Interview guides · Aptyx

Aptyx Software Engineer Interview Questions & Guide 2026

Software engineer prep for Aptyx: design exercises weighted toward the constraints of product and business analytics work.

The 60-second brief

At Aptyx, the Software Engineer role is pivotal to the intersection of advanced manufacturing and precision engineering. While the title reflects software development capabilities, the role is deeply embedded within the R&D and production ecosystems, focusing on high-stakes environments like Precision Extrusion, Catheter R&D, and ePTFE development. You are not just writing code; you are building the digital infrastructure that enables the creation of life-critical medical components.

The title spans product, platform and infrastructure work, and which of those the seat actually is decides whether design or algorithms carries more weight in your preparation. The posting rarely settles it; what the team is on call for usually does.

Focus areas

Choose indexes from the query's access path
Bound every outbound call with a timeout
Paginate large result sets with keyset cursors

Overview

The work at Aptyx is characterized by high complexity and a focus on technical rigor. You will collaborate with cross-functional teams to translate engineering requirements into scalable, reliable systems. Whether you are optimizing data workflows for R&D processes or developing tools to support material science breakthroughs, your contribution directly impacts the efficiency and quality of Aptyx products. This position is ideal for engineers who thrive in a fast-paced environment and are motivated by the tangible, real-world impact of their software solutions.

Where to focus your preparation

Treat the sections below as preparation areas and confirm the format with your recruiter.

Preparation focus

No round sequence has been reported for this company, so work the categories below and confirm the format with your recruiter.

What to demonstrate

Breadth across SQL, experimentation and product reasoning
Ability to state assumptions before choosing a method

How to prepare

Drill the practice exercises below and time yourself
Prepare three quantified stories about decisions you drove

Where candidates lose points

Assuming an isolation level prevents the anomaly you actually have

Isolation levels are named by the SQL standard but implemented differently, so any claim about one is only true of a named engine. PostgreSQL defaults to READ COMMITTED, where every statement takes a fresh snapshot, so two statements inside one transaction can legitimately disagree about the same row. Its REPEATABLE READ is snapshot isolation: it removes non-repeatable and phantom reads but permits write skew, where two transactions each read a set, each conclude their own write is safe, both commit, and the combined result violates a constraint that no single row expresses. Only SERIALIZABLE closes that, and it closes it by aborting a transaction with a serialization failure (SQLSTATE 40001), which means the guarantee is theoretical unless the application has a retry loop. InnoDB's REPEATABLE READ is a different mechanism again - plain SELECTs read a consistent snapshot while locking reads and writes see the latest committed row - so a read-modify-write inside one transaction can act on a value that the transaction's own earlier SELECT never returned.

Letting a slow dependency consume unbounded concurrency

The failure that takes a service down is usually not an error but a delay. A dependency answering in thirty seconds instead of fifty milliseconds holds each request's worker or connection six hundred times longer, and since required concurrency is arrival rate times latency, a fleet sized for sixty in-flight requests now needs thirty-six thousand to sustain the same rate - so it queues, and requests whose clients have already abandoned them still occupy resources. Retries make it precisely worse: a policy of three attempts triples the load on a dependency at the exact moment it is least able to serve, which is how one slow dependency becomes an outage of everything sharing that pool. Containment is four specific things - a timeout on every outbound call shorter than the caller's remaining budget, a bounded pool per dependency so one cannot starve the others, backoff with full jitter rather than a fixed delay so retries do not resynchronise, and a circuit that stops sending once the failure rate makes an attempt pointless.

Never running a concrete value through the code

Trace one small input and one edge input by hand, index by index, out loud. Re-reading your own code catches design mistakes; walking a real value through it catches the off-by-one, the uninitialised accumulator and the loop that never advances.

Issuing one query per row of a result set

Fetch related rows in a single batched query keyed by the ids you already hold, or join them into the original query. A per-row round trip multiplies network latency by the row count, and it looks perfectly fine against the ten rows in your development database.

Questions & practice

Diff a projection against the primary without per-row point reads

Hard · reconciliation · range hashing · throttling

The listing projection has drifted and some rows show a stale version. The primary holds 40,000,000 resource rows across 12,000 tenants while serving 1,200 writes and 14,000 reads per second. The obvious repair, reading each resource row and comparing its version against the projection, is correct and would eventually finish. Explain precisely why it is unacceptable here, then give a diff that finds the differing rows, state its complexity, and make it safe to run against a live primary. Replication lag is usually under 100 ms and is not bounded.

Approach

Quantify the naive cost rather than calling it slow: 40,000,000 point reads at even 0.5 ms each is over five hours serialised, and the only lever is concurrency, which is exactly what you cannot spend. The primary's pool is sized for the write path, and 40,000,000 random reads evict the buffer cache that sustains the 85 percent cache hit rate, so the audit degrades the system it is auditing.
Replace random access with one ordered pass per side. Both sides can be read in (tenant_id, resource_id) order, which is a sequential scan on each and a merge join in O(n) time and O(1) memory. For a dense diff that is the whole answer, and it reads the primary once instead of 40,000,000 times.
For the expected sparse case, compare range hashes instead of rows: partition the key space, compute per range an order-independent aggregate over hash(resource_id, version), compare aggregates, and descend only into ranges that differ. With d differing rows and branching factor B, at most d ranges mismatch per level, so the drill-down examines O(d log_B(n/d)) ranges and reads full rows only in mismatching leaves.
Aggregate with a sum modulo 2^64 or a multiset hash, never XOR. XOR is order-independent but self-cancelling, so two rows wrong in the same way, or a row duplicated on one side, leave the range aggregate matching and the range is declared clean.
Pin the comparison to a point in time or it reports lag as drift: consider only rows whose updated_at is older than now minus a lag margin, and re-check each candidate mismatch individually before repairing. At 1,200 writes per second a diff without this reports thousands of false positives, and an unattended repairer would then overwrite live rows with stale values.
Make the run resumable and throttled: batch by range key, persist the last completed range, and watch a signal such as replica lag or primary CPU, pausing rather than pressing on. A reconciliation that cannot be stopped and resumed gets killed halfway and restarted from zero, which is how a repair becomes an incident.

Follow-up

The diff reports 900 stale rows. How do you decide between patching those rows and rebuilding the projection from resource_revision?
Same job, but the projection lives in a search index that cannot be scanned in key order. What changes?
How would you run this continuously at low cost instead of only as incident response?

Merge partitioned event streams into one ordered feed with bounded lateness

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

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

Approach

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

Follow-up

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

Collapse a redelivered event batch into per-aggregate high-water marks

Easy · hashing · at-least-once · aggregation

You drain a batch of up to 5,000,000 events, each (aggregate_id BIGINT, aggregate_version INT, event_type, payload). The log guarantees order within one aggregate only; the batch merges 64 partitions, and a relay failover has redelivered a range, so an older version for an aggregate can appear after a newer one. Given a map of last_applied_version per aggregate, produce the events worth applying, at most one per (aggregate_id, version), plus the count discarded. Target O(n) time. State the memory for 2,000,000 distinct aggregates and what you do when it does not fit.

Approach

One pass, one hash map from aggregate_id to the highest version kept, and a discard counter. An event whose version is at or below last_applied_version for its aggregate is dropped without further work, which is the whole reason the event carries its version rather than a delta. O(n) expected time, O(d) space in distinct aggregates.
Keep the maximum, never the last occurrence. The redelivered range means the final appearance of an aggregate in the batch can be an older version than one seen earlier in the same batch, so last-wins applies stale state over newer state and the projection regresses with no error anywhere.
Cost the memory instead of calling it large: an 8-byte key plus a 4-byte version is 12 bytes of payload, and an open-addressed table held at a 0.7 load factor costs roughly 17 bytes per entry before per-slot metadata, so 2,000,000 aggregates is tens of megabytes in a native layout and several times that in a runtime that boxes both key and value.
If the distinct set exceeds memory, partition on hash(aggregate_id) mod P and reduce each partition independently. Every event for one aggregate hashes to the same partition, so the per-partition result is exact and the merge is concatenation rather than a second reduction.
Reject sorting the batch by (aggregate_id, version) as the default. It is O(n log n) and buys nothing, because max is associative and commutative and needs no ordering; sorting earns its cost only when the downstream consumer must receive the events in order rather than a per-aggregate winner.
Separate the two mechanisms out loud: in-batch deduplication does not make the consumer idempotent, because the same event redelivered tomorrow arrives in a different batch entirely. The projection write itself still has to be keyed on (aggregate_id, version).

Worked solution (20 min)

Write the pass: look up last_applied_version, skip if the event's version is not greater, otherwise upsert into the keep-map only when the incoming version exceeds the version already held, incrementing the discard counter on every skip.
Hand-trace one aggregate whose events arrive as v5, v3, v4, v5 with last_applied_version = 2, and confirm the output holds v5 once while the counter reads 3.
Compute the table footprint for 2,000,000 entries at 12 bytes of payload and a 0.7 load factor, then state the multiplier for a runtime that boxes keys and values.
Add the hash-partitioning fallback and say in one sentence why the per-partition results need no cross-partition merge logic.

Expected result: A single-pass O(n) reduction keyed on aggregate_id that keeps the maximum version rather than the last occurrence, O(d) space with the byte cost stated for 2,000,000 aggregates, a hash-partitioning fallback keyed on aggregate_id, and an explicit statement that batch-local deduplication does not replace a projection write keyed on (aggregate_id, version).

Verify it yourself

Shuffling the batch order changes nothing in the output, which is the test that separates max-wins from last-wins.
Feeding the same batch twice produces an identical keep-set and only doubles the discard counter.
Every emitted event has a version strictly greater than the supplied last_applied_version for its aggregate.

Follow-up

The payload is a patch rather than a snapshot, so applying only the highest version loses the intermediate changes. What changes in your reduction?
How do you detect that version 7 arrived while version 6 was never delivered, and what should the consumer do about the gap?
Two events for one aggregate carry the same version with different payloads. Which one is wrong, and how would you find out?

Hold a per-tenant active cap against concurrent creates

Hard · write skew · isolation levels · constraints

A tenant on the standard plan may hold at most 50 resources with status='active'. The create handler runs SELECT count(*) FROM resource WHERE tenant_id = $1 AND status = 'active', compares to 50, then inserts. Two creates arrive 3 ms apart on different instances and the tenant lands at 51. Name the anomaly, say whether PostgreSQL 16 READ COMMITTED or REPEATABLE READ prevents it and why, then give an implementation that holds the cap at READ COMMITTED with the exact statements. Finally, say what changes when the cap is 'at most one running export per tenant' on job_run.

Approach

Name it: write skew. The two transactions read an overlapping set and write disjoint rows, so there is no row-level conflict for the engine to detect and each commit is individually legal.
Rule out the levels precisely. READ COMMITTED takes a fresh snapshot per statement and takes no lock on the counted rows, so both see 49. PostgreSQL's REPEATABLE READ is snapshot isolation: it removes non-repeatable reads and phantoms within the snapshot but still admits write skew, because the anomaly is not a re-read of a changed row, it is a read of a set that a concurrent transaction invalidates. Only SERIALIZABLE closes it, by tracking the read dependency and aborting one transaction with SQLSTATE 40001 — a guarantee that exists only if the application re-runs the whole transaction from the read.
Convert the set predicate into a single-row conflict: keep tenant.active_resource_count and run UPDATE tenant SET active_resource_count = active_resource_count + 1 WHERE tenant_id = $1 AND active_resource_count < 50 in the same transaction as the INSERT. Zero affected rows is the cap, returned as 409. The row lock serialises the decision at any isolation level, and contention is bounded to one tenant's row — which is also the fair-scheduling unit, unlike a global counter that would convoy every tenant behind one row.
State the cost you just took on: a counter is a second source of truth that can drift, so every path that changes status must adjust it inside the same transaction, and a periodic reconciliation has to exist, with resource_revision as the authority for what the count should have been.
For the job case the invariant is expressible per row, so let the database hold it: a partial unique index on job_run (tenant_id, job_type) WHERE status IN ('queued','running') makes a second running export unwritable and the loser takes 23505, mapped to 409. That is strictly better than a counter — no drift, no reconciliation — and it is available only because the cap is one rather than fifty.
Add the retry discipline each route demands: under SERIALIZABLE both 40001 and deadlock 40P01 are retryable and the retry must re-execute the read, while under READ COMMITTED with the counter nothing retries, because the conflict is reported to the caller rather than raised as an error.

Follow-up

A resource moves from archived back to active. Which statements change, and what breaks if the counter update and the status change land in different transactions?
The cap becomes plan-dependent and a plan can change mid-month. Where does the number 50 live, and who reads it?
How do you detect after the fact that the counter drifted, without locking the table?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

resource carries version INT NOT NULL DEFAULT 1. resource_revision holds revision_id, resource_id, version, actor_user_id, change_kind, patch JSONB, request_id, created_at with UNIQUE (resource_id, version). outbox_event holds aggregate_type, aggregate_id, aggregate_version, event_type, payload, status. A PUT carries the version the client read. Write the exact statements for the single transaction that applies the edit, records the revision and enqueues 'resource.updated', and give the handler's branch on zero affected rows. Then say what PostgreSQL 16 does under READ COMMITTED when two of these updates hit one row at once.

Approach

One transaction, three writes, no network call inside it: UPDATE resource SET title = $3, version = version + 1, updated_at = now() WHERE resource_id = $1 AND tenant_id = $4 AND version = $2; then INSERT the resource_revision row at version $2 + 1; then INSERT the outbox_event row at the same aggregate_version. The event goes to a table rather than a broker because no transaction spans both.
Branch on the affected-row count before doing anything else. Zero has three causes — stale version, wrong tenant, row gone — so re-read once and map to 409 carrying the current version, or 404 for an id outside the caller's tenant, which also stops the endpoint confirming that another tenant's id exists.
State the engine behaviour instead of assuming it. Under READ COMMITTED the second UPDATE blocks on the row lock, and when the first commits PostgreSQL re-evaluates the WHERE clause against the newly committed row, so the version predicate now fails and the statement reports zero rows. Under REPEATABLE READ the identical collision raises SQLSTATE 40001 instead, so the handler must fold both shapes into one conflict response.
Keep UNIQUE (resource_id, version) even though the predicate already serialises writers. It is what makes a lost update unwritable if any other path ever reaches the revision table, and it converts a logic bug into 23505 rather than into a silently missing history row.
Refuse to auto-retry the whole PUT. A retry re-reads the winner's state and reapplies an intent formed against data that no longer exists — the silent overwrite the version token was added to detect. Return the conflict; merge field-wise only if the patches are provably disjoint.
Note that now() is the transaction timestamp in PostgreSQL, so resource.updated_at, the revision's created_at and the outbox row share one instant, which is what later makes reconciliation between the three tables unambiguous.

Worked solution (25 min)

Write the three statements plus the rowcount branch and confirm they sit in one BEGIN/COMMIT with no outbound call between them.
Run two clients that both read version 7 and apply their updates 5 ms apart; assert one 200 and one 409.
Assert resource.version = 8, exactly one resource_revision row at version 8, and one outbox_event row at aggregate_version 8.
Repeat at REPEATABLE READ and record the different failure shape (SQLSTATE 40001) the handler must also map to 409.
Delete the version predicate and re-run: both writes commit and the first edit disappears with no error raised anywhere.

Expected result: Exactly one of the two commits. version = 8 with one revision row and one outbox row at version 8, the loser gets 409 carrying version 8, and no state exists in which the resource moved but the revision or the outbox row is missing.

Verify it yourself

COUNT(*) of resource_revision for that resource equals resource.version, with no gap in the version sequence.
Kill the process between the UPDATE and the outbox INSERT — the transaction rolls back and no event describes a change that never happened.
Under a 50-way race, resource.version equals the number of 2xx responses rather than the number of requests.

Follow-up

A client sends the version it read ten minutes ago and the resource has moved three versions. What is in your 409 so it can resolve the conflict without a full re-fetch?
Two editors, two disjoint fields, no overlap. Does your answer still refuse the second write, and should it?
Every write now touches a second hot table. How do you keep the outbox insert and its partial index from becoming the write bottleneck at 1.2k writes/second?

How do you approach the design of a system that requires high-precision data inputs?

Medium

Approach

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

Follow-up

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

How do you ensure your code is scalable and maintainable for long-term R&D projects?

Medium

Approach

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

Follow-up

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

Describe a situation where you had to pivot your technical approach due to changing project requirements.

Medium

Approach

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

Follow-up

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

Choose what to break when replication lag reaches forty seconds

Hard · replication lag · read-after-write · consistency vs availability · load shedding

Reads are served from two replicas: 14k requests/second, about 85% absorbed by cache, so roughly 2.1k reads/second reach the database. Writes go to the primary at 1.2k/second. A tenant's backfill drives replication lag from under 100 ms to 40 seconds and it is still climbing. Sessions that have just written are pinned to the primary. Decide, endpoint class by endpoint class, whether to serve stale, fail, or route to the primary, and justify each choice with the load it adds to the primary. Then state what you would have built beforehand.

Approach

Establish blast radius before cause, because mitigation and diagnosis have different deadlines. The decisive arithmetic is what happens if the database reads move to the primary: 2.1k reads/second on top of 1.2k writes/second roughly triples its operation count, on the node already absorbing the backfill that caused this. Reads and writes are not equal in cost, so treat that as an argument against a blanket move rather than as a capacity model - but it is enough to rule out routing everything to the primary.
Classify endpoints by what staleness costs, not by how important they feel. Reads whose staleness is invisible - listings, search, counters - stay on the replica and return the watermark so the client can tell. Reads that immediately follow that same session's write keep their primary pin, which is a small bounded slice of traffic rather than the whole 2.1k/second. Reads that feed a decision with a side effect - authorisation, quota, the read half of a read-modify-write - must not be stale at all, because a 40-second-old permission row is the stale-permission failure wearing a different costume; those go to the primary or fail.
Shed instead of queueing. If the must-be-fresh class alone exceeds the primary's headroom, refuse its lowest-value slice with 503 and a retry-after. A request queued behind a saturated primary holds a connection for a client that has already given up, and the retry storm that follows is what turns degradation into an outage. Bound the connection pool per role so the read fallback cannot consume the write path's connections - that bulkhead is the single decision that determines whether writes survive the next ten minutes.
Attack the cause in parallel, since it is the one thing that can be stopped. The backfill is the load generator. A backfill that reads replication lag as its throttle signal and pauses above a threshold would have made this a non-event, with batch sizes small enough that each batch's write volume is a fraction of what a replica can apply per second. That is most of the answer to what should have existed beforehand.
Name the mechanism you would prefer over session pinning. Capture the write position at commit and require the read path to be at or past it: compare the primary's pg_current_wal_lsn() at commit time against the replica's pg_last_wal_replay_lsn(), and fall back to the primary only for the specific request that is ahead of the replica. Session pinning is the cheap approximation and it over-pins - every read in the window goes to the primary whether or not it needed to, which is a share of the cost being paid right now.

Worked solution (35 min)

List the endpoints in three buckets - staleness invisible, staleness visible to the writer only, staleness unsafe - and attach the share of the 2.1k reads/second each bucket carries.
Compute the primary's operation count under each routing option and mark which options are arithmetically available.
Write the pin rule and its window, then the shed rule: which slice, what status code, what retry-after.
Write the backfill's throttle predicate against a measured lag value, including its pause threshold and resume condition.

Expected result: Staleness-invisible reads stay on the replica and return a watermark. Reads following the session's own write keep a bounded primary pin. Reads that gate a side effect go to the primary or fail, with the lowest-value slice shed as 503 plus retry-after if headroom runs out. Connection pools are bounded per role so the read fallback cannot starve writes. The backfill is throttled on measured replication lag. The durable fix is a commit-position token compared against the replica's replay position, replacing blanket session pinning.

Verify it yourself

The traffic routed to the primary under the plan is stated as a rate, not as a category.
No element of the plan increases the write path's connection usage.
The backfill throttle references a measured lag value and has an explicit resume condition.
Every endpoint left on a replica returns something with which a client can detect staleness.

Follow-up

Lag returns to normal in nine minutes. Which mitigation do you remove first, and which one stays permanently?
A user reports their change did not save, and the write committed. Trace the path that produces that report and name the signal that would have shown it before the report arrived.
The replica is 40 seconds behind but otherwise healthy. Do you take it out of rotation? What does that do to the other replica's lag?

Exports duplicate a row range about once a week

Hard · leases · clock skew · idempotency · fencing

Roughly once a week an export writes a file containing a duplicated range of rows. The affected job_run rows show attempt = 1, status = succeeded, one started_at, and a lease_owner naming a different host from the one whose logs show the job starting. Leases last 30 seconds and are heartbeated every 10 from inside the handler; lease_expires_at is computed on the worker and compared against the database's now(). Find the mechanism, and give a fix that holds even if you cannot fix the clocks.

Approach

Start from the fact that eliminates the obvious answer. attempt = 1 means no retry was recorded, so this is not a re-run after failure; two workers ran the same row concurrently and the takeover path never touched the counter. lease_owner naming a host other than the one that started the job is the same statement from the other side.
Enumerate the mechanisms that cause a premature takeover, then find the signal that separates them. Either the lease genuinely expired because the heartbeat did not fire, which is what happens when the heartbeat runs on the handler's own thread and the handler makes a long blocking call, or it only appeared expired because two clocks disagree, since lease_expires_at is written from the worker's clock and evaluated against the database's. The discriminator is the distribution: incidents clustered on the longest exports indict the heartbeat, incidents clustered on one host indict skew. Measure both, and measure each host's offset against the database directly.
Read the reclaim query precisely. In PostgreSQL now() is transaction start time, not statement time, so a reclaimer holding a long transaction compares against an older timestamp than expected; clock_timestamp() is the statement-time function. This is worth ruling in or out before you redesign anything, because it changes which rows look expired.
Remove the second clock rather than trying to synchronise it. Issue and extend the lease in the database, with lease_expires_at = now() + interval '30 seconds' in both the claim and the heartbeat, so exactly one clock is ever compared and worker skew stops mattering to this predicate.
Accept that a lease can still expire under a slow worker, because a lease cannot distinguish slow from dead, and fence the work. Carry a monotonically increasing lease generation and make every write the handler performs conditional on still holding it, as UPDATE ... WHERE job_run_id = $1 AND lease_owner = $2 AND lease_generation = $3, so a displaced worker's writes affect zero rows and it aborts instead of duplicating.
Make the handler's writes idempotent independently of all that: give each exported chunk a natural key of (job_run_id, batch_start) with a unique constraint so a second copy conflicts rather than appends, move the heartbeat off the handler's thread, and increment attempt on takeover so the event is visible in a metric.

Follow-up

The displaced worker has already streamed half the file to object storage. What makes that side effect safe to repeat?
You now count takeovers. What alert fires on that counter, and at what threshold?
What breaks if you simply raise the lease to five minutes?

Suggested practice checklist

For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.

Day 1 — Rebuild the primitives by implementing them

Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.

Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.

Day 2 — Arrays under an invariant: two pointers, sliding window, binary search

Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.

Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.

Day 3 — Sorting, heaps, and the greedy argument that has to be proved

Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.

Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.

Day 4 — Recursion, memoisation, and the step to a table

Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.

Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.

Day 5 — Graphs, where most of the work is choosing the traversal

Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.

Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.

Day 6 — One day for everything that is not an algorithm

Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.

Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.

Day 7 — Solve out loud, under time

Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.

Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.

What the behavioural round is actually measuring

Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.

Can you describe a time you had to troubleshoot a complex technical issue in a production-critical environment?

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
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.

Follow-up

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

What is your experience with integrating software solutions with hardware or manufacturing processes?

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?

Tell callers you do not own that their integration breaks

Medium · deprecation · compatibility · stakeholders

A field in a write endpoint's response must change shape. You own the endpoint; you do not own the four internal callers or the outbound webhook consumers who read it. Describe a deprecation you were responsible for: what you shipped first, how you established who was actually reading the field, the window you gave and what set its length, what you did about the consumer who never moved, and how you decided removal was safe. Name the signal you used, not the announcement you sent.

Approach

Establish the reader set empirically rather than from a wiki of owners: per-field usage counters keyed by principal, or access logs attributed to a consumer. State the blind spot of whichever you pick, since a consumer that reads the field only on a monthly job will not appear in a week of logs.
Ship additive first. Populate the new field alongside the old one so no reader is forced to move, which is also what keeps a rolling deploy safe, because old and new instances answer the same requests at the same time and a rollback must still find the old shape present.
Set the window from the slowest legitimate consumer's release cadence, not from your calendar, and decide separately what to do for a consumer with no release process at all, such as an external webhook endpoint you can only email.
Convert silence into evidence before you rely on it: a short, low-traffic removal window that makes a still-dependent consumer fail visibly and loudly while you are watching, rather than at three in the morning after you have moved on.
State the removal criterion as a measurement with a duration attached, such as observed reads at zero across a full billing cycle, and keep the change reversible for one release after removal.

Follow-up

How would you detect a consumer that reads the field only during a monthly export?
One caller refuses to move and has a commercial relationship behind it. What changes in your plan and what does not?
After removal, what makes the change irreversible, and how long before you cross that line?