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ACONEXT Holding Software Engineer Interview Questions & Guide 2026

System design practice for ACONEXT Holding software engineer candidates: reach a data model and an API contract, not a box diagram.

The 60-second brief

As a Software Engineer at ACONEXT Holding, you are at the intersection of high-stakes engineering and complex industrial application. ACONEXT Holding operates as a critical partner in sectors ranging from automotive and robotics to defense and software-defined radio. Your role is not merely to write code, but to engineer robust, scalable, and precise solutions that power mission-critical systems.

Browser-facing seats are not covered by algorithm practice. Scope in state ownership, what the page does on a slow or failed request, and how you would diagnose something that renders correctly but feels slow.

Focus areas

Bound every outbound call with a timeout
Detect concurrent edits instead of losing writes
Choose indexes from the query's access path

Overview

Whether you are working on Embedded Systems, PLM Solutions, or Defence Systems, your contributions directly impact the reliability and performance of advanced technological products. You will join a team that values technical depth, systematic thinking, and the ability to bridge the gap between abstract requirements and physical reality. This role is ideal for engineers who thrive on technical rigor and seek to solve challenges that have real-world, high-consequence implications.

The ACONEXT Holding interview process

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

Technical Background Screening

What this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.

What to demonstrate

Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly

How to prepare

Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live

In-Depth Technical Discussions

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

Team Member Interaction

When a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.

What to demonstrate

Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened

How to prepare

Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure

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.

Assuming an isolation level prevents the anomaly you actually have

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

Check-then-act on shared state

Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.

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

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.

Worked solution (20 min)

Define the bucket struct and the advance step: take floor(finished_at_ms / 1000), compare with the ring's current second, zero min(delta, 60) buckets forward, then write into the new head.
Trace a destination that receives 5 attempts, goes silent for 90 seconds, then receives one more, and confirm the rate is computed from one attempt rather than six.
Compute total memory for 40,000 destinations at 60 buckets of two 4-byte counters, and state what changes if the window widens to 300 seconds.
Write the open rule as a single predicate combining the minimum-attempt floor with the rate threshold.

Expected result: A per-destination ring of 60 one-second (attempts, failures) buckets advanced lazily for amortised O(1) cost, roughly 480 bytes of counters per destination and about 20 to 25 MB for 40,000 of them, an LRU bound on live rings, and an open rule requiring both a minimum attempt count in the window and a rate above threshold.

Verify it yourself

After 60 seconds of silence a destination's window holds zero attempts and cannot open a circuit.
Memory measured with one hot destination and 39,999 idle ones is indistinguishable from the reverse, which the deque version would fail.
Replaying the same attempt stream with the clock shifted by an hour yields the same final rate.

Follow-up

The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
How would you make the window survive a process restart, and is it worth the cost?

Collapse a redelivered event batch into per-aggregate high-water marks

Easy · hashing · at-least-once · aggregation

You drain a batch of up to 5,000,000 events, each (aggregate_id BIGINT, aggregate_version INT, event_type, payload). The log guarantees order within one aggregate only; the batch merges 64 partitions, and a relay failover has redelivered a range, so an older version for an aggregate can appear after a newer one. Given a map of last_applied_version per aggregate, produce the events worth applying, at most one per (aggregate_id, version), plus the count discarded. Target O(n) time. State the memory for 2,000,000 distinct aggregates and what you do when it does not fit.

Approach

One pass, one hash map from aggregate_id to the highest version kept, and a discard counter. An event whose version is at or below last_applied_version for its aggregate is dropped without further work, which is the whole reason the event carries its version rather than a delta. O(n) expected time, O(d) space in distinct aggregates.
Keep the maximum, never the last occurrence. The redelivered range means the final appearance of an aggregate in the batch can be an older version than one seen earlier in the same batch, so last-wins applies stale state over newer state and the projection regresses with no error anywhere.
Cost the memory instead of calling it large: an 8-byte key plus a 4-byte version is 12 bytes of payload, and an open-addressed table held at a 0.7 load factor costs roughly 17 bytes per entry before per-slot metadata, so 2,000,000 aggregates is tens of megabytes in a native layout and several times that in a runtime that boxes both key and value.
If the distinct set exceeds memory, partition on hash(aggregate_id) mod P and reduce each partition independently. Every event for one aggregate hashes to the same partition, so the per-partition result is exact and the merge is concatenation rather than a second reduction.
Reject sorting the batch by (aggregate_id, version) as the default. It is O(n log n) and buys nothing, because max is associative and commutative and needs no ordering; sorting earns its cost only when the downstream consumer must receive the events in order rather than a per-aggregate winner.
Separate the two mechanisms out loud: in-batch deduplication does not make the consumer idempotent, because the same event redelivered tomorrow arrives in a different batch entirely. The projection write itself still has to be keyed on (aggregate_id, version).

Follow-up

The payload is a patch rather than a snapshot, so applying only the highest version loses the intermediate changes. What changes in your reduction?
How do you detect that version 7 arrived while version 6 was never delivered, and what should the consumer do about the gap?
Two events for one aggregate carry the same version with different payloads. Which one is wrong, and how would you find out?

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?

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

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.
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.
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.
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.
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.
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)

Write the three statements plus the rowcount branch and confirm they sit in one BEGIN/COMMIT with no outbound call between them.
Run two clients that both read version 7 and apply their updates 5 ms apart; assert one 200 and one 409.
Assert resource.version = 8, exactly one resource_revision row at version 8, and one outbox_event row at aggregate_version 8.
Repeat at REPEATABLE READ and record the different failure shape (SQLSTATE 40001) the handler must also map to 409.
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?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.

Approach

Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.
Answer the two negatives explicitly. owner_user_id alone is not a left prefix of the new index, so it degrades to a full scan of the index at best. Ordered by updated_at, the query still seeks on the (tenant, owner) pair but must sort, because only created_at is ordered within that pair.
Measure both sides with EXPLAIN (ANALYZE, BUFFERS) and compare estimated against actual rows at the lowest node — a 2M-versus-200 misestimate there is usually what chose the plan, and adding an index will not fix a statistics problem.

Follow-up

90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
A dashboard runs this for 40 owners in one page load. What changes about the design?
How do you roll this index out on a table taking 1.2k writes/second, and what does it cost on every insert from then on?

Explain your process for debugging a critical failure in a deployed production environment.

Medium

Approach

Work from the requirement backwards to the design.
Clarify what is being asked and what a complete answer contains.
State your assumptions explicitly before working the problem.

Follow-up

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

How do you manage memory and concurrency in high-performance embedded systems?

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.
State your assumptions explicitly before working the problem.

Follow-up

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

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

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.
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.
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.
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.
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.
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)

Write the request schema with the per-item client id, and the response schema with per-item status and a top-level failed count.
Write the atomicity decision and the sentence of justification that names the lock cost or the resend cost.
Define both key levels and trace a resend of a half-applied batch through them, item by item.
List the whole-request rejections with their status codes and limits.
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?

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

Roughly ninety minutes on weeknights with one longer weekend block. The plan cuts scope rather than compressing everything, on the assumption that one thing finished per night beats four half-started.

Day 1 — Fix the scope and take a cold baseline

Read the role description and write the three things the loop will almost certainly test, then write an explicit not-doing list and keep it visible all week.
Take one twenty-five-minute coding problem and one fifteen-minute design prompt cold, and write the single sentence naming what blocked each, because those two sentences decide where the remaining evenings go.
Set the week's rule: one thing finished every night, including the night you only have forty minutes.

Deliverable: A one-page scope with a not-doing list and two cold attempts, each carrying one sentence on what blocked it.

Day 2 — One pattern, written three times from blank

Choose the single pattern most likely to appear in your loop and write it three times from an empty file rather than editing the previous attempt.
On the third pass, write the invariant as a comment before the loop body and the complexity before the first line of code.
Stop at ninety minutes even if the third version is imperfect, and write the one thing you would fix given another hour.

Deliverable: Three independent implementations of the same pattern plus a note on what changed between them.

Day 3 — One design, only to the depth you can defend

Take one system shape and go only as far as requirements, interface and data model, refusing to draw a box you could not survive a follow-up about.
Attach one number to each non-functional requirement, deriving it rather than asserting it, and write the assumption the number rests on.
Write the one tradeoff you are choosing against and the observation that would make you reverse it.

Deliverable: One design at interface-and-schema depth with derived numbers and one written reversible tradeoff.

Day 4 — Only the fundamentals you will have to defend

Write, in under two hundred words each, the answers to the two questions that follow almost any implementation: why this structure and not the obvious alternative, and what happens to this code at a hundred times the input.
Write what an index actually costs: faster lookups on the indexed columns against a write that now maintains a second structure, plus the cases where the planner declines to use it anyway, low selectivity, or a predicate wrapping the column in a function.
Delete any answer you cannot deliver aloud in under a minute, since an answer that needs reading is not an answer you have.

Deliverable: Three written answers, each under two hundred words and each timed aloud.

Day 5 — Your own work, timed

Write a ninety-second and a four-minute version of your main project and time both aloud rather than reading them.
Prepare the two follow-ups that always come: what you would do differently, and how you knew it worked.
Put one number in the first sentence and be ready to say exactly where it came from and what it excludes.

Deliverable: Two timed narratives with one defensible number in the opening line.

Day 6 — The one full rehearsal, in the weekend block

Run a sixty-minute mock covering a coding round and a design round in one sitting with no break, because sustained attention is the thing evenings have not trained.
Immediately afterwards, and before hearing any feedback, write the three moments you lost the thread.
Spend the rest of the block only on those three moments, and on nothing you merely feel shaky about.

Deliverable: Mock notes naming three failure moments with a specific fix written under each.

Day 7 — Taper

Write the twenty-minute warm-up you will actually do on the morning: one problem you can already solve from a blank file, one design you can narrate, and nothing you have never seen.
Re-read only your own notes from this week and open no new material.
Write the logistics down: the editor or shared document you will be working in, whether execution and lookups are permitted, and the sentence you will use when you do not know something.

Deliverable: A one-page card holding the design structure, the project numbers, and the logistics.

Risky changes and the reversal path

For anything that touched live traffic, be ready to say how you would have undone it: a flag, a staged rollout, dual writes with the old path still authoritative. Once the old column is dropped or the source rows are overwritten there is no reverse, so name what you kept a copy of and for how long.

Tell me about a time you had to explain a complex technical issue to a non-technical stakeholder.

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

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

How do you handle disagreements regarding technical architecture within your team?

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.
Name the disagreement and how you resolved it with evidence.

Follow-up

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

How do you prioritize features when faced with conflicting technical requirements?

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Name the disagreement and how you resolved it with evidence.
Close with what you would do differently, concretely.

Follow-up

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

Describe your experience working in a regulated environment where safety and compliance were paramount.

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.
Close with what you would do differently, concretely.

Follow-up

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