Interview guides · Okx

Okx Software Engineer Interview Questions & Guide 2026

Code reading practice for Okx software engineer prep: work out what unfamiliar code does before judging it.

The 60-second brief

A Software Engineer at OKX is responsible for building and scaling some of the most highly trafficked systems in the cryptocurrency and Web3 industries. OKX operates a global, high-frequency digital asset exchange that requires sub-millisecond execution speeds, robust security protocols, and financial-grade reliability. Engineers here do not just write standard web applications; they design and maintain high-throughput trading engines, low-latency APIs, decentralized wallet architectures, and complex blockchain integrations that process billions of dollars in daily transaction volume.

For a design discussion, a catalogue of architectures is worth less than the ability to turn a vague requirement into a data model and an API contract. A box diagram with no schema under it collapses at the first follow-up question.

Focus areas

Evolve APIs without breaking pinned SDK clients
Scope every query and cache key by tenant
Bound blast radius with per-tenant concurrency limits

Overview

The impact of this role is immediate and global. A minor latency spike or system bottleneck can have massive financial implications for millions of users worldwide. Consequently, software engineers at OKX work on highly complex distributed systems, optimizing database operations, managing massive message queues, and refining multi-threaded execution models. Because the crypto market operates 24/7/365, system resilience, fault tolerance, and zero-downtime deployments are central to every engineering decision.

Whether you are working on core backend services, mobile platforms, or Web3 products, you will be expected to solve highly ambiguous technical challenges. The engineering culture is fast-paced, deeply technical, and highly pragmatic. Successful engineers at OKX are those who possess a strong foundation in computer science fundamentals, enjoy optimizing system performance at the operating system and network levels, and can adapt quickly to the rapidly evolving blockchain landscape.

The Okx interview process

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

Online Assessment

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

Initial Screening Call

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

What to demonstrate

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

How to prepare

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

Technical Evaluation Phase

Input bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.

What to demonstrate

Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply

How to prepare

For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last

Interviews with Leadership

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

Serialising a tenant's writes through select ... for update on a single counter row

It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.

Holding money in a floating-point type, or rounding it more than once

Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.

Starting work without saying what you are about to spend time on

State the plan before executing it: the approach, roughly how long it will take, and what you intend to leave hand-waved. That gives the interviewer a chance to redirect you in ten seconds rather than watching you spend fifteen minutes on the wrong sub-problem.

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

Given an array of integers, find the maximum sum of a contiguous subarray using a sliding window approach.

Medium · data structures and algorithms

Approach

State the target complexity and say which constraint rules the naive version out.
Name the brute-force solution and its complexity before improving on it.
Walk one small example through your approach before writing the whole thing.

Follow-up

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

Solve a medium-difficulty LeetCode problem involving binary trees or graph traversal, ensuring optimal time complexity.

Medium · data structures and algorithms

Approach

Walk one small example through your approach before writing the whole thing.
Restate the input: its shape, its size, and what is guaranteed about it.
Name the brute-force solution and its complexity before improving on it.

Follow-up

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

Explain the internal mechanics of a Hash Map and how it handles hash collisions.

Medium · data structures and algorithms

Approach

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

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?

Discuss the differences between process-level threading and lightweight coroutines (e.g., Go goroutines or Rust async tasks).

Medium · languages, concurrency and fundamentals

Approach

Reach for the cheapest primitive that closes the race, not the broadest lock.
Say what the runtime actually does before reasoning about the code.
Name what is shared across threads and what owns each piece of state.

Follow-up

How would you prove the race exists rather than suspect it?
Where could this allocate more than you expect?

What are memory orderings in multi-threaded programming, and why are they important for low-latency development?

Medium · languages, concurrency and fundamentals

Approach

Say what the runtime actually does before reasoning about the code.
Distinguish a value from a reference to it, and say which one you handed out.
Name what is shared across threads and what owns each piece of state.

Follow-up

Where could this allocate more than you expect?
What happens if two callers reach this at the same time?

Fold a deduplicated usage stream into hourly rollups

Easy · aggregation · deduplication · watermarks · exact-arithmetic

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

Approach

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

Worked solution (25 min)

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

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

Verify it yourself

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

Follow-up

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

Enforce a concurrent-run quota that survives simultaneous requests

Medium · isolation levels · write skew · quotas · lock contention

A plan allows at most 20 concurrently running rows in job_run per tenant. The table holds run_id, tenant_id, workspace_id, status (queued, leased, running, succeeded, failed, timed_out, cancelled, lost), lease_token, leased_until, started_at and finished_at. Today the service runs select count(*) from job_run where tenant_id = $1 and status = 'running', compares the result to 20, then inserts. Under load a tenant exceeds the cap by exactly the number of concurrent requests. Name the anomaly, say which isolation levels do and do not prevent it, and give a version that holds, as SQL.

Approach

Name it: write skew. Each transaction reads a predicate (the count of running rows), neither modifies what the other read, and both then insert rows that jointly violate an invariant no single row expresses. Read committed permits it. So does repeatable read, because snapshot isolation's first-updater-wins check fires only on conflicting row updates, and these are inserts touching disjoint rows.
Enumerate the fixes with their real costs. SERIALIZABLE works: PostgreSQL's SSI tracks the predicate read and aborts one transaction with SQLSTATE 40001, which obliges the caller to retry and makes the abort rate rise with contention on a hot tenant. Folding the predicate into the write as insert ... select ... where (select count(*) ...) < 20 narrows the race to the statement's snapshot but does not close it under read committed.
Give the version that holds at read committed: serialise on a row both transactions must touch. update tenant_concurrency set running = running + 1 where tenant_id = $1 and running < 20 returning running updates zero rows when the cap is reached, and zero rows is the rejection. This works because at read committed a blocked UPDATE re-evaluates its WHERE clause against the newly committed row; at repeatable read the same statement raises a serialisation error instead, so the isolation level changes the calling contract.
State the cost you just bought. That row is now a per-tenant serialisation point, so admission throughput for the tenant is bounded by one divided by the lock hold time; at a 2 ms hold that is roughly 500 admissions/second. Keep the critical section to the single UPDATE, with no network call or scheduling decision inside the transaction, and decrement in the same transaction that writes the terminal status.
Close the leak the status enum implies: a run can end as lost, so a crashed worker otherwise consumes a slot forever. Reconcile on a schedule against status = 'running' and leased_until < now(), and treat the counter as a fast path over job_run, which stays the system of record.

Follow-up

Write the retry loop for the SERIALIZABLE version. What does the caller see when it keeps aborting, and what bounds the retries?
Two regions each keep a counter. What is the effective cap, and what does admission do when the counter store is unreachable?
The cap changes mid-flight on a plan upgrade. Do running jobs get killed, and what does the counter row look like during the change?

Explain why the metering dashboard scans every daily partition

Medium · indexing · partition pruning · sargability · tenant skew

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

Approach

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

Worked solution (30 min)

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

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

Verify it yourself

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

Follow-up

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

How does the HTTPS protocol work? Explain the handshaking process, specifically focusing on how symmetric and asymmetric encryption are utilized.

Medium · system design and architecture

Approach

State the consistency you need, and where you are willing to be stale.
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.

Follow-up

What would you drop to keep the system up under load?
What breaks first when traffic grows ten times?

Design a high-throughput, low-latency order matching engine for a crypto exchange.

Medium · system design and architecture

Approach

Name the read and write paths separately; they rarely have the same bottleneck.
Choose a partition key and say what query it makes expensive.
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?
What breaks first when traffic grows ten times?

How would you design a distributed rate limiter to protect public-facing APIs from abuse?

Medium · system design and architecture

Approach

Choose a partition key and say what query it makes expensive.
Name the read and write paths separately; they rarely have the same bottleneck.
State the consistency you need, and where you are willing to be stale.

Follow-up

What would you drop to keep the system up under load?
What breaks first when traffic grows ten times?

In a high-concurrency system, how do you prevent cache stampede (cache breakdown) and cache penetration?

Medium

Approach

Say what you would check first and why it is the highest-information step.
State your assumptions explicitly before working the problem.
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?

An idempotent create endpoint that returns a one-time secret

Easy · idempotency · race conditions · api design

POST /v1/api-keys inserts a tenant_api_key row (tenant_id, workspace_id, name, key_prefix, secret_hash, scopes, status, auth_version, created_at, expires_at) and returns the plaintext secret exactly once, since only its SHA-256 is stored. Write volume is tens per second and clients retry on timeout. Design the idempotency mechanism: what the key is scoped by, where the record lives, its retention, what happens when the same key arrives with a different body, what happens when a retry arrives while the first request is still in flight, and what a replay returns for the secret.

Approach

Scope the key by tenant, not globally: uniqueness is on (tenant_id, idempotency_key), or one tenant's key collides with another's and the second caller receives a stored response for a request it never sent. Store a fingerprint of the request alongside it - method, path and a hash of the canonicalised body - so a mismatch is detectable.
Let the unique constraint decide the race instead of application logic. In the same transaction as the credential insert, run INSERT INTO idempotency_record (tenant_id, key, request_fingerprint, status) VALUES (...) ON CONFLICT DO NOTHING RETURNING id; no returned row means this is a replay, and the existing record is then read. A select-then-insert here loses to itself under concurrency in exactly the way this endpoint is meant to prevent.
Write the three replay outcomes as a decision table rather than as prose: same fingerprint and completed returns the stored response; same fingerprint and still in flight returns 409 with Retry-After, without blocking and without executing; different fingerprint returns 422, because replaying a stored response for a mutated body would tell the caller a credential was created for parameters it never sent.
Handle the secret as the part that makes this endpoint different from an ordinary idempotent create. The plaintext cannot be regenerated from secret_hash, so either the stored response holds it - making the idempotency record a secret at rest whose retention is now the secret's exposure window - or a replay returns the key metadata without the secret and the documentation says a lost response is resolved by listing keys and revoking the orphan. The second is the safer default precisely because credentials are listable and revocable.
Set retention from the client's retry budget, not from a round number: the record must outlive the SDK's maximum total retry duration, so twenty-four hours is defensible if that budget is minutes. After expiry the key is reusable and a very late retry creates a second credential, which is acceptable here only because the object is listable and revocable, and would not be for an unlistable side effect. Expire with a scheduled delete on an index over created_at.

Worked solution (20 min)

Write the DDL for the idempotency record, including the unique constraint that makes the concurrent case impossible rather than unlikely.
Write the four outcomes - fresh, replay-completed, replay-in-flight, fingerprint-mismatch - as a decision table with the HTTP status for each.
Decide what a replay returns for the plaintext secret and write the exact sentence the API reference has to carry about it.
Pick a retention and justify it from the client library's own maximum retry duration rather than from a round number.

Expected result: Unique (tenant_id, idempotency_key); the record inserted in the same transaction as the credential with ON CONFLICT DO NOTHING deciding the winner; 200 with the stored response, 409 for in-flight, 422 for a fingerprint mismatch; a documented rule that the plaintext secret is not replayed and a lost response is resolved by listing and revoking; retention exceeding the SDK retry budget, for example twenty-four hours.

Verify it yourself

Two concurrent requests with the same key must leave exactly one row in tenant_api_key. If the logic is select-then-insert, they will not.
A retry with a mutated body must not return 200. If it does, the fingerprint is not being compared or is not covering the fields that matter.
The retention you chose must be strictly longer than the client library's maximum total retry duration. Look that number up rather than assuming it.

Follow-up

Two requests with the same key arrive concurrently. Show the exact statements and say which one loses, and how it finds out.
The client receives a timeout, retries, and gets 409 in-flight. What should the client library do next, and for how long?
How does the design change if the created object is not listable - a one-off payout, say - so an orphan cannot be found afterwards?

Webhook workers leak until OOM and drop in-flight deliveries

Medium · memory leak · eviction · leases

webhook-delivery workers grow from 400 MB to a 2 GB limit over about 36 hours, are OOM-killed, restart, and repeat. Each restart abandons in-flight attempts, so webhook_delivery rows sit in in_flight until their leases expire and the backlog spikes. The live set measured after a forced full collection also grows. The fleet serves tens of thousands of subscriptions, several thousand of which have been failing for weeks. Give an ordered checklist, the measurement separating retention from fragmentation, and the fix.

Approach

Separate the two failure shapes with one measurement: track resident set size against the live set after a forced full collection. A live set that climbs monotonically is retention; a flat live set under a rising RSS is fragmentation, off-heap or native allocation, or an allocator that never returns pages. The stated symptom puts this in the first category, which rules out allocator tuning as a fix.
Characterise the curve rather than the total. Growth linear in uptime implies an unbounded structure keyed by something that keeps arriving; step growth implies buffering a large object. Correlate the slope against event rate and separately against the count of distinct subscriptions seen, because those two diverge and only one of them will fit.
Diff two heap snapshots an hour apart by retained size grouped by dominant root, not by allocation count, which is dominated by short-lived objects and will point at the wrong thing.
Expect a per-subscription map with no eviction: circuit-breaker or backoff state created on first failure and never removed, so the retained set grows with endpoints that have ever failed, and the several thousand permanently dead endpoints hold theirs forever.
Fix in two places. Bound the in-memory structure with a size-capped LRU or a TTL keyed on last use, and move state that must survive a restart onto the subscription or webhook_delivery row, since the worker holding it in memory is exactly why a restart loses it.
Repair the second-order damage separately, because it will outlive the leak: workers claim by compare-and-set with leased_until, so a bounded lease returns in_flight rows to pending on a known schedule, and a graceful shutdown releases leases instead of waiting them out.

Follow-up

The backlog spike after a restart is itself a thundering herd against customer endpoints. What stops the recovery from becoming a second incident?
Suppose the live set had been flat while RSS still climbed. Name two causes and the measurement that separates them.
How would you size the LRU, and what does a miss on an evicted circuit-breaker entry cost a customer whose endpoint is down?

Suggested practice checklist

For someone fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.

Day 1 — Measure before reasoning

Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.

Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.

Day 2 — References, copies, and the bugs they produce

Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.

Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.

Day 3 — Types, once, in a language that checks them

Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.

Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.

Day 4 — Concurrency, starting with what actually runs at the same time

Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.

Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.

Day 5 — Debugging as a procedure rather than an instinct

Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.

Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.

Day 6 — Tests that catch the bug you are about to write

Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.

Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.

Day 7 — Debug something broken, out loud

Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.

Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.

The bug you decided not to fix

Engineers over-index on what they repaired. A stronger answer covers something you knowingly left broken: the alert you tuned down, the data inconsistency you documented instead of chasing, the cleanup you deferred past two quarters. Give the reasoning and the condition that would have reopened it, so it reads as a decision and not as neglect.

Tell me about a time you had to learn a completely new technology or domain on short notice to deliver a critical feature.

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

Ship metered billing with a named deduplication horizon

Medium · technical debt · deduplication · deadline pressure · detectors

Metered billing must be on in three weeks. usage_event is partitioned daily, so its unique index must include the partition key and deduplicates only within a day: a producer retry that crosses midnight, or a replay run a week later, gets through. A cross-partition dedup store is two weeks you do not have. Describe shipping with debt you named in advance: what you shipped, what you wrote down, the detector you added, the trigger and date for paying it off, and what you would have refused to ship under the same pressure.

Approach

Show you can separate the two kinds of debt, because that distinction is what the question actually probes. Debt that costs engineering time later is shippable on a deadline. Debt that silently corrupts a number a customer gets charged for is not shippable unless the corruption is detectable, and detectability is the whole negotiation.
Make the exposure narrow and measured rather than gestural. The hole is duplicates whose occurrences straddle a UTC day boundary, plus any replay older than partition retention. Measure it before arguing about it: how often an idempotency_key recurs at all, and the distribution of the gap between first and last occurrence. If the ninety-ninth percentile of that gap is four minutes, the residual risk is a small band around midnight and you can say so numerically.
Add the detector before the feature, not after. A nightly job counting keys that appear in more than one partition is one grouped scan over recent partitions, and it converts a silent overcount into a page. State what it costs to run and what it fires on.
Buy the cheap half of the real fix immediately: extend partition retention so the dedup horizon exceeds the producer's maximum retry window plus the longest replay you intend to support. That reframes retention as a correctness parameter rather than a storage cost, which is the sentence you need on record before someone optimises the bill.
Make repayment mechanical instead of aspirational: a dated entry with a named owner, plus a threshold that pulls the date forward — first detector hit above N events, or first customer dispute. Debt with a trigger gets paid; debt with only a date does not.
Answer the second half honestly by naming what you would refuse under identical pressure: the sealing path, because a sealed row is frozen and a wrong number there stops being a bug and becomes an adjustment line, a dispute and an audit question.

Follow-up

The detector fires on forty duplicate events for one tenant, and two of their invoices have already sealed. What happens next?
Whom did you tell that the billing numbers had a known hole, and in what words?
Finance asks you to cut storage by shortening partition retention. What do you say, and to whom?

Argue against failing open when the control plane is unreachable

Hard · revocation · cache staleness · fail-open vs fail-closed · influence

The gateway caches credential-to-context decisions with a sixty-second TTL. A design proposal says that when control-plane reads fail, pods should keep serving from expired entries indefinitely so a control-plane outage never becomes a product outage. You believe that converts every revocation into an unbounded one. Describe a design you argued against while it was still a live proposal: what you measured or modelled to make the case, what you conceded, who decided, and what happened afterwards. Say what would have changed your mind before the decision, not after it.

Approach

Reframe it from a values argument into a bounded-staleness argument. Both sides already accept the cache; the disagreement is only about the ceiling on how long a revoked credential keeps authorising. Put a number on the table — serve stale for up to fifteen minutes, then fail closed — and make the other side argue against a number rather than against a principle.
Bring arithmetic rather than adjectives: the rate of revocations with revoked_reason in ('suspected_leak','auth_version_bump'), the observed distribution of control-plane unavailability, and the product of the two, which is expected requests served by revoked credentials per outage-hour. At 30k requests/second the unbounded version is not a subtle exposure and the number says so.
Concede the strong half of the opposing case first, because that is what buys you the room: failing closed turns one service's outage into a total outage across three regions, and a control plane doing tens of writes per second is not engineered to the gateway's availability target. A proposal you have not steelmanned reads as reflex.
Propose the asymmetry that usually resolves this: stale entitlements cost bounded money (a quota fifteen minutes out of date over-serves by a computable amount), while a stale revocation costs unbounded access. Split the cached decision by what it authorises, give the two halves different staleness ceilings, and let the entitlement half fail open while the revocation half fails closed.
State the propagation dependency plainly, since it is the part that is missed: validity is also derived from the principal's auth_version, so password reset and sign-out-everywhere flow through this same cache. A design that bounds staleness for explicit revocation and not for auth_version bumps has only fixed half of it.
Say who decided, and what you did afterwards in either outcome: write the decision down with its number and a review date, and instrument the exposure you were worried about so the next round of the argument is settled by data instead of by seniority.

Follow-up

Publish-subscribe invalidation is lossy under a partition, and a TTL is the only hard bound. What TTL do you pick, and what does it cost you at 30k requests/second?
The key was revoked because it was found in a public repository. Does your answer change, and where does that urgency live in the design?
You lost the argument and six weeks later the failure you predicted happens. What do you say in the review, and what do you not say?