Interview guides · Acushnet

Acushnet Software Engineer Interview Questions & Guide 2026

Take-home style practice for Acushnet software engineer candidates: a small service, a fixed budget, a written tradeoff.

The 60-second brief

As a Software Engineer at Acushnet, you are at the intersection of world-class manufacturing and cutting-edge technology. Acushnet is the home of iconic brands like Titleist, and your work directly influences the precision, quality, and innovation behind the equipment used by the world's best athletes. This role is not just about writing code; it is about building the digital infrastructure that maintains the company’s reputation for excellence in the golf industry.

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 every outbound call with a timeout
Paginate large result sets with keyset cursors
Make every write idempotent under retry

Overview

You will contribute to high-impact projects that bridge the gap between software development and physical production. Whether you are working on Central Quality & Quality Control systems or supporting CNC/CAM project engineering, your contributions will directly impact product consistency and operational efficiency. The environment is collaborative and rooted in a deep understanding of manufacturing processes, making this an ideal role for engineers who enjoy seeing their technical solutions materialize in real-world, high-stakes environments.

Familiarize yourself with how Acushnet manufactures its products. Watching company videos on their production facilities and ball plants will provide you with essential context that demonstrates your genuine interest and business acumen during interviews.

The Acushnet interview process

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

Video Screening

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

What to demonstrate

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

How to prepare

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

On-Site Interview

Nobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.

What to demonstrate

Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience

How to prepare

Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed

Where candidates lose points

Running a schema change as though the lock lasts as long as the statement

In PostgreSQL an ALTER TABLE that needs an ACCESS EXCLUSIVE lock must first wait for every open transaction touching that table, and while it waits, later queries needing a conflicting lock queue behind it rather than overtaking it. A DDL statement that would execute in milliseconds, issued while a thirty-second analytics query is open, therefore stalls all traffic on that table for thirty seconds: the outage length is set by the longest open transaction, not by the change. The defences are specific and worth knowing by name - set lock_timeout low and retry rather than queue, add columns without a volatile default so no table rewrite occurs (from version 11 a non-volatile default is a metadata-only change), build indexes with CREATE INDEX CONCURRENTLY while accepting that it cannot run inside a transaction block and leaves an invalid index behind if it fails, and add constraints as NOT VALID followed by a separate VALIDATE CONSTRAINT, which takes a weaker lock.

Assuming an isolation level prevents the anomaly you actually have

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

Retrying a write that is not safe to repeat

A timeout tells you nothing about whether the server applied the write, so a blind retry of a create or a charge can duplicate it. Either make the operation idempotent, with a caller-supplied key the server deduplicates on or a conditional update, or do not retry it; and use exponential backoff with jitter so the retries of many clients do not synchronise into a second outage.

Finishing a solution without stating its complexity

Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.

Questions & practice

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

Hard · reconciliation · range hashing · throttling

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

Approach

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

Follow-up

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

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.

Follow-up

The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
How would you make the window survive a process restart, and is it worth the cost?

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.

Approach

Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.

Worked solution (30 min)

Write the edge-loading query with the tenant predicate on both endpoints and state what it does with a cross-tenant edge.
Implement iterative Tarjan with an explicit stack and confirm on a three-node cycle that it emits one component of size three.
Build the transpose restricted to the visited set and mark every node with an in-edge from outside it as refused, carrying the referrer id.
Run Kahn over the condensation and verify the emitted order against the referrer-before-referenced rule.
Size the CSR arrays for 2,000,000 nodes and 8,000,000 edges and compare against a boxed adjacency map.

Expected result: An iterative O(V+E) traversal over a tenant-scoped CSR subgraph, SCC condensation so cycles archive atomically as one component, a transpose-based refusal list naming the external referrer for each blocked resource, and a Kahn topological order over the condensation, with recursion replaced by an explicit stack because of graph depth rather than style.

Verify it yourself

A two-node cycle with no external referrers archives as one unit, and adding a single external referrer moves both nodes to the refusal list.
No resource appears in the archive order before something that references it, except within its own component.
A counter on edge visits matches the loaded edge count exactly, confirming each edge is traversed once.

Follow-up

The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?

Denormalise tenant onto revisions and backfill it live

Hard · denormalisation · online migration · backfill

resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.

Approach

Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.
Step three, backfill: batch by primary key rather than by created_at so the cursor is dense and resumable — UPDATE resource_revision rr SET tenant_id = r.tenant_id FROM resource r WHERE r.resource_id = rr.resource_id AND rr.revision_id > $1 AND rr.revision_id <= $1 + 5000 AND rr.tenant_id IS NULL — committing per batch and persisting the cursor. Throttle on replica replay lag and on dead-tuple count, since each batch writes 5,000 new row versions. Run the backfill before the index exists so those updates can stay HOT.
Step four, index then enforce then contract: CREATE INDEX CONCURRENTLY (cannot run inside a transaction block, scans the table twice, waits on open transactions, and leaves an INVALID index to drop concurrently if it fails); ADD CONSTRAINT ... CHECK (tenant_id IS NOT NULL) NOT VALID, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE, after which SET NOT NULL uses the validated check instead of re-scanning on PostgreSQL 12 and later. Only then move the audit reads onto the column and, in a later deploy, delete the join path.

Worked solution (40 min)

Write the five steps as separate scripts and state, for each, the lock mode it acquires and the deploy it pairs with.
On a 20M-row copy, run the ADD COLUMN while a 30-second transaction holds a lock on the table, and record how long unrelated queries queue behind it.
Run the batched backfill at 5,000 rows, kill it mid-run, restart from the persisted cursor, and confirm no row is processed twice and none is skipped.
Build the index concurrently under concurrent write load, then add the CHECK ... NOT VALID, VALIDATE it and SET NOT NULL, timing each.
Compare the audit-feed plan before and after: join-and-filter versus an index seek with no Sort.

Expected result: ADD COLUMN completes in milliseconds once it holds the lock, but with a 30-second transaction open it stalls all traffic on that table for the full 30 seconds unless lock_timeout cuts it off. The backfill resumes exactly at the persisted revision_id. After VALIDATE, SET NOT NULL returns without a table scan. The audit feed drops from scanning the tenant's history to reading about 50 rows.

Verify it yourself

SELECT count(*) FROM resource_revision WHERE tenant_id IS NULL returns 0, re-checked after the dual-writer has been live for a full deploy cycle.
A sampled join confirms rr.tenant_id = r.tenant_id across 100k random revisions; the composite foreign key makes a mismatch unwritable from then on.
pg_stat_activity shows no transaction older than one batch's duration while the backfill runs.
EXPLAIN of the audit feed shows an Index Scan with no Sort node.

Follow-up

The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
How do you verify the backfill actually finished, given rows are still being inserted while it runs?
A resource must now be movable between tenants. What does that do to the composite foreign key and to the revisions already written?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

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

Approach

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

Follow-up

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

Specify the outbound webhook contract a customer endpoint can verify

Medium · webhooks · hmac · at-least-once · ordering

Egress delivery sends resource.created, resource.updated and resource.archived to roughly 40k customer endpoints at about 1.5k deliveries/second, with a per-destination concurrency cap of 4 and a 10 second connect-plus-read timeout, at-least-once from the relay. The receiver is code you do not control, often behind a proxy that may re-encode JSON. Specify the payload envelope, how the receiver verifies authenticity and rejects replays, what ordering it may assume, how it deduplicates, which status codes it should return and what each means to us, and the retry, circuit and replay behaviour.

Approach

Send the fact, not a pointer: event_id, event_type, aggregate_type, aggregate_id, aggregate_version, occurred_at and the payload copied from outbox_event. A receiver told only that resource 42 changed has to call back, and the call back returns a newer state than the event describes - that is precisely how a consumer applies changes out of order while believing it is following them.
Sign the exact bytes on the wire: HMAC-SHA256 over the timestamp, a separator, and the raw body, in a header, compared in constant time before the body is parsed. Instruct receivers to verify the raw bytes, never a re-serialised copy, since a proxy that reorders keys or changes whitespace invalidates a signature computed over re-encoded JSON. Reject a timestamp outside a tolerance - five minutes is typical - and keep a short seen-set of event_ids, because a capture replayed inside the tolerance still verifies.
Support key rotation by sending several signature values in one header during the overlap window and publishing that any one that verifies is sufficient. Without that rule a key rotation is a synchronised outage across every destination at once.
State the guarantee as it actually is: at-least-once and unordered, including within a single aggregate_id. The outbox is ordered per aggregate at the source, but the relay holds up to 4 requests in flight per destination and backs off each retry independently, so a version 6 delivery that needs one retry lands after a version 7 that succeeded on its first attempt. Publishing per-aggregate ordering we do not enforce is worse than publishing none, because a receiver that trusts it overwrites new state with old. Receivers deduplicate on event_id and apply conditionally - UPDATE ... WHERE aggregate_version < :incoming, or the same comparison under a row lock on the aggregate - which discards stale redeliveries and stays correct when two deliveries for one aggregate are processed at the same time. A duplicate and an out-of-order delivery then have the same outcome: no-op.
Define the response contract in the receiver's terms. 2xx means durably accepted, and accepting means writing it down - a receiver that does its processing inline inside our 10 second timeout will be retried while it is still working. 4xx is permanent and we stop. 429, 5xx and timeouts are retried with capped exponential backoff and full jitter within the per-destination concurrency cap, and a circuit opens after a run of consecutive failures and half-opens on a schedule, so a permanently dead endpoint costs one probe per interval instead of a growing queue.
Give them a recovery path that does not involve us holding a queue for everyone: a visible parked state, and a replay endpoint keyed by event_id or time range, so a receiver that was down for two hours asks for the gap on its own schedule.

Worked solution (30 min)

Write the envelope fields and mark which are needed for deduplication and which for discarding a stale delivery.
Write the signature string construction and the receiver's verification steps in order, including the constant-time compare and the tolerance check.
Write the receiver's idempotent apply: dedupe on event_id, then the version-conditional write that discards stale and concurrently delivered redeliveries.
Write the status-to-action table from our side: which codes retry, which stop, and what opens the circuit.
Describe the parked state and the replay call, including what the customer must supply.

Expected result: A self-contained signed envelope, verification over raw bytes with a replay window and a seen-set, an at-least-once guarantee that states delivery is unordered even within one aggregate, a version-conditional apply, a status table that distinguishes permanent from transient, and a customer-driven replay path.

Verify it yourself

Delivering the same event twice leaves the receiver's state identical to delivering it once.
Delivering version 7 and then version 6 for one aggregate leaves version 7 applied.
A destination that always times out costs one probe per circuit interval rather than a retry queue.

Follow-up

A customer insists an event never arrived. What do you check, in order, and which table settles the argument?
A receiver cannot store a version and asks us to deliver their events in order. What would the relay have to give up to offer that, and what does one failing event then do to the rest of their stream?

Convert the listing endpoint from offset pages to stable cursors

Medium · pagination · cursors · index access path

GET /v1/resources returns a tenant's resources newest-updated first, today with page and per_page, backed by index (tenant_id, status, updated_at DESC, resource_id DESC). Callers are a browser feed and a nightly sync job that walks every page. Users report items appearing twice or vanishing between pages, and page 400 is slow. Design the cursor contract: what the cursor contains and how it is encoded, the exact WHERE and ORDER BY, what happens when a row's updated_at changes mid-walk, how a client detects the end, and what the sync job does when a cursor is rejected.

Approach

Separate the two defects, because they have different fixes. Cost: OFFSET n makes the engine produce and discard n rows, so price grows with page depth rather than page size and page 400 pays for 400 pages of work. Correctness: while the set shifts, rows cross the offset boundary and are skipped or repeated, and nothing in the response lets the client detect it.
Write the seek: WHERE tenant_id = $1 AND status = $2 AND (updated_at, resource_id) < ($k, $id) ORDER BY updated_at DESC, resource_id DESC LIMIT n. The tie-break is not decoration - updated_at is not unique, and two rows sharing a timestamp across a page boundary reintroduce exactly the skip this was adopted to remove. PostgreSQL seeks the composite index on the row-value comparison directly; on an engine that does not optimise a row constructor, expand it into the equivalent OR form or the plan quietly degrades to a scan.
Encode the cursor as an opaque token carrying the sort key, the id, and a fingerprint of the filter and sort order, signed or at minimum validated. A cursor replayed against a different sort or filter must be a 400 with its own code, not a silently wrong page - the sync job cannot notice the difference otherwise.
State the guarantee precisely rather than generously. Keyset is stable against inserts and deletes elsewhere in the set, because the position is a value and not a count. It is not a snapshot: updated_at is mutable, so a row that is edited during the walk re-sorts and may be seen twice or not at all. If the sync job needs exactly-once coverage, order on an immutable key such as (created_at, resource_id), or walk resource_revision by revision_id and treat updated_at as data.
Define termination and limits in the response, not in the client's inference: fetch n+1 rows, return n, and emit next_cursor only when the extra row existed. Absence of next_cursor is the sole end signal, because a page can legitimately come back short when rows are filtered after retrieval. Cap n and document the cap rather than honouring per_page=10000.
Migrate without a flag day: keep page and per_page working, add the cursor, count usage per credential, and remove the offset path only once the sync job's traffic on it is zero.

Follow-up

The client wants a total count and the ability to jump to page 400. What can you honestly offer instead, and what does each option cost?
How do you paginate backwards, and what does that require of the index?

Edge instances grow 400 MB per hour until the nightly restart

Hard · memory leak · heap analysis · metric cardinality · gc

Edge API instances start at 700 MB resident and grow about 400 MB/hour; a nightly rolling restart has hidden it for weeks. Growth continues unchanged when request rate halves overnight, p99 degrades in the last hours before an instance is recycled, and heap used immediately after a forced full GC rises monotonically. The service holds no product state. Name the discriminating measurement that separates the plausible causes, give the most likely cause, and give the fix and how you would verify it.

Approach

Separate resident memory from live heap first, because they fail differently. Resident size can grow from fragmentation, native buffers or thread stacks while the heap is flat; heap used after a full GC rising monotonically is the measurement that says objects are reachable and not being released. You already have it, so this is retention, not fragmentation, and that closes off half the candidate list.
Use the rate's independence from traffic as the discriminator. Growth that continues at half the request rate rules out per-request objects that are merely slow to collect and points at a structure that grows with distinct values observed rather than with call volume. Write the candidates that have that property: a metrics registry keyed on a high-cardinality label, an unevicted cache, an interner, a per-key lock map.
Take two heap snapshots an hour apart and diff by retained size, reading the dominator tree, not by allocation count or instance count. Expect one root holding a map with millions of entries, then follow the reference chain to the code that inserts and never removes. Allocation profilers point at churn, which is the wrong signal here.
The candidate that fits this service is an observability label carrying an identifier, such as a request path recorded before templating so that /v1/resources/48213 becomes its own metric series. That grows with distinct ids seen, is independent of rate, and explains the late p99 degradation, since GC cost rises with the size of the live set.
Fix by bounding cardinality at the source: template the path to /v1/resources/{id} before it becomes a label, move tenant id from a label to a log field or an exemplar, and cap the registry with a bounded map that evicts. Add a cardinality ceiling that fails loudly in a lower environment rather than growing quietly in production.
Verify with a soak rather than a restart. Hold one instance out of the nightly recycle for 48 hours with the fix and compare post-GC heap and series count against an unfixed control taking the same traffic.

Follow-up

Post-GC heap is now flat but resident size still creeps. What are you looking at, and does it matter?
How would you have detected this before an OOM, given the nightly restart masked the trend?
That label is what makes one dashboard useful. How do you keep the dashboard and lose the leak?

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.

Disagreement needs an outcome you did not choose

Conflict answers where you were right and everyone came round are the weakest ones. Stronger: the evidence you went and collected, what would have changed your mind, and what you did in the weeks after the call went against you. Implementing a design you argued against, properly, is a specific and checkable behaviour.

Describe a situation where you had to prioritize conflicting project requirements.

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
Name the disagreement and how you resolved it with evidence.
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?

How do you handle feedback from non-technical stakeholders?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

Explain a time you solved a problem in a group.

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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