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Appetiser Apps Software Engineer Interview Questions & Guide 2026

Take-home style practice for Appetiser Apps software engineer candidates: a small service, a fixed budget, a written tradeoff.

The 60-second brief

As a Software Engineer at Appetiser Apps, you are a critical contributor to the end-to-end development of high-impact digital products. The role is centered on building scalable, user-centric applications, requiring you to bridge the gap between complex business requirements and elegant technical execution. You will work within a dynamic environment where your code directly influences the experience of diverse users across various industries.

Prepare in one language you know well enough to debug in rather than the one you think reads best. Under a clock an unfamiliar language costs you standard-library lookups and iteration mechanics, and that time comes out of your thinking budget, not your typing budget.

Focus areas

  • Scope every query and cache key by tenant
  • Keep money in integer minor units
  • Build at-least-once pipelines with explicit deduplication horizons

Overview

This position is inherently collaborative, placing you in teams where you will tackle challenges ranging from API architecture and frontend performance to backend optimization. You aren't just writing code; you are solving specific, real-world problems. Success here requires a blend of technical precision, a proactive approach to refactoring, and the ability to communicate technical concepts clearly to both developers and non-technical stakeholders.

The Appetiser Apps interview process

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

Initial Interaction

When a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.

What to demonstrate

  • Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
  • Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
  • Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
  • Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened

How to prepare

  • Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
  • Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
  • Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure

Technical Evaluations

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

Demonstrate Skills

You cannot drill a format you do not know, so put the preparation into material that travels. Three pieces of your own work, each rehearsed until you can take a follow-up you did not anticipate, will carry a conversation or a code walkthrough equally well. Specificity is what separates that from filler. A number needs its definition before it means anything: a p99 is over some window and measured at some hop, and a server-side figure excludes the queueing and network time a client would see. The number you cannot qualify is the one to leave out.

What to demonstrate

  • Whether your examples carry detail only someone who did the work would hold, such as what the binding constraint actually was, which alternative you rejected and why it was worse, and what you measured on each side of the change
  • Whether a number survives one follow-up, meaning you can say what it was measured over and whether it moved because of your change or merely alongside it
  • Whether a failure is described with the specific change that followed it, rather than a lesson stated in general terms
  • Whether your part in a team effort is stated accurately, including what other people did

How to prepare

  • Write a page on each of three projects covering the constraint, the option you rejected, the measurement before and after, and what went wrong. Cut any line you cannot take a follow-up on, since you are writing the parts you will be pressed on rather than a summary.
  • Recover the real figures while you still have access: request volume, data size, latency with its percentile and window, team size, timeline. Note where each came from, whether a dashboard, a design document or memory, and mark the estimates so you can say which they are out loud.
  • Take your weakest project story to someone who works in a different area and have them ask why four times in succession. The point where you run out of answer is the part to go and re-read before the round.

Where candidates lose points

One shared connection pool for every tenant and every query class

A single tenant with a large table and a missing index can occupy every connection with slow queries, and every other tenant then waits in connection acquisition -- a queue invisible in database metrics, because the database itself looks healthy while the application starves. Containment is bulkheads: separate pools or per-tenant concurrency caps for interactive requests, background jobs and exports, a statement timeout low enough that a pathological query dies before it accumulates, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks indefinitely. One caveat worth knowing in advance: if a transaction-pooling proxy sits in front of the database, session-scoped behaviour changes, so session-level advisory locks and settings applied outside a transaction do not survive the way they do on a direct connection.

Holding money in a floating-point type, or rounding it more than once

Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.

Sharing mutable state with no stated owner

Say which thread, request or task owns each mutable structure, and what protects it when the answer is more than one: a lock, a queue that hands ownership across, or an immutable copy per reader. A structure documented as safe for concurrent reads is usually not safe for a concurrent write alongside those reads.

Designing for a scale nobody asked for

Ask for request rate, data size, read-to-write ratio and expected growth, then size the simplest option first; one relational instance on current hardware covers a large share of real workloads. Reaching for shards, queues and a cache tier before any number has been quoted reads as pattern-matching rather than judgement.

Questions & practice

Seal an hour under late data with bounded memory

Hard · watermarks · late-data · quantile-sketch · conditional-write

Metering ingest reads 256 partitions at 10,000 to 40,000 events/second. Events carry occurred_at and ingested_at, and during a producer replay the gap between them is hours. Seal each UTC hour once no more than 50 parts per million of that hour's eventual quantity can still arrive, using memory that does not grow with the size of the replay. Define the watermark, the lateness parameter and how you measure it, the structure holding open hours, and the write that performs the seal. State what an idle partition does to your watermark.

Approach

  1. Two clocks, two jobs. Bucket by occurred_at, because that is the hour the customer is billed for, and advance the watermark on ingested_at, because that is what the fold has consumed and what source_max_ingested_at records. Conflating them is what makes late data invisible.
  2. The global watermark is the min over partitions of each partition's committed ingested_at, not the max: the fold is trustworthy only as far as the slowest partition. The consequence is that one idle partition pins the watermark forever and nothing seals, so an idle partition must promote its watermark to wall clock after a stated idle timeout, and that timeout becomes a correctness parameter, because a partition that is slow rather than idle gets sealed past.
  3. Choose the lateness L from the measured distribution of ingested_at - occurred_at, weighted by quantity rather than by event count. The target is 50 ppm of the hour's quantity, and a replay is rare in events while carrying disproportionate mass, so an event-weighted quantile picks an L that is comfortably wrong at exactly the moment it matters.
  4. Measure that quantile in bounded memory. A Greenwald-Khanna summary gives epsilon-approximate quantiles in O((1/epsilon) log(epsilon n)) space; a t-digest costs more per merge but has relative error that tightens at the tails, which is the half of the distribution you are reading at p99.99. Keep a separate summary per tenant class, because one tenant's batch importer is not the population.
  5. Hold open hours in a min-heap keyed by hour_start. When the watermark advances, pop every hour with hour_end + L < W and seal it: O(log H_open) per advance and O(1) amortised per event to touch its bucket. Memory is open hours multiplied by distinct (tenant, workspace, sku) keys, so cap the number of simultaneously open hours and spill the oldest into usage_rollup_hourly as status='open' with a revision bump. While an hour is open the row is upsertable, so the store is your overflow.
  6. The seal itself is a conditional write: update ... set status='sealed', sealed_at=now() where status='open' returning .... Two sealers race on every restart, and the loser must see zero rows and stop rather than write a second value. After the seal, an event for that hour is not an upsert but an adjustment, and source_max_ingested_at is what proves it arrived afterwards.

Follow-up

  • A replay starts during the sealing window for a period you are about to close. What do you do, and what is the customer-visible consequence of each option?
  • Your measured quantity-weighted p99.99 lateness is six hours and the invoice must be issued at 02:00 UTC on the first. How do you reconcile those two numbers?
  • How would you detect that L has drifted before it costs you an hour's quantity?

Schedule ordered webhook retries with a heap of subscription queues

Medium · heap · scheduling · backoff · head-of-line-blocking

Design the in-memory scheduler for webhook delivery. Up to 20 million rows sit in status pending or failed_retryable across 200,000 subscriptions, each row carrying next_attempt_at and attempt_count, and each endpoint having a circuit breaker. Deliveries for one subscription must be attempted in order, so at most one attempt per subscription may be in flight. Support due(now), complete(delivery, outcome) and insert(delivery) in O(log S), where S is the subscription count rather than the delivery count. Give the backoff formula you schedule retries with.

Approach

  1. Key the global heap by subscription, not by delivery. Each subscription owns a FIFO of its due deliveries in event order; the heap holds one entry per eligible subscription, keyed by its head's next_attempt_at. That is 200,000 heap entries instead of 20 million, and it makes the one-in-flight rule structural rather than a check somebody can forget.
  2. due(now): peek the minimum. If its key is in the future, sleep until then instead of spinning. Otherwise pop it, move the subscription into an in-flight set, and do not re-push it. A subscription absent from the heap cannot be dispatched twice, which is precisely how ordering is preserved.
  3. complete: on success, drop the head and re-push the subscription keyed by its new head, or leave it out when the queue empties. On a retryable failure, increment attempt_count and set next_attempt_at = now + uniform(0, min(cap, base * 2^attempt)), sampled uniformly across the whole interval. That is full jitter; deterministic backoff re-synchronises the herd you just created.
  4. Circuit breaker: park the subscription in a second heap keyed by its half-open time, so an endpoint dead for six hours costs one heap entry and zero attempts rather than consuming worker slots. Admit exactly one probe at half-open and close the breaker only on its success.
  5. Say the price of the ordering guarantee out loud. One in-flight attempt per subscription means an endpoint answering in 10 seconds drains at 0.1 deliveries/second however many workers you run, and its backlog grows until it recovers. If the customer does not need order, allow k in flight and document delivery as unordered; that is the trade, and it is a product decision.
  6. All three operations are O(log S) with O(S) resident heap memory and the queues themselves backed by the store. The database-backed equivalent is a partial index on (subscription_id, next_attempt_at) where status in ('pending','failed_retryable') claimed with FOR UPDATE SKIP LOCKED, and the write-back must be fenced on lease_token so a worker that stalled and resumed cannot overwrite a newer attempt.

Worked solution (30 min)

  1. Define the four structures explicitly: queues: subscription_id -> deque[delivery], ready: min-heap of (next_attempt_at, subscription_id), inflight: set[subscription_id], breaker: min-heap of (half_open_at, subscription_id).
  2. Write down the invariant you will assert after every operation: a subscription appears in at most one of ready, inflight and breaker, never in two.
  3. Implement due, complete and insert, then simulate 200,000 subscriptions with Zipf-distributed queue depths totalling 20 million deliveries.
  4. Add one endpoint that always times out after 10 seconds and one that always answers in 20 ms, then measure the fast endpoint's throughput with and without the per-endpoint breaker.
  5. Instrument heap size across the run.

Expected result: Heap size stays at or below 200,000 regardless of the 20 million deliveries. The fast endpoint's throughput is unaffected by the dead one once the breaker trips. The dead endpoint's deliveries accumulate in their own deque and cost exactly one heap entry.

Verify it yourself

  • The asserted invariant never fires under concurrent workers.
  • Replaying the same delivery set twice produces identical per-subscription ordering, even though the interleaving across subscriptions differs.
  • Replacing full jitter with fixed backoff makes retry attempts cluster visibly in a histogram of attempt timestamps.

Follow-up

  • One subscription has 4 million queued deliveries. What stops it from starving the other 199,999, and what does your heap look like under that load?
  • A customer requests redelivery of last Tuesday's events. Where do those rows enter your structure, and what keeps them from reordering live traffic?
  • The process restarts. How much state do you rebuild, and what stops every subscription from being attempted in the same second?

Hold a tenant to a trailing sixty-second request limit

Medium · sliding-window · two-pointer · rate-limiting · tenant-skew

The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.

Approach

  1. Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request.
  2. Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
  3. Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate, prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact.
  4. Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (tokens, last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual.
  5. Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
  6. Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.

Follow-up

  • One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
  • Quotas rather than rate limits: the check is select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes.
  • How do you return an accurate Retry-After from the exact algorithm without a second scan?

Paginate a tenant's delivery export without skipping rows

Easy · keyset pagination · indexing · multi-tenancy

A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.

Approach

  1. Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
  2. Replace the position with a value predicate over a stable, unique, indexed ordering: where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window.
  3. Index (tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit.
  4. Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
  5. Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.

Worked solution (20 min)

  1. Load about 50k deliveries for one tenant, then walk them with the OFFSET query while a writer inserts 10 rows/second, collecting every returned delivery_id.
  2. Compare the distinct ids collected against the set of ids that existed when the walk started, and record the shortfall.
  3. Repeat the walk with the keyset query and confirm every pre-existing id is returned exactly once.
  4. Run explain (analyze, buffers) on page 1 and page 500 of each form and compare shared buffer hits.

Expected result: A keyset query using the row comparison (created_at, delivery_id) < ($2, $3) with order by created_at desc, delivery_id desc limit 100, backed by (tenant_id, created_at, delivery_id) scanned backwards; plus the statement that OFFSET drops rows under concurrent insert and costs time proportional to the offset, and that updated_at is disqualified because it mutates on retry.

Verify it yourself

  • Buffer counts for page 500 are the same order of magnitude as page 1.
  • Every id present before the walk is returned exactly once when only inserts run concurrently.
  • The query plan shows no Sort node above the index scan.
  • Two rows sharing a created_at value to the microsecond are both returned across the page boundary.

Follow-up

  • The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added created_at <= $snapshot bound, and a materialised export table.
  • A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
  • The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?

Model credential revocation so history survives the delete

Easy · soft delete · partial indexes · check constraints · audit history

tenant_api_key stores key_id, tenant_id, workspace_id, name, key_prefix, secret_hash, scopes text[], status (active, revoked, expired, compromised), auth_version, created_at, expires_at, last_used_at, revoked_at, revoked_reason. Rotation inserts a new row and revocation never deletes, because an incident review asks which credential served a request last quarter. Write the constraints that enforce: a label is unique only among a tenant's live keys, revoked_at and status can never disagree, and scopes is never empty. Then write the authentication lookup predicate, and name one column in this table that must stay out of it.

Approach

  1. Reach for a partial unique index rather than a plain UNIQUE: create unique index on tenant_api_key (tenant_id, name) where revoked_at is null. Any number of revoked rows may share a label, the live namespace stays unique per tenant, and the revoked majority is not in the index at all, so it stays small on a table that only grows.
  2. Tie the nullable timestamp to the enum so the two cannot drift: check ((revoked_at is not null) = (status in ('revoked','compromised'))) and check ((revoked_at is null) = (revoked_reason is null)). A revocation that records no reason is the one an incident review cannot use.
  3. Write the emptiness check as check (cardinality(scopes) > 0), not array_length(scopes, 1) > 0. array_length returns NULL for an empty array, a CHECK constraint passes when its expression is NULL, so the array_length version accepts exactly the value it was written to reject.
  4. Make the lookup a single index probe with every liveness condition inside it: where secret_hash = $1 and revoked_at is null and (expires_at is null or expires_at > now()) and auth_version = $2, backed by a unique index on secret_hash. Nothing is filtered in application code, so there is no path that forgets a clause.
  5. Keep last_used_at out of that predicate. It is written asynchronously and is allowed to lag by a minute, so it is a usage signal; feeding it into an authorisation decision makes the decision depend on a write that may be late, batched away or lost.
  6. Flag the modelling smell while you are here: expired is derivable from expires_at < now(), so storing it as a status obliges a job to keep it true and guarantees the column is wrong between the expiry instant and that job's next run. Derive it in the predicate; keep the stored status for states that are decisions rather than clock readings.

Follow-up

  • Rotation issues a replacement while the old key stays live for a 30-day overlap. What does the uniqueness rule become, and what does the UI show to tell two same-named keys apart?
  • A password reset bumps the principal's auth_version. No row in this table changed. How does the next request fail, and what query counts how many keys that bump just killed?
  • A key turns up in a public repository. Which columns let you find it, and what do you write to the row?

Describe Laravel and Vue JS in your own words.

Medium

Approach

  1. State your assumptions explicitly before working the problem.
  2. Work from the requirement backwards to the design.
  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?

Explain what Middleware is and how it functions.

Medium

Approach

  1. Say what you would check first and why it is the highest-information step.
  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?

Specify webhook signature verification a customer can implement

Easy · webhooks · hmac · replay protection

The webhook-delivery service signs each payload before POSTing it to a customer endpoint. Write the signature specification a customer implements in their own language: the header format, exactly which bytes are signed, the algorithm, how replay is bounded, and how a signing secret rotates without a delivery gap. Then write the verification steps the customer performs, in order, including what they compare and what they return on failure. Constraint: most customers reach for their web framework's parsed JSON body by default. Deliverable: the spec section plus reference pseudocode.

Approach

  1. Sign the concatenation of the timestamp and the raw body, t + "." + body, and emit a header of the form t=<unix seconds>,v1=<hex>. The timestamp has to be inside the MAC, or an attacker re-stamps a captured body and the tolerance window buys nothing.
  2. Require the raw request bytes. A framework that parses JSON and re-serialises it changes key order, whitespace and number formatting, so the spec must tell the customer to capture the body before the parser runs and give the middleware note for each common framework.
  3. Use HMAC-SHA256, not sha256(secret || body): SHA-256 is a Merkle-Damgard construction, so the naive form admits length extension. Require a constant-time comparison as well, since a short-circuiting byte compare leaks the expected prefix under repeated probing.
  4. Bound replay in two layers: reject when |now - t| exceeds a stated tolerance such as 300 seconds, then deduplicate on the event identifier header. The tolerance is what makes the customer's dedup store finite rather than unbounded.
  5. Rotate by allowing two live secrets and emitting both signatures in one header (v1=<old>,v1=<new>); the customer accepts if any candidate matches, so neither side needs an instantaneous cutover. A failed verification returns 400 and the body is not processed.

Worked solution (15 min)

  1. Write the header grammar and one real example line with a plausible timestamp and hex digest.
  2. Write the signed string construction explicitly as a byte concatenation, and add the sentence telling the customer where in their framework to obtain the raw body.
  3. Write the five verification steps in order: extract t and candidates, check the tolerance, recompute the HMAC over t + '.' + raw body, compare in constant time against each candidate, then deduplicate on the event identifier.
  4. Add the rotation paragraph: two active secrets, both signatures sent, overlap window stated in the dashboard.
  5. State the failure response and the fact that the payload is not processed, plus what the sender does with that 400.

Expected result: A spec where the signed input is the timestamp joined to the exact raw bytes, the algorithm is HMAC-SHA256 compared in constant time, replay is bounded by a stated tolerance plus event-id deduplication, and rotation is an overlap of two accepted signatures rather than a cutover.

Verify it yourself

  • The timestamp is inside the signed string, not merely sent alongside it.
  • The spec names raw bytes and says where the customer gets them before parsing.
  • Rotation requires no coordinated instant on either side.
  • The reason for HMAC over sha256(secret || body) is stated as length extension, not as a preference.

Follow-up

  • A customer's verification passes locally and fails in production behind a proxy that re-encodes the response body. Where do you look first?
  • Why sign with a per-endpoint secret rather than the tenant's API key?

A rare job-run overwrite that logging makes disappear

Hard · fencing tokens · leases · concurrency

About one job run in fifty thousand records billable_seconds matching no observed sandbox lifetime, and a few rows show worker_id changing after finished_at was already set. It does not reproduce: debug logging around the terminal write made it vanish for two weeks before it returned. Runs last from 200 ms to 30 minutes, the lease is 60 seconds and is renewed while a run executes. Give an ordered checklist, the mechanism, and a fix that makes the illegal write impossible rather than merely rarer.

Approach

  1. Mine the evidence instead of chasing a repro: select rows where updated_at is later than finished_at, or where a terminal status was written twice, and join them to attempt history to recover both writer identities. The defect has already happened tens of times and the rows are the recording.
  2. State the signature before measuring it. If the cause is a lease that expired while the original worker was stalled, affected runs should cluster where the gap between the last renewal and the terminal write exceeds the lease, and should correlate with worker pause metrics rather than with workload shape.
  3. Read the disappearance honestly. Logging inside the window changed the timing and lowered the probability; it is evidence about how narrow the window is, not a fix. Reproduce by widening the window on purpose, shortening the lease and injecting a pause between sandbox exit and the terminal write, rather than by adding more instrumentation.
  4. Name the mechanism precisely: the lease expires during a stall such as a long garbage-collection pause or a brief partition, the run is re-dispatched, and the original worker then wakes and writes its terminal state over the new attempt's row. A lease alone cannot stop this, because the check and the write are separated by the stall.
  5. Fix by fencing the write itself: UPDATE job_run SET status = $2, finished_at = $3, billable_seconds = $4 WHERE run_id = $1 AND status = 'running' AND lease_token = $5, with zero rows affected interpreted as having been fenced rather than as success. The token lives on the row so the store arbitrates, not the worker's memory.
  6. Keep the state machine honest: a retry inserts a new row pointing at parent_run_id rather than resetting the old one, and a run whose worker vanished terminates as lost with billable_seconds null, because recording failure asserts an outcome nobody observed and then bills and retries on that assertion.

Follow-up

  • The supervisor also emits a usage event on completion. What does the fenced worker do about the event it already emitted, and how does metering absorb it?
  • Lease renewal is itself a network call. What happens when a renewal times out, and how does the worker decide whether it still holds the lease?
  • Why is lengthening the lease past the longest legitimate run the wrong lever, and what breaks if you do it anyway?

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.

Review and mentoring: what you let through

Counting review comments or mentees proves nothing. The useful version is a specific change you approved with a reservation you stated, or one you blocked and the delay that cost. Say which standard you were holding and why it was worth the friction. A mentoring story needs the thing the other person can now do without you.

What do you know about us?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

What is your vision for the company?

Medium · behavioural and engineering judgement

Approach

  1. Give the blast radius: what could have broken, and what you measured.
  2. Pick a story where you made the decision, not one where you watched it.
  3. Name the disagreement and how you resolved it with evidence.

Follow-up

  • How did you know your change caused the improvement?
  • What would you do differently if you ran that again?

How do you address the communication gap between you and a teammate when working in a remote position?

Medium · behavioural and engineering judgement

Approach

  1. Name the disagreement and how you resolved it with evidence.
  2. State the situation in two sentences and spend the rest on the reasoning.
  3. Pick a story where you made the decision, not one where you watched it.

Follow-up

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

Describe one time you received criticism from your manager; how did you receive it?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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