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affinis AG Software Engineer Interview Questions & Guide 2026

affinis AG software engineer prep on operations: decide what you would need to see to know a release is going wrong.

The 60-second brief

As a Software Engineer at affinis AG, you are at the intersection of complex enterprise architecture and strategic business transformation. The role is critical to the firm's mission of delivering high-impact digital solutions, particularly within the ecosystems of Adobe Experience Platform (AEP), Adobe Campaign, and Microsoft Dynamics 365 Business Central. You aren't just writing code; you are architecting solutions that allow clients to streamline operations, manage customer data, and drive meaningful digital engagement.

Browser-facing seats are not covered by algorithm practice. Scope in state ownership, what the page does on a slow or failed request, and how you would diagnose something that renders correctly but feels slow.

Focus areas

Bound blast radius with per-tenant concurrency limits
Build at-least-once pipelines with explicit deduplication horizons
Make every write idempotent under client retries

Overview

This position demands a blend of deep technical expertise and a consultative mindset. Because affinis AG functions as a high-level partner to its clients, you will often find yourself bridging the gap between raw technical requirements and long-term business goals. Success in this role requires the ability to navigate complex software landscapes, troubleshoot intricate system architectures, and maintain a high standard of quality in every deployment.

The affinis AG interview process

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

Company Introduction

You cannot drill a format you do not know, so put the preparation into material that travels. Three pieces of your own work, each rehearsed until you can take a follow-up you did not anticipate, will carry a conversation or a code walkthrough equally well. Specificity is what separates that from filler. A number needs its definition before it means anything: a p99 is over some window and measured at some hop, and a server-side figure excludes the queueing and network time a client would see. The number you cannot qualify is the one to leave out.

What to demonstrate

Whether your examples carry detail only someone who did the work would hold, such as what the binding constraint actually was, which alternative you rejected and why it was worse, and what you measured on each side of the change
Whether a number survives one follow-up, meaning you can say what it was measured over and whether it moved because of your change or merely alongside it
Whether a failure is described with the specific change that followed it, rather than a lesson stated in general terms
Whether your part in a team effort is stated accurately, including what other people did

How to prepare

Write a page on each of three projects covering the constraint, the option you rejected, the measurement before and after, and what went wrong. Cut any line you cannot take a follow-up on, since you are writing the parts you will be pressed on rather than a summary.
Recover the real figures while you still have access: request volume, data size, latency with its percentile and window, team size, timeline. Note where each came from, whether a dashboard, a design document or memory, and mark the estimates so you can say which they are out loud.
Take your weakest project story to someone who works in a different area and have them ask why four times in succession. The point where you run out of answer is the part to go and re-read before the round.

Candidate Background Discussion

An unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.

What to demonstrate

Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating

How to prepare

Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes

Mutual Discovery

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

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.

One shared connection pool for every tenant and every query class

A single tenant with a large table and a missing index can occupy every connection with slow queries, and every other tenant then waits in connection acquisition -- a queue invisible in database metrics, because the database itself looks healthy while the application starves. Containment is bulkheads: separate pools or per-tenant concurrency caps for interactive requests, background jobs and exports, a statement timeout low enough that a pathological query dies before it accumulates, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks indefinitely. One caveat worth knowing in advance: if a transaction-pooling proxy sits in front of the database, session-scoped behaviour changes, so session-level advisory locks and settings applied outside a transaction do not survive the way they do on a direct connection.

A queue or buffer with no bound

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

Check-then-act on shared state

Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.

Questions & practice

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.

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?

Fold a deduplicated usage stream into hourly rollups

Easy · aggregation · deduplication · watermarks · exact-arithmetic

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

Approach

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

Worked solution (25 min)

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

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

Verify it yourself

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

Follow-up

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

Explain why the metering dashboard scans every daily partition

Medium · indexing · partition pruning · sargability · tenant skew

usage_event is range-partitioned daily on ingested_at and holds tenant_id, workspace_id, environment, sku, quantity numeric(20,6), occurred_at and ingested_at. The only relevant index is on (occurred_at). A dashboard runs select sku, sum(quantity) from usage_event where tenant_id = $1 and date_trunc('hour', occurred_at) >= $2 and environment = 'production' group by sku, and EXPLAIN shows a sequential scan of every partition. Give each distinct reason, rewrite the predicate so an index can serve it, propose the index, and state the write cost its column order adds.

Approach

Separate the three causes rather than blaming one. First, date_trunc('hour', occurred_at) wraps the column, so the predicate is not sargable against a btree on the bare column. Second, pruning keys off ingested_at while the query constrains occurred_at, so no partition can be excluded. Third, even made sargable, (occurred_at) is not tenant-leading, so for one tenant among thousands the scan reads the whole time range and discards nearly all of it.
Rewrite the bound carefully, because the obvious rewrite is only conditionally equivalent. date_trunc('hour', x) >= $2 equals x >= $2 only when $2 is already hour-aligned; for an arbitrary $2 it means x >= date_trunc('hour', $2) + interval '1 hour'. Normalise the parameter in the caller and leave the column bare.
Restore pruning with a second, redundant predicate on the partition key: ingested_at >= $2 - interval '<late-data horizon>'. State both sides of it. It prunes to a handful of partitions, and it silently omits any event whose ingest lagged past that horizon, which is precisely what a producer replay produces. Either document the horizon as a stated bound, or partition on occurred_at and move the problem into the dedup window instead.
Propose (tenant_id, occurred_at) include (sku, quantity) per partition. A partial index where environment = 'production' mostly saves size rather than selectivity, since production dominates the three environments; take it if non-production is a meaningful share and skip it otherwise.
Price the write path honestly. At roughly 250M rows/day each extra index is another insert plus WAL per row, and a tenant-leading key scatters inserts across one hot leaf per active tenant instead of appending to a single rightmost leaf, so page dirtying and random I/O both rise. An INCLUDE payload widens every leaf entry and enlarges the index accordingly.
Add the index-only-scan caveat before someone reports it as a regression: on a freshly appended table the visibility map is not yet set for recent pages, so the INCLUDE columns still cost heap fetches until autovacuum has been through, and the newest hour is exactly the data the dashboard reads.

Worked solution (30 min)

Build 30 daily partitions with skewed tenants, one holding about 40% of the rows, then ANALYZE.
Run explain (analyze, buffers) on the original query and record how many partitions were scanned and the rows removed by filter.
Apply the rewritten predicate and the index, re-run, and confirm the plan lists only the partitions inside the ingested_at bound.
Re-run with $2 set to a non-hour-aligned timestamp and confirm the rewritten and original predicates return identical rows.
Insert an event with ingested_at six hours past occurred_at and check whether the pruning predicate excludes it.

Expected result: Three causes named separately: a non-sargable date_trunc, pruning on ingested_at against a predicate on occurred_at, and a non-tenant-leading index. A rewrite to occurred_at >= <hour-normalised $2> plus a redundant ingested_at >= $2 - horizon, an index on (tenant_id, occurred_at) including sku and quantity, a per-row write cost stated, and the late-data omission the pruning predicate introduces stated as a known bound.

Verify it yourself

The plan lists only partitions inside the ingested_at bound.
Rows removed by filter on the index scan is small relative to rows returned.
The rewritten and original predicates agree for both hour-aligned and non-aligned bounds.
A deliberately late-ingested event is visible or omitted exactly as the stated horizon predicts.

Follow-up

CREATE INDEX CONCURRENTLY is not supported on a partitioned parent. Give the sequence that gets this index onto 400 existing partitions without blocking ingest.
One tenant holds 200 times the median row count and the dashboard still times out for them with the index in place. What changes?
Should this read hit usage_rollup_hourly instead? State what that costs in freshness and what the watermark lets you promise.

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.

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?

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.

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?

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?

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 a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.

Day 1 — Numbers before diagrams

Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.

Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.

Day 2 — One system, from requirements to schema

Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.

Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.

Day 3 — The consistency you are actually buying

Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.

Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.

Day 4 — Failure is the design

For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.

Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.

Day 5 — Scaling the hot path

Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.

Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.

Day 6 — Keep the coding hand in, at the bar that applies to you

Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.

Deliverable: One problem hardened to review standard, with its test list and one written review comment.

Day 7 — Defend it while being interrupted

Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.

Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.

Outage answers are graded on instruments

Nobody is scoring your stamina at three in the morning. What carries weight is which signal told you something was wrong, what you measured before touching anything, what you rolled back versus what you fixed forward, and why you picked one. 'We restarted it and it went away' is a story about not knowing.

Can you walk us through your professional background and key milestones in your career?

Medium · behavioural and engineering judgement

Approach

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.
Name the disagreement and how you resolved it with evidence.

Follow-up

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

Why do you want to transition from your current position to affinis AG?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Name the disagreement and how you resolved it with evidence.
Close with what you would do differently, concretely.

Follow-up

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

How do you handle collaborative environments where team dynamics shift?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Close with what you would do differently, concretely.
Name the disagreement and how you resolved it with evidence.

Follow-up

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

What draws you to the specific technologies we use, such as Adobe or Microsoft Dynamics?

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.
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?