Interview guides · AMP Limited

AMP Limited Software Engineer Interview Questions & Guide 2026

Timed coding practice for AMP Limited software engineer candidates: full attempts under a clock, then scored against a rubric.

The 60-second brief

As a Software Engineer at AMP Limited, you are joining a pivotal institution in the Australian financial services landscape. The work you perform is foundational, supporting critical systems that manage superannuation, investment, and financial planning for thousands of clients. Whether you are building robust backend infrastructure or crafting intuitive digital interfaces, your contributions directly influence the stability and scalability of AMP Limited’s product ecosystem.

Scope in the operational half of the job when the seat carries on-call. Rollout, rollback and what you would put on a dashboard belong inside a design answer rather than after it, and an answer that never reaches them reads as someone who has built systems but not run them.

Focus areas

Make every write idempotent under retry
Detect concurrent edits instead of losing writes
Paginate large result sets with keyset cursors

Overview

This role is designed for engineers who are passionate about hands-on technical craftsmanship. AMP Limited prioritizes individuals who want to be "in the code," solving complex engineering challenges rather than moving toward management. You will be part of a high-performing, professional environment that values technical depth, collaborative problem-solving, and a pragmatic approach to delivering reliable financial software solutions.

The organization maintains a strong hybrid work culture, particularly at their headquarters in the Quay Quarter Tower in Sydney, where collaboration is a core component of the team dynamic.

The AMP Limited interview process

AMP Limited candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Initial Conversation

An unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.

What to demonstrate

Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating

How to prepare

Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes

Technical Assessment

The same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.

What to demonstrate

In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes

How to prepare

Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first

Where candidates lose points

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.

Choosing an index from the columns a query mentions rather than from how it filters and orders

A composite B-tree index on (a, b, c) can be seeked only as a left prefix: equality on a, then equality on b, then a range or an ordering on c. A query that filters on b alone cannot seek into it at all and at best gets a full scan of the index; a query that filters a and ranges on b gets no benefit from c, because the index is only sorted by c within a fixed (a, b) pair. The practical consequence is that one index per column is close to useless for multi-predicate queries while a single correctly ordered composite index turns a scan into a lookup. The ordering half is what gets missed: if the index cannot satisfy the ORDER BY, the database must read every matching row and sort before the limit can apply, so a LIMIT 20 over a million matching rows still reads a million rows.

Answering a debugging question with a guess instead of a bisection

Give a procedure that halves the search space at each step: confirm the symptom reproduces, establish the last known-good version, input or timestamp, then bisect over commits, over the data, or over the layers of the request path. A plausible cause with no way to confirm it is the same move whether it happens to be right or wrong, which is why it scores nothing.

Finishing a solution without stating its complexity

Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.

Questions & practice

Identify the heaviest tenants in a five-minute window under memory pressure

Medium · top-k · heavy hitters · streaming

The edge service handles about 3,000 requests per second across roughly 50,000 tenants, peaking near 9,000. Expose the 50 heaviest tenants by request count over the trailing five minutes so limits can be tightened before one tenant's backfill starves the fleet. You may not retain five minutes of raw records. Give the exact solution and its memory, then the bounded-memory approximation with its error stated as a formula, and say which you would ship and at what tenant cardinality that choice changes.

Approach

Do the exact version first, because it is affordable at this cardinality: a ring of 300 one-second counters per tenant, advanced lazily, is 1,200 bytes of counters per tenant and roughly 60 to 90 MB for 50,000 tenants with overhead. Carry a running total and subtract the bucket you overwrite so a window read is O(1) rather than 300 adds.
Extract the top 50 with a size-k min-heap over the tenant sums: O(d log k) for d tenants, against O(d log d) to sort them all. Maintaining the heap continuously instead of on query requires a tenant-to-heap-index map, because incrementing a count already inside the heap means sifting from a known position, and without that map you rebuild the heap on every request.
State the approximation precisely rather than gesturing at sketches. Misra-Gries with m counters retains every item whose true count exceeds N/(m+1), and each retained count underestimates the truth by at most N/(m+1). With m = 1,000 and N = 900,000 requests in the window the error is roughly 900 requests, which is fine for spotting a tenant sending 50,000 and useless for ranking two tenants 200 apart.
Say what breaks when the window slides: Misra-Gries and Space-Saving are insert-only and cannot be decremented as records age out. The workable construction is one summary per sub-window, say ten seconds, with 30 summaries merged at query time, and the merged error is the sum of the per-summary errors, so the bound degrades linearly in the number of sub-windows.
Choose and defend it: at 50,000 tenants the exact rings cost under 100 MB in a process that already holds more, so ship exact. Keep the sketch for the case that actually motivates it, a per-principal or per-IP key where cardinality runs to millions and is not bounded by anything you control.
Raise the fleet problem before it is asked: each of 20 to 40 instances sees only its share, and the top 50 of one shard is not the top 50 of the fleet. Either aggregate counts centrally or accept that a per-instance threshold multiplied by instance count is the limit you are really enforcing.

Follow-up

The heaviest tenant is heavy because of one export job rather than user traffic. Should the limiter treat those as the same tenant?
Two tenants sit tied at the boundary of the top 50. Does your answer flap, and does the flapping matter?
You switch to per-principal keys and cardinality goes to 10 million. Walk through what changes.

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.

Approach

Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.

Follow-up

The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?

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

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.
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.
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.
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.
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.
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)

Write the heap comparator on (occurred_at, event_id) and the per-partition head refill.
Write the watermark computation and the emit-loop condition, then list which buffered events are held at a chosen instant.
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.
Add the idle-partition marker and trace the watermark with one silent partition, both with and without the marker.
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?

Replace offset paging on the resource feed with keyset

Easy · keyset pagination · composite index · sort order

resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.

Approach

Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.

Follow-up

The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
What does the cursor do when the row it points at has since been deleted?

Find version gaps and relay lag with window functions

Medium · window functions · event tables · frame semantics

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

Approach

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

Worked solution (30 min)

Write the three queries against seven days of data and confirm each returns without error.
In a scratch copy, delete one middle event for a single aggregate and confirm the gap query names that aggregate and the versions either side.
Run a running total over ungrouped rows ordered by date_trunc('second', created_at), once with the default frame and once with ROWS, and record where the two series diverge.
Compare the DISTINCT ON and row_number() plans on the same data and record rows-read for each.

Expected result: The gap query returns one row per hole with the version on each side; the daily query returns seven rows whose running total ends at the seven-day count; the newest-per-aggregate query returns exactly one row per aggregate_id present in the window. The RANGE running total is higher than the ROWS one at every tied second.

Verify it yourself

The per-day counts sum to COUNT(*) over the same predicate.
Re-running the gap query after reinserting the deleted event returns zero rows.
count(*) of the newest-per-aggregate result equals count(DISTINCT aggregate_id) over the same filter.

Follow-up

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

Describe your current tech stack and the architecture of the projects you have led.

Medium

Approach

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

Follow-up

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

Explain how you would approach a full-stack task involving both frontend and backend integration.

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?

Design the async export contract a client can resume safely

Hard · async operations · idempotency · leases

A tenant asks for a CSV of every resource. The work runs for minutes on the worker fleet through a job_run row carrying a lease, an attempt count and a dedupe_key, far past the edge's 400 ms budget. Callers are a browser that polls and a script that walks away and checks later. Specify what the submit call returns, the operation resource and its states, how a duplicate submit is handled, how a client learns about completion, what cancellation means given that a lease can expire mid-run, and how the result is fetched and when it expires.

Approach

Split the API in two. Submit returns 202 with an operation id and a location to poll, and never blocks on the work. The operation is a real resource with its own lifecycle - queued, running, succeeded, failed, cancelled - plus attempt, a monotonic progress figure, and a terminal error drawn from the same code taxonomy the synchronous endpoints use, so a client needs one error vocabulary rather than two.
Deduplicate at submit using job_run.dedupe_key, unique over (job_type, dedupe_key) while status is 'queued' or 'running': a repeat submit of the same logical export returns 200 with the existing operation instead of 202 with a new one, and the partial index deliberately permits a legitimate re-run once the first has finished. Pair it with the request's idempotency key so an HTTP-level retry of the submit is exact rather than merely similar.
Tell the poller how to poll: Retry-After on the polling response, a minimum interval enforced at the edge, and a documented maximum lifetime after which an operation is reaped. Polling is the contract of record; the webhook is the fast path, and both must lead to the same terminal state, so a client that receives the completion event and then polls anyway sees no contradiction.
Be exact about cancellation. A cancel request records intent; it cannot stop work already executing. The handler reads the flag at checkpoints, and because a lease expires on a clock that cannot distinguish a dead worker from a slow one, a second copy may start after the cancel was recorded - so the handler re-reads the flag immediately after claiming the lease. 'cancelled' becomes terminal only when no lease is outstanding; reporting it earlier shows a client a stopped job while a worker is still writing output.
Make the handler safe to run twice, because the lease guarantees that it will be. Write output to a deterministic object key derived from the operation id so a second copy overwrites its own work instead of appending a second file, and record completion with a conditional update that only the copy holding the current lease can win.
Treat result fetch as a separate authorised read: a short-lived signed URL, the tenant checked when it is issued rather than only when the file was produced, and a documented retention after which the operation remains terminal but the bytes are gone - a state the client must be able to tell apart from a failure.

Worked solution (40 min)

Write the submit request and its two possible responses, 202 for new and 200 for a duplicate, and the dedupe_key construction.
Draw the operation state machine, marking which transitions a client may observe and which are terminal.
Write the cancellation sequence across a lease expiry, showing where the second copy reads the flag.
Define the output key, the completion update's predicate, and why both are needed for a double run.
Specify result fetch: URL lifetime, authorisation point, retention, and the distinct response once the bytes are gone.

Expected result: A 202-plus-operation contract with deduplicated submits, a documented poll interval alongside an optional completion webhook, cancellation defined as recorded intent with a terminal state gated on lease ownership, an idempotent handler writing to a deterministic key, and a separately authorised expiring result.

Verify it yourself

Submitting the same export twice while the first is running produces one job_run row and one operation id.
Killing a worker mid-export and letting the lease expire yields one output file, not two, and one terminal transition.
A client that receives the completion webhook and then polls sees the same terminal state and the same result location.

Follow-up

An operation has said 'running' for 40 minutes and the worker is gone. What does the client see, and which columns in job_run decide that?
Two tenants each submit 50 exports at once. What in this contract stops one of them delaying the other?
The customer wants the export emailed instead. What changes, and what becomes harder to make exactly-once?

p99 jumped on one listing filter while p50 stayed flat

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

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

Approach

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

Follow-up

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

Suggested practice checklist

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

Day 1 — Coding, one pass at shallow depth

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

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

Day 2 — Design, one pass at shallow depth

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

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

Day 3 — Fundamentals and the practical rounds

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

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

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

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

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

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

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

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

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

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

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

Day 7 — Reassemble the loop

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

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

The estimate and the day you knew

A slipped date is only a bad story if you sat on it. What matters is what you believed when you gave the number, the signal that told you it was wrong, how many days passed before you said so, and what you cut rather than asking for more time. Scope you defended counts as much as scope you dropped.

How do you handle situations where a project is stalled or facing critical blockers?

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
Name the disagreement and how you resolved it with evidence.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

How do you communicate technical constraints to non-technical stakeholders?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
State the situation in two sentences and spend the rest on the reasoning.
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?
How did you know your change caused the improvement?

Tell us about a time you had to pivot your technical approach due to changing requirements.

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
Close with what you would do differently, concretely.
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?

Describe a time you faced a significant obstacle in a project; what specific steps did you take to move forward?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
State the situation in two sentences and spend the rest on the reasoning.
Pick a story where you made the decision, not one where you watched it.

Follow-up

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