Interview guides · Mactores

Mactores Software Engineer Interview Questions & Guide 2026

Timed coding practice for Mactores software engineer candidates: full attempts under a clock, then scored against a rubric.

The 60-second brief

As a Software Engineer at Mactores, you are at the intersection of complex cloud architecture and high-impact digital transformation. You are not just writing code; you are building scalable, resilient solutions that leverage the full power of cloud ecosystems to solve real-world enterprise challenges. This role is critical to the company’s mission of accelerating customer success through cloud-native innovation.

The behavioural round is a technical round in narrative form. Prepare it by collecting specifics you actually owned, such as a design you argued against, an incident you diagnosed, or a decision you later reversed, rather than by rehearsing phrasing.

Focus areas

Scope every query and cache key by tenant
Make every write idempotent under client retries
Bound blast radius with per-tenant concurrency limits

Overview

You will find yourself working in dynamic environments where deep technical expertise in AWS services, automation, and system design is paramount. Whether you are optimizing existing infrastructure or architecting new cloud-native applications, your work directly influences the reliability and performance of systems that power client businesses. This position is ideal for engineers who thrive on technical complexity and enjoy bridging the gap between abstract requirements and robust, production-ready software.

The Mactores interview process

Mactores candidates report 4 rounds · ≈ 3-5 weeks. The stages below are what candidates describe, not a published process.

Initial Screening Call

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

Cognitive Ability Assessment

Input bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.

What to demonstrate

Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply

How to prepare

For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last

Technical Rounds

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

Discussion with Hiring Manager

Expect the estimate to be probed harder than the outcome. A manager asking how long something will take is checking whether your number covers the parts that reliably go wrong: migrating the rows that already exist, and the second service that has to change at the same time. A single confident number is the weak answer. The strong one states what the estimate assumes, names the unknown that could double it, and says what you would cut first if the date were fixed. Scope is the variable you control; hours are not.

What to demonstrate

Whether an estimate arrives with its assumptions attached, including what it excludes and how much of it is waiting on review, environments or another team
Whether you can name work you cut and the person you told, since a cut nobody heard about surfaces later as a surprise rather than a decision
How you order two tasks that both look urgent, and whether that order survives being told the deadline moved in by a week
Whether you separate what must be done before launch from what must be done before the thing can be left running unattended

How to prepare

Take the last three things you shipped and write down, from memory, the estimate you gave and the time it actually took. Bring the largest miss and why it missed; a candidate who has never missed has never committed to a date.
Against your current queue, write who is blocked until each item lands. Anything with nobody behind it becomes your worked example of what you would drop and how you would say so.
Rehearse the fixed-date answer aloud: which slice ships, which slice is deferred and written down somewhere it will be found, and the one thing you would refuse to cut because removing it makes the release unsafe rather than smaller.

Where candidates lose points

Treating a timed-out write as a failed write

A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.

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.

Treating a network call as though it were a local function call

A remote call can be slow, fail, or return after you stopped waiting, so name the timeout, the retry policy, and what the caller sees while the dependency is down. A call with no timeout turns one slow dependency into an exhausted thread or connection pool in every service upstream of it.

Not asking what the system looks like if it dies halfway through

For any multi-step write, say what state remains if the process stops between step two and step three, and what brings it back: a single transaction, a saga with compensating actions, an outbox, or a reconciliation job. Partial failure is routine at any real call volume, so 'that shouldn't happen' is an answer with nothing behind it.

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?

Find peak concurrent sandbox usage from run intervals

Medium · sweep-line · intervals · concurrency-caps · null-semantics

Given up to 5 million job_run rows for one tenant over one day, with run_id, started_at, finished_at, status and wall_clock_limit_seconds, report the maximum number of sandboxes running at once, the earliest instant that maximum is reached, and the first run_id that would breach a per-tenant cap of C. started_at is null while a run is queued; finished_at is null both for runs still executing and for runs in status lost. Treat a run as occupying [started_at, finished_at). Give the complexity and state how you handle each null.

Approach

Turn each run into two sweep events, (started_at, +1) and (end, -1), then sort the 2n events by timestamp with -1 ordered before +1 at equal timestamps. That tie-break is what makes the interval half-open, so a run finishing at 10:00:00 and one starting at 10:00:00 never overlap.
Decide each null out loud before sweeping, because each choice moves the answer. A null started_at means queued and contributes nothing. A null finished_at with status running or leased is clipped to the window end. Status lost has no observed end at all, so clip it at started_at + wall_clock_limit_seconds on the grounds that the supervisor owns the timeout, and record that you did. The table's check (finished_at is null or started_at is not null) guarantees you never see an end without a start.
Sweep once, maintaining a running counter, the maximum, and the timestamp at which the maximum was first attained (update peak_at only on a strict increase, or you will report the last such instant instead of the earliest). Capture the first run_id whose +1 takes the counter to C+1 during the same sweep rather than in a second pass.
Complexity: O(n log n) dominated by the sort, O(n) space. If rows already arrive ordered by started_at, a min-heap of end times gives O(n log k) time and O(k) space with k the peak concurrency, which is the better shape when the rows come from an index scan on (tenant_id, started_at).
If second resolution is acceptable, counting-sort the endpoints into an 86,400-slot delta array and prefix-sum it: O(n + T) time and O(T) space, which beats the comparison sort at 5 million rows. It answers only at second granularity, so state which resolution the cap is defined in.

Worked solution (20 min)

Write the null policy as three lines of prose first, one per case, and keep them beside the output.
Emit 2n endpoint tuples (timestamp, delta, run_id) and sort on the key (timestamp, delta) so -1 precedes +1.
Sweep, tracking cur, peak, peak_at updated only on a strict increase, and the first run_id whose +1 takes cur to C+1.
Build a fixture with two runs where one ends exactly when the next starts, three genuinely overlapping runs, one run with a null finished_at and status running, and one with status lost and a 300-second wall_clock_limit_seconds.
Re-run with every timestamp shifted by a constant and confirm the peak is unchanged while peak_at shifts by the same constant.

Expected result: Peak is 3, from the overlapping trio. The exact-handoff pair yields a peak of 1, not 2. peak_at is the start instant of the third overlapping run. The lost run occupies exactly [started_at, started_at + 300s) under the stated policy.

Verify it yourself

Setting peak_at on >= rather than > moves the reported instant to the last time the peak was attained. Flip it and watch the fixture change.
Sorting by timestamp alone, without the delta tie-break, raises the reported peak on the handoff fixture.
The deltas over the whole sweep sum to zero once every open-ended run has been clipped.

Follow-up

Now report peak concurrency per tenant for 10,000 tenants from one globally sorted stream. What changes about memory and about the sort?
The cap has to be enforced at dispatch rather than reported afterwards. What does the admission check look like, and where does it race?
How would you answer 'peak concurrency within any 5-minute window' without re-sorting?

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.

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?

Rebuild an hourly rollup with deduplication and late-arrival accounting

Hard · window functions · deduplication · event time vs ingest time · watermarks

From usage_event (event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at), produce the values usage_rollup_hourly should hold for one tenant over one day: per (workspace_id, sku, hour_start) the deduplicated quantity_sum, event_count and source_max_ingested_at, bucketed by occurred_at. Duplicates share (tenant_id, idempotency_key). Also report, per hour, the running total across the day and the share of quantity that arrived more than two hours after the hour began. Write the query, and state which duplicates a daily unique index cannot catch.

Approach

Deduplicate in its own CTE before any aggregation, because a SUM cannot be un-summed: row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id) = 1. Include the tiebreaker. Without it the surviving row is non-deterministic when two duplicates share an ingested_at, and a rollup described as deterministically recomputable then disagrees with itself between runs.
Bucket on occurred_at and nothing else, and pin the timezone explicitly. date_trunc('hour', timestamptz) truncates in the session's TimeZone setting, so the same query run by a session set to a non-UTC zone buckets differently; use the three-argument date_trunc('hour', occurred_at, 'UTC') on PostgreSQL 16 or later, or date_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC' before that. Filter environment = 'production' explicitly, since metering covers three environments and billing covers one.
Aggregate to the grain with sum(quantity), count(*) and max(ingested_at). The last is not decoration: it is the watermark the row consumed up to, and without it there is no way to prove afterwards what a number did and did not include.
Compute the late share inside the dedup-and-aggregate step as a conditional aggregate, sum(quantity) filter (where ingested_at > hour_start + interval '2 hours'), then divide by the hour's total. Compute the running total as a window over the already aggregated rows: sum(quantity_sum) over (partition by workspace_id, sku order by hour_start rows between unbounded preceding and current row). Running either over raw rows puts the duplicates back.
Answer the index question exactly. The unique constraint is on (ingested_day, tenant_id, idempotency_key), because a unique index on a partitioned table must contain the partition key. It therefore deduplicates only within one ingest day and admits a duplicate whose retry crosses midnight or whose replay runs a week later. That is why this CTE dedups across the whole window being recomputed, and why the dedup horizon is a correctness parameter rather than a retention cost.
Keep the numeric type all the way through. quantity is numeric so the sums are exact; a cast to double precision anywhere in this pipeline reintroduces drift that surfaces only as a few unreconcilable cents per tenant per month, long after the query is out of anyone's mind.

Follow-up

A dispute forces the same recompute over 40 days for one tenant. What changes about the dedup CTE's memory use and the chosen plan, and what would you do about it?
Two runs a minute apart return different quantity_sum values for an hour that is already closed. Give two mechanisms that produce that, and the single query that distinguishes them.
Express the same rollup incrementally so it does not re-scan the day each time the watermark advances. What does the incremental version stop being able to answer?

Enforce a concurrent-run quota that survives simultaneous requests

Medium · isolation levels · write skew · quotas · lock contention

A plan allows at most 20 concurrently running rows in job_run per tenant. The table holds run_id, tenant_id, workspace_id, status (queued, leased, running, succeeded, failed, timed_out, cancelled, lost), lease_token, leased_until, started_at and finished_at. Today the service runs select count(*) from job_run where tenant_id = $1 and status = 'running', compares the result to 20, then inserts. Under load a tenant exceeds the cap by exactly the number of concurrent requests. Name the anomaly, say which isolation levels do and do not prevent it, and give a version that holds, as SQL.

Approach

Name it: write skew. Each transaction reads a predicate (the count of running rows), neither modifies what the other read, and both then insert rows that jointly violate an invariant no single row expresses. Read committed permits it. So does repeatable read, because snapshot isolation's first-updater-wins check fires only on conflicting row updates, and these are inserts touching disjoint rows.
Enumerate the fixes with their real costs. SERIALIZABLE works: PostgreSQL's SSI tracks the predicate read and aborts one transaction with SQLSTATE 40001, which obliges the caller to retry and makes the abort rate rise with contention on a hot tenant. Folding the predicate into the write as insert ... select ... where (select count(*) ...) < 20 narrows the race to the statement's snapshot but does not close it under read committed.
Give the version that holds at read committed: serialise on a row both transactions must touch. update tenant_concurrency set running = running + 1 where tenant_id = $1 and running < 20 returning running updates zero rows when the cap is reached, and zero rows is the rejection. This works because at read committed a blocked UPDATE re-evaluates its WHERE clause against the newly committed row; at repeatable read the same statement raises a serialisation error instead, so the isolation level changes the calling contract.
State the cost you just bought. That row is now a per-tenant serialisation point, so admission throughput for the tenant is bounded by one divided by the lock hold time; at a 2 ms hold that is roughly 500 admissions/second. Keep the critical section to the single UPDATE, with no network call or scheduling decision inside the transaction, and decrement in the same transaction that writes the terminal status.
Close the leak the status enum implies: a run can end as lost, so a crashed worker otherwise consumes a slot forever. Reconcile on a schedule against status = 'running' and leased_until < now(), and treat the counter as a fast path over job_run, which stays the system of record.

Worked solution (25 min)

Seed a tenant with 19 running rows, then fire 8 concurrent sessions each running the select-then-insert, and count the resulting running rows.
Repeat at REPEATABLE READ and confirm the count still exceeds 20.
Repeat at SERIALIZABLE, count the 40001 aborts, and note that without a retry loop those requests fail rather than queue.
Implement the atomic counter UPDATE, re-run the 8-way test, and confirm exactly 20 running rows with zero over-admissions.
Kill a worker mid-run, let the lease expire, and check whether the slot comes back without intervention.

Expected result: Write skew identified; read committed and repeatable read both permit it; serialisable prevents it at the price of 40001 retries; the working read-committed version is update tenant_concurrency set running = running + 1 where tenant_id = $1 and running < 20 returning running, whose zero-row result is the rejection, plus a reconciliation path for runs that terminate as 'lost'.

Verify it yourself

At 8x concurrency against a cap of 20, running rows never exceed 20.
The rejection returns zero rows rather than raising, so it is distinguishable from an error in the caller.
Killing a worker does not permanently consume a slot.
The counter and select count(*) from job_run where status = 'running' agree after reconciliation.

Follow-up

Write the retry loop for the SERIALIZABLE version. What does the caller see when it keeps aborting, and what bounds the retries?
Two regions each keep a counter. What is the effective cap, and what does admission do when the counter store is unreachable?
The cap changes mid-flight on a plan upgrade. Do running jobs get killed, and what does the counter row look like during the change?

How would you design a solution using S3, Lambda, IAM roles, API Gateway, and VPC?

Medium · system design and architecture

Approach

State the consistency you need, and where you are willing to be stale.
Name the failure you are designing for, then the recovery path.
Choose a partition key and say what query it makes expensive.

Follow-up

What breaks first when traffic grows ten times?
How does this behave when that dependency is down for an hour?

Can you explain how different AWS services interact in a real-world scenario?

Medium · system design and architecture

Approach

State the consistency you need, and where you are willing to be stale.
Fix the scope first: who calls this, how often, and what they do when it fails.
Name the failure you are designing for, then the recovery path.

Follow-up

How does this behave when that dependency is down for an hour?
What breaks first when traffic grows ten times?

Can you describe the projects you have worked on and the specific challenges you faced?

Medium

Approach

Work from the requirement backwards to the design.
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?

What are your preferred methods for troubleshooting performance issues?

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

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

An idempotent create endpoint that returns a one-time secret

Easy · idempotency · race conditions · api design

POST /v1/api-keys inserts a tenant_api_key row (tenant_id, workspace_id, name, key_prefix, secret_hash, scopes, status, auth_version, created_at, expires_at) and returns the plaintext secret exactly once, since only its SHA-256 is stored. Write volume is tens per second and clients retry on timeout. Design the idempotency mechanism: what the key is scoped by, where the record lives, its retention, what happens when the same key arrives with a different body, what happens when a retry arrives while the first request is still in flight, and what a replay returns for the secret.

Approach

Scope the key by tenant, not globally: uniqueness is on (tenant_id, idempotency_key), or one tenant's key collides with another's and the second caller receives a stored response for a request it never sent. Store a fingerprint of the request alongside it - method, path and a hash of the canonicalised body - so a mismatch is detectable.
Let the unique constraint decide the race instead of application logic. In the same transaction as the credential insert, run INSERT INTO idempotency_record (tenant_id, key, request_fingerprint, status) VALUES (...) ON CONFLICT DO NOTHING RETURNING id; no returned row means this is a replay, and the existing record is then read. A select-then-insert here loses to itself under concurrency in exactly the way this endpoint is meant to prevent.
Write the three replay outcomes as a decision table rather than as prose: same fingerprint and completed returns the stored response; same fingerprint and still in flight returns 409 with Retry-After, without blocking and without executing; different fingerprint returns 422, because replaying a stored response for a mutated body would tell the caller a credential was created for parameters it never sent.
Handle the secret as the part that makes this endpoint different from an ordinary idempotent create. The plaintext cannot be regenerated from secret_hash, so either the stored response holds it - making the idempotency record a secret at rest whose retention is now the secret's exposure window - or a replay returns the key metadata without the secret and the documentation says a lost response is resolved by listing keys and revoking the orphan. The second is the safer default precisely because credentials are listable and revocable.
Set retention from the client's retry budget, not from a round number: the record must outlive the SDK's maximum total retry duration, so twenty-four hours is defensible if that budget is minutes. After expiry the key is reusable and a very late retry creates a second credential, which is acceptable here only because the object is listable and revocable, and would not be for an unlistable side effect. Expire with a scheduled delete on an index over created_at.

Worked solution (20 min)

Write the DDL for the idempotency record, including the unique constraint that makes the concurrent case impossible rather than unlikely.
Write the four outcomes - fresh, replay-completed, replay-in-flight, fingerprint-mismatch - as a decision table with the HTTP status for each.
Decide what a replay returns for the plaintext secret and write the exact sentence the API reference has to carry about it.
Pick a retention and justify it from the client library's own maximum retry duration rather than from a round number.

Expected result: Unique (tenant_id, idempotency_key); the record inserted in the same transaction as the credential with ON CONFLICT DO NOTHING deciding the winner; 200 with the stored response, 409 for in-flight, 422 for a fingerprint mismatch; a documented rule that the plaintext secret is not replayed and a lost response is resolved by listing and revoking; retention exceeding the SDK retry budget, for example twenty-four hours.

Verify it yourself

Two concurrent requests with the same key must leave exactly one row in tenant_api_key. If the logic is select-then-insert, they will not.
A retry with a mutated body must not return 200. If it does, the fingerprint is not being compared or is not covering the fields that matter.
The retention you chose must be strictly longer than the client library's maximum total retry duration. Look that number up rather than assuming it.

Follow-up

Two requests with the same key arrive concurrently. Show the exact statements and say which one loses, and how it finds out.
The client receives a timeout, retries, and gets 409 in-flight. What should the client library do next, and for how long?
How does the design change if the created object is not listable - a one-off payout, say - so an orphan cannot be found afterwards?

A rare job-run overwrite that logging makes disappear

Hard · fencing tokens · leases · concurrency

About one job run in fifty thousand records billable_seconds matching no observed sandbox lifetime, and a few rows show worker_id changing after finished_at was already set. It does not reproduce: debug logging around the terminal write made it vanish for two weeks before it returned. Runs last from 200 ms to 30 minutes, the lease is 60 seconds and is renewed while a run executes. Give an ordered checklist, the mechanism, and a fix that makes the illegal write impossible rather than merely rarer.

Approach

Mine the evidence instead of chasing a repro: select rows where updated_at is later than finished_at, or where a terminal status was written twice, and join them to attempt history to recover both writer identities. The defect has already happened tens of times and the rows are the recording.
State the signature before measuring it. If the cause is a lease that expired while the original worker was stalled, affected runs should cluster where the gap between the last renewal and the terminal write exceeds the lease, and should correlate with worker pause metrics rather than with workload shape.
Read the disappearance honestly. Logging inside the window changed the timing and lowered the probability; it is evidence about how narrow the window is, not a fix. Reproduce by widening the window on purpose, shortening the lease and injecting a pause between sandbox exit and the terminal write, rather than by adding more instrumentation.
Name the mechanism precisely: the lease expires during a stall such as a long garbage-collection pause or a brief partition, the run is re-dispatched, and the original worker then wakes and writes its terminal state over the new attempt's row. A lease alone cannot stop this, because the check and the write are separated by the stall.
Fix by fencing the write itself: UPDATE job_run SET status = $2, finished_at = $3, billable_seconds = $4 WHERE run_id = $1 AND status = 'running' AND lease_token = $5, with zero rows affected interpreted as having been fenced rather than as success. The token lives on the row so the store arbitrates, not the worker's memory.
Keep the state machine honest: a retry inserts a new row pointing at parent_run_id rather than resetting the old one, and a run whose worker vanished terminates as lost with billable_seconds null, because recording failure asserts an outcome nobody observed and then bills and retries on that assertion.

Follow-up

The supervisor also emits a usage event on completion. What does the fenced worker do about the event it already emitted, and how does metering absorb it?
Lease renewal is itself a network call. What happens when a renewal times out, and how does the worker decide whether it still holds the lease?
Why is lengthening the lease past the longest legitimate run the wrong lever, and what breaks if you do it anyway?

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.

Scope is read from what you decided

Team size, service count and tickets closed say very little. Seniority shows in the decision you owned: what you chose not to build, which constraint you traded away, whose objection you had to resolve before anything could move. A large project where you executed someone else's plan is a small story.

How do you handle complex problem-solving in a time-constrained environment?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Pick a story where you made the decision, not one where you watched it.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

What are your experiences with Cloud computing and open-source cloud technologies?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Close with what you would do differently, concretely.
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?
What did you decide not to do, and why?

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?