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Helix Linear Technologies Software Engineer Interview Questions & Guide 2026

Helix Linear Technologies software engineer prep on databases: indexes, query plans, and the anomalies an isolation level still permits.

The 60-second brief

The Software Engineer role—often categorized within the organization as a Quality CMM Programmer—is a foundational position at Helix Linear Technologies. You will play a critical role in bridging the gap between high-precision manufacturing and digital quality assurance. By developing and maintaining programs for Coordinate Measuring Machines (CMM), you ensure that Helix Linear Technologies' linear motion components meet the rigorous dimensional tolerances they require.

The shape of the workload matters more for your prep than the industry label does. Read-heavy serving, write-heavy ingestion and scheduled batch processing have different binding constraints and fail in different places, so find out which one the team lives in before picking design topics.

Focus areas

Make every write idempotent under client retries
Build at-least-once pipelines with explicit deduplication horizons
Evolve APIs without breaking pinned SDK clients

Overview

This role is ideal for engineers who appreciate the intersection of software logic and physical hardware. You are not just writing code; you are building the validation framework that guarantees the reliability of components used in mission-critical applications. Your work directly impacts product quality, operational efficiency, and the company's reputation for engineering excellence.

While the role title may vary, be prepared to discuss how your programming logic applies to physical measurement systems and quality control standards.

The Helix Linear Technologies interview process

Helix Linear Technologies candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Initial Screening

The person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.

What to demonstrate

Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural

How to prepare

Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one

Professional Dialogue

Because the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.

What to demonstrate

Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
Whether you can be redirected in the middle of an answer without restarting it from the beginning
Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it

How to prepare

Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing

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.

Answering a debugging question with a guess instead of a bisection

Give a procedure that halves the search space at each step: confirm the symptom reproduces, establish the last known-good version, input or timestamp, then bisect over commits, over the data, or over the layers of the request path. A plausible cause with no way to confirm it is the same move whether it happens to be right or wrong, which is why it scores nothing.

Optimising an axis nobody named

Ask which resource is actually scarce here: wall-clock latency, throughput, memory footprint, cost per request, or engineering time. Shaving a constant factor off an in-memory step is wasted effort when the same function makes a blocking remote call inside the loop.

Questions & practice

Locate a billing reconciliation gap without rescanning ninety million events

Hard · reconciliation · dimensional-bisection · watermarks · hypothesis-testing

A tenant's sealed invoice total is 0.4% below the sum of its raw usage_event rows for the period. That tenant has 90 million events over 30 days in a table partitioned daily on ingested_at, and its rollups carry source_max_ingested_at, revision and sealed_at. Recomputing all 30 days from raw is correct, and you are not going to do it. Give the procedure that locates the divergent (workspace, sku, hour) cell, the cost of each probe, and the one query you run before any of it.

Approach

Run the free query first. Sum raw quantity for the period restricted to ingested_at <= source_max_ingested_at of the sealed rollups, and compare that against the unrestricted sum. The rollup stores the watermark precisely so this can be answered without a scan. If the whole 0.4% sits above the watermark, nothing is broken: it is late data, it becomes an adjustment line, and the investigation ends in one query.
Only if the gap survives that test do you bisect, and you bisect by dimension rather than by rows. Compare 30 per-day totals, then inside the offending day compare the 6 SKUs, then the workspaces, then the 24 hours. That is roughly 30 + 6 + W + 24 grouped probes, each an indexed range scan over one daily partition for one tenant, against O(N) per attempt for the naive re-fold.
Quantify why naive is not merely slow but unusable mid-incident: at a generous 200,000 rows/second sequential, 90 million rows is about 7.5 minutes per attempt, you will want ten attempts, and every one competes for I/O on the same partitions live ingest is writing. The diagnostic worsens the backlog it is diagnosing.
Before fetching each comparison, state what it would look like under each hypothesis. Two adjacent hours off by equal and opposite amounts is occurred_at versus ingested_at bucketing. A whole day offset by exactly N hours is a timezone applied at the wrong layer. A gap confined to one SKU in one workspace is an environment filter. The same (tenant_id, idempotency_key) present in two ingested_day partitions is the dedup horizon losing a retry that crossed midnight.
Make the next bisection cheap by storing the aggregate you keep recomputing. A per-(tenant_id, ingested_day) count and quantity checksum turns step two from thirty probes into one read, and it is the same number the reconciliation job already produces.
Whatever you find, the sealed period does not change value. The correction is an adjustment line pointing at the line it reverses, carrying its own source_rollup_watermark, because the original invoice is the evidence of what the customer was charged.

Follow-up

The gap is 0.4% in one direction on one day and 0.4% the other way the next day. What does that shape rule in, and what does it rule out?
How do you distinguish a duplicate from a restatement, given revision and recomputed_at on the rollup?
Ingest is still running while you investigate. What makes your two numbers comparable at all?

Schedule ordered webhook retries with a heap of subscription queues

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

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

Approach

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

Worked solution (30 min)

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

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

Verify it yourself

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

Follow-up

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

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.

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?

Model credential revocation so history survives the delete

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

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

Approach

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

Worked solution (20 min)

Create the table with all three constraints on a scratch database and insert two revoked rows sharing (tenant_id, name); the partial index should accept both.
Insert a second live row with that same name and confirm the violation names the partial index.
Run update tenant_api_key set revoked_at = now() leaving status = 'active' and confirm the CHECK rejects it; then try insert ... scopes = '{}' against both the cardinality and the array_length forms and note that only one rejects it.
Run explain (analyze, buffers) on the lookup predicate for a live key and confirm an index scan on secret_hash with rows removed by filter equal to zero.

Expected result: A partial unique index on (tenant_id, name) where revoked_at is null; check ((revoked_at is not null) = (status in ('revoked','compromised'))); check ((revoked_at is null) = (revoked_reason is null)); check (cardinality(scopes) > 0); a single-probe lookup on secret_hash plus revoked_at, expires_at and auth_version; last_used_at excluded, with a note that 'expired' should be derived rather than stored.

Verify it yourself

Two revoked keys with the same name insert cleanly; a second live one does not.
The array_length form of the emptiness check accepts an empty array and the cardinality form rejects it.
No statement anywhere in the design deletes or updates a revoked row.
Revoking a key touches exactly one row and takes effect on the next lookup with no application-side filtering.

Follow-up

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

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.

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 is your process for validating that a new program meets engineering drawings?

Medium

Approach

State your assumptions explicitly before working the problem.
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.

Follow-up

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

How do you approach debugging a program that produces inconsistent measurement results?

Medium

Approach

Work from the requirement backwards to the design.
Clarify what is being asked and what a complete answer contains.
State your assumptions explicitly before working the problem.

Follow-up

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

Leasing sandboxed job runs without double execution

Medium · leases · fencing · state machines · metering

job-runner leases work from a queue and executes customer workloads in sandboxes with CPU, memory, wall-clock and egress limits. Run durations span 200 ms to 30 minutes, thousands are concurrent, and job_run.status must reach exactly one terminal state among succeeded, failed, timed_out, cancelled and lost. A worker can pause 45 seconds for garbage collection, or be briefly partitioned, after its sandbox has exited but before it writes the outcome. Design the lease duration, the fencing, and timeout ownership, and state how billable_seconds is computed for a run whose worker never returns.

Approach

Resolve the lease-duration conflict rather than picking a compromise. A lease shorter than the longest legitimate run re-dispatches work that is still executing; a lease long enough for a 30-minute run leaves a crashed 200 ms run undetected for half an hour. Use a short lease - about 30 seconds - renewed by a heartbeat every 10 seconds while the supervisor is alive, each renewal pushing leased_until to now() + 30 s. Detection latency is then one lease, not one heartbeat: a dead worker's last successful heartbeat landed at most 10 s before it died, so leased_until expires 20 to 30 s after the death, plus whatever interval the reaper scans on. What the 3:1 ratio of lease to heartbeat buys is headroom - one or two missed heartbeats change nothing, and a stall only becomes a re-dispatch once it outlives the lease remaining when it began, between 20 and 30 s here. Both costs follow from those two numbers: heartbeat write load proportional to concurrent runs (3,000 runs at a 10-second interval is about 300 updates/second), and any pause longer than the lease - the 45-second GC - re-dispatching a run that is perfectly healthy.
Fence the outcome write so the store, not the worker's memory, arbitrates: UPDATE job_run SET status = $1, finished_at = $2, exit_code = $3 WHERE run_id = $4 AND status = 'running' AND lease_token = $5, and require exactly one row affected. Zero rows means this worker was fenced while it was paused, and its correct behaviour is to discard the result, not to retry - the retry is how a resumed worker overwrites a newer attempt.
Make the status machine monotone with a check constraint or a trigger so no terminal row can move, and model a retry as a new row with parent_run_id set and attempt incremented rather than a reset of the old one. That is what keeps how many times did this actually execute answerable afterwards and keeps each attempt's resource usage attributable to itself.
Put the wall-clock timeout in the supervisor, never in the workload: a customer-supplied program asked to time itself out will not. Distinguish timed_out, which the supervisor observed and caused, from lost, which nobody observed at all, and leave exit_code null unless the supervisor actually saw the process exit. Recording lost as failed asserts an outcome no one witnessed and then bills and retries on that assertion.
Decide billable_seconds for the unobserved case explicitly, because it cannot be derived. It is not now() - started_at, which bills queue and pause time the customer never consumed. The two defensible policies are leaving it null while status = 'lost' and billing nothing, or flooring it at the last heartbeat's observed running time as a stated lower bound; pick one and write it down. Add a per-tenant concurrency cap on sandbox slots so one tenant cannot occupy the whole fleet while these questions are being answered.

Worked solution (30 min)

Draw the timeline: last successful heartbeat at t, sandbox exits at t, worker pauses from t to t+45 s, so leased_until = t+30 s and the lease expires there. Mark where the second worker starts and where the first worker's write arrives, then say how short the pause would have had to be to change nothing.
Write the fenced terminal update and state the expected rows-affected in both the healthy case and the fenced case.
Size the heartbeat: 3,000 concurrent runs at a 10-second interval is about 300 row updates/second on job_run - decide whether that lands on the same table and what it contends with.
Write the billable_seconds rule for each terminal status and say which status leaves it null.

Expected result: A short lease of roughly 30 seconds renewed by a 10-second heartbeat, so detection latency is one lease - 20 to 30 seconds after a worker dies, plus the reaper's scan interval - independent of the 200 ms to 30 minute run spread, with one or two missed heartbeats tolerated rather than fatal; a terminal write fenced by AND status = 'running' AND lease_token = $x where zero rows affected means discard; retries as new rows with parent_run_id; a supervisor-owned wall-clock kill that separates timed_out from lost; billable_seconds left null while status = 'lost', or floored at the last heartbeat as a stated policy, never inferred from wall-clock.

Verify it yourself

In your timeline the fenced write must affect zero rows. If it affects one, the lease alone was doing the work and the token adds nothing.
At 3,000 concurrent runs and a 10-second heartbeat, check 300 updates/second against the other write load on that table before accepting the interval; if it is the dominant writer, the heartbeat belongs somewhere else. Check the ratio too: at lease = 3 x interval a lost heartbeat is survivable, whereas at lease = interval every scheduling hiccup is a double execution.
State whether a duplicate execution is billed, then confirm that summing billable_seconds across a parent and its retry matches that stated policy rather than matching it by accident.

Follow-up

A worker returns from a 45-second GC pause and writes succeeded. Walk through what the database does statement by statement.
Heartbeats are now a few hundred writes a second against job_run. How do you keep that off the path that matters, and what do you lose by moving it?
The re-dispatched workload has an external side effect the first attempt already performed. What does your design owe the customer, and what can it not fix?

Gateway p99 spikes on a five-minute cadence

Medium · cache stampede · ttl · invalidation

edge-gateway caches each credential-to-authorisation-context decision for five minutes. p99 sits at 6 ms except for a spike to 900 ms roughly every five minutes, worst in the region with the most pods, and control-plane CPU and read latency rise in step with it. The error rate stays near zero. Customers are told a revoked credential stops authorising within 60 seconds. Give the ordered checklist that identifies the mechanism, and a fix that removes the spike without weakening the 60-second bound.

Approach

Test periodicity before anything else: take the spike timestamps modulo the TTL in seconds. A tight cluster at a fixed offset means expiry phase, while traffic-driven spikes scatter.
Overlay pod start times. Entries filled at first request inherit the phase of the pod that filled them, so a cohort of pods deployed together expires together and the amplitude should track cohort size rather than tenant count.
Separate a herd from a capacity shortfall by measuring control-plane requests per second during a spike against baseline. A stampede shows a step of roughly (pods x hot keys) for one interval with hit rate collapsing to near zero, not a gradual climb that would indicate the dependency is simply undersized.
Apply three independent controls: randomise each key's TTL by a factor drawn uniformly from something like 0.8 to 1.0 so cohorts de-phase; coalesce concurrent misses per key per pod so exactly one refresh is in flight; and serve the stale value while that refresh runs so a miss costs the stale read rather than the dependency's queue.
Bound staleness against the published contract rather than against comfort: serve-stale is admissible only up to the 60-second revocation bound, so the TTL floor and the stale window together must stay inside it, and the published invalidation must delete the entry rather than schedule a refresh.
Decide in advance what a miss does when the control plane is unreachable, because that is now the only uncached path: failing closed converts a dependency outage into a total outage, while extending stale service past the bound breaks the revocation promise. Pick one and configure it explicitly.

Follow-up

Publish-subscribe invalidation is lossy under a partition. Given that, what actually enforces the 60-second bound, and what number would you put in the contract if asked to defend it?
One tenant's key is hot enough that a single pod's coalesced refresh still matters. What changes?
Would a shared cache tier in front of the control plane help or shift the problem, and what new failure does it add?

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.

The bug you decided not to fix

Engineers over-index on what they repaired. A stronger answer covers something you knowingly left broken: the alert you tuned down, the data inconsistency you documented instead of chasing, the cleanup you deferred past two quarters. Give the reasoning and the condition that would have reopened it, so it reads as a decision and not as neglect.

Describe your experience with CMM software or similar automated inspection tools.

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 did you decide not to do, and why?
How did you know your change caused the improvement?

Why are you interested in the intersection of software and precision manufacturing?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Pick a story where you made the decision, not one where you watched it.
Close with what you would do differently, concretely.

Follow-up

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

How do you handle feedback regarding technical specifications or programming errors?

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Give the blast radius: what could have broken, and what you measured.
Close with what you would do differently, concretely.

Follow-up

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

Where do you see yourself in 5 years?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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