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Essense Of Australia Software Engineer Interview Questions & Guide 2026

Software engineer interview prep for Essense Of Australia: original coding, system design and debugging practice, each with a worked solution.

The 60-second brief

As a Software Engineer at Essense Of Australia, you work between bridal fashion and the digital experiences that connect the company's global brand with brides-to-be. You build the digital infrastructure that lets the company's design-led products reach a diverse, international audience. Your work affects how the company's users interact with its collections, manage their journey, and engage with its retail partners.

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
Choose indexes from the query's access path
Make every write idempotent under retry

Overview

This role requires a unique balance of technical precision and creative problem-solving. The engineering team at Essense Of Australia, a fashion-industry company, is tasked with maintaining high-performing, scalable, and intuitive web platforms that reflect the brand's elegance and quality. You will be expected to contribute to complex technical challenges, collaborate across departments, and keep the company's digital presence in line with the quality of the gowns it creates.

The Essense Of Australia interview process

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

Initial Screening

Half of this call is the part candidates treat as small talk: start date, notice period, work authorisation and its timing, location and time zone, on-call, and the number. Those are what kill offers late, after several engineers have each spent a day. Surfacing a hard constraint now costs you nothing and occasionally buys you something, since a loop compressed to fit a competing deadline can usually only be arranged if it is asked for early. The common failure is deflecting the compensation question twice, then discovering at offer stage that the band never reached your number.

What to demonstrate

Whether your hard constraints are compatible with the role before a loop gets booked: earliest start, notice period, what authorisation you hold and when it needs action, days on site, willingness to carry a pager
Whether you give a compensation range with something behind it, such as current total compensation or a competing timeline, rather than leaving the band untested
Whether your stated timeline is real, since a competing deadline raised now is something scheduling can sometimes work around and the same deadline raised at offer stage usually is not

How to prepare

Write each constraint down in one line before the call and state them as facts rather than negotiating them live under a question you were not expecting
Set your range from two or three current data points for that level and location, and name the structure you are quoting in, so the number is comparable to the one they are holding
If another process is running, say where it stands and by when, and ask directly whether this loop can be scheduled inside that window

Peer-Level Technical Conversation

Most of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.

What to demonstrate

Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
Whether a failing case is isolated and explained before any edit is made to the code

How to prepare

From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.

High-Level 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

Discussion with Leadership

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

Choosing an index from the columns a query mentions rather than from how it filters and orders

A composite B-tree index on (a, b, c) can be seeked only as a left prefix: equality on a, then equality on b, then a range or an ordering on c. A query that filters on b alone cannot seek into it at all and at best gets a full scan of the index; a query that filters a and ranges on b gets no benefit from c, because the index is only sorted by c within a fixed (a, b) pair. The practical consequence is that one index per column is close to useless for multi-predicate queries while a single correctly ordered composite index turns a scan into a lookup. The ordering half is what gets missed: if the index cannot satisfy the ORDER BY, the database must read every matching row and sort before the limit can apply, so a LIMIT 20 over a million matching rows still reads a million rows.

Letting a slow dependency consume unbounded concurrency

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

Abandoning working code to chase the optimal solution

Get the straightforward version correct, state its complexity, and only then optimise, keeping the working version until the faster one passes the same cases. A correct quadratic solution with a stated path to linear beats a half-written optimal one that never ran.

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.

Questions & practice

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

Easy · hashing · at-least-once · aggregation

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

Approach

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

Follow-up

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

Merge partitioned event streams into one ordered feed with bounded lateness

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

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

Approach

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

Follow-up

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

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.

Worked solution (20 min)

Define the bucket struct and the advance step: take floor(finished_at_ms / 1000), compare with the ring's current second, zero min(delta, 60) buckets forward, then write into the new head.
Trace a destination that receives 5 attempts, goes silent for 90 seconds, then receives one more, and confirm the rate is computed from one attempt rather than six.
Compute total memory for 40,000 destinations at 60 buckets of two 4-byte counters, and state what changes if the window widens to 300 seconds.
Write the open rule as a single predicate combining the minimum-attempt floor with the rate threshold.

Expected result: A per-destination ring of 60 one-second (attempts, failures) buckets advanced lazily for amortised O(1) cost, roughly 480 bytes of counters per destination and about 20 to 25 MB for 40,000 of them, an LRU bound on live rings, and an open rule requiring both a minimum attempt count in the window and a rate above threshold.

Verify it yourself

After 60 seconds of silence a destination's window holds zero attempts and cannot open a circuit.
Memory measured with one hot destination and 39,999 idle ones is indistinguishable from the reverse, which the deque version would fail.
Replaying the same attempt stream with the clock shifted by an hour yields the same final rate.

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?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

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

Approach

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

Worked solution (25 min)

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

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

Verify it yourself

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

Follow-up

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

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.

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?

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.

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?

Make the resource write endpoint safe for retrying clients

Easy · idempotency · unique constraints · retries

The edge API accepts POST /v1/resources at 1.2k writes/second peak under a 400 ms p99 budget, and clients retry on timeout. idempotency_key has PRIMARY KEY (tenant_id, idempotency_key) plus request_fingerprint CHAR(64), state in ('in_flight','succeeded','failed'), response_status, response_body, resource_id, locked_until and expires_at. Specify the exact sequence the handler runs: what the second request does when it arrives while the first is still executing, what a retry carrying a different body receives, what happens when a process dies mid-request, and how the table is kept from growing without bound.

Approach

Start from the race rather than the happy path. Insert the key row first, in its own short transaction, with state 'in_flight' and locked_until set to now() plus the request deadline. The unique constraint on (tenant_id, idempotency_key) is the arbitration mechanism: exactly one caller commits the insert, and every other caller takes a unique violation (SQLSTATE 23505) and becomes a follower. Reading the table and then inserting is a check-then-act race that both callers pass, and they pass it most often under the load that generates the retries.
Give the follower a branch for every state it can observe, with no fall-through to doing the work: fingerprint mismatch means the key is being reused for a different request, so reject with 422 and never replay the stored response; 'succeeded' replays response_status and response_body verbatim; 'failed' allows a fresh attempt; 'in_flight' with locked_until in the future is answered with 409 and a retry hint, or a short bounded wait, because the only alternative is executing the effect twice.
Commit the effect and the record of the effect together. The resource insert, its resource_revision row, its outbox_event row and the transition of the key row to 'succeeded' all happen in one transaction, so no crash can leave the work done and the key still 'in_flight'. The earlier insert is deliberately a separate transaction: it has to be visible to a concurrent caller before the work starts, which an uncommitted row is not.
Close the wedged-key path. A process that dies after the insert leaves a row that would otherwise block that key forever, which is what locked_until exists for. Reclaim with a single conditional statement, UPDATE ... SET state='in_flight', locked_until=now()+interval WHERE ... AND state='in_flight' AND locked_until < now(), and treat a zero rowcount as losing the reclaim. Two statements that read and then update reintroduce the original race one layer down.
Size the retention before it becomes an incident. At the 1.2k/second peak a 24-hour window is on the order of 100 million rows, and the table grows with traffic rather than with data, so the sweep is load-bearing. Batched deletes driven by an index on expires_at keep it bounded. Range-partitioning by day and dropping whole partitions is cheaper, but PostgreSQL requires every partition-key column to appear in the primary key, so the key becomes (tenant_id, idempotency_key, created_at) and a retry that straddles the boundary no longer collides with its original. State that cost rather than meeting it later.

Worked solution (20 min)

Draw a timeline with the retry arriving 150 ms into a 900 ms first attempt, and mark the instant each row becomes visible to the other transaction.
Write the insert-first statement, then a five-row branch table for the follower: fingerprint mismatch, 'succeeded', 'failed', live 'in_flight', expired 'in_flight' - one action each.
Write the single conditional UPDATE that reclaims an expired in_flight row, and argue why two concurrent reclaimers cannot both see a non-zero rowcount.
Compute rows per day at peak, name the index the sweep uses, and check the sweep rate against the insert rate.

Expected result: The key row is inserted at 'in_flight' in its own transaction before any work; the unique violation identifies the follower. The follower replays the stored response on 'succeeded', returns 422 on a fingerprint mismatch, and waits or returns 409 on a live 'in_flight', never performing the work itself. The effect, its revision row, its outbox row and the transition to 'succeeded' commit in one transaction. Expired in_flight rows are reclaimed by one conditional UPDATE, and expired keys are swept in batches.

Verify it yourself

The retry creates no second resource row, whether it arrives during the first attempt or after it.
A retry with the same key and a changed body is rejected rather than served the first attempt's response.
Killing the process between the key insert and the commit leaves a key that is reclaimable within locked_until rather than one that is stuck.
For create operations, every key row in state 'succeeded' carries a resource_id that exists.

Follow-up

The stored response body averages 200 KB and this table is now the largest in the database. What do you store instead, and what does a replay return once the body has been pruned?
A client library generates a fresh idempotency key on every attempt. What breaks, which layer should have caught it, and does the server have any defence?
The first attempt succeeded but its response was lost, and the client retries 30 hours later, after the key expired. What does the second attempt do, and is that acceptable?

Read latency spikes on a sixty-second sawtooth

Medium · cache stampede · single-flight · ttl · tail latency

The cached listing read path serves about 14k reads/second at an 85% hit rate. p99 sits at 35 ms for 57 seconds, jumps to 900 ms for 3, and repeats. During each spike the primary shows several hundred identical listing queries starting within the same millisecond, all carrying one large tenant's id. Cache entries use a 60-second TTL. Give the mechanism, the ordered checks, the fix, and the correctness hazard your fix must not introduce.

Approach

Match the period to a configured number before theorising about load. A spike every 60 seconds against a 60-second TTL is an entry expiring, and you confirm it by correlating spike timestamps with the entry's write time rather than with the traffic curve. If the period had matched a cron or a GC interval instead, this is a different investigation.
Establish the concurrency of the miss. Several hundred identical queries in one millisecond means the miss path has no coalescing: every request that arrives between expiry and repopulation recomputes. The herd size is that key's arrival rate times its recompute time, so at 1.2k reads/second for the hot key and a 250 ms recompute you expect about 300 concurrent misses, which matches what is observed.
Add single-flight on the miss path so one caller per key recomputes under a short-lived lock while the rest wait for its result. Prefer stale-while-revalidate where the read tolerates it: return the expired value immediately and refresh asynchronously, which removes the latency spike rather than serialising it into a queue of waiters.
De-synchronise the keys. Write TTLs with jitter, for example 60 seconds plus or minus 10%, so a deploy or a mass invalidation does not align every key on the same second and turn a per-key herd into a fleet-wide one.
Name the hazard the fix must not introduce. Serving a stale listing is acceptable only because the API reports the projection watermark, and a reader that loaded the old value before a write can repopulate the entry after the invalidation, so the bounded TTL is what actually caps staleness rather than the delete. Keep read-after-write pinned to the primary for the writing session regardless.
Verify on miss concurrency, not hit rate. The hit rate barely moves, because the herd is one miss multiplied; the number that must change is distinct origin queries per key per minute.

Follow-up

The same sawtooth appears on a key that is invalidated on write rather than expired. Is that the same bug?
How does your answer change if the recompute takes 4 seconds instead of 250 ms?
What exactly does a client see during a stale-while-revalidate window, and how does the watermark let them tell?

Suggested practice checklist

Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.

Day 1 — Diagnostic, scored before you study anything

Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.

Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.

Day 2 — Largest gap: find the boundary rather than the subject

Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
Re-attempt one of them from blank four hours later with nothing open.

Deliverable: A sub-skill map with the two blocking sub-skills circled.

Day 3 — Drill the blocking sub-skill by repeating the shape

Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
Have someone else read your one-sentence rule and find the precondition you left out.

Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.

Day 4 — Second gap, plus maintenance on the strongest area

Run the same sub-skill decomposition on the second-largest gap in half the time.
Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.

Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.

Day 5 — The gap that is not a skill

Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.

Deliverable: Two recordings with a counted reduction in time-to-first-question.

Day 6 — Retest under day-one conditions

Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
Write down which single block you would still lose the offer on.

Deliverable: A second scored rubric placed beside the first, with one named remaining risk.

Day 7 — Full loop under interview conditions

Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.

Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.

The estimate and the day you knew

A slipped date is only a bad story if you sat on it. What matters is what you believed when you gave the number, the signal that told you it was wrong, how many days passed before you said so, and what you cut rather than asking for more time. Scope you defended counts as much as scope you dropped.

Describe a situation where you had to learn a new technology or tool quickly to complete a project.

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Name the disagreement and how you resolved it with evidence.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

Can you provide examples of your work that demonstrate the characteristics we value as a company?

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?
What would you do differently if you ran that again?

What motivates you to produce high-quality code in a fast-paced environment?

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
Name the disagreement and how you resolved it with evidence.
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?

Tell me about a time you had to navigate a disagreement with a team member; how did you resolve it?

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

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