Interview guides · AI Tech Start-up

AI Tech Start-up Software Engineer Interview Questions & Guide 2026

Software engineer prep for AI Tech Start-up: design exercises weighted toward the constraints of B2B software and infrastructure work.

The 60-second brief

As a Software Engineer at AI Tech Start-up, you are at the forefront of bridging the gap between cutting-edge Large Language Models (LLMs) and complex industrial reality. You will not just be writing code; you will be architecting systems that transform how global manufacturing businesses operate. By building agentic AI workflows, you are directly responsible for automating high-stakes industrial procurement and sales processes.

Find out what you will be typing into. A shared plain-text editor with no autocomplete, compiler or test runner changes what you have to hold in your head, and practising inside your own configured environment hides exactly that gap.

Focus areas

Bound blast radius with per-tenant concurrency limits
Evolve APIs without breaking pinned SDK clients
Keep money in integer minor units

Overview

This role is uniquely challenging because it requires both deep technical rigor and a tangible product mindset. You will be expected to work across the full stack, utilizing Python and Next.js to deliver end-to-end features. Beyond the keyboard, you will engage directly with the domain, including site visits to manufacturing facilities to observe the real-world impact of the software you build. If you thrive in environments that move fast, value technical precision, and prioritize solving "real" problems over theoretical research, this position offers significant strategic influence.

The AI Tech Start-up interview process

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

Initial Screening

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

What to demonstrate

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

How to prepare

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

Technical Deep Dives

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

What to demonstrate

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

How to prepare

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

Behavioral Assessments

Your first answer is not really what is scored. It buys the follow-up questions, and those decide the round. An interviewer with fifteen minutes takes one thread and pushes on it four or five times, so a story you can only tell at a single level of detail collapses under the third why. That is an argument for fewer stories known deeply rather than one prepared per prompt. Four or five pieces of work you can still explain down to the code you changed and the argument you had about it will cover nearly anything asked in this round.

What to demonstrate

Whether a story holds as the questioning moves from what you did to why that instead of the alternative, and then to what you would change knowing what you know now
Whether you can re-cut a project to answer the question actually asked rather than delivering a rehearsed block that answers an adjacent one
Whether your level of detail is chosen rather than habitual: going down to the schema when the question is about the data model, staying out of it when the question is about the person who disagreed with you

How to prepare

Pick four projects and write the chain out four levels deep for each: what you did, why that, why not the alternative, and what would have to be true for the alternative to have won. Where you cannot reach the fourth level, you have a placeholder rather than a story
Have someone ask why three times in a row on a single thread with nothing else added, and mark the point where you start repeating a sentence you already said. That point is where the interviewer stops learning anything
Build a one-page index instead of an answer bank: the common prompts in this round (disagreement, a failure that was yours, thin requirements, a deadline you missed, work you inherited) mapped to which of your four projects you would use for each, so the choosing is done now rather than while an interviewer waits

Final Team Interviews

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

Treating a timed-out write as a failed write

A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.

Paginating a growing table with limit and offset

Two unrelated defects share the idiom. Correctness: rows inserted or deleted between page requests shift the window, so a consumer walking an export skips rows and sees others twice, which for a customer-facing sync is silent data loss rather than an error anyone notices. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N times the page size and a deep page on a large table degrades from milliseconds to seconds. Keyset pagination over a stable, unique, indexed ordering -- where (created_at, id) < ($1, $2) order by created_at desc, id desc limit $3 -- is constant-cost per page and immune to shifting, on the precondition that the cursor columns never change value for a row, which disqualifies updated_at as a cursor.

Assuming the input fits in memory

Ask how large the input is in bytes before committing to an in-memory algorithm; beyond that point the options are a single streaming pass, an external sort with bounded buffers, or a sketch that trades exactness for constant memory. An algorithm that assumes random access to the whole input is a different algorithm from one that sees each element once.

Sorting when the problem never required a total order

Match the algorithm to the guarantee actually needed: the top k comes from a size-k heap in O(n log k) time and O(k) space, distinctness needs a set rather than an ordering, and a small bounded integer key range admits a linear counting pass. A full O(n log n) sort is the right default only when you genuinely need everything in order.

Questions & practice

Find peak concurrent sandbox usage from run intervals

Medium · sweep-line · intervals · concurrency-caps · null-semantics

Given up to 5 million job_run rows for one tenant over one day, with run_id, started_at, finished_at, status and wall_clock_limit_seconds, report the maximum number of sandboxes running at once, the earliest instant that maximum is reached, and the first run_id that would breach a per-tenant cap of C. started_at is null while a run is queued; finished_at is null both for runs still executing and for runs in status lost. Treat a run as occupying [started_at, finished_at). Give the complexity and state how you handle each null.

Approach

Turn each run into two sweep events, (started_at, +1) and (end, -1), then sort the 2n events by timestamp with -1 ordered before +1 at equal timestamps. That tie-break is what makes the interval half-open, so a run finishing at 10:00:00 and one starting at 10:00:00 never overlap.
Decide each null out loud before sweeping, because each choice moves the answer. A null started_at means queued and contributes nothing. A null finished_at with status running or leased is clipped to the window end. Status lost has no observed end at all, so clip it at started_at + wall_clock_limit_seconds on the grounds that the supervisor owns the timeout, and record that you did. The table's check (finished_at is null or started_at is not null) guarantees you never see an end without a start.
Sweep once, maintaining a running counter, the maximum, and the timestamp at which the maximum was first attained (update peak_at only on a strict increase, or you will report the last such instant instead of the earliest). Capture the first run_id whose +1 takes the counter to C+1 during the same sweep rather than in a second pass.
Complexity: O(n log n) dominated by the sort, O(n) space. If rows already arrive ordered by started_at, a min-heap of end times gives O(n log k) time and O(k) space with k the peak concurrency, which is the better shape when the rows come from an index scan on (tenant_id, started_at).
If second resolution is acceptable, counting-sort the endpoints into an 86,400-slot delta array and prefix-sum it: O(n + T) time and O(T) space, which beats the comparison sort at 5 million rows. It answers only at second granularity, so state which resolution the cap is defined in.

Follow-up

Now report peak concurrency per tenant for 10,000 tenants from one globally sorted stream. What changes about memory and about the sort?
The cap has to be enforced at dispatch rather than reported afterwards. What does the admission check look like, and where does it race?
How would you answer 'peak concurrency within any 5-minute window' without re-sorting?

Locate a billing reconciliation gap without rescanning ninety million events

Hard · reconciliation · dimensional-bisection · watermarks · hypothesis-testing

A tenant's sealed invoice total is 0.4% below the sum of its raw usage_event rows for the period. That tenant has 90 million events over 30 days in a table partitioned daily on ingested_at, and its rollups carry source_max_ingested_at, revision and sealed_at. Recomputing all 30 days from raw is correct, and you are not going to do it. Give the procedure that locates the divergent (workspace, sku, hour) cell, the cost of each probe, and the one query you run before any of it.

Approach

Run the free query first. Sum raw quantity for the period restricted to ingested_at <= source_max_ingested_at of the sealed rollups, and compare that against the unrestricted sum. The rollup stores the watermark precisely so this can be answered without a scan. If the whole 0.4% sits above the watermark, nothing is broken: it is late data, it becomes an adjustment line, and the investigation ends in one query.
Only if the gap survives that test do you bisect, and you bisect by dimension rather than by rows. Compare 30 per-day totals, then inside the offending day compare the 6 SKUs, then the workspaces, then the 24 hours. That is roughly 30 + 6 + W + 24 grouped probes, each an indexed range scan over one daily partition for one tenant, against O(N) per attempt for the naive re-fold.
Quantify why naive is not merely slow but unusable mid-incident: at a generous 200,000 rows/second sequential, 90 million rows is about 7.5 minutes per attempt, you will want ten attempts, and every one competes for I/O on the same partitions live ingest is writing. The diagnostic worsens the backlog it is diagnosing.
Before fetching each comparison, state what it would look like under each hypothesis. Two adjacent hours off by equal and opposite amounts is occurred_at versus ingested_at bucketing. A whole day offset by exactly N hours is a timezone applied at the wrong layer. A gap confined to one SKU in one workspace is an environment filter. The same (tenant_id, idempotency_key) present in two ingested_day partitions is the dedup horizon losing a retry that crossed midnight.
Make the next bisection cheap by storing the aggregate you keep recomputing. A per-(tenant_id, ingested_day) count and quantity checksum turns step two from thirty probes into one read, and it is the same number the reconciliation job already produces.
Whatever you find, the sealed period does not change value. The correction is an adjustment line pointing at the line it reverses, carrying its own source_rollup_watermark, because the original invoice is the evidence of what the customer was charged.

Worked solution (35 min)

Reproduce the shape locally: generate 2 million events for one synthetic tenant over 5 days, fold them, then inject three defects, namely 0.2% of events bucketed by ingested_at, a handful of duplicate idempotency_key values whose retries cross midnight, and a block of events ingested after the seal.
Write the watermark-bounded query first and record how much of the gap it explains on its own.
Bisect by day, then SKU, then hour, recording the probe count and the rows each probe touches.
For each located cell, write down the predicted signature before querying it, then check whether the data matches the prediction.
Time a full re-fold of the 2 million rows and extrapolate to 90 million.

Expected result: The watermark-bounded query accounts for the post-seal block entirely and removes it from the investigation. The ingested_at bucketing appears as adjacent hours off by equal and opposite amounts. The midnight-crossing duplicates appear as one (tenant_id, idempotency_key) present in two ingested_day partitions. Bisection reaches each cell in fewer than 70 probes.

Verify it yourself

The three injected defects sum to the original gap with no unexplained remainder.
Each probe's row count is bounded by one day-partition for one tenant, not by the period.
Removing the source_max_ingested_at bound from the first query makes a gap appear, and grow, while ingest runs, even after every defect is removed.

Follow-up

The gap is 0.4% in one direction on one day and 0.4% the other way the next day. What does that shape rule in, and what does it rule out?
How do you distinguish a duplicate from a restatement, given revision and recomputed_at on the rollup?
Ingest is still running while you investigate. What makes your two numbers comparable at all?

Schedule ordered webhook retries with a heap of subscription queues

Medium · heap · scheduling · backoff · head-of-line-blocking

Design the in-memory scheduler for webhook delivery. Up to 20 million rows sit in status pending or failed_retryable across 200,000 subscriptions, each row carrying next_attempt_at and attempt_count, and each endpoint having a circuit breaker. Deliveries for one subscription must be attempted in order, so at most one attempt per subscription may be in flight. Support due(now), complete(delivery, outcome) and insert(delivery) in O(log S), where S is the subscription count rather than the delivery count. Give the backoff formula you schedule retries with.

Approach

Key the global heap by subscription, not by delivery. Each subscription owns a FIFO of its due deliveries in event order; the heap holds one entry per eligible subscription, keyed by its head's next_attempt_at. That is 200,000 heap entries instead of 20 million, and it makes the one-in-flight rule structural rather than a check somebody can forget.
due(now): peek the minimum. If its key is in the future, sleep until then instead of spinning. Otherwise pop it, move the subscription into an in-flight set, and do not re-push it. A subscription absent from the heap cannot be dispatched twice, which is precisely how ordering is preserved.
complete: on success, drop the head and re-push the subscription keyed by its new head, or leave it out when the queue empties. On a retryable failure, increment attempt_count and set next_attempt_at = now + uniform(0, min(cap, base * 2^attempt)), sampled uniformly across the whole interval. That is full jitter; deterministic backoff re-synchronises the herd you just created.
Circuit breaker: park the subscription in a second heap keyed by its half-open time, so an endpoint dead for six hours costs one heap entry and zero attempts rather than consuming worker slots. Admit exactly one probe at half-open and close the breaker only on its success.
Say the price of the ordering guarantee out loud. One in-flight attempt per subscription means an endpoint answering in 10 seconds drains at 0.1 deliveries/second however many workers you run, and its backlog grows until it recovers. If the customer does not need order, allow k in flight and document delivery as unordered; that is the trade, and it is a product decision.
All three operations are O(log S) with O(S) resident heap memory and the queues themselves backed by the store. The database-backed equivalent is a partial index on (subscription_id, next_attempt_at) where status in ('pending','failed_retryable') claimed with FOR UPDATE SKIP LOCKED, and the write-back must be fenced on lease_token so a worker that stalled and resumed cannot overwrite a newer attempt.

Follow-up

One subscription has 4 million queued deliveries. What stops it from starving the other 199,999, and what does your heap look like under that load?
A customer requests redelivery of last Tuesday's events. Where do those rows enter your structure, and what keeps them from reordering live traffic?
The process restarts. How much state do you rebuild, and what stops every subscription from being attempted in the same second?

Paginate a tenant's delivery export without skipping rows

Easy · keyset pagination · indexing · multi-tenancy

A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.

Approach

Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
Replace the position with a value predicate over a stable, unique, indexed ordering: where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window.
Index (tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit.
Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.

Worked solution (20 min)

Load about 50k deliveries for one tenant, then walk them with the OFFSET query while a writer inserts 10 rows/second, collecting every returned delivery_id.
Compare the distinct ids collected against the set of ids that existed when the walk started, and record the shortfall.
Repeat the walk with the keyset query and confirm every pre-existing id is returned exactly once.
Run explain (analyze, buffers) on page 1 and page 500 of each form and compare shared buffer hits.

Expected result: A keyset query using the row comparison (created_at, delivery_id) < ($2, $3) with order by created_at desc, delivery_id desc limit 100, backed by (tenant_id, created_at, delivery_id) scanned backwards; plus the statement that OFFSET drops rows under concurrent insert and costs time proportional to the offset, and that updated_at is disqualified because it mutates on retry.

Verify it yourself

Buffer counts for page 500 are the same order of magnitude as page 1.
Every id present before the walk is returned exactly once when only inserts run concurrently.
The query plan shows no Sort node above the index scan.
Two rows sharing a created_at value to the microsecond are both returned across the page boundary.

Follow-up

The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added created_at <= $snapshot bound, and a materialised export table.
A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?

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?

What are your best practices for unit testing and ensuring code quality?

Medium

Approach

Say what you would check first and why it is the highest-information step.
State your assumptions explicitly before working the problem.
Work from the requirement backwards to the design.

Follow-up

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

What is the difference between a no-code and a low-code platform?

Medium

Approach

Clarify what is being asked and what a complete answer contains.
Work from the requirement backwards to the design.
Say what you would check first and why it is the highest-information step.

Follow-up

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

How do you approach the design of software architecture when building new features?

Medium

Approach

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

Follow-up

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

Build a resumable usage export the customer can reconcile against

Medium · pagination · cursors · late data

Customers reconcile invoices against usage_event (event_id uuid, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at, source_service, request_id), partitioned daily on ingested_at. The current export is ?page=N&per_page=1000 ordered by occurred_at, and a customer syncing hourly reports rows that never appear in their export but do appear on their invoice. Design the replacement: the ordering, the cursor's contents, the index it requires, and the rule deciding where a page stops. State what the client does on a timeout and on a cursor older than retention.

Approach

Separate the two defects in the current shape. Offset makes the database produce and discard N * per_page rows, so page cost grows linearly and a deep page degrades from milliseconds to seconds. Concurrent inserts also shift the window between requests, so a walker skips rows with no error raised anywhere, which for a customer sync is silent data loss.
Order by ingestion, not by occurrence. During a replay events arrive hours out of occurred_at order, so a consumer holding an occurred_at high-water mark can never see a late event that falls below it; (ingested_at, event_id) is the only ordering under which 'everything after my cursor' is a complete statement.
Page with a row comparison: where tenant_id = $1 and (ingested_at, event_id) > ($2, $3) order by ingested_at, event_id limit $4, backed by an index on (tenant_id, ingested_at, event_id). That seeks directly to the resume point, so every page costs the same regardless of depth.
Trail the head of the table. With ingested_at defaulting to now(), which is transaction start time, a long insert transaction receives an earlier timestamp and becomes visible after a reader has already passed it. Cap each page at ingested_at <= now() - delta, with delta larger than the longest write transaction as bounded by statement_timeout and idle_in_transaction_session_timeout, or the export skips exactly the rows written under load.
Make the cursor opaque and self-describing: base64 of the timestamp, the event id and a fingerprint of the filters, rejected when the filters differ from the current request. Return 410 with a cursor_expired code once the cursor's partition has been dropped, so the client restarts from a known time instead of resuming into a hole.
Keep a page a pure GET with no server-side consumption, so a timeout is resolved by retrying the identical cursor.

Worked solution (30 min)

Construct the failing case on paper: an event with occurred_at at 09:00 ingested at 14:00, and a consumer that read up to 10:00 at 11:00.
Write the keyset query with the row comparison and the exact index it needs, then say which column of the index each predicate uses.
Add the trailing-head predicate and pick delta from a named timeout setting rather than a round number.
Define the cursor's encoded contents and the two error cases: filter mismatch and expired partition, with their status codes.
Write the client's algorithm in four lines: request, persist cursor after processing the page, retry the same cursor on timeout, restart from a time on 410.

Expected result: Keyset pagination over (ingested_at, event_id) with an index on (tenant_id, ingested_at, event_id), a trailing head predicate sized above the longest write transaction, an opaque filter-bound cursor, and 410 for a cursor past retention.

Verify it yourself

A late event with an old occurred_at is still delivered by the export.
Page cost is independent of how many pages have already been read.
The cursor persists across a client restart and survives retrying the same page.
The tie-break column is unique, so no row can be skipped or repeated at a page boundary.

Follow-up

The customer asks for a total count alongside the first page. What do you offer instead, and why is an exact count both expensive here and wrong by the time it is read?
How would you let a customer re-read a window they have already consumed without giving up the forward-only cursor?

One tenant's counter writes stall the whole connection pool

Medium · lock contention · connection pool · multi-tenancy

A change that made a per-tenant usage counter correct now produces site-wide latency whenever one large tenant writes: unrelated endpoints time out waiting for a connection while database CPU stays low and no statement is slow. The change wraps the counter update in a transaction that takes SELECT ... FOR UPDATE on one row, calls an external pricing service, then updates and commits. Give an ordered checklist, the arithmetic that bounds that tenant's write rate, and three repairs with the cost each one accepts.

Approach

Separate waiting from working. Low database CPU alongside high application latency points at a queue, so instrument connection-acquisition wait separately from query execution time; that queue forms in the application and is invisible in database metrics, which is why the database looks healthy throughout.
Confirm the lock rather than assuming it: sample waiting sessions and group by wait event, relation and tuple. Contention concentrated on one tuple belonging to one tenant is the signature; a deadlock would instead show the database aborting transactions after its detection timeout, which is not happening here.
Do the arithmetic out loud. Throughput on a serialised row is one divided by the lock hold time, and the hold spans the external call, so a 20 ms pricing call caps that tenant near 50 writes per second no matter how many pods run. Every waiter also holds a pooled connection while it queues, so the shared pool drains and unrelated tenants fail at acquisition.
Repair one: shrink the critical section to a single statement with the price resolved before the transaction opens. Cost is a stale price for the duration of one request and a second round trip; benefit is a hold time measured in the database's own execution time.
Repairs two and three change where the contention lives rather than how long it is held. Sharding the counter into per-(tenant, bucket) rows and summing on read multiplies write throughput by the shard count, at the cost of an aggregate on every read and a shard count you must size against the largest tenant rather than the median. Accumulating in memory and flushing periodically removes the per-write round trip entirely, paid for with a bounded loss window on crash, which is acceptable for a rate limiter and not for a billing counter.
Contain independently of which repair wins: a separate pool or per-tenant concurrency cap for this write class, a statement timeout low enough that a pathological query dies before it accumulates waiters, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks.

Follow-up

What would a genuine deadlock look like here, which two code paths would produce one, and how does the database's response differ from what you observed?
If a transaction-pooling proxy sits in front of the database, which of your three repairs changes behaviour, and what stops working that would have worked on a direct connection?
The counter also enforces a quota. Why is SELECT the count and then INSERT still wrong after you have fixed the contention?

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.

Review and mentoring: what you let through

Counting review comments or mentees proves nothing. The useful version is a specific change you approved with a reservation you stated, or one you blocked and the delay that cost. Say which standard you were holding and why it was worth the friction. A mentoring story needs the thing the other person can now do without you.

Describe a time you had to make a difficult technical decision. What was your reasoning?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Close with what you would do differently, concretely.
Pick a story where you made the decision, not one where you watched it.

Follow-up

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

How do you handle ambiguity when requirements are not fully defined?

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
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.

Follow-up

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

What do you look for in a team and how do you contribute to a positive engineering culture?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Pick a story where you made the decision, not one where you watched it.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

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