At Buildsafe Australia, the Software Engineer role maintains the digital systems that support the company's safety solutions. Buildsafe Australia works in fall prevention and height safety and relies on its systems to manage complex logistics, compliance tracking, and operational efficiency. You will be responsible for keeping the company's infrastructure resilient, scalable, and able to support a high-growth environment.
The position is about more than writing code: it connects engineering requirements with the practical, safety-critical needs of Buildsafe Australia's field teams. You will work in a collaborative, fast-paced environment where your technical decisions directly affect the safety and productivity of the company's nationwide operations. Whether you are optimizing internal systems or scaling the company's infrastructure, your work shapes how Buildsafe Australia delivers on every job site.
Initial Screening
reportedThe person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.
What to demonstrate
- Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
- Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
- Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural
How to prepare
- Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
- Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
- Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one
Technical Deep Dives
reportedInput 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
Behavioral Assessments
reportedWhat you say here is written down by each interviewer and compared afterwards, so the unit of evaluation is a claim someone else could check, not a well-told narrative. Two things make a story checkable: detail only a participant would hold, and a clean line around which part was yours. Vague ownership is the usual failure and it is usually accidental, because engineers say we about the team's work and we about their own, so the thing they personally built disappears into the plural. Name the part you wrote, and name who did the rest.
What to demonstrate
- Whether your details are ones a participant would hold and an observer would not: the constraint that ruled out the obvious approach, the first attempt that failed, the person who objected and on what grounds
- Whether ownership survives a direct question, since a follow-up to we decided is routinely who decided, and an answer that stays plural at that point is read as the work belonging to someone else
- Whether the numbers you quote are ones you would say identically to a former colleague with the dashboard open
How to prepare
- Go through each story replacing every we with either I or a named role (the on-call engineer, the reviewer, the other team) and check the story still holds together. Wherever it stops making sense you have found a part you cannot actually speak to
- Open the artefacts for two of your stories, the pull request, the design doc, the incident notes, and read them for dates and figures you have been rounding in the retelling. Correct your version to match
- For each story write the single sentence you would least want repeated to a former teammate, then either make it accurate or take it out
Final Assessment
reportedNobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.
What to demonstrate
- Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
- Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
- Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
- Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience
How to prepare
- Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
- For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
- Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed
PracHub editorial advice for the preparation topics above.
Checking a quota with a select and then writing
Under read-committed isolation, two concurrent transactions both observe a count below the limit and both insert, so the limit is exceeded by exactly the concurrency. Repeatable read does not rescue it either: it provides a stable snapshot, and this is write skew, which snapshot isolation permits by design. The options are serialisable isolation, which detects the conflict and aborts one transaction with a serialisation failure and therefore obliges the caller to retry; a single statement with the predicate inside the write; or a constraint that makes the surplus insert fail outright. The reason this pattern survives review is that it is correct in every test that runs one request at a time.
Treating a timed-out write as a failed write
A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.
Reading the constraints as preamble rather than as part of the problem
The bounds are usually there to eliminate the obvious approach: n up to 10^5 makes an O(n^2) scan roughly 10^10 operations, far outside any per-test time budget, and an input larger than memory rules out loading it at all. When a bound is not given, ask for it, then say out loud which approach it kills.
Hardcoding to the sample inputs
Solve the stated problem rather than the two examples; special-casing a literal to make a sample pass is obvious immediately and reads as either a misunderstanding or an attempt to fake progress. If you genuinely cannot generalise yet, say which part is a stub and what would replace it.
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.
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?
Locate a billing reconciliation gap without rescanning ninety million events
A tenant's sealed invoice total is 0.4% below the sum of its raw usage_event rows for the period. That tenant has 90 million events over 30 days in a table partitioned daily on ingested_at, and its rollups carry source_max_ingested_at, revision and sealed_at. Recomputing all 30 days from raw is correct, and you are not going to do it. Give the procedure that locates the divergent (workspace, sku, hour) cell, the cost of each probe, and the one query you run before any of it.
Approach
- Run the free query first. Sum raw quantity for the period restricted to
ingested_at <= source_max_ingested_atof the sealed rollups, and compare that against the unrestricted sum. The rollup stores the watermark precisely so this can be answered without a scan. If the whole 0.4% sits above the watermark, nothing is broken: it is late data, it becomes an adjustment line, and the investigation ends in one query. - Only if the gap survives that test do you bisect, and you bisect by dimension rather than by rows. Compare 30 per-day totals, then inside the offending day compare the 6 SKUs, then the workspaces, then the 24 hours. That is roughly 30 + 6 + W + 24 grouped probes, each an indexed range scan over one daily partition for one tenant, against O(N) per attempt for the naive re-fold.
- Quantify why naive is not merely slow but unusable mid-incident: at a generous 200,000 rows/second sequential, 90 million rows is about 7.5 minutes per attempt, you will want ten attempts, and every one competes for I/O on the same partitions live ingest is writing. The diagnostic worsens the backlog it is diagnosing.
- Before fetching each comparison, state what it would look like under each hypothesis. Two adjacent hours off by equal and opposite amounts is
occurred_atversusingested_atbucketing. A whole day offset by exactly N hours is a timezone applied at the wrong layer. A gap confined to one SKU in one workspace is an environment filter. The same(tenant_id, idempotency_key)present in twoingested_daypartitions is the dedup horizon losing a retry that crossed midnight. - Make the next bisection cheap by storing the aggregate you keep recomputing. A per-
(tenant_id, ingested_day)count and quantity checksum turns step two from thirty probes into one read, and it is the same number the reconciliation job already produces. - Whatever you find, the sealed period does not change value. The correction is an adjustment line pointing at the line it reverses, carrying its own
source_rollup_watermark, because the original invoice is the evidence of what the customer was charged.
Worked solution 35 min
- Reproduce the shape locally: generate 2 million events for one synthetic tenant over 5 days, fold them, then inject three defects, namely 0.2% of events bucketed by
ingested_at, a handful of duplicateidempotency_keyvalues whose retries cross midnight, and a block of events ingested after the seal. - Write the watermark-bounded query first and record how much of the gap it explains on its own.
- Bisect by day, then SKU, then hour, recording the probe count and the rows each probe touches.
- For each located cell, write down the predicted signature before querying it, then check whether the data matches the prediction.
- Time a full re-fold of the 2 million rows and extrapolate to 90 million.
Follow-up
- The gap is 0.4% in one direction on one day and 0.4% the other way the next day. What does that shape rule in, and what does it rule out?
- How do you distinguish a duplicate from a restatement, given
revisionandrecomputed_aton the rollup? - Ingest is still running while you investigate. What makes your two numbers comparable at all?
Find the join that inflates every invoice total
invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.
Approach
- Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row:
sum(l.amount_minor)is the true total multiplied by the attempt count, andcount(p.attempt_id)is attempts times lines. Three lines and two attempts report double the money and six attempts. - Reject the reflex repair.
count(distinct p.attempt_id)does fix the count, because attempt_id is unique.sum(distinct l.amount_minor)does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match. - Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
- Keep invoices with no attempts by making the attempt branch a LEFT JOIN with
coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about. - Push each filter to its own grain:
where l.voided_at is nullbelongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants. - Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.
Follow-up
- Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with
filter (where ...)be better? - Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
- The same report is needed per tenant per month. What index makes the line CTE cheap?
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?
How do you balance the need for new features with the necessity of sys…
How do you balance the need for new features with the necessity of system stability?
Approach
- Fix the scope first: who calls this, how often, and what they do when it fails.
- State the consistency you need, and where you are willing to be stale.
- Name the failure you are designing for, then the recovery path.
Follow-up
- What would you drop to keep the system up under load?
- How does this behave when that dependency is down for an hour?
What methods do you use to ensure data security and system integrity?
What methods do you use to ensure data security and system integrity?
Approach
- State the consistency you need, and where you are willing to be stale.
- Fix the scope first: who calls this, how often, and what they do when it fails.
- Choose a partition key and say what query it makes expensive.
Follow-up
- What would you drop to keep the system up under load?
- What breaks first when traffic grows ten times?
Explain how you handle deployment cycles and minimize downtime during …
Explain how you handle deployment cycles and minimize downtime during updates.
Approach
- Fix the scope first: who calls this, how often, and what they do when it fails.
- Choose a partition key and say what query it makes expensive.
- Name the read and write paths separately; they rarely have the same bottleneck.
Follow-up
- How does this behave when that dependency is down for an hour?
- What breaks first when traffic grows ten times?
How do you approach troubleshooting a critical system failure under pr…
How do you approach troubleshooting a critical system failure under pressure?
Approach
- Fix the scope first: who calls this, how often, and what they do when it fails.
- Choose a partition key and say what query it makes expensive.
- State the consistency you need, and where you are willing to be stale.
Follow-up
- What would you drop to keep the system up under load?
- What breaks first when traffic grows ten times?
Metering ingest that survives a six-hour producer replay
metering-ingest consumes usage events at-least-once - 250M/day, 10-40k/second at peak - and folds them into usage_rollup_hourly keyed (tenant_id, workspace_id, sku, hour_start). usage_event is partitioned daily on ingested_at with unique (ingested_day, tenant_id, idempotency_key). A producer outage ends in a six-hour replay that re-sends events already ingested, some of whose originals crossed midnight. Design the consumer: partitioning, where the acknowledgement sits relative to the commit, the deduplication horizon and its storage cost, and how the rollup watermark advances. Nothing may be double-counted and nothing may be silently dropped.
Approach
- Choose the acknowledgement position deliberately and name what each choice costs. Acknowledging after the fold commits makes the consumer at-least-once: a crash between the two replays the batch and produces duplicates, which are ordinary and absorbable. Acknowledging first makes it at-most-once: a crash between the two drops revenue with no error raised anywhere and no way to detect it later. Take at-least-once and design everything downstream to absorb duplicates.
- Put the dedup and the fold in one transaction so there is no window between them. Insert the batch into usage_event with ON CONFLICT DO NOTHING, take the rows actually inserted, and fold only those into usage_rollup_hourly with an upsert on (tenant_id, workspace_id, sku, hour_start) bucketed by occurred_at, not ingested_at. A batch of about 2,000 rows is one round trip and one index probe per event.
- Attack the partition-key flaw head on: the unique index includes ingested_day because a unique index on a partitioned table must contain the partition key, so the same (tenant_id, idempotency_key) re-sent after midnight is a different index entry and passes. Deduplicate instead against a store keyed (tenant_id, idempotency_key) with no date component, whose horizon exceeds the producer's maximum retry window plus the longest replay you intend to support. At 14 days that is 250M x 14 = 3.5 billion keys, which is a dedicated key-value store, not a larger index on the same table. The alternative - partitioning usage_event on (tenant_id, occurred_day) so the natural key is stable - fixes dedup but loses pruning on ingest time and makes retention by dropping partitions awkward.
- Partition the consumer by hash of tenant_id so one tenant's replay stalls only its own partitions, and give replay traffic a separate lower-priority lane so live ingest keeps its latency. The cost is explicit: that tenant's watermark lags while the replay drains, and everything gated on the watermark waits for it.
- Define the watermark as a property of committed work, not of wall-clock time: per partition it is the largest occurred_at such that every event with a smaller occurred_at has committed, and the sealing decision uses the minimum across partitions. Record source_max_ingested_at on every rollup row so any number can prove what it did and did not include, and keep restatement legal only while status = 'open' - after sealed_at the value is frozen and a late event becomes an invoice adjustment instead.
Worked solution 40 min
- Write the consumer loop in pseudocode with the acknowledgement after the commit, then annotate each line with what is lost or duplicated if the process dies exactly there.
- Size the dedup store: events/day x horizon_days keys, bytes per key including the tenant prefix, and the resulting memory or disk. Compare that cost against simply extending retention on the partitioned table and say why the latter does not fix the problem.
- Take one event ingested at 23:59:58 and replayed at 00:00:04 and work out its fate under (a) the partitioned unique index alone and (b) the separate dedup store.
- Write the per-partition watermark formula, then what the seal uses, then what a single stalled partition does to sealing.
Follow-up
- The dedup store is lost entirely. What can you still guarantee, and how do you rebuild it from what remains?
- A replay delivers events for an hour that is already sealed. Trace exactly what happens to them, row by row.
- One partition is stuck on a poison message, so the minimum-across-partitions watermark never advances and no tenant can be sealed. What is your escape hatch and what does it cost in correctness?
Regional error rate explodes after a dependency merely slows
A control-plane read replica in one region degrades from 4 ms to 120 ms. Within ninety seconds that region's gateway error rate rises from 0.01% to 40% and its p99 becomes bimodal, one mode near the old p99 and one at the client timeout. The other two regions are unaffected. The gateway retries control-plane reads three times with exponential backoff and no jitter. Give an ordered checklist that separates trigger from amplifier, the offered-load arithmetic, and the controls that break the loop.
Approach
- Split the incident into three questions before touching a control: what started it, what amplified it, and what would make recovery slow. Here they are the replica slowdown, the retry policy interacting with queueing, and a synchronised unjittered herd at recovery. They are different mechanisms and each needs its own control.
- Read the distribution rather than the mean. A bimodal p99 with one mode pinned at the client timeout is two populations, not one degraded path; split latency by cache hit and miss and confirm the fast mode is hits and the timeout mode is misses that reached the replica.
- Do the load arithmetic. Three retries turn one client request into up to four upstream requests, so offered load reaches roughly 4x on a dependency that is already slower, and it arrives at the worst moment. With utilisation approaching one, queueing delay grows superlinearly, which is why a 30x latency increase upstream does not produce a 30x increase downstream, it produces timeouts.
- Break the loop with controls that bound offered load rather than with more attempts: a concurrency limit on the control-plane client so at most N calls are in flight and the remainder fail fast, a circuit breaker scoped per dependency and region, and a retry budget capping retries at a small fraction of base traffic so amplification has a ceiling that does not depend on how many clients are retrying.
- Add full jitter to whatever retries survive, sleeping uniformly in [0, min(cap, base x 2^attempt)], so attempts de-correlate instead of arriving in waves aligned to the moment of failure.
- Decide the unreachable-dependency behaviour in advance, because it is the actual product decision underneath: serving from an expired credential cache keeps the product available while extending a revoked key's life past the stated bound, and failing closed converts a dependency degradation into a total outage. State the mode and the staleness number rather than letting the timeout choose.
Follow-up
- The replica recovers. Describe what happens in the first ten seconds with your controls in place versus without them.
- Which single metric would have paged before the error rate moved, and why is upstream latency by itself not it?
- Requests that fail fast under the concurrency limit still need an answer. What does the gateway return, and what does it do to the usage event it would otherwise have emitted?
Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Diagnostic, scored before you study anything
- Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
- Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
- Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.
Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Largest gap: find the boundary rather than the subject
- Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
- Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
- Re-attempt one of them from blank four hours later with nothing open.
Deliverable: A sub-skill map with the two blocking sub-skills circled.
Practice prompt ↗Practice prompt ↗03Drill the blocking sub-skill by repeating the shape
- Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
- State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
- Have someone else read your one-sentence rule and find the precondition you left out.
Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.
Practice prompt ↗Practice prompt ↗04Second gap, plus maintenance on the strongest area
- Run the same sub-skill decomposition on the second-largest gap in half the time.
- Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
- Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.
Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.
Practice prompt ↗Practice prompt ↗Worked solution ↗05The gap that is not a skill
- Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
- Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
- Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.
Deliverable: Two recordings with a counted reduction in time-to-first-question.
Practice prompt ↗Practice prompt ↗06Retest under day-one conditions
- Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
- For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
- Write down which single block you would still lose the offer on.
Deliverable: A second scored rubric placed beside the first, with one named remaining risk.
Practice prompt ↗Practice prompt ↗07Full loop under interview conditions
- Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
- Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
- Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.
Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.
Practice prompt ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Engineers over-index on what they repaired. A stronger answer covers something you knowingly left broken: the alert you tuned down, the data inconsistency you documented instead of chasing, the cleanup you deferred past two quarters. Give the reasoning and the condition that would have reopened it, so it reads as a decision and not as neglect.
Give an example of a time you identified an inefficiency in a process …
Give an example of a time you identified an inefficiency in a process and implemented a solution.
Approach
- Give the blast radius: what could have broken, and what you measured.
- Pick a story where you made the decision, not one where you watched it.
- 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?
How do you handle conflict or differing opinions when collaborating wi…
How do you handle conflict or differing opinions when collaborating with non-technical stakeholders?
Approach
- 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.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
Estimate a tenant-leading index migration you have never run
Someone needs a date. usage_event carries an index on (occurred_at) and needs (tenant_id, occurred_at); the largest tenant holds roughly a hundred times the median tenant's rows, the table is partitioned daily with years of retention, and you have never run a migration on a table this large. Give an estimate you would defend: how you decompose the work, the two or three numbers you would go and measure first, the range and confidence you state, and what you commit to when the person asking needs a single date today.
Approach
- Refuse the bare number and then give one anyway, in the form that is actually useful: a range plus the measurement that collapses it. 'Four to eleven days; one afternoon building this index on a restored copy of the largest partition takes that to within a day' is an answer, while 'it depends' is not.
- Decompose by failure mode rather than into equal chunks, because that is where estimates go wrong. On a partitioned parent you create the index ON ONLY the parent, build each partition's index with CREATE INDEX CONCURRENTLY, then ALTER INDEX ... ATTACH PARTITION, at which point the parent index becomes valid. CONCURRENTLY does not block writes but scans each partition twice, waits out older transactions, cannot run inside a transaction block, and on failure leaves an invalid index you must drop concurrently and retry.
- Name the two unknowns that dominate and price them: build time on one restored partition of realistic size, and whether the planner actually chooses the new index for the skewed tenant, since selectivity for a tenant holding most of the rows is a different question from selectivity for the median tenant. Both are half-day measurements against a replica, and both are cheaper than being wrong by a week.
- State the assumptions the range is conditional on, because that is what makes a slip a re-estimate instead of a credibility event: no partition above a stated row count, one concurrent build at a time so it does not compete with ingest for I/O, and an ingest backlog that can absorb the added write amplification while both indexes exist.
- Budget the step nobody budgets: verification and the old index's removal. Dropping the old index is fast, but deciding it is safe to drop means confirming no plan still uses it, and that confirmation waits on real traffic across a full weekly cycle rather than on your patience.
- Answer the single-date request honestly. Commit to a date for the first checkpoint — the measured build number from the replica — and to re-estimating on that date, and say plainly what you are not committing to yet. A date with a scheduled re-estimate is worth more to the asker than a confident wrong one, and you should say why in those words.
Follow-up
- The concurrent build fails half way through the largest partition. What is the state of the database and what do you do next?
- Your estimate slips by sixty percent. Which assumption broke, and at what point would you have known?
- The person asking needs the date for a customer commitment. Does your answer change?
- 01
Give an example of a time you identified an inefficiency in a process and implemented a solution.
- 02
How do you handle conflict or differing opinions when collaborating with non-technical stakeholders?
- 03
Someone needs a date. usage_event carries an index on (occurred_at) and needs (tenant_id, occurred_at); the largest tenant holds roughly a hundred times the median tenant's rows, the table is partitioned daily with years of retention, and you have never run a migration on a table this large. Give an estimate you would defend: how you decompose the work, the two or three numbers you would go and measure first, the range and confidence you state, and what you commit to when the person asking needs a single date today.
Is this an official Buildsafe Australia interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Buildsafe Australia. Rounds and questions reflect what candidates have reported, not a process Buildsafe Australia has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How should I prepare for the technical portions of the interview?
Focus on your past work. Be prepared to explain the architecture of the systems you have built or maintained and be ready to defend your design choices against potential failure points.
PracHub interview research ↗What is the company culture like for engineers?
Candidates report an emphasis on pragmatism, safety, and continuous improvement. Interviewers are reported to favor engineers interested less in the latest tools than in building systems that work reliably under pressure.
PracHub interview research ↗What is the typical timeline for the hiring process?
The process usually spans a few weeks. The process is described as prioritizing quality and fit, with enough time for you to meet the key team members you would be working with.
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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