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Apex Software Engineer Interview Questions & Guide 2026

Software engineer prep for Apex: design exercises weighted toward the constraints of B2B software and infrastructure work.

The 60-second brief

As a Software Engineer at Apex, you will design, develop, and support custom software solutions that solve critical business problems across complex technical environments. You operate at the intersection of robust backend data structures and responsive user interfaces, building secure, scalable applications tailored to internal teams and high-stakes operations. Your work directly impacts how systems handle demanding workloads, ensuring that internal stakeholders and external clients rely on fast, fault-tolerant infrastructure.

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

Focus areas

Make every write idempotent under client retries
Bound blast radius with per-tenant concurrency limits
Evolve APIs without breaking pinned SDK clients

Overview

This role requires deep technical fluency in database architecture, integration frameworks, and rapid application development tools. Whether you are building and maintaining high-performance data systems, optimizing complex queries, or implementing robust interfaces, your contributions drive operational efficiency. You will collaborate closely with cross-functional partners, translating ambiguous business needs into clean, maintainable technical designs while navigating non-routine problem spaces with autonomy and precision.

Expect an environment characterized by high expectations, rapid iteration, and intellectual curiosity. values engineers who can balance rapid prototyping with enterprise-grade stability, maintaining strict coding standards while pivoting quickly to capture new business opportunities. You will take ownership of features from conception through deployment, ensuring that your code remains reliable, well-tested, and fully optimized for scale.

The Apex interview process

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

Initial Screening

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

What to demonstrate

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

How to prepare

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

Technical Interviews

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

What to demonstrate

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

How to prepare

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

Team Engagement

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

Serialising a tenant's writes through select ... for update on a single counter row

It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.

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

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

Assuming the input fits in memory

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

A queue or buffer with no bound

Every producer-consumer boundary needs a capacity and a policy for reaching it: block the producer, shed load, or drop the oldest entry. Unbounded buffering converts a temporary slowdown into memory exhaustion and hides the backpressure signal that would have revealed the consumer was falling behind.

Questions & practice

Hold a tenant to a trailing sixty-second request limit

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

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

Approach

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-After from the exact algorithm without a second scan?

Seal an hour under late data with bounded memory

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

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

Approach

Two clocks, two jobs. Bucket by occurred_at, because that is the hour the customer is billed for, and advance the watermark on ingested_at, because that is what the fold has consumed and what source_max_ingested_at records. Conflating them is what makes late data invisible.
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 with hour_end + L < W and seal it: O(log H_open) per advance and O(1) amortised per event to touch its bucket. Memory is open hours multiplied by distinct (tenant, workspace, sku) keys, so cap the number of simultaneously open hours and spill the oldest into usage_rollup_hourly as status='open' with a revision bump. While an hour is open the row is upsertable, so the store is your overflow.
The seal itself is a conditional write: update ... set status='sealed', sealed_at=now() where status='open' returning .... Two sealers race on every restart, and the loser must see zero rows and stop rather than write a second value. After the seal, an event for that hour is not an upsert but an adjustment, and source_max_ingested_at is what proves it arrived afterwards.

Follow-up

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

Fold a deduplicated usage stream into hourly rollups

Easy · aggregation · deduplication · watermarks · exact-arithmetic

You are given one day of usage_event rows, up to 250 million, each carrying event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at and ingested_at. Produce usage_rollup_hourly cells keyed (tenant_id, workspace_id, sku, hour_start) with quantity_sum, event_count and source_max_ingested_at. An event counts once per (tenant_id, idempotency_key). The rollup grain has no environment column, so state your filter. One pass. Give your time and space bounds, and say what the deduplication actually costs in memory.

Approach

Bucket on occurred_at, never ingested_at: hour_start = date_trunc('hour', occurred_at at time zone 'UTC'). The two columns answer different questions. occurred_at says which hour the customer is billed for; ingested_at says how current the fold is. Using the second for the first makes late data invisible instead of correctable.
The fold is trivial and the deduplication is the entire cost, so price it before designing anything clever. An exact set over (tenant_id, idempotency_key) at 250M entries, stored as a 16-byte 128-bit hash in an open-addressed table at 0.7 load factor, needs about 357M slots at 16 bytes each, roughly 5.7 GB. The fix is partitioning by hash(tenant_id) % P so each shard holds 1/P of the set and no tenant's keys straddle shards.
Rule out a Bloom filter as a replacement, in the right direction: a false positive reports 'already seen' for an event never seen, so you drop a real event and lose revenue with no error raised. It is usable only as a negative pre-filter in front of the exact set, where a miss is conclusive and a hit must fall through to the real lookup.
Accumulate in scaled integers, not binary floating point. numeric(20,6) admits values below 10^14, so one event scaled to micro-units can reach 10^20, past int64's 9.22 x 10^18; use a 128-bit or arbitrary-precision accumulator unless you first bound the per-event maximum. binary64 represents integers exactly only to 2^53, about 9.01 x 10^15, and cannot represent 0.1 at all, so two runs that sum in different orders disagree.
Carry source_max_ingested_at = max(ingested_at) over the events folded into each cell, and count event_count over accepted, post-dedup events. Without that watermark there is no way to prove later what a number did and did not include, which is the first question any reconciliation asks.
State the environment filter explicitly, because the rollup grain cannot record it. A fold that quietly includes staging bills non-production traffic; one that quietly excludes it loses a cost signal. Production-only is the billing answer, and either way it belongs in the job name and the output metadata. Complexity: O(n) time, O(distinct dedup keys) space, dominated by the dedup set rather than by the cells.

Worked solution (25 min)

Write both key tuples down before any code: dedup key (tenant_id, idempotency_key), cell key (tenant_id, workspace_id, sku, hour_start), with hour_start derived from occurred_at in UTC.
Build a 10,000-row fixture containing one event duplicated three times under the same idempotency_key, two events sharing an idempotency_key across different tenant_id values, one event whose occurred_at is two hours before its ingested_at, and one staging event inside an otherwise production cell.
Fold it and assert each of those four expectations separately rather than eyeballing a grand total.
Re-run with the input shuffled and diff the output files.
Size the dedup set for 250M keys using the load-factor arithmetic and write the number down next to the fixture.

Expected result: The triplicate contributes one event and its quantity once. The two same-key, different-tenant events both count, because the dedup key is the pair. The late event lands in the hour of its occurred_at while that cell's source_max_ingested_at advances to the later timestamp. The staging event is included or excluded per the stated filter and never silently.

Verify it yourself

Shuffled input produces byte-identical cells, including event_count and source_max_ingested_at.
The sum of quantity_sum across all cells equals the sum of quantity over the distinct dedup keys, computed independently.
Swapping the accumulator to binary64 makes a fixture of 0.1-valued quantities disagree with the exact sum. That is a failing test, not a rounding note.

Follow-up

A producer retries at 23:59:59 and the retry lands at 00:00:01. The unique index on the daily-partitioned table must include the partition key. What gets double-counted, and what is the smallest change that fixes it?
The consumer acknowledges its batch before committing the fold. Which failure loses revenue now, and which arrangement duplicates instead?
What makes a re-run over the same day produce byte-identical rollups?

Migrate a live partitioned event table without blocking ingest

Hard · online migration · ddl locks · expand and contract · partitioning

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 ... type rewrites 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) and environment with its default on the parent. Both are catalogue-only but both take a brief ACCESS EXCLUSIVE that cascades to partitions, so run each with lock_timeout set 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), then validate constraint (SHARE UPDATE EXCLUSIVE, scans while reads and writes continue), then set 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.

Paginate a tenant's delivery export without skipping rows

Easy · keyset pagination · indexing · multi-tenancy

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

Approach

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

Worked solution (20 min)

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

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

Verify it yourself

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

Follow-up

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

What strategies do you use to manage data mapping, conversion, and validation during system migrations?

Medium · api and integration design

Approach

Design the error taxonomy before the success shape; callers branch on it.
State how the contract changes without breaking existing clients.
Separate accepted, pending, failed and confirmed; they are different facts.

Follow-up

What does a partial failure look like to the caller?
How does a client discover it is on an old version of this contract?

How do you design robust inbound and outbound integrations using REST or SOAP web services?

Medium · api and integration design

Approach

Say who the caller is and what they do when the call fails halfway.
Design the error taxonomy before the success shape; callers branch on it.
State how the contract changes without breaking existing clients.

Follow-up

How does a client discover it is on an old version of this contract?
What happens if the caller retries after a timeout?

How would you simulate an IMU on the ground within a specialized testing environment?

Medium · api and integration design

Approach

Design the error taxonomy before the success shape; callers branch on it.
State how the contract changes without breaking existing clients.
Define the identity of a request so a retry cannot double-apply it.

Follow-up

What does a partial failure look like to the caller?
How does a client discover it is on an old version of this contract?

1–2 sentences introducing the category and what it tests.

Medium · api and integration design

Approach

Separate accepted, pending, failed and confirmed; they are different facts.
Define the identity of a request so a retry cannot double-apply it.
Say who the caller is and what they do when the call fails halfway.

Follow-up

What does a partial failure look like to the caller?
How does a client discover it is on an old version of this contract?

Explain the role of unit testing in validating individual components and units of code in isolation.

Medium

Approach

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

Follow-up

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

What is polymorphism, and how is it implemented?

Medium

Approach

State your assumptions explicitly before working the problem.
Say what you would check first and why it is the highest-information step.
Work from the requirement backwards to the design.

Follow-up

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

Sealing a billing period against late-arriving usage

Medium · idempotency · immutable ledger · rounding · conditional writes

billing seals a tenant's period once the metering watermark passes the period end, prices the sealed rollups against the plan (included allowance, tier boundaries, negotiated discount) and writes invoice_line_item rows: quantity numeric(28,6), unit_price_micros bigint, amount_minor bigint, currency char(3), source_rollup_watermark timestamptz. The job is re-run after failures and two workers may attempt the same tenant. Events legitimately arrive with occurred_at inside the period and ingested_at after it. Specify the seal transition, the idempotency key for every write including the payment-processor call, the rounding position and mode, and the fate of a post-seal event.

Approach

Make sealing a conditional write rather than a read followed by a write: UPDATE usage_rollup_hourly SET status = 'sealed', sealed_at = now() WHERE tenant_id = $1 AND hour_start >= $2 AND hour_start < $3 AND status = 'open' RETURNING rollup_id. Two concurrent sealers cannot both win because the loser's predicate no longer matches, and the loser learns it lost from an empty result rather than from a lock timeout.
Gate the seal on the watermark, not on the clock - the period closes when the fold is trustworthy past period_end - but cap the wait, because an unsealed period blocks the whole billing run. Seal anyway after a stated maximum (six hours past period end is a reasonable default) and let whatever arrives afterwards become an adjustment. The trade-off is named in one line: adjustments are cheap, a missed billing cycle is not.
Make every write converge under re-run. Line items are idempotent on (invoice_id, sku, rate_tier) with ON CONFLICT DO UPDATE permitted only while the invoice is draft; the payment-processor call carries (tenant_id, billing_period_start) as its idempotency key, because a timeout there is an unknown outcome, not a failed one, and a plain retry is how a customer gets charged twice.
Keep the arithmetic exact and round exactly once. quantity stays numeric, the rate is an integer unit_price_micros in millionths of a minor unit because per-request prices are genuinely below a cent, and amount_minor = round_half_even(quantity x unit_price_micros / 1,000,000) is computed once per line and stored. Tiering walks the boundaries in order emitting one line per (sku, rate_tier) with tier 0 as the included allowance at price zero; the invoice total is the sum of stored amount_minor values and is never recomputed downstream from quantity and rate.
Handle the post-seal event as a forward-only correction. The sealed rollup is frozen, so the difference between the sealed value and the restated value becomes a new invoice_line_item with kind = 'adjustment' and voided_by_line_id pointing at the line it reverses, carrying its own source_rollup_watermark. Nothing is edited in place, because the original line is the only evidence of what the customer was actually charged and is precisely what a dispute, a refund or an audit asks to see.

Worked solution (35 min)

Write the seal as a single conditional UPDATE ... WHERE status = 'open' RETURNING and state exactly what the losing worker sees.
Price one line by hand: quantity 4,318,904.250000 at unit_price_micros 1200 gives 5,182.6851 minor units and 5,183 after a single half-even rounding. Then do quantity 2,500,000.000000 at unit_price_micros 1, which lands exactly on 2.5 and rounds to 2 under half-even but 3 under half-up.
Take four hundred synthetic lines with fractional minor units and compute the total two ways - sum of per-line rounded amounts, and a single rounding of the summed exact amounts - then record the gap.
Trace one event with occurred_at inside the period and ingested_at two days after the seal all the way to the row that eventually reflects it.

Expected result: Seal as a conditional UPDATE from status = 'open' gated on the metering watermark with a bounded maximum wait; line writes idempotent on (invoice_id, sku, rate_tier) and the processor call idempotent on (tenant_id, billing_period_start); amount_minor rounded once per line with half-even; late data becoming an adjustment line that points at the line it reverses and records its own watermark, with no in-place edit anywhere.

Verify it yourself

Re-running the entire billing job on an already-invoiced period must produce identical line rows and must not call the payment processor a second time. If it charges again, the idempotency key was on the retry rather than on the business fact.
Over four hundred lines with uniformly distributed fractional parts the two totals should differ by a handful of minor units - standard deviation around sqrt(400/12), about 5.8, with a worst case of 200. If your two totals match exactly, the test data has no fractional parts and has not exercised the trap.
Every line carries a source_rollup_watermark, and two lines for the same SKU with different watermarks should explain their difference without opening any other table.

Follow-up

Two workers start the same tenant's seal a millisecond apart and the winner crashes after sealing but before writing any line. What does the second worker observe, and is the resulting invoice correct?
Show the divergence between rounding each line and rounding the total once over four hundred lines, and say which direction it goes.
A customer disputes a charge from two quarters ago. Which rows answer it, and which single design decision made that answer possible?

Metering partition crash-loops and the sealing watermark freezes

Medium · poison message · at-least-once · dead-letter queue

One metering-ingest partition has stopped advancing. Lag grows linearly, the consumer restarts about every 40 seconds, and the same offset appears in every startup log while other partitions stay healthy. Events are committed in batches of a few thousand and the acknowledgement follows the commit. Sealing is six hours away and source_max_ingested_at for that partition's tenants is frozen. Give an ordered checklist, a containment action available within minutes, and the durable fix, saying what each does to exactly-once accounting.

Approach

Distinguish a poison record from a capacity problem in one measurement: compare the offset and the exception across restarts. An identical pair every time is deterministic failure on one record, whereas a throughput problem still advances the offset between crashes.
Read the record from a separate consumer group so the bytes can be inspected without perturbing the stuck consumer, then classify the defect: schema violation, a quantity failing the non-negative check, a null workspace, an unmappable SKU enum, or a payload past a size limit. That classification decides whether this is a producer bug or a missing consumer guard.
Account for batch granularity before acting. With commits of a few thousand, one bad record fails thousands of good ones, so the blast radius is the batch. Halve the batch around the offset to isolate the record, or move to per-record error isolation so the radius becomes the record.
Contain by diverting that record to a dead-letter store with its raw bytes and offset, then resume. This is safe here precisely because the acknowledgement follows the commit: the good records from the failed batch are re-consumed and absorbed by the uniqueness check on (tenant_id, idempotency_key) rather than counted twice.
Make the fix durable with per-record error isolation, a bounded poison counter, and an alert on dead-letter rate rather than on lag alone, since lag only reveals this after the sealing margin has already been eaten.
Check the horizon before replaying anything. The unique index lives on a daily-partitioned table and therefore includes the partition key, so it deduplicates within a day only; a replay landing on a later ingest day needs the separate dedup store or it double-counts into a tenant's bill.

Follow-up

Move the acknowledgement before the commit and describe exactly what is lost and what is duplicated in each of the two crash windows.
Sealing is in six hours and the partition will not drain in time. What do you seal on, and what does the invoice have to record so the difference is explainable later?
A producer replays two weeks of events next month. Which part of your fix stops holding, and what is the dedup horizon you would actually configure?

Suggested practice checklist

For someone fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.

Day 1 — Measure before reasoning

Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.

Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.

Day 2 — References, copies, and the bugs they produce

Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.

Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.

Day 3 — Types, once, in a language that checks them

Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.

Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.

Day 4 — Concurrency, starting with what actually runs at the same time

Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.

Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.

Day 5 — Debugging as a procedure rather than an instinct

Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.

Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.

Day 6 — Tests that catch the bug you are about to write

Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.

Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.

Day 7 — Debug something broken, out loud

Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.

Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.

Risky changes and the reversal path

For anything that touched live traffic, be ready to say how you would have undone it: a flag, a staged rollout, dual writes with the old path still authoritative. Once the old column is dropped or the source rows are overwritten there is no reverse, so name what you kept a copy of and for how long.

How do you approach collaboration with functional teams and non-technical stakeholders?

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
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?
How did you know your change caused the improvement?

What brings you to our company, and why do you want to work here specifically?

Medium · behavioural and engineering judgement

Approach

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.
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?

Unblock an engineer on a job run that finished twice

Easy · mentoring · fencing tokens · lease expiry · debugging method

An engineer two weeks into the team brings you a job_run row showing status succeeded with an exit_code written by a worker declared dead ten minutes earlier; the retry attempt also shows succeeded. They have spent a day adding logging and are no closer. You have twenty minutes and you do not want to take the keyboard. Describe how you unblock someone: the question you ask first, what you let them find themselves, the concept you name and when, and how you check the next day that they own the fix rather than having watched you produce it.

Approach

Ask what they expect rather than what they see: which statement set status to succeeded, and what did it check before writing? That question points directly at the update's WHERE clause, which is where the answer lives, and it costs them nothing to answer, so it does not read as a test.
Let them build the timeline themselves from the row: queued_at, started_at, leased_until, finished_at and worker_id, on both the original run and the retry. Two different worker_ids with a lease expiry between them tells the whole story, and they will see it before you say it.
Name the concept once the evidence has earned it. A lease bounds time; it does not prevent a write. The store has to reject a stale writer, which means the update carries a fencing token the row compares — update job_run set status = 'succeeded' where run_id = $1 and lease_token = $2 and status = 'running' — and a long garbage-collection pause or a brief partition is enough to produce what they are looking at.
Point at the second, less obvious half and let them decide it: 'lost' exists in the status enum precisely so a run whose worker vanished is not recorded as failed, because failed asserts an outcome nobody observed and the system then bills and retries on that assertion. Ask them what these two rows should have said.
Leave them with the next step rather than the patch — a test that kills the first worker after the sandbox exits and before the row is written — and say when you are available again, so the offer is real rather than polite.
Check ownership the next day by what they produced, not by asking if it went well: a test that reproduces the window proves they understood it; a test that only asserts the new WHERE clause proves they copied it. Ask them to explain it to a third person and listen for whether the explanation is theirs.

Follow-up

They propose a longer lease instead of a token. What do you say, and what breaks when legitimate runs last thirty minutes?
How can you tell whether your explanation landed or they simply deferred to you?
The same engineer hits a variant of this next month. What did you fail to teach the first time?