimising an axis nobody named

Ask which resource is actually scarce here: wall-clock latency, throughput, memory footprint, cost per request, or engineering time. Shaving a constant factor off an in-memory step is wasted effort when the same function makes a blocking remote call inside the loop.

Questions & practice

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

  1. Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
  2. Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
  3. State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
  4. Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
  5. Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
  6. Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.

Follow-up

  • The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
  • A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
  • How would you make the window survive a process restart, and is it worth the cost?

Merge partitioned event streams into one ordered feed with bounded lateness

Hard · k-way merge · watermarks · out-of-order streams

The read-model service consumes 64 log partitions carrying about 4,000 events per second in total. Each partition is ordered within itself, but partitions drift by up to 30 seconds, and the activity feed must present a tenant's events in occurred_at order. Produce the merge. State its complexity, the buffer it requires in events and in bytes, what happens when one partition is idle, and what you do with an event that arrives after you have already emitted its position. Payloads average 1 KB.

Approach

  1. Merge with a min-heap over the 64 partition heads keyed on (occurred_at, event_id): O(log P) per event and O(n log P) overall. The tie-break on event_id is what makes the output deterministic when two partitions carry the same millisecond, which matters because the feed is paginated and a non-deterministic order reorders pages under the reader.
  2. Emitting the heap head is only correct once every partition has produced everything up to that timestamp, so the emit condition is a watermark: the minimum across partitions of the highest occurred_at seen, less the allowed lateness. Events are held until the watermark passes them, which is what turns individually ordered streams into a jointly ordered one.
  3. Size the buffer from the lateness rather than guessing: 4,000 events per second times 30 seconds is 120,000 buffered events, and at 1 KB each about 120 MB of heap. That number is the real price of the ordering guarantee and belongs in front of whoever asked for it.
  4. Handle the idle partition explicitly, because it fails the feed rather than corrupting it: a partition with no traffic never advances its own maximum, so the watermark freezes and output stops entirely. Either every partition emits a periodic idle marker carrying the broker's current time, or the watermark falls back to wall clock for a partition silent beyond a threshold.
  5. Choose the late-event policy from what the projection is keyed on. The projection upserts on (aggregate_id, aggregate_version) and discards a version it has already applied, so a late event is safe to apply out of order and correctness never depended on the merge at all. Apply it, recompute the affected feed page, and count lateness so the 30-second budget can be re-derived from data rather than folklore.
  6. Say what the merge does not buy: ordering is guaranteed within one aggregate by the log's partitioning, and no watermark makes the cross-aggregate order authoritative. Two events from different aggregates in the same millisecond have no true order, so the feed's order is a presentation choice that must be stable rather than correct.

Worked solution (35 min)

  1. Write the heap comparator on (occurred_at, event_id) and the per-partition head refill.
  2. Write the watermark computation and the emit-loop condition, then list which buffered events are held at a chosen instant.
  3. Compute the buffer at 4,000 events per second, 30 seconds and 1 KB per event, and state what fraction of a worker's heap that represents.
  4. Add the idle-partition marker and trace the watermark with one silent partition, both with and without the marker.
  5. Write the late-event path and name the key that makes applying it safe.

Expected result: A 64-way min-heap merge at O(n log P) with a deterministic (occurred_at, event_id) comparator, a watermark of the per-partition minimum less 30 seconds gating emission, a stated buffer of 120,000 events and roughly 120 MB, idle markers so a silent partition cannot freeze the watermark, and a late-event policy justified by the projection's idempotency on (aggregate_id, aggregate_version).

Verify it yourself

  • Feeding the partitions in an adversarial interleaving produces the same output sequence as feeding them in timestamp order.
  • Silencing one partition stalls output without the idle marker and does not stall it with the marker.
  • Applying a late event twice leaves the projection byte-identical, which is what permits the out-of-order path to exist.

Follow-up

  • The lateness budget is raised to five minutes. What is the new buffer, and what besides memory changes?
  • The consumer restarts. Where does it resume from, and what does the feed look like for the first 30 seconds?
  • One partition is ten minutes behind because its producer is slow. Do you stall the feed or emit without it?

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

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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.

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?

Find version gaps and relay lag with window functions

Medium · window functions · event tables · frame semantics

outbox_event holds event_id, aggregate_type, aggregate_id, aggregate_version, event_type, payload, status ('pending','published','dead'), attempts, created_at, published_at. A projection is missing rows and you must decide whether the relay skipped events or the consumer dropped them. Write three queries over the last seven days: one listing every aggregate_id whose published aggregate_version sequence has a hole, one giving per-day counts with a running total, and one returning the newest published event per aggregate. For each, say where the window function is evaluated relative to WHERE and LIMIT. PostgreSQL 16.

Approach

  1. Gaps: compute lead(aggregate_version) OVER (PARTITION BY aggregate_id ORDER BY aggregate_version) in a subquery, then filter next_version <> aggregate_version + 1 in the outer query. Window functions are evaluated after WHERE, GROUP BY and HAVING and before the outer ORDER BY and LIMIT, so the predicate cannot sit in the same WHERE clause and PostgreSQL 16 has no QUALIFY.
  2. Say what the seven-day filter does to the answer: it truncates every partition, so the first row per aggregate has no predecessor inside the window and a hole spanning the boundary is invisible. Widen the window, or join to resource.version as the authority for the true maximum.
  3. Running total: SELECT date_trunc('day', created_at) AS d, count() AS n, sum(count()) OVER (ORDER BY date_trunc('day', created_at) ROWS UNBOUNDED PRECEDING). An aggregate inside a window call is legal because grouping runs before windowing. The grouping key is unique per row here so ROWS and RANGE agree, but write the frame anyway — over ungrouped rows with tied timestamps the default RANGE frame pulls in every peer row and the total jumps.
  4. Newest per aggregate: DISTINCT ON (aggregate_id) ... ORDER BY aggregate_id, aggregate_version DESC is the cheap PostgreSQL-only form when an index matches that order; row_number() OVER (PARTITION BY aggregate_id ORDER BY aggregate_version DESC) = 1 is the portable form and needs a subquery for the same evaluation-order reason as the gap query.
  5. Interpret rather than report: no gaps plus a normal p95 of published_at - created_at points at the consumer; gaps or a fat lag tail point at the relay; rows still 'pending' with attempts > 0 point at neither, because they never left the database.
  6. Be explicit that the partial index on (created_at, event_id) WHERE status = 'pending' does not serve any of these — they read published rows. Name the index a recurring monitor would need, and say why a query run twice a year may not deserve one.

Worked solution (30 min)

  1. Write the three queries against seven days of data and confirm each returns without error.
  2. In a scratch copy, delete one middle event for a single aggregate and confirm the gap query names that aggregate and the versions either side.
  3. Run a running total over ungrouped rows ordered by date_trunc('second', created_at), once with the default frame and once with ROWS, and record where the two series diverge.
  4. Compare the DISTINCT ON and row_number() plans on the same data and record rows-read for each.

Expected result: The gap query returns one row per hole with the version on each side; the daily query returns seven rows whose running total ends at the seven-day count; the newest-per-aggregate query returns exactly one row per aggregate_id present in the window. The RANGE running total is higher than the ROWS one at every tied second.

Verify it yourself

  • The per-day counts sum to COUNT(*) over the same predicate.
  • Re-running the gap query after reinserting the deleted event returns zero rows.
  • count(*) of the newest-per-aggregate result equals count(DISTINCT aggregate_id) over the same filter.

Follow-up

  • Relay failover redelivers events. Does a duplicate break the gap query, and how would you detect one from this table alone?
  • Turn the gap check into a continuous monitor rather than a query someone runs after an incident. What does it watch?
  • The consumer claims it never received event 4,812,006. What do you look at, in what order?

Replace offset paging on the resource feed with keyset

Easy · keyset pagination · composite index · sort order

resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.

Approach

  1. Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
  2. Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
  3. Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
  4. Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
  5. State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
  6. Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.

Follow-up

  • The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
  • Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
  • What does the cursor do when the row it points at has since been deleted?

What experience do you have with PLC programming or industrial automation software?

Medium

Approach

  1. State your assumptions explicitly before working the problem.
  2. Say what you would check first and why it is the highest-information step.
  3. Clarify what is being asked and what a complete answer contains.

Follow-up

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

How do you troubleshoot a software-driven refrigeration unit that is failing to maintain temperature?

Medium

Approach

  1. Clarify what is being asked and what a complete answer contains.
  2. Work from the requirement backwards to the design.
  3. State your assumptions explicitly before working the problem.

Follow-up

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

Evolve the resource contract without breaking integrations you cannot upgrade

Hard · versioning · backward compatibility · deprecation

GET /v1/resources/{id} returns status from a four-value enum ('draft','active','archived','deleted'), a numeric version, and the body inline. Consumers are a browser app you deploy and roughly 300 server-side integrations, some untouched for two years, that switch exhaustively on status and parse ids as JSON numbers. You must add a 'pending_review' status, move bodies over 256 KB to a body_ref pointer, and expose per-field change history from resource_revision. Specify the compatibility policy, the wire changes, how both generations are served, and the evidence that lets you remove the old shape.

Approach

  1. Write the policy first and date it: fields are added, never retyped or repurposed; consumers ignore unknown fields; an unknown enum value maps to a documented fallback; nothing is removed until telemetry shows no caller reads it. Then say the uncomfortable part out loud - v1 shipped without the unknown-value rule, so 300 running integrations have no fallback, and no server change can install one into code that is already deployed.
  2. That single fact forces per-request negotiation rather than a server-side default. Keep one internal model and select a serialiser from an explicit version in the request, and default a caller that sends nothing to the oldest supported version. Defaulting to the newest is the change that breaks every integration that never asked for anything, on the day you ship.
  3. Downgrade 'pending_review' for old callers to the nearest state they already handle, 'draft', and state the loss explicitly: those integrations cannot see review state and will treat the resource as editable. If that is unacceptable for one integration, the remedy is moving it to the new version, not a cleverer projection - there is no mapping that invents a state the client has no code for.
  4. Make the body change additive. Old callers keep body inline; the new shape adds body_ref and a size field, and resources over the limit are served to old callers by resolving the pointer server-side or by refusing with a documented code, chosen once and published. Never repurpose body to carry the pointer: a client that renders it shows a storage key to a user, and that failure is silent, where a missing field would have been loud. While you are here, serialise BIGINT ids as strings in the new shape - a browser parsing JSON numbers gets IEEE-754 doubles, exact only to 2^53 - and treat that as its own breaking change requiring the same negotiation, not a quiet fix.
  5. Add change history as a separate sub-resource, GET /v1/resources/{id}/revisions, keyset-paginated over (resource_id, version) rather than as an array inside the resource. A field added to a hot response is paid for by every caller including those that never read it, and an unbounded array inside a cached object destroys the size assumptions the cache was configured with.
  6. Retire on evidence rather than on a date alone: count requests per negotiated version per credential, publish a Sunset header (RFC 8594) with the removal date and a link to the migration, contact the credentials still on the old version, then answer 410 Gone once it is removed. Keep each version's serialiser under snapshot tests so a refactor cannot change v1's bytes by accident.

Worked solution (40 min)

  1. Write and date the compatibility policy as five rules, then mark which of them v1 callers cannot honour.
  2. Choose the negotiation mechanism and the default for an unversioned request, and justify the default in one sentence.
  3. Write the downgrade table: new state to old state, new body shape to old body shape, and what is lost in each direction.
  4. Specify the revisions sub-resource: its URL, its cursor, and why it is not a field on the resource.
  5. Write the sunset plan: the per-version per-credential metric, the header, the lead time, and the terminal status code.

Expected result: A dated additive policy, version negotiation over one internal model with the oldest version as the default, a downgrade table with its losses named, history as a paginated sub-resource, and a removal gated on per-credential version telemetry rather than a calendar entry.

Verify it yourself

  • A two-year-old integration receives a resource that is really in 'pending_review' and neither crashes nor renders a blank state.
  • Deleting the legacy body field from the internal model breaks exactly one serialiser and no business logic.
  • Every v1 response is byte-stable under snapshot tests after an unrelated refactor.

Follow-up

  • An old integration submits a status transition while the resource is really in 'pending_review'. What does the write path accept, and what does it reject?
  • Two years on you want to delete the v1 serialiser. What evidence makes that safe, and who must be contacted before it happens?
  • How would you test against a two-year-old integration rather than against today's source?

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

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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

For a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.

Day 1 — Numbers before diagrams

  • Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
  • Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
  • Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.

Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.

Day 2 — One system, from requirements to schema

  • Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
  • Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
  • Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.

Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.

Day 3 — The consistency you are actually buying

  • Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
  • Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
  • Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.

Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.

Day 4 — Failure is the design

  • For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
  • Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
  • Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
  • Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.

Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.

Day 5 — Scaling the hot path

  • Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
  • Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
  • Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
  • Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.

Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.

Day 6 — Keep the coding hand in, at the bar that applies to you

  • Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
  • Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
  • Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.

Deliverable: One problem hardened to review standard, with its test list and one written review comment.

Day 7 — Defend it while being interrupted

  • Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
  • Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
  • Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.

Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.

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.

Describe a time you had to optimize an existing system to improve energy efficiency.

Medium · behavioural and engineering judgement

Approach

  1. Name the disagreement and how you resolved it with evidence.
  2. Give the blast radius: what could have broken, and what you measured.
  3. Close with what you would do differently, concretely.

Follow-up

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

Describe a time you had to explain a complex technical issue to a non-technical client.

Medium · behavioural and engineering judgement

Approach

  1. State the situation in two sentences and spend the rest on the reasoning.
  2. Close with what you would do differently, concretely.
  3. Pick a story where you made the decision, not one where you watched it.

Follow-up

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

What steps do you take to ensure your work adheres to industry safety and compliance standards?

Medium · behavioural and engineering judgement

Approach

  1. Pick a story where you made the decision, not one where you watched it.
  2. Name the disagreement and how you resolved it with evidence.
  3. 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 would you do differently if you ran that again?

How do you handle high-pressure situations where a system failure requires an immediate fix?

Medium · behavioural and engineering judgement

Approach

  1. Name the disagreement and how you resolved it with evidence.
  2. Pick a story where you made the decision, not one where you watched it.
  3. 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?
  • How did you know your change caused the improvement?