Grammarly Machine Learning Engineer Interview Questions & Guide 2026

The 60-second brief

This guide covers what a Machine Learning Engineer at Grammarly is expected to do and how to prepare for the interview.

Ask how many rounds there are and what each one is before you plan your weeks, because the composition is what your hours should follow. A loop with three coding rounds and one short design conversation deserves a different split from the reverse, and whoever schedules it will usually say plainly which it is.

Focus areas

• Choose indexes from the query's access path
• Trace a symptom to a mechanism under load
• Paginate large result sets with keyset cursors

The Grammarly interview process

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

Phone Screen

The title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.

What to demonstrate

• Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
• Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
• Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year

How to prepare

• Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
• Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
• Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter

Technical Interviews

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

Behavioral Interviews

This round is deciding whether a change you make without supervision can be allowed to reach production. It is scored on what you knew at the moment you decided, not on how it turned out, so a story that opens with the result and works backwards reads as luck retold as judgement. Say what the options were, what you did not know, what you did to shrink the unknown before committing, and what you accepted as the worst plausible case. The detail that separates answers is a bound: how many users, how much data, and for how long, if you had been wrong.

What to demonstrate

• Whether the reasoning you give was available at the time you decided rather than after the result came in, since a story whose deciding evidence arrived later describes an outcome and not a judgement
• Whether you can put units on the exposure (users, rows, minutes of degraded service) and whether the containment you chose actually bounded it: a canary bounds the request path it fronts, while a background job writing to a shared table reaches every user regardless of which version served their requests
• Whether the reversal path existed before you shipped or was improvised during the incident, and whether it restores state or only stops further damage

How to prepare

• For your three largest changes, write down the one thing you would have had to be wrong about for it to fail, and what your best estimate of it was on the day you shipped. If you never held an estimate, that is the gap the follow-up questions will find
• Write the undo procedure for one of those changes as it existed at the time, then mark which steps restore data and which only stop new damage. Turning a flag off or reverting a deploy ends the new writes; rows already written come back only from a copy you kept, and a dropped column comes back empty unless something outside the schema holds the values
• Rehearse one story from the decision point forward and stop before the outcome, then have someone ask what you would do next. If the story only works with the ending attached, it is an anecdote rather than a decision you can defend

Final Round with Leadership

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

Assuming an isolation level prevents the anomaly you actually have

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

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.

Saying 'eventually consistent' without naming the anomaly a user would see

Describe the concrete symptom you are choosing to accept: the author reloads and their own comment is missing for two seconds, or two devices show different balances for a minute. The class of consistency model is a technical label; the tolerable anomaly is the actual product decision.

A queue or buffer with no bound

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

Questions & practice

Explain overfitting, normalization, and regularization.

Medium · machine learning fundamentals

Approach

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

Follow-up

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

These questions assess your ability to write clean, efficient code and translate machine learning concepts into working implementations under live interview conditions.

Medium · machine learning fundamentals

Approach

1. Name the simplest model that could work and what would make you move past it.
2. Say how you would validate it, and where leakage could enter the split.
3. 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?

How do you evaluate the convergence and generalization capabilities of a complex model?

Medium · machine learning fundamentals

Approach

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

Follow-up

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

How do you handle class imbalance in large-scale classification datasets?

Medium · machine learning fundamentals

Approach

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

Follow-up

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

Write a function to handle complex string manipulation and combinatorics for text normalization.

Medium · coding and algorithms

Approach

1. Walk one small example through your approach before writing the whole thing.
2. Restate the input: its shape, its size, and what is guaranteed about it.
3. State the target complexity and say which constraint rules the naive version out.

Follow-up

• Which test case would catch an off-by-one here?
• What is the worst case, and how likely is it on real data?

Interviewers use these questions to explore how you collaborate, manage timelines, handle workplace relationships, and navigate ambiguous project requirements.

Medium · coding and algorithms

Approach

1. State the target complexity and say which constraint rules the naive version out.
2. Name the brute-force solution and its complexity before improving on it.
3. Choose the data structure from the access pattern, not from familiarity.

Follow-up

• What is the worst case, and how likely is it on real data?
• Which test case would catch an off-by-one here?

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

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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)

1. 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.
2. 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.
3. 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.
4. 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

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. Note that now() is the transaction timestamp in PostgreSQL, so resource.updated_at, the revision's created_at and the outbox row share one instant, which is what later makes reconciliation between the three tables unambiguous.

Follow-up

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

Hold a per-tenant active cap against concurrent creates

Hard · write skew · isolation levels · constraints

A tenant on the standard plan may hold at most 50 resources with status='active'. The create handler runs SELECT count(*) FROM resource WHERE tenant_id = $1 AND status = 'active', compares to 50, then inserts. Two creates arrive 3 ms apart on different instances and the tenant lands at 51. Name the anomaly, say whether PostgreSQL 16 READ COMMITTED or REPEATABLE READ prevents it and why, then give an implementation that holds the cap at READ COMMITTED with the exact statements. Finally, say what changes when the cap is 'at most one running export per tenant' on job_run.

Approach

1. Name it: write skew. The two transactions read an overlapping set and write disjoint rows, so there is no row-level conflict for the engine to detect and each commit is individually legal.
2. Rule out the levels precisely. READ COMMITTED takes a fresh snapshot per statement and takes no lock on the counted rows, so both see 49. PostgreSQL's REPEATABLE READ is snapshot isolation: it removes non-repeatable reads and phantoms within the snapshot but still admits write skew, because the anomaly is not a re-read of a changed row, it is a read of a set that a concurrent transaction invalidates. Only SERIALIZABLE closes it, by tracking the read dependency and aborting one transaction with SQLSTATE 40001 — a guarantee that exists only if the application re-runs the whole transaction from the read.
3. Convert the set predicate into a single-row conflict: keep tenant.active_resource_count and run UPDATE tenant SET active_resource_count = active_resource_count + 1 WHERE tenant_id = $1 AND active_resource_count < 50 in the same transaction as the INSERT. Zero affected rows is the cap, returned as 409. The row lock serialises the decision at any isolation level, and contention is bounded to one tenant's row — which is also the fair-scheduling unit, unlike a global counter that would convoy every tenant behind one row.
4. State the cost you just took on: a counter is a second source of truth that can drift, so every path that changes status must adjust it inside the same transaction, and a periodic reconciliation has to exist, with resource_revision as the authority for what the count should have been.
5. For the job case the invariant is expressible per row, so let the database hold it: a partial unique index on job_run (tenant_id, job_type) WHERE status IN ('queued','running') makes a second running export unwritable and the loser takes 23505, mapped to 409. That is strictly better than a counter — no drift, no reconciliation — and it is available only because the cap is one rather than fifty.
6. Add the retry discipline each route demands: under SERIALIZABLE both 40001 and deadlock 40P01 are retryable and the retry must re-execute the read, while under READ COMMITTED with the counter nothing retries, because the conflict is reported to the caller rather than raised as an error.

Worked solution (35 min)

1. Reproduce with two sessions that both count 49, both insert and both commit, at READ COMMITTED and then at REPEATABLE READ; record the final active count for each.
2. Repeat both sessions at SERIALIZABLE and record which SQLSTATE the loser receives and at which statement it is raised.
3. Implement the counter form and run a 20-way concurrent create against a tenant sitting at 45 active resources.
4. Implement the partial unique index for the job case and race 20 enqueues of the same export.

Expected result: READ COMMITTED and REPEATABLE READ both finish at 51. SERIALIZABLE finishes at 50 with one transaction aborted at 40001 — usually at commit, though SSI may raise it at an earlier statement. The counter form finishes at exactly 50: of the twenty creates, the five that take the tenant from 45 to 50 return 201 and the remaining fifteen match zero rows and return 409, with no error raised. The job index admits one running export and rejects the rest with 23505.

Verify it yourself

• count(*) WHERE status='active' never exceeds the cap across 50 randomised concurrent runs.
• Remove the active_resource_count < 50 predicate and the same test must start failing; if it still passes, the race is not being exercised.
• At SERIALIZABLE, confirm every participating session is SERIALIZABLE — one session at a lower level and the guarantee is gone.
• After each run, the counter equals the real count of active rows.

Follow-up

• A resource moves from archived back to active. Which statements change, and what breaks if the counter update and the status change land in different transactions?
• The cap becomes plan-dependent and a plan can change mid-month. Where does the number 50 live, and who reads it?
• How do you detect after the fact that the counter drifted, without locking the table?

Design a machine learning system to verify an existing spell checking API and improve user acceptance of recommendations.

Medium · ml system design and production

Approach

1. Separate the offline training path from the online serving path.
2. Name what you would monitor after launch and what triggers a retrain.
3. Fix the product goal and the online metric before choosing any model.

Follow-up

• What happens when a feature is missing at serving time?
• How would you roll the new model out safely?

Optimize an inference pipeline to reduce latency for real-time text processing.

Medium · ml system design and production

Approach

1. Separate the offline training path from the online serving path.
2. Say where features come from at serving time and how they match training.
3. Name what you would monitor after launch and what triggers a retrain.

Follow-up

• What happens when a feature is missing at serving time?
• How would you detect drift before the metric drops?

This category evaluates your ability to architect scalable, resilient end-to-end machine learning systems that handle high throughput and stringent latency requirements.

Medium · system design

Approach

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

Follow-up

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

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

1. 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.
2. 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.
3. 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.
4. 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.
5. 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)

1. 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.
2. 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.
3. 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.
4. 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?

p99 jumped on one listing filter while p50 stayed flat

Medium · query planning · composite index · percentiles · tail latency

After a release that added an owner_user_id filter to the resource listing, p99 rose from 90 ms to 1.9 s while p50 stayed at 40 ms. Traffic and row counts are unchanged. resource carries the index (tenant_id, status, updated_at DESC, resource_id DESC). The new query filters tenant_id and owner_user_id, orders by updated_at DESC, resource_id DESC, and takes 20 rows. On PostgreSQL, explain the shape of the regression, prove it from a query plan, and give the index you would add.

Approach

1. Start from the shape. A flat p50 with a moved p99 means a subset of requests changed cost, not all of them, so the first job is naming the subset. Bucket the endpoint's latency by the tenant's row count; the natural hypothesis is that large tenants are a small share of requests and all of the tail.
2. Get the plan for the new query on a large tenant with EXPLAIN (ANALYZE, BUFFERS). Expect an index scan over the tenant's range, a filter discarding most of it, then a Sort feeding the Limit, possibly reporting Sort Method: external merge Disk. Read actual rows on the scan node, not estimated.
3. Explain why the existing index cannot serve it. A composite B-tree is seekable only as a left prefix, and with no equality predicate on status the scan cannot treat updated_at as an ordering, because rows in the tenant's range are ordered by status first. Everything matching must be read and sorted before LIMIT 20 can apply, so a tenant with 400,000 rows pays 400,000 rows to return 20.
4. Add (tenant_id, owner_user_id, updated_at DESC, resource_id DESC). Equality on the first two columns leaves the index ordered by updated_at within that pair, so the plan becomes an index scan that stops after 20 rows with no Sort node. PostgreSQL can scan a B-tree backwards, so the DESC markers matter only if the two sort columns ever disagree in direction; keeping them explicit documents the order the keyset cursor depends on.
5. Price the fix. This is a fourth index on a table taking about 1.2k writes/second, and every insert and version bump maintains it. Justify it against the query it serves, and check whether it makes an existing index redundant, which here it does not, since the original still serves the status-filtered default listing.
6. Re-measure per tenant-size bucket rather than in aggregate. A fleet-wide p99 can improve while the largest tenant is still on the old plan.

Follow-up

• The endpoint paginates with OFFSET. What does page 500 cost with your index, and what does the keyset version cost?
• How would you have caught this before release, given that a 10,000-row seed database produces the same plan shape at an unnoticeable cost?
• If a fourth index were unacceptable on write grounds, what else could serve this query?

Suggested practice checklist

Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.

Day 1 — Coding, one pass at shallow depth

• Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
• For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
• Fix nothing today. The value of the pass is the unfixed record.

Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.

Day 2 — Design, one pass at shallow depth

• Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
• After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
• Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.

Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.

Day 3 — Fundamentals and the practical rounds

• Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
• Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
• Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.

Deliverable: Eight scored short answers and one written reading of unfamiliar code.

Day 4 — The rounds that are about you, and the map

• Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
• Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
• Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.

Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.

Day 5 — First chosen area, to the depth you skipped

• Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
• After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
• Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.

Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.

Day 6 — Second chosen area, where the gap is coverage rather than speed

• Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
• Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
• Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.

Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.

Day 7 — Reassemble the loop

• Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
• Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
• Reduce the week to one page holding only the rules you can state without reading them.

Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.

Disagreement needs an outcome you did not choose

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

Discuss your academic background, career trajectory, and future professional goals.

Medium · behavioural and collaboration

Approach

1. Pick a story where you made the decision, not one where you watched it.
2. Name the disagreement and how you resolved it with evidence.
3. Close with what you would do differently, concretely.

Follow-up

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

Unblock an engineer without taking the keyboard

Easy · mentoring · leases · at-least-once

A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.

Approach

1. Ask before diagnosing, and ask for things answerable from data they already have: the attempt count on the job rows that produced duplicates, the handler's observed duration against its lease expiry, and whether the duplicate rows share a natural key that a unique constraint could have caught.
2. Teach the shape rather than the answer. A lease cannot distinguish a dead worker from a slow one, so a handler that outruns its lease is running twice by design, and deploys deliver the other half by killing handlers mid-run on every rollout. Both of their candidate theories produce identical duplicate rows, which is why the evidence has to come from timings rather than from argument.
3. Hand over a checklist they execute: a natural key on every write the handler performs so the second copy collides rather than appends, the record of intent written before any external effect, a lease heartbeat while running, and the metric that shows it working.
4. Keep ownership with them deliberately. Pair on the first write, then step back; if you finish it yourself you have closed one ticket and left the same person stuck on the next redelivery.
5. Close on the systemic gap that let two days pass, which is usually a missing dashboard for attempt counts or an undocumented at-least-once contract, and fix that rather than only the bug.

Follow-up

• How would you distinguish a genuine double-delivery from a lease expiry using only the data already stored?
• Their handler calls an external endpoint before recording that it did. What do you tell them to change first?
• What do you do the third time the same person brings you the same class of bug?

Estimate work you have never done and defend the range

Hard · estimation · backfills · expand-contract

You are asked to estimate a change you have never attempted: add a column to a 100-million-row table, populate it, move reads across, and drop the old shape. Give a range with the assumptions that generate it, including batch size, the signal your backfill throttles on, and wall-clock hours, and name the three unknowns that would move the number most. Then describe a real estimate you gave under comparable ignorance: how you expressed its uncertainty, what you committed to, and how wrong you turned out to be.

Approach

1. Decompose into independently deployable steps before estimating anything: add the column nullable, write both shapes, backfill in batches, verify, move reads, stop writing the old shape, drop it. That is four deploys spread over days, and the calendar estimate is dominated by them rather than by the loop's runtime.
2. Do the arithmetic aloud for the part that has arithmetic in it: batch size times number of batches times per-batch duration, at a write rate the primary can absorb alongside roughly 1.2k writes per second of production traffic. The loop is throttled by replication lag and lock waits, not by how fast it can issue statements.
3. Price the schema step by its lock rather than its statement duration. In PostgreSQL an ALTER TABLE taking ACCESS EXCLUSIVE waits for every open transaction on that table while later queries queue behind it, so a millisecond change issued during a thirty-second analytics query stalls that table for thirty seconds. Adding a nullable column with a non-volatile default avoids a rewrite from version 11; a new index wants CREATE INDEX CONCURRENTLY, which cannot run inside a transaction block and leaves an invalid index behind if it fails.
4. Express the answer as a range whose endpoints each trace to a stated assumption, then name the cheapest experiment that collapses it, which is almost always running one real batch against the real table and multiplying.
5. Commit to a checkpoint rather than a completion date: the day you report a measured number from that first batch. That is a promise you can keep under uncertainty, and it is what the asker actually needs in order to plan.

Follow-up

• How do you verify the backfill genuinely finished, given rows written by production traffic while it ran?
• Where does the backfill resume from after a worker is killed mid-batch, and what makes that resume point trustworthy?
• Your first batch comes back ten times slower than assumed. What do you tell the person waiting on the estimate, and when?