As a Software Engineer at Acuiti Labs, you will play a pivotal role in designing and implementing robust technical solutions that drive the company’s mission. You will work at the intersection of complex software development and cutting-edge technologies, contributing to products that require high precision and scalability. This position is critical to the organization’s ability to deliver value to its clients through sophisticated engineering practices.
The work environment at Acuiti Labs is defined by professionalism and a focus on technical excellence. Whether you are working on AI-driven initiatives or core infrastructure, you will be expected to demonstrate a deep understanding of programming fundamentals and a commitment to high-quality code. This is an ideal role for engineers who thrive in collaborative, growth-oriented settings and who are eager to solve challenging problems in a fast-paced environment.
Initial Screening
reportedHalf 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 Assessment
reportedWhat 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
Professional Fit Discussion
reportedBecause the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.
What to demonstrate
- Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
- Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
- Whether you can be redirected in the middle of an answer without restarting it from the beginning
- Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it
How to prepare
- Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
- Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
- Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing
PracHub editorial advice for the preparation topics above.
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.
Paginating a growing table with limit and offset
Two unrelated defects share the idiom. Correctness: rows inserted or deleted between page requests shift the window, so a consumer walking an export skips rows and sees others twice, which for a customer-facing sync is silent data loss rather than an error anyone notices. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N times the page size and a deep page on a large table degrades from milliseconds to seconds. Keyset pagination over a stable, unique, indexed ordering -- where (created_at, id) < ($1, $2) order by created_at desc, id desc limit $3 -- is constant-cost per page and immune to shifting, on the precondition that the cursor columns never change value for a row, which disqualifies updated_at as a cursor.
Comparing floating-point values for equality, or holding money in them
Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.
Retrying a write that is not safe to repeat
A timeout tells you nothing about whether the server applied the write, so a blind retry of a create or a charge can duplicate it. Either make the operation idempotent, with a caller-supplied key the server deduplicates on or a conditional update, or do not retry it; and use exponential backoff with jitter so the retries of many clients do not synchronise into a second outage.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Hold a tenant to a trailing sixty-second request limit
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
- 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. - 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.
- 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. - 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. - 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.
- 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-Afterfrom the exact algorithm without a second scan?
Order a job dependency graph and find its critical path
A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.
Approach
- Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
- Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
- Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order:
earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E). - Second pass in reverse topological order for
latest_finish, thenslack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain.slack[v] = 0is exactly the statement that some longest path runs through v; equivalently, the longest path through v has lengthT - slack[v]. - The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by
min(d, T - L_avoid(v)), whereL_avoid(v)is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, whilemin(10, 15 - 14)predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan ismax(T - d, L_avoid(v)). - Compute
L_avoid(v)the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, sinceL_avoid(v)only ever matters through that max: setduration[v] := 0, recompute the makespan asT0(v) = max(T - duration[v], L_avoid(v)), and the gain ismin(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.
Worked solution 30 min
- Build four fixtures. A: 12 jobs, two branches of 100 s and 95 s that share no job. B: fixture A plus one back edge. C: two disjoint paths tied at 100 s. D: the shared-prefix case, one job of 10 s feeding a 5 s job and a 4 s job, so the longest path is 15 s and the runner-up is 14 s.
- Run Kahn; on fixture B confirm it emits fewer than V nodes, then run the residual-subgraph DFS and print the actual cycle.
- Compute
earliest_finishforward andlatest_finishbackward, and list the zero-slack set for each fixture. - For each zero-slack job v, recompute the makespan with
duration[v] := 0to getT0(v), and record both the correct boundT - T0(v)and the wrong one,T - second_longest_path, side by side. - Apply the shortening for real (20 s off the critical branch of A, 10 s off the shared prefix of D) and diff the recomputed makespan against each prediction.
Follow-up
- Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
- Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
- Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?
Seal an hour under late data with bounded memory
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
- Two clocks, two jobs. Bucket by
occurred_at, because that is the hour the customer is billed for, and advance the watermark oningested_at, because that is what the fold has consumed and whatsource_max_ingested_atrecords. Conflating them is what makes late data invisible. - 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. - 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. - 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.
- Hold open hours in a min-heap keyed by
hour_start. When the watermark advances, pop every hour withhour_end + L < Wand 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 intousage_rollup_hourlyasstatus='open'with arevisionbump. While an hour is open the row is upsertable, so the store is your overflow. - 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, andsource_max_ingested_atis 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?
Migrate a live partitioned event table without blocking ingest
usage_event is range-partitioned daily on ingested_at, holds roughly 250M rows per day across 400 live partitions, and is written at 10-40k rows/second. Two changes are required: quantity must move from double precision to numeric(20,6), and a new environment column must become NOT NULL with a default of 'production'. Ingest cannot stop. Give the ordered plan, naming for each step the lock it takes, what that lock blocks, and roughly how long it is held. Identify the one step that cannot be rolled back cleanly once traffic depends on it.
Approach
- Classify the two changes before planning anything. Adding a column with a non-volatile default has been metadata-only since PostgreSQL 11, so it is cheap. Changing double precision to numeric is not binary-coercible, so
alter column ... typerewrites every partition under ACCESS EXCLUSIVE and rebuilds its indexes; on this volume that is hours of blocked ingest and is simply not an option, which is why the plan is expand-and-contract rather than one statement. - Expand: add
quantity_numeric numeric(20,6)andenvironmentwith its default on the parent. Both are catalogue-only but both take a brief ACCESS EXCLUSIVE that cascades to partitions, so run each withlock_timeoutset to a second or two and retry on failure. A queued ACCESS EXCLUSIVE request blocks every reader behind it, which is how a metadata-only change turns into an outage. - Dual-write: deploy producer code that populates both columns on every insert, and leave it running before anything reads the new column. This is the step that cannot be reverted cleanly. Once readers depend on quantity_numeric, reverting the writer leaves rows with a null there, and the gap is only discoverable by re-reading the old column, which the readers have stopped doing.
- Backfill older partitions in batches keyed on the primary key, oldest first, committing every few thousand rows with a pause between batches, and skipping the partition still receiving writes until it rotates. Each batch is an ordinary UPDATE taking row locks only. The cost is bloat and WAL rather than blocking, so watch dead tuples and let autovacuum keep pace instead of wrapping 400 partitions in one transaction.
- Make NOT NULL cheap with the three-step form:
add constraint ... check (environment is not null) not valid(brief ACCESS EXCLUSIVE, no scan), thenvalidate constraint(SHARE UPDATE EXCLUSIVE, scans while reads and writes continue), thenset not null, which from PostgreSQL 12 uses the validated check and skips its own full scan. Do this per partition, then on the parent. - Switch and contract: move reads to the new column behind a flag, verify over a full period that both columns agree on freshly written rows, drop the old column (metadata-only), and only then remove the dual-write. Any index on the new column goes on with CREATE INDEX CONCURRENTLY per partition, since CIC is not supported on a partitioned parent: create the parent index with ONLY, build each child concurrently, then ALTER INDEX ... ATTACH PARTITION until the parent index becomes valid.
Follow-up
- A CREATE INDEX CONCURRENTLY fails halfway through the partition list. What state is the table in, how do you detect it, and what do you run?
- The producer computes quantity itself. What happens to a request already in flight when the dual-write deploy lands, and does it matter?
- Give two queries that prove the backfill is complete: one cheap enough to run every minute, one authoritative.
Rebuild an hourly rollup with deduplication and late-arrival accounting
From usage_event (event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at), produce the values usage_rollup_hourly should hold for one tenant over one day: per (workspace_id, sku, hour_start) the deduplicated quantity_sum, event_count and source_max_ingested_at, bucketed by occurred_at. Duplicates share (tenant_id, idempotency_key). Also report, per hour, the running total across the day and the share of quantity that arrived more than two hours after the hour began. Write the query, and state which duplicates a daily unique index cannot catch.
Approach
- Deduplicate in its own CTE before any aggregation, because a SUM cannot be un-summed:
row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id) = 1. Include the tiebreaker. Without it the surviving row is non-deterministic when two duplicates share an ingested_at, and a rollup described as deterministically recomputable then disagrees with itself between runs. - Bucket on occurred_at and nothing else, and pin the timezone explicitly.
date_trunc('hour', timestamptz)truncates in the session's TimeZone setting, so the same query run by a session set to a non-UTC zone buckets differently; use the three-argumentdate_trunc('hour', occurred_at, 'UTC')on PostgreSQL 16 or later, ordate_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC'before that. Filterenvironment = 'production'explicitly, since metering covers three environments and billing covers one. - Aggregate to the grain with
sum(quantity),count(*)andmax(ingested_at). The last is not decoration: it is the watermark the row consumed up to, and without it there is no way to prove afterwards what a number did and did not include. - Compute the late share inside the dedup-and-aggregate step as a conditional aggregate,
sum(quantity) filter (where ingested_at > hour_start + interval '2 hours'), then divide by the hour's total. Compute the running total as a window over the already aggregated rows:sum(quantity_sum) over (partition by workspace_id, sku order by hour_start rows between unbounded preceding and current row). Running either over raw rows puts the duplicates back. - Answer the index question exactly. The unique constraint is on (ingested_day, tenant_id, idempotency_key), because a unique index on a partitioned table must contain the partition key. It therefore deduplicates only within one ingest day and admits a duplicate whose retry crosses midnight or whose replay runs a week later. That is why this CTE dedups across the whole window being recomputed, and why the dedup horizon is a correctness parameter rather than a retention cost.
- Keep the numeric type all the way through. quantity is numeric so the sums are exact; a cast to double precision anywhere in this pipeline reintroduces drift that surfaces only as a few unreconcilable cents per tenant per month, long after the query is out of anyone's mind.
Worked solution 40 min
- Generate a day of events for one tenant across three workspaces and two SKUs, inject 2% duplicates sharing an idempotency_key with a later ingested_at, and inject 1% whose ingested_at is six hours past occurred_at.
- Write the dedup CTE alone and confirm its row count drops by exactly the injected duplicate count.
- Aggregate and compare quantity_sum against a ground truth computed outside SQL from the generator's own records.
- Add the two window and filtered aggregates, and confirm the final hour's running total equals the ungrouped day total for each (workspace, sku).
- Run the whole query twice over static data and diff the two outputs.
- Re-run with the session TimeZone set to a non-UTC zone and confirm hour_start does not move.
Follow-up
- A dispute forces the same recompute over 40 days for one tenant. What changes about the dedup CTE's memory use and the chosen plan, and what would you do about it?
- Two runs a minute apart return different quantity_sum values for an hour that is already closed. Give two mechanisms that produce that, and the single query that distinguishes them.
- Express the same rollup incrementally so it does not re-scan the day each time the watermark advances. What does the incremental version stop being able to answer?
What are the primary differences between memory management in the lang…
What are the primary differences between memory management in the languages you are most proficient in?
Approach
- Clarify what is being asked and what a complete answer contains.
- Say what you would check first and why it is the highest-information step.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Describe the logic behind your approach to the coding challenge provid…
Describe the logic behind your approach to the coding challenge provided during the screening.
Approach
- Clarify what is being asked and what a complete answer contains.
- State your assumptions explicitly before working the problem.
- Say what you would check first and why it is the highest-information step.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Per-tenant rate limiting that holds across one hundred pods
The same gateway enforces each tenant's plan rate limit in requests/second and its monthly quota. Traffic is heavily skewed: a handful of tenants exceed all others combined, while a small tenant's requests may all land on one of the 120 pods. Both limits must hold in aggregate rather than per pod, per credential or per region, and the check may add at most 2 ms at p99. Specify the algorithm, where the state lives, the exact operation performed per request, and the behaviour when the counter store is unreachable - separately for the rate limit and the quota.
Approach
- Do the arithmetic on per-pod buckets before discarding them, and be precise about which half fails. Giving each pod rate/N keeps the sustained aggregate correct - 120 buckets of 5 rps sum to the 600 rps limit - but a tenant whose traffic all lands on one pod is throttled at 5 rps, a hundred and twentieth of what it pays for. The over-admission comes from burst depth, not sustained rate: if each pod carries a full bucket depth B, the aggregate burst is 120B against a limit that intended to allow B.
- Move the rate limit to a shared counter accessed in exactly one atomic round trip - a token bucket or GCRA as a single server-side script returning allow/deny plus a retry-after - never a read followed by a write. Keep the key tenant-scoped and region-local; a globally shared counter costs a cross-region round trip that alone exceeds the 2 ms budget, so shard the global limit per region in proportion to observed regional share and accept that a tenant whose traffic shifts regions is briefly throttled until the shards rebalance.
- Cut round trips for the largest tenants with leased tokens: a pod leases 50 at a time and spends them locally, reducing counter round trips by 50x at the cost of bounded over-admission of pods x lease_size, which is 6,000 requests in the worst case here. Shrink the lease for tenants with small limits, where that bound would swamp the limit itself.
- Treat the quota as a different problem, not a longer window. It is a count that other in-flight requests are changing, so select-the-usage-then-insert is write skew: read-committed permits it and repeatable read permits it too, because snapshot isolation is exactly what allows two transactions to read a consistent count and both write. Enforce it inside one statement - an atomic increment whose returned value is compared, or UPDATE ... SET used = used + $1 WHERE used + $1 <= limit RETURNING - or with a constraint that makes the surplus write fail.
- Write the unavailability policy per limit rather than one policy for both. The rate limit fails open to a conservative local bucket, because it protects capacity and failing closed turns a counter blip into a total outage. The quota fails open for tenants whose last known usage was well below the cap and closed for those at or above it, since that is the only decision with revenue attached, and reconciles from the usage ledger on recovery - which is why the monthly counter must be re-seedable from that ledger rather than being the only copy.
Worked solution 30 min
- For a 600 rps limit across 120 pods, compute the sustained rate and the aggregate burst under local buckets of depth B, then the throughput of a tenant whose traffic all lands on one pod.
- Write the shared-counter operation as a single atomic script: its inputs, its return value, and the number of network round trips per request.
- Size a lease of 50 tokens: compute the worst-case over-admission as pods x lease_size and the reduction in round trips, then decide the smallest limit at which the lease is still acceptable.
- Write the two unavailability policies side by side and the reconciliation step that runs when the counter store returns.
Follow-up
- One tenant is 40% of all traffic and its counter key lands on a single shard. What do you change, and what does that cost in accuracy?
- Clients receive 429s and retry. What stops the well-behaved ones from synchronising into a herd?
- A tenant reports being throttled below its limit. Which evidence do you produce, and does your design emit it today?
Gateway p99 spikes on a five-minute cadence
edge-gateway caches each credential-to-authorisation-context decision for five minutes. p99 sits at 6 ms except for a spike to 900 ms roughly every five minutes, worst in the region with the most pods, and control-plane CPU and read latency rise in step with it. The error rate stays near zero. Customers are told a revoked credential stops authorising within 60 seconds. Give the ordered checklist that identifies the mechanism, and a fix that removes the spike without weakening the 60-second bound.
Approach
- Test periodicity before anything else: take the spike timestamps modulo the TTL in seconds. A tight cluster at a fixed offset means expiry phase, while traffic-driven spikes scatter.
- Overlay pod start times. Entries filled at first request inherit the phase of the pod that filled them, so a cohort of pods deployed together expires together and the amplitude should track cohort size rather than tenant count.
- Separate a herd from a capacity shortfall by measuring control-plane requests per second during a spike against baseline. A stampede shows a step of roughly (pods x hot keys) for one interval with hit rate collapsing to near zero, not a gradual climb that would indicate the dependency is simply undersized.
- Apply three independent controls: randomise each key's TTL by a factor drawn uniformly from something like 0.8 to 1.0 so cohorts de-phase; coalesce concurrent misses per key per pod so exactly one refresh is in flight; and serve the stale value while that refresh runs so a miss costs the stale read rather than the dependency's queue.
- Bound staleness against the published contract rather than against comfort: serve-stale is admissible only up to the 60-second revocation bound, so the TTL floor and the stale window together must stay inside it, and the published invalidation must delete the entry rather than schedule a refresh.
- Decide in advance what a miss does when the control plane is unreachable, because that is now the only uncached path: failing closed converts a dependency outage into a total outage, while extending stale service past the bound breaks the revocation promise. Pick one and configure it explicitly.
Follow-up
- Publish-subscribe invalidation is lossy under a partition. Given that, what actually enforces the 60-second bound, and what number would you put in the contract if asked to defend it?
- One tenant's key is hot enough that a single pod's coalesced refresh still matters. What changes?
- Would a shared cache tier in front of the control plane help or shift the problem, and what new failure does it add?
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.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Coding, 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.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Design, 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.
Practice prompt ↗Practice prompt ↗03Fundamentals 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.
Practice prompt ↗Practice prompt ↗04The 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.
Practice prompt ↗Practice prompt ↗Worked solution ↗05First 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.
Practice prompt ↗Practice prompt ↗06Second 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.
Practice prompt ↗Practice prompt ↗07Reassemble 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.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Team size, service count and tickets closed say very little. Seniority shows in the decision you owned: what you chose not to build, which constraint you traded away, whose objection you had to resolve before anything could move. A large project where you executed someone else's plan is a small story.
Walk us through your resume and highlight the projects where you faced…
Walk us through your resume and highlight the projects where you faced the most significant technical hurdles.
Approach
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
How do you balance speed of delivery with the need for high-quality, b…
How do you balance speed of delivery with the need for high-quality, bug-free software?
Approach
- Close with what you would do differently, concretely.
- State the situation in two sentences and spend the rest on the reasoning.
- 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?
- How did you know your change caused the improvement?
How do you handle feedback regarding your code or design choices?
How do you handle feedback regarding your code or design choices?
Approach
- Pick a story where you made the decision, not one where you watched it.
- Give the blast radius: what could have broken, and what you measured.
- 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?
What interests you about the specific engineering challenges we are so…
What interests you about the specific engineering challenges we are solving here?
Approach
- Close with what you would do differently, concretely.
- Pick a story where you made the decision, not one where you watched it.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
- 01
Walk us through your resume and highlight the projects where you faced the most significant technical hurdles.
- 02
How do you balance speed of delivery with the need for high-quality, bug-free software?
- 03
How do you handle feedback regarding your code or design choices?
- 04
What interests you about the specific engineering challenges we are solving here?
Is this an official Acuiti Labs interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Acuiti Labs. Rounds and questions reflect what candidates have reported, not a process Acuiti Labs has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long does the entire interview process usually take?
The process typically spans a few weeks, often concluding in roughly 20 days. The team values transparency, so keep an open line of communication with your recruiter regarding your status.
PracHub interview research ↗What is the most important thing to focus on for preparation?
Focus on your technical fundamentals—specifically coding skills in your primary language and your ability to explain your previous work. Being able to articulate the "why" behind your code is just as important as the code itself.
PracHub interview research ↗How difficult is the technical assessment?
The difficulty is generally considered average, provided you have a solid grasp of your chosen programming language and core data structures. Focus on writing clean, logical code rather than attempting to memorize complex algorithms.
PracHub interview research ↗Is the culture at Acuiti Labs collaborative?
Yes, the environment is highly professional and emphasizes clear communication and teamwork. They value engineers who are transparent about their work and open to feedback.
PracHub interview research ↗Sources & methodology 3 sources ↗
No official company page is cited. Rounds and questions come from candidate reports and PracHub editorial material; each source shows the date it was read.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
PracHub page · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-22
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