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Compass Machine Learning Engineer Interview Questions & Guide 2026

Compass Machine Learning Engineer interview guide: reported rounds, practice questions with worked solutions and a 7-day prep plan.

The 60-second brief

As a Machine Learning Engineer at Compass, you play a pivotal role in enhancing the company’s ability to leverage data-driven insights to refine its real estate technology platform. This position is integral to developing algorithms and models that inform user experiences, improve operational efficiencies, and drive strategic business decisions. By applying advanced machine learning techniques, you will directly impact the quality of services offered to clients and enhance decision-making processes for agents, buyers, and sellers alike.

If the loop includes an asynchronous take-home, treat it as a code review of you rather than as a puzzle: structure, what you chose to test, and what you wrote down about the constraint you were working under. Hold the stated time box and say what you would have done with more of it.

Focus areas

Bound every outbound call with a timeout
Choose indexes from the query's access path
Make every write idempotent under retry

Overview

The complexity and scale at which you will work are significant. You will engage with vast datasets, employing cutting-edge technologies to solve challenging problems that directly influence Compass’s competitive edge in the market. Your contributions will not only help in optimizing existing features but will also be crucial in shaping innovative products that cater to an evolving industry landscape. This role requires both technical acumen and creativity, providing an exciting opportunity to be at the forefront of technology in real estate.

The Compass interview process

Compass candidates report 5 rounds · ≈ 4-6 weeks. The stages below are what candidates describe, not a published process.

Technical Phone Screens

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

What to demonstrate

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

How to prepare

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

Onsite Interviews

Where the day includes a partner from product, design or data, that conversation is weighted like the technical ones and prepared for least. They are deciding one thing: whether having you in the room makes their decisions cheaper. That means options with costs attached, not implementation detail and not "it depends". An estimate someone can plan against — a range, the assumption that would push it to the high end, and what you would drop to hit the low one — is worth more than a confident single number, which everyone present already knows is wrong.

What to demonstrate

Whether an estimate comes as a range with the assumption most likely to break it, and states what a specific scope cut would actually buy
Whether a technical constraint is handed over as a choice with consequences on their side, rather than as a verdict they have no standing to argue with
Whether you establish what decision is on the table before proposing anything
Whether risk is raised while it can still change the plan, with the trigger that would confirm it, instead of reported afterwards as a slip

How to prepare

Take a project that shipped late and write the two-sentence warning you could have given three weeks earlier, naming what you would have needed decided at that point
Rehearse one estimate out loud until it arrives in three parts: the range, the single assumption that would blow it, and the smallest thing you would cut to protect the date
Rewrite an objection you have actually made — the "we can't do that" version — as two options with their costs, so the choice ends up with the person who owns it

Technical Coding Assessments

What this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.

What to demonstrate

Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly

How to prepare

Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live

Machine Learning Discussions

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

Problem-Solving Evaluation

What to demonstrate

Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly

How to prepare

Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live

Where candidates lose points

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.

Paginating with LIMIT/OFFSET over a set that changes while the client is reading it

OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.

Optimising an axis nobody named

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

Choosing a schema before the access patterns are known

Write the queries first, with their filters, sort orders, cardinalities and which ones sit on the latency-critical path, then design tables and indexes to serve them. An index nothing queries still costs write throughput and storage, and a hot query with no supporting index becomes a full scan that only hurts once the table is big.

Questions & practice

How would you implement a decision tree from scratch?

Medium · machine learning fundamentals

Approach

Name the simplest model that could work and what would make you move past it.
Pick the metric from the cost of each error type, not from habit.
Say how you would validate it, and where leakage could enter the split.

Follow-up

What changes if the classes are heavily imbalanced?
How would you know the model is overfitting?

What is overfitting, and how can it be mitigated?

Medium · machine learning fundamentals

Approach

Say how you would validate it, and where leakage could enter the split.
Name the simplest model that could work and what would make you move past it.
State the learning problem: the label, the unit of prediction and how the model is used.

Follow-up

What changes if the classes are heavily imbalanced?
Where could label leakage enter this setup?

Discuss the end-to-end process of deploying a machine learning model.

Medium · machine learning fundamentals

Approach

Say how you would validate it, and where leakage could enter the split.
Pick the metric from the cost of each error type, not from habit.
Name the simplest model that could work and what would make you move past it.

Follow-up

How would you know the model is overfitting?
Where could label leakage enter this setup?

Describe how you would approach a situation where your model is underperforming.

Medium · machine learning fundamentals

Approach

Pick the metric from the cost of each error type, not from habit.
State the learning problem: the label, the unit of prediction and how the model is used.
Name the simplest model that could work and what would make you move past it.

Follow-up

Where could label leakage enter this setup?
How would you know the model is overfitting?

Solve a problem related to time complexity and space complexity.

Medium · coding and algorithms

Approach

Choose the data structure from the access pattern, not from familiarity.
Restate the input: its shape, its size, and what is guaranteed about it.
State the target complexity and say which constraint rules the naive version out.

Follow-up

How does this change if the input no longer fits in memory?
What is the worst case, and how likely is it on real data?

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).

Worked solution (20 min)

Write the pass: look up last_applied_version, skip if the event's version is not greater, otherwise upsert into the keep-map only when the incoming version exceeds the version already held, incrementing the discard counter on every skip.
Hand-trace one aggregate whose events arrive as v5, v3, v4, v5 with last_applied_version = 2, and confirm the output holds v5 once while the counter reads 3.
Compute the table footprint for 2,000,000 entries at 12 bytes of payload and a 0.7 load factor, then state the multiplier for a runtime that boxes keys and values.
Add the hash-partitioning fallback and say in one sentence why the per-partition results need no cross-partition merge logic.

Expected result: A single-pass O(n) reduction keyed on aggregate_id that keeps the maximum version rather than the last occurrence, O(d) space with the byte cost stated for 2,000,000 aggregates, a hash-partitioning fallback keyed on aggregate_id, and an explicit statement that batch-local deduplication does not replace a projection write keyed on (aggregate_id, version).

Verify it yourself

Shuffling the batch order changes nothing in the output, which is the test that separates max-wins from last-wins.
Feeding the same batch twice produces an identical keep-set and only doubles the discard counter.
Every emitted event has a version strictly greater than the supplied last_applied_version for its aggregate.

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?

Find version gaps and relay lag with window functions

Medium · window functions · event tables · frame semantics

outbox_event holds event_id, aggregate_type, aggregate_id, aggregate_version, event_type, payload, status ('pending','published','dead'), attempts, created_at, published_at. A projection is missing rows and you must decide whether the relay skipped events or the consumer dropped them. Write three queries over the last seven days: one listing every aggregate_id whose published aggregate_version sequence has a hole, one giving per-day counts with a running total, and one returning the newest published event per aggregate. For each, say where the window function is evaluated relative to WHERE and LIMIT. PostgreSQL 16.

Approach

Gaps: compute lead(aggregate_version) OVER (PARTITION BY aggregate_id ORDER BY aggregate_version) in a subquery, then filter next_version <> aggregate_version + 1 in the outer query. Window functions are evaluated after WHERE, GROUP BY and HAVING and before the outer ORDER BY and LIMIT, so the predicate cannot sit in the same WHERE clause and PostgreSQL 16 has no QUALIFY.
Say what the seven-day filter does to the answer: it truncates every partition, so the first row per aggregate has no predecessor inside the window and a hole spanning the boundary is invisible. Widen the window, or join to resource.version as the authority for the true maximum.
Running total: SELECT date_trunc('day', created_at) AS d, count() AS n, sum(count()) OVER (ORDER BY date_trunc('day', created_at) ROWS UNBOUNDED PRECEDING). An aggregate inside a window call is legal because grouping runs before windowing. The grouping key is unique per row here so ROWS and RANGE agree, but write the frame anyway — over ungrouped rows with tied timestamps the default RANGE frame pulls in every peer row and the total jumps.
Newest per aggregate: DISTINCT ON (aggregate_id) ... ORDER BY aggregate_id, aggregate_version DESC is the cheap PostgreSQL-only form when an index matches that order; row_number() OVER (PARTITION BY aggregate_id ORDER BY aggregate_version DESC) = 1 is the portable form and needs a subquery for the same evaluation-order reason as the gap query.
Interpret rather than report: no gaps plus a normal p95 of published_at - created_at points at the consumer; gaps or a fat lag tail point at the relay; rows still 'pending' with attempts > 0 point at neither, because they never left the database.
Be explicit that the partial index on (created_at, event_id) WHERE status = 'pending' does not serve any of these — they read published rows. Name the index a recurring monitor would need, and say why a query run twice a year may not deserve one.

Follow-up

Relay failover redelivers events. Does a duplicate break the gap query, and how would you detect one from this table alone?
Turn the gap check into a continuous monitor rather than a query someone runs after an incident. What does it watch?
The consumer claims it never received event 4,812,006. What do you look at, in what order?

Stop tag and share joins from fanning out a page

Medium · join fan-out · semi-joins · many-to-many

resource_tag is (resource_id, tag_id) with PK (resource_id, tag_id); resource_share is (resource_id, shared_with_user_id, permission). The tagged-and-shared listing inner-joins resource to both, filters tenant_id, tag_id = ANY($2) and shared_with_user_id = $3, orders by updated_at DESC and takes 50. Pages come back with fewer than 50 distinct resources and the total in the header is far too high. Explain the row multiplication, rewrite both the page query and the count query so each is correct, and name the index each one needs. PostgreSQL 16.

Approach

Do the arithmetic against the predicates that are actually there. An inner join emits one row per matching child row, and both joins are filtered: tag_id = ANY($2) admits only the requested tags, shared_with_user_id = $3 admits one user's share rows. So a resource holding three of the requested tags and shared with $3 once yields three rows, not one — the multiplier is its count of matching tags times its share rows for that single user, and that second factor is 1 unless the table admits duplicate (resource_id, shared_with_user_id) pairs. LIMIT 50 then limits rows rather than resources, and COUNT(*) counts pairs — the header is the product, not the population.
Reject DISTINCT as the fix. It deduplicates after the product has been built, so the planner must materialise and sort the fanned-out set before the LIMIT can apply, and it leaves any SUM or AVG in the same select list wrong.
Rewrite both filters as semi-joins, keeping resource as the only row source: AND EXISTS (SELECT 1 FROM resource_tag rt WHERE rt.resource_id = r.resource_id AND rt.tag_id = ANY($2)) and the same shape against resource_share. A semi-join stops at the first match per resource and preserves the driving index order, so ORDER BY updated_at DESC, resource_id DESC LIMIT 50 still stops after 50 rows.
Count with the same predicates and no join at all: SELECT count(*) FROM resource r WHERE r.tenant_id = $1 AND r.status = 'active' AND EXISTS (...) AND EXISTS (...). Nothing multiplies a resource, so the number is the population.
Attach the tags for display after the page has been cut — LEFT JOIN LATERAL (SELECT array_agg(rt.tag_id) FROM resource_tag rt WHERE rt.resource_id = p.resource_id) ON TRUE over the 50 returned rows. Aggregate over the page, never over the tenant.
Index both directions and say which query each serves: PK (resource_id, tag_id) serves the lateral lookup, (tag_id, resource_id) serves the EXISTS probe by tag, and resource_share needs (shared_with_user_id, resource_id) for the same reason. An index covering one direction only leaves the other as a scan.

Worked solution (30 min)

Build a tenant where each resource carries 0-5 tags from a 20-tag vocabulary and is shared with 0-4 distinct users, then bind $2 to three tags and $3 to a user holding shares on about half the resources. Run the joined query and compare its row count to the distinct resource count on page one.
Run COUNT(*) on the joined shape and on the EXISTS shape and compare both to a ground truth computed from distinct ids; then give $3 a second permission row on 10% of resources and record which of the two counts moves.
EXPLAIN both page queries and compare rows-read plus the presence of a Sort or HashAggregate node above the join.
Add (tag_id, resource_id), re-run the EXISTS probe, and record the plan change on the inner side.

Expected result: The EXISTS page returns exactly 50 distinct resources and reads about 50 driving rows with no aggregate above the join. The EXISTS count matches the ground truth. The joined COUNT(*) is larger by roughly the average number of matching tags per qualifying resource — the tags it holds from $2, not its total tag count — times its share rows for $3, which is 1 until the duplicate permission rows exist and rises with them afterwards. The EXISTS count is unchanged by those duplicates.

Verify it yourself

The page's resource_id list has no duplicates and is 50 long whenever at least 50 rows match.
count(*) from the rewrite equals count(DISTINCT r.resource_id) over the joined shape.
EXPLAIN shows a semi-join node on the inner side rather than a full join followed by a unique or aggregate step.

Follow-up

The filter changes from 'any of these tags' to 'all of these tags'. Rewrite it and state what it costs relative to the ANY form.
A resource can be shared with the same user twice under different permissions. Does your count change, and should it?
Where does the correct total come from when the tenant holds 4M resources and the header must not cost 200 ms?

How would you design a recommendation system for a real estate platform?

Medium · ml system design and production

Approach

Separate the offline training path from the online serving path.
Say where features come from at serving time and how they match training.
Fix the product goal and the online metric before choosing any model.

Follow-up

How would you roll the new model out safely?
How would you detect drift before the metric drops?

What considerations would you take into account for data pipeline architecture?

Medium · system design

Approach

Fix the scope first: who calls this, how often, and what they do when it fails.
Name the failure you are designing for, then the recovery path.
Choose a partition key and say what query it makes expensive.

Follow-up

How does this behave when that dependency is down for an hour?
What would you drop to keep the system up under load?

Given a dataset with missing values, how would you approach the problem?

Medium

Approach

State your assumptions explicitly before working the problem.
Say what you would check first and why it is the highest-information step.
Clarify what is being asked and what a complete answer contains.

Follow-up

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

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?

One customer endpoint stalls deliveries to every other destination

Medium · bulkhead · timeouts · saturation · retry amplification

The egress service delivers about 1.5k webhooks/second across 40,000 destinations, with a per-destination concurrency cap of 4 and a 10-second connect-plus-read timeout. Throughput falls to 300/second, queue depth climbs, and p99 delivery latency for unaffected destinations goes from 200 ms to minutes, while the error rate barely moves. One tenant holds 900 destination rows whose URLs share a hostname that now answers in 9.5 seconds. Explain the mechanism with the arithmetic, then give the containment in the order you would apply it.

Approach

Look at saturation before errors. A flat error rate with collapsing throughput says nothing is failing, things are waiting, so the first signal to pull is in-flight request count or pool wait time rather than the error counter. This is the distinction that decides the whole investigation.
Group in-flight work by resolved host, not by destination id. The cap is keyed per destination row, so 900 rows sharing one hostname buy 3,600 concurrent slots against a single host, each held for 9.5 seconds. The bulkhead was never a bulkhead for that host, and grouping by the wrong dimension is why the dashboard looked healthy.
Do the arithmetic in both directions. Required concurrency is arrival rate times latency, so 1.5k/second at 200 ms needs about 300 in flight, which is entirely consumed by 3,600 slow slots; conversely whatever concurrency is left sustains rate equals concurrency divided by 9.5 seconds, which is the 300/second you are seeing. Matching both numbers is what promotes this from a plausible story to the mechanism.
Explain why the circuit breaker never helped. It opens on consecutive failures, and a 9.5-second response inside a 10-second timeout is a success. Slow is not failing, so an error-rate breaker cannot see this; you need a slow-call ratio, a deadline propagated from the caller's remaining budget, or a concurrency limiter.
Contain in order: park the offending host so the shared pool drains, add a per-resolved-host concurrency cap alongside the per-destination one, give slow hosts their own queue so they cannot occupy the general pool, derive the timeout from the delivery deadline rather than a round number, and check the retry policy is not tripling load on a host that is already slow. Use backoff with full jitter so retries do not resynchronise on recovery.
State the invariant you are restoring: one tenant's endpoints degrade only that tenant's deliveries. That is a property to load-test for, not to assume from a config value.

Follow-up

The host recovers to 80 ms. How long does the queue take to drain, and what does the drain do to the recovered host?
Where should the 10-second timeout number actually come from?
If that tenant had one destination row instead of 900, would the cap of 4 have saved you? What would you measure to be sure?

Suggested practice checklist

For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.

Day 1 — Rebuild the primitives by implementing them

Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.

Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.

Day 2 — Arrays under an invariant: two pointers, sliding window, binary search

Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.

Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.

Day 3 — Sorting, heaps, and the greedy argument that has to be proved

Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.

Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.

Day 4 — Recursion, memoisation, and the step to a table

Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.

Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.

Day 5 — Graphs, where most of the work is choosing the traversal

Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.

Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.

Day 6 — One day for everything that is not an algorithm

Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.

Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.

Day 7 — Solve out loud, under time

Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.

Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.

The failure that was yours

Keep one story where the bad call was yours rather than a dependency's or a manager's. Name the check that would have caught it, whether you added that check afterwards, and whether it has fired since. Answers that route blame outward end the conversation early; answers that end in a guardrail someone still relies on tend to open it up.

Describe a time when you had to work collaboratively with a team to achieve a goal.

Medium · behavioural and collaboration

Approach

Give the blast radius: what could have broken, and what you measured.
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

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

Turn a code review disagreement into a decision

Easy · code review · optimistic concurrency · communication

A colleague's change updates a row with UPDATE resource SET version = version + 1 WHERE resource_id = $1 AND version = $2 and treats an affected-row count of zero as a successful no-op. You read that as a silently lost update; they think returning 200 is friendlier to clients than returning a conflict. Describe how you have handled a review disagreement of this shape: what goes in the comment, when you leave the thread, and who decides. Then write the comment you would leave here, in under 80 words.

Approach

Sort the disagreement before writing anything. A silently discarded write is a correctness claim about data; the choice between 409 and 412 is taste. Only the first justifies blocking a merge, and saying which one you are doing is most of the value of the comment.
Make the claim reproducible in the comment itself with an interleaving rather than a principle: A reads version 7, B reads version 7, B commits version 8, A's predicate matches zero rows, A is told it succeeded and A's edit is gone.
Offer the alternative with its cost attached: return 409 carrying the current version and the revision that won, so the client can re-read and re-apply. Note that automatic retry is not the fix, because a retry re-reads the winner's state and reapplies an intent formed against data that no longer exists.
Apply an escalation rule you can state: two round trips on the thread, then a call, and the service's owner decides rather than the reviewer. A reviewer who cannot be overruled is a bottleneck with extra steps.
Close in writing wherever the decision lands, so the next reader finds the reasoning in the code or the ticket instead of in a collapsed review thread.

Follow-up

Where would you put the test that fails if someone reintroduces the swallowed zero rowcount?
The author says clients cannot handle a 409. How do you check whether that is true?
How do you handle the same review comment when the author is more senior than you and in a hurry?

Tell callers you do not own that their integration breaks

Medium · deprecation · compatibility · stakeholders

A field in a write endpoint's response must change shape. You own the endpoint; you do not own the four internal callers or the outbound webhook consumers who read it. Describe a deprecation you were responsible for: what you shipped first, how you established who was actually reading the field, the window you gave and what set its length, what you did about the consumer who never moved, and how you decided removal was safe. Name the signal you used, not the announcement you sent.

Approach

Establish the reader set empirically rather than from a wiki of owners: per-field usage counters keyed by principal, or access logs attributed to a consumer. State the blind spot of whichever you pick, since a consumer that reads the field only on a monthly job will not appear in a week of logs.
Ship additive first. Populate the new field alongside the old one so no reader is forced to move, which is also what keeps a rolling deploy safe, because old and new instances answer the same requests at the same time and a rollback must still find the old shape present.
Set the window from the slowest legitimate consumer's release cadence, not from your calendar, and decide separately what to do for a consumer with no release process at all, such as an external webhook endpoint you can only email.
Convert silence into evidence before you rely on it: a short, low-traffic removal window that makes a still-dependent consumer fail visibly and loudly while you are watching, rather than at three in the morning after you have moved on.
State the removal criterion as a measurement with a duration attached, such as observed reads at zero across a full billing cycle, and keep the change reversible for one release after removal.

Follow-up

How would you detect a consumer that reads the field only during a monthly export?
One caller refuses to move and has a commercial relationship behind it. What changes in your plan and what does not?
After removal, what makes the change irreversible, and how long before you cross that line?