Kasada Software Engineer Interview Questions & Guide 2026

The 60-second brief

A Software Engineer at Kasada is at the forefront of the fight against sophisticated automated threats. You are responsible for engineering solutions that protect global brands from malicious bots, scrapers, and account takeover attacks. The work is inherently complex, requiring a deep understanding of web security, obfuscation techniques, and high-performance systems that operate in real-time.

Allocate preparation against your weakest link rather than your favourite topic. A strong algorithm habit usually comes with weak out-loud explanation of tradeoffs, and years of shipping usually come with rusty from-scratch implementation under a clock.

Focus areas

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

Overview

This role is critical to the Kasada mission of making the internet safer by rendering automated attacks ineffective. You will work on products that must remain invisible to users while being incredibly resilient against adversarial attempts to reverse-engineer or bypass security controls. Success in this role requires a blend of creative problem-solving and rigorous technical execution, as you are constantly navigating an evolving landscape of cyber threats.

Where to focus your preparation

Treat the sections below as preparation areas and confirm the format with your recruiter.

Preparation focus

No round sequence has been reported for this company, so work the categories below and confirm the format with your recruiter.

What to demonstrate

• Breadth across SQL, experimentation and product reasoning
• Ability to state assumptions before choosing a method

How to prepare

• Drill the practice exercises below and time yourself
• Prepare three quantified stories about decisions you drove

Where candidates lose points

Shipping a migration and the code that depends on it as a single change

During any rolling deploy, and for as long as a rollback remains possible, old and new code execute against the same schema at the same time. A migration that drops or renames a column breaks every instance that has not restarted yet, and code that requires a column the migration has not applied breaks every instance that restarted early. The discipline is expand then contract: add the new column nullable, write both shapes, backfill in batches, move reads across once the backfill is verified, and only then stop writing the old shape and drop it - four deploys, usually spread over days. It feels disproportionate until the first rollback, at which point it is the only reason the previous version still runs.

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

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

Treating a network call as though it were a local function call

A remote call can be slow, fail, or return after you stopped waiting, so name the timeout, the retry policy, and what the caller sees while the dependency is down. A call with no timeout turns one slow dependency into an exhausted thread or connection pool in every service upstream of it.

Arguing past a hint

When the interviewer asks what happens for a particular input or floats a different data structure, stop and take it seriously; it is almost always a correction rather than idle curiosity. Talking over it converts a recoverable wrong turn into a data point about how you handle review.

Questions & practice

Can you explain your approach to identifying and mitigating automated bot traffic?

Medium · data structures and algorithms

Approach

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

Follow-up

• Which test case would catch an off-by-one here?
• How does this change if the input no longer fits in memory?

How would you use the Chrome Developer Console to debug a complex web request?

Medium · data structures 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

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

Merge partitioned event streams into one ordered feed with bounded lateness

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

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

Approach

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

Worked solution (35 min)

1. Write the heap comparator on (occurred_at, event_id) and the per-partition head refill.
2. Write the watermark computation and the emit-loop condition, then list which buffered events are held at a chosen instant.
3. Compute the buffer at 4,000 events per second, 30 seconds and 1 KB per event, and state what fraction of a worker's heap that represents.
4. Add the idle-partition marker and trace the watermark with one silent partition, both with and without the marker.
5. Write the late-event path and name the key that makes applying it safe.

Expected result: A 64-way min-heap merge at O(n log P) with a deterministic (occurred_at, event_id) comparator, a watermark of the per-partition minimum less 30 seconds gating emission, a stated buffer of 120,000 events and roughly 120 MB, idle markers so a silent partition cannot freeze the watermark, and a late-event policy justified by the projection's idempotency on (aggregate_id, aggregate_version).

Verify it yourself

• Feeding the partitions in an adversarial interleaving produces the same output sequence as feeding them in timestamp order.
• Silencing one partition stalls output without the idle marker and does not stall it with the marker.
• Applying a late event twice leaves the projection byte-identical, which is what permits the out-of-order path to exist.

Follow-up

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

Stop tag and share joins from fanning out a page

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

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

Approach

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

Follow-up

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

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.

Worked solution (25 min)

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

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

Verify it yourself

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

Follow-up

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

Specify the outbound webhook contract a customer endpoint can verify

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

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

Approach

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

Follow-up

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

Design the bulk write endpoint a migration script retries blindly

Medium · batch api · partial failure · idempotency

A customer's migration script pushes 2 million resources through POST /v1/resources:batch, up to 500 items per call, and retries any call that errors or times out. Within a call some items fail validation, some collide with rows that already exist, and some succeed. Specify the request and response shape, whether a batch is atomic or per-item, how idempotency works for the call and for each item, the status code for a mixed outcome, the size and item-count limits with their error codes, and exactly what the script does after a timeout mid-batch.

Approach

1. Choose atomicity deliberately and price it. All-or-nothing means one transaction holding locks for the whole batch on a primary already absorbing about 1.2k writes/second, which bounds batch size by lock duration, and it turns one bad row into 499 rejections the script must resend. Per-item partial success is the right default for a migration, and the contract's job is then to make a partial outcome impossible to miss.
2. Require a client-supplied id on every item and echo it in every result. Deriving a per-item key from the array index breaks the first time the script resends a batch with the failures removed: the indices shift, previously-succeeded items acquire new keys, and they are created a second time.
3. Key the effects at two levels. The call's Idempotency-Key covers an exact resend of the same bytes; per-item keys of (tenant_id, client_item_id) make a partially-applied batch safe to resend whole. Resending an identical batch must reproduce the same per-item results, not 500 conflicts the script has to interpret.
4. Answer a mixed outcome with one status plus per-item detail: 200, or 207 borrowed from WebDAV if you prefer it - document whichever you pick - carrying an array of client_item_id, per-item status, and either resource_id or an error code from the same taxonomy the single-item endpoint uses. Reserve 4xx for the request as a whole: unparseable body, too many items, payload over the limit (413). Put a failed count at the top level so that even a script checking only the cheapest thing cannot conclude success while rows were dropped.
5. Cap the request before doing any of it: item count, total bytes, and concurrent batches per tenant, since one tenant's migration otherwise consumes write capacity everyone shares. Anything that cannot finish inside the request deadline belongs on job_run behind a 202, not in a synchronous call that will time out halfway.
6. Script behaviour after a timeout: the outcome is unknown and no results were received, so resend the identical batch with the same keys and read the results. Never resend 'only the items I have no result for' - a timeout yields no results at all, and that rule silently means resend everything anyway.

Worked solution (30 min)

1. Write the request schema with the per-item client id, and the response schema with per-item status and a top-level failed count.
2. Write the atomicity decision and the sentence of justification that names the lock cost or the resend cost.
3. Define both key levels and trace a resend of a half-applied batch through them, item by item.
4. List the whole-request rejections with their status codes and limits.
5. Write the script's timeout rule and the pacing it should apply between calls.

Expected result: A per-item batch with client-supplied item ids, per-item idempotency plus a call-level key, a mixed-outcome response that a status-only check cannot misread, explicit request caps, and a resend rule that is correct when no results were received.

Verify it yourself

• Resending an identical batch twice leaves exactly the same set of resource rows as sending it once.
• Resending a batch with the successful items removed does not create duplicates of anything.
• A caller that reads only the top-level fields can still tell that some items failed.

Follow-up

• At 500 items per call, what is the wall-clock time for 2 million rows, and what pacing do you publish so the migration does not become an incident?
• One item in every batch fails with the same code. How does the script discover that without a human reading logs?

Listing latency scales with page size, not with filters

Easy · n+1 · latency · query batching

The tenant listing endpoint reads resource filtered by tenant_id and status, ordered by updated_at DESC, and returns each row plus the owner's display name from app_user and the actor of that resource's latest resource_revision. p99 is 55 ms at 10 rows per page and 1.4 s at 200. Database telemetry shows 401 statements per request, each under 1 ms, and nothing in the slow-query log. Diagnose the cause and give the fix, stating the statement count per request and the p99 you expect afterwards.

Approach

1. Read the counters before forming a theory. 401 statements for 200 rows is one driver query plus two per row, and sub-millisecond execution with an empty slow-query log rules out a bad plan. The time is round trips, which is why it is invisible in every per-query metric and scales with rows returned rather than with filter selectivity.
2. Name the two per-row statements from their normalised text: a single-row app_user lookup by user_id, and a resource_revision lookup by resource_id ordered by version DESC LIMIT 1. Confirm by dropping those two response fields and watching the statement count fall to one. That locates the calls in the serialisation layer, not the repository.
3. Check that the arithmetic accounts for the whole gap. Measure one round trip to the replica in isolation; 400 trips at roughly 3 ms of network plus 0.2 ms of execution is about 1.3 s on top of a 55 ms baseline, which matches. If the multiplication had fallen short, the N+1 would only be part of the story and you would keep looking.
4. Batch both lookups. Collect owner_user_ids and resource_ids from the driver query, then issue WHERE tenant_id = $1 AND user_id = ANY($2) for the users, and PostgreSQL's SELECT DISTINCT ON (resource_id) ... WHERE resource_id = ANY($2) ORDER BY resource_id, version DESC for the latest revision, which the UNIQUE (resource_id, version) index serves directly. On an engine without DISTINCT ON, use a lateral join or a row_number window. Three statements per request at any page size.
5. Keep the tenant predicate in the batched query. The per-row version was implicitly scoped because its ids came from tenant-scoped rows; a batched user_id = ANY(...) with no tenant_id is an unscoped read that behaves correctly only as long as the id list is trustworthy.
6. Re-measure at 10, 50 and 200 rows and confirm the statement count is constant. Latency should now track bytes returned.

Follow-up

• The page size is capped at 200 today. What breaks first if it is raised to 2,000, and is it still this bug?
• How do you stop the next N+1 from reaching production, given that no individual query is slow and the endpoint's tests pass?
• The latest-revision actor is only used to render an avatar. Make the case for denormalising it onto resource, and name the write anomaly that introduces.

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.

Describe a time you had to optimize a piece of code for high-throughput performance.

Medium · behavioural and engineering judgement

Approach

1. Pick a story where you made the decision, not one where you watched it.
2. State the situation in two sentences and spend the rest on the reasoning.
3. 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?

How do you handle asynchronous operations in JavaScript environments?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

• What would you do differently if you ran that again?
• What did you decide not to do, and why?

Ship under a deadline and bound the debt you chose

Medium · pagination · technical debt · tradeoffs

You have four days to ship a tenant-facing listing endpoint. The version you would defend uses keyset pagination over (tenant_id, status, updated_at DESC, resource_id DESC); the version you can finish uses LIMIT/OFFSET with no matching index. Describe a deadline call you actually made of this shape: what you shipped, what you knowingly deferred, how you bounded the damage with a mechanism rather than an intention, and the specific numeric condition that would force the follow-up. Name who you told and where you wrote it down.

Approach

1. Name the deferred failure precisely instead of calling it slow. OFFSET n makes the database produce and discard n rows, so cost grows with page depth; without an index matching the sort, every matching row is read and sorted before the limit applies; and rows inserted between two page fetches shift across the boundary so items are skipped or repeated with nothing in the response to signal it.
2. Bound the blast radius with something mechanical rather than a promise: cap maximum page depth, cap page size, restrict the endpoint to one internal caller, or keep it behind a flag. State which failure each cap removes and which it leaves standing.
3. Attach a number to the trigger and wire it to an alarm: the first tenant crossing N resources, or the endpoint's p99 crossing its share of the 400 ms budget, so the debt announces itself instead of waiting to be remembered.
4. Write it where the next engineer looks, which is the code and the ticket, not a chat message: what was deferred, why, the cap, and the trigger.
5. Report what actually happened in your real example, including the case where the trigger never fired and the debt was correctly never repaid.

Follow-up

• At what page depth does the offset version breach your latency budget, given your page size and row counts?
• What breaks first when you switch to keyset pagination later, and what does a client holding an old page token see?
• Who would have overruled you if you had asked for two more days, and did you ask?