Interview guides · ArianeGroup
ArianeGroup Software Engineer Interview Questions & Guide 2026
Testing practice for ArianeGroup software engineer candidates: the boundaries worth covering, and why flakiness is a real defect.
The 60-second brief
As a Software Engineer at ArianeGroup, you are at the heart of Europe’s sovereign access to space. This role is not merely about writing code; it is about engineering robust, high-reliability software solutions that support the design, development, and operation of complex aerospace systems, including the Ariane 6 launcher. Your work directly impacts the safety, precision, and mission success of cutting-edge space technology.
The title spans product, platform and infrastructure work, and which of those the seat actually is decides whether design or algorithms carries more weight in your preparation. The posting rarely settles it; what the team is on call for usually does.
Focus areas
- Bound every outbound call with a timeout
- Paginate large result sets with keyset cursors
- Choose indexes from the query's access path
Overview
You will contribute to critical infrastructure, ranging from human-machine interfaces (HMI) to complex simulation and data processing environments. This position requires a unique blend of technical rigor and industrial awareness. Whether you are working on ground systems or embedded software, your contributions help ensure that the software powering ArianeGroup's products meets the extreme standards of the aerospace industry.
This is a high-stakes, intellectually stimulating environment where attention to detail is paramount. You will collaborate with multidisciplinary teams of engineers, physicists, and project managers. Success in this role requires not just proficiency in software development, but a genuine passion for the space sector and an ability to navigate the complex, regulated landscape of a major aerospace leader.
The ArianeGroup interview process
ArianeGroup candidates report 4 rounds · ≈ 3-5 weeks. The stages below are what candidates describe, not a published process.
Initial Screening
Before anything technical happens, someone has to decide which rung of the ladder your loop is calibrated to, and that decision sets the bar for every round after it. It comes from how you describe scope, not from your title, because titles do not convert cleanly between companies. The weak version of the answer is team size and years. The strong version names the largest change you shipped where nobody reviewed the design, what would have broken if you had been wrong, and what you were paged for. Get the level said out loud on this call, because the range and the loop both follow from it.
What to demonstrate
- Whether the scope in your own account maps onto a level the team actually has an opening at, so a mismatch ends the process cheaply rather than after four interviewers have spent a day
- Whether your title needs re-mapping: the same word describes very different amounts of independent decision-making at a twenty-person company and a ten-thousand-person one
- Whether your compensation expectation can be filled at that level in the structure the role pays in, which is why the number gets asked for before any engineer is scheduled
How to prepare
- Write down two changes from the last two years: the largest one you designed with nobody reviewing the design, and the largest one where someone more senior did. Lead with the first when scope comes up, and be ready to say which parts of the second were yours
- Ask which level the loop is calibrated to and what changes at the level above it, then plan your weeks from that answer rather than from the posting
- Settle a total-compensation range beforehand with the split named, base against bonus against equity and its vesting period, so a question about numbers gets a number instead of the word market
In-Depth Interviews
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
Technical Evaluation
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
Cultural Fit Assessment
Input bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.
What to demonstrate
- Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
- Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
- Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
- Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply
How to prepare
- For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
- For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
- Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last
Where candidates lose points
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.
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.
Saying 'eventually consistent' without naming the anomaly a user would see
Describe the concrete symptom you are choosing to accept: the author reloads and their own comment is missing for two seconds, or two devices show different balances for a minute. The class of consistency model is a technical label; the tolerable anomaly is the actual product decision.
Sharing mutable state with no stated owner
Say which thread, request or task owns each mutable structure, and what protects it when the answer is more than one: a lock, a queue that hands ownership across, or an immutable copy per reader. A structure documented as safe for concurrent reads is usually not safe for a concurrent write alongside those reads.
Questions & practice
Find overlapping job attempts and peak concurrency from lease records
Medium · sweep line · intervals · leases
A day of job_run history yields about 50,000,000 attempt records: (job_run_id, job_type, attempt, started_at, finished_at which is NULL when the worker died, lease_expires_at). Leases expire on a clock, so a job that outran its lease ran twice. Produce (a) every job_run_id whose attempts overlapped in wall-clock time and (b) the peak number of simultaneously running attempts per job_type with the minute it occurred. Target O(n log n). State how you treat a NULL finished_at and what clock skew does to your answer.
Approach
- Define the interval before sorting anything: an attempt occupies [started_at, COALESCE(finished_at, lease_expires_at)). finished_at is observed and lease_expires_at is only a promise, so every attempt without a finish contributes an estimate and the whole result is a lower bound on overlap rather than an exact count.
- For peak concurrency, sweep: emit 2n endpoints, sort by (timestamp, kind) with ends ordered before starts at equal timestamps, then walk the sequence maintaining a counter per job_type and record each type's maximum with its timestamp. O(n log n) dominated by the sort, O(n) space, or O(1) extra if the sort is external and the walk streams.
- For overlap detection, do not compare attempts pairwise. A single global sort by (job_run_id, started_at) gives both the grouping and the order; within a group, keep the maximum end seen so far and report an overlap exactly when the next start is less than that running maximum, which is one linear pass after the sort.
- Half-open intervals matter and are easy to get wrong: with closed intervals an attempt ending at the same millisecond another begins reads as concurrency two, and across 50,000,000 records that artefact swamps the real signal.
- State the clock caveat: started_at and finished_at are written by different workers, so under skew of a few hundred milliseconds an apparent overlap shorter than that bound is not evidence. Filter reported overlaps by a minimum duration, or prefer timestamps written by whichever component heartbeats the lease.
- Scale the sort rather than assuming it fits: the sweep emits two endpoints per attempt, so 50,000,000 records become 100,000,000 endpoints, and at roughly 24 bytes each, an 8-byte timestamp plus a 4-byte job_type plus a kind flag padded to alignment, that is about 2.4 GB of sort keys before any scratch space. Either push the ordering into the database behind an index on (job_type, started_at) or run an external merge sort in chunks; the overlap pass sorts n records rather than 2n, so it is the cheaper of the two.
Worked solution (30 min)
- Write the interval derivation with the COALESCE and state in one line which of the two end sources is observed and which is assumed.
- Write the concurrency sweep: the endpoint tuples, the sort key including the end-before-start tie-break, and the per-job_type counter.
- Hand-trace four attempts of one job, two disjoint and two overlapping by three seconds, and confirm the overlap detector fires exactly once.
- Add the skew filter as a minimum overlap duration, state the value you chose, and justify it from how the timestamps are written.
Expected result: One global sort by (job_run_id, started_at) driving a running-maximum-end check for overlaps, plus a sweep over 2n endpoints with ends ordered before starts for per-job_type peak concurrency, both O(n log n), with half-open intervals, NULL finished_at falling back to lease_expires_at, and overlaps shorter than the clock-skew bound excluded.
Verify it yourself
- Two attempts touching at exactly the same instant report concurrency 1, not 2.
- A job with a single attempt never appears in the overlap output regardless of its lease.
- The reported peak timestamp for a job_type lies inside at least one of that type's intervals.
Follow-up
- A handler is not idempotent and you have found 400 overlapping jobs. Which of them actually caused damage, and what would you query to find out?
- Peak concurrency for one job_type is 4 against a configured cap of 4. Is the cap working, or is the data hiding attempts that never started?
- How would you compute both answers incrementally as records arrive rather than in a daily batch?
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?
Canonicalise a request body into a stable idempotency fingerprint
Medium · parsing · canonicalisation · hashing
idempotency_key.request_fingerprint is a SHA-256 over the method, path and canonicalised body, and a retry whose fingerprint differs must be rejected with 422 rather than served the stored response. Write the canonicaliser. Bodies are JSON up to 256 KB nested at most 32 levels; clients vary key order, whitespace and unicode escaping, and some send 64-bit ids as JSON numbers. Produce a deterministic byte string such that semantically identical bodies match and any semantic difference does not. State your complexity and name two normalisations you refuse to perform.
Approach
- Parse once into a tree, then re-serialise under fixed rules: object keys sorted, array order preserved, one escaping convention, no insignificant whitespace. Parsing is O(n) and sorting keys is O(k log k) per object, so O(n log n) overall with O(depth) stack, and the 32-level cap is enforced during parsing because hostile nesting is how a canonicaliser becomes a stack overflow.
- Sort keys by their UTF-8 bytes and say why the obvious implementation is wrong in some runtimes: a default string comparison that orders by UTF-16 code units places surrogate pairs, meaning code points from U+10000 up, below U+E000 to U+FFFF, which is not UTF-8 byte order, so two services written in different languages disagree on the same document.
- Do not re-encode numbers through a double. IEEE-754 binary64 represents integers exactly only up to 2^53, so normalising a 19-digit id through a float changes it, and 1 against 1.0 cannot be reconciled without deciding whether they are the same value. Preserve the literal token, and require ids as strings at the API boundary if you want them comparable.
- Reject duplicate keys rather than picking one. JSON permits them and parsers disagree, most keeping the last, so any choice you make ties the fingerprint to a parser detail that the code handling the request does not necessarily share.
- Frame the hash preimage so concatenation cannot collide: delimit or length-prefix the method, path and body, otherwise one request's fields can be rearranged into another request with the same byte stream and the same fingerprint.
- Name the refusals and their consequence: no case folding, no dropping of null-valued keys, no Unicode normalisation. Each makes two different requests fingerprint alike, and the resulting failure is the worst one this table has, since the second request is answered with the first one's stored response and its effect never happens.
Follow-up
- A client sends the same logical request with an extra field your API ignores. Same key, different fingerprint, so you return 422. Is that the right answer?
- Where does the fingerprint get computed relative to request decompression and the body-size limit?
- The endpoint takes 1,000 requests per second with 256 KB bodies. What does hashing cost, and does it belong at the edge or in the core service?
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.
Worked solution (25 min)
- Load 2M resource rows across 5k owners in one tenant, run the query under EXPLAIN (ANALYZE, BUFFERS), and record the node reading the most rows plus the Sort Method line.
- Create (tenant_id, owner_user_id, created_at DESC) concurrently and re-run, confirming the Sort node is gone and actual rows fall to about 20.
- Run the two negative cases and capture the plan for each.
- Re-run the original tenant listing query to confirm the new index has not displaced the index that query depends on.
Expected result: With (tenant_id, owner_user_id, created_at DESC) the plan is an Index Scan with no Sort, roughly 20 rows and a handful of buffers, sub-millisecond. owner_user_id alone still scans; ORDER BY updated_at seeks on the pair and then sorts that owner's rows.
Verify it yourself
- Actual rows at the leaf node is on the order of the LIMIT, not the tenant's row count.
- No Sort node appears above the index scan.
- Buffers read stays constant as one owner's row count grows; if it grows with it, the ORDER BY is not being satisfied by the index.
- pg_stat_user_indexes shows idx_scan climbing on the new index and not on the listing index for this endpoint.
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?
Hold a per-tenant active cap against concurrent creates
Hard · write skew · isolation levels · constraints
A tenant on the standard plan may hold at most 50 resources with status='active'. The create handler runs SELECT count(*) FROM resource WHERE tenant_id = $1 AND status = 'active', compares to 50, then inserts. Two creates arrive 3 ms apart on different instances and the tenant lands at 51. Name the anomaly, say whether PostgreSQL 16 READ COMMITTED or REPEATABLE READ prevents it and why, then give an implementation that holds the cap at READ COMMITTED with the exact statements. Finally, say what changes when the cap is 'at most one running export per tenant' on job_run.
Approach
- Name it: write skew. The two transactions read an overlapping set and write disjoint rows, so there is no row-level conflict for the engine to detect and each commit is individually legal.
- Rule out the levels precisely. READ COMMITTED takes a fresh snapshot per statement and takes no lock on the counted rows, so both see 49. PostgreSQL's REPEATABLE READ is snapshot isolation: it removes non-repeatable reads and phantoms within the snapshot but still admits write skew, because the anomaly is not a re-read of a changed row, it is a read of a set that a concurrent transaction invalidates. Only SERIALIZABLE closes it, by tracking the read dependency and aborting one transaction with SQLSTATE 40001 — a guarantee that exists only if the application re-runs the whole transaction from the read.
- Convert the set predicate into a single-row conflict: keep tenant.active_resource_count and run UPDATE tenant SET active_resource_count = active_resource_count + 1 WHERE tenant_id = $1 AND active_resource_count < 50 in the same transaction as the INSERT. Zero affected rows is the cap, returned as 409. The row lock serialises the decision at any isolation level, and contention is bounded to one tenant's row — which is also the fair-scheduling unit, unlike a global counter that would convoy every tenant behind one row.
- State the cost you just took on: a counter is a second source of truth that can drift, so every path that changes status must adjust it inside the same transaction, and a periodic reconciliation has to exist, with resource_revision as the authority for what the count should have been.
- For the job case the invariant is expressible per row, so let the database hold it: a partial unique index on job_run (tenant_id, job_type) WHERE status IN ('queued','running') makes a second running export unwritable and the loser takes 23505, mapped to 409. That is strictly better than a counter — no drift, no reconciliation — and it is available only because the cap is one rather than fifty.
- Add the retry discipline each route demands: under SERIALIZABLE both 40001 and deadlock 40P01 are retryable and the retry must re-execute the read, while under READ COMMITTED with the counter nothing retries, because the conflict is reported to the caller rather than raised as an error.
Follow-up
- A resource moves from archived back to active. Which statements change, and what breaks if the counter update and the status change land in different transactions?
- The cap becomes plan-dependent and a plan can change mid-month. Where does the number 50 live, and who reads it?
- How do you detect after the fact that the counter drifted, without locking the table?
What is your position regarding the military activities of the company?
Medium
Approach
- Work from the requirement backwards to the design.
- 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?
Can you explain the TRL (Technology Readiness Level) scale?
Medium
Approach
- Say what you would check first and why it is the highest-information step.
- Clarify what is being asked and what a complete answer contains.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Convert the listing endpoint from offset pages to stable cursors
Medium · pagination · cursors · index access path
GET /v1/resources returns a tenant's resources newest-updated first, today with page and per_page, backed by index (tenant_id, status, updated_at DESC, resource_id DESC). Callers are a browser feed and a nightly sync job that walks every page. Users report items appearing twice or vanishing between pages, and page 400 is slow. Design the cursor contract: what the cursor contains and how it is encoded, the exact WHERE and ORDER BY, what happens when a row's updated_at changes mid-walk, how a client detects the end, and what the sync job does when a cursor is rejected.
Approach
- Separate the two defects, because they have different fixes. Cost: OFFSET n makes the engine produce and discard n rows, so price grows with page depth rather than page size and page 400 pays for 400 pages of work. Correctness: while the set shifts, rows cross the offset boundary and are skipped or repeated, and nothing in the response lets the client detect it.
- Write the seek: WHERE tenant_id = $1 AND status = $2 AND (updated_at, resource_id) < ($k, $id) ORDER BY updated_at DESC, resource_id DESC LIMIT n. The tie-break is not decoration - updated_at is not unique, and two rows sharing a timestamp across a page boundary reintroduce exactly the skip this was adopted to remove. PostgreSQL seeks the composite index on the row-value comparison directly; on an engine that does not optimise a row constructor, expand it into the equivalent OR form or the plan quietly degrades to a scan.
- Encode the cursor as an opaque token carrying the sort key, the id, and a fingerprint of the filter and sort order, signed or at minimum validated. A cursor replayed against a different sort or filter must be a 400 with its own code, not a silently wrong page - the sync job cannot notice the difference otherwise.
- State the guarantee precisely rather than generously. Keyset is stable against inserts and deletes elsewhere in the set, because the position is a value and not a count. It is not a snapshot: updated_at is mutable, so a row that is edited during the walk re-sorts and may be seen twice or not at all. If the sync job needs exactly-once coverage, order on an immutable key such as (created_at, resource_id), or walk resource_revision by revision_id and treat updated_at as data.
- Define termination and limits in the response, not in the client's inference: fetch n+1 rows, return n, and emit next_cursor only when the extra row existed. Absence of next_cursor is the sole end signal, because a page can legitimately come back short when rows are filtered after retrieval. Cap n and document the cap rather than honouring per_page=10000.
- Migrate without a flag day: keep page and per_page working, add the cursor, count usage per credential, and remove the offset path only once the sync job's traffic on it is zero.
Worked solution (25 min)
- Write the old and new queries side by side and state the rows examined for page 400 under each.
- Define the cursor payload field by field, including the filter and sort fingerprint, and choose the encoding.
- Write the end-of-results rule and the cap, then the 400 response for a cursor that does not match the current query.
- Construct the mid-walk edit case: a row updated between page two and page three, and say exactly what the client sees.
- Write the deprecation plan for page and per_page, including the signal that says removal is safe.
Expected result: A keyset query that seeks the existing composite index with no sort step, an opaque cursor bound to its filter and sort order, a single documented end condition, and an explicit statement that mutable sort keys make the walk non-snapshot.
Verify it yourself
- The plan for a page fetch shows an index seek and no sort, and reads about n rows regardless of depth.
- Two rows sharing an updated_at value that straddle a boundary are both returned exactly once.
- A cursor issued under one filter and replayed under another is rejected rather than served.
Follow-up
- The client wants a total count and the ability to jump to page 400. What can you honestly offer instead, and what does each option cost?
- How do you paginate backwards, and what does that require of the index?
Listing latency scales with page size, not with filters
Easy · n+1 · latency · query batching
The tenant listing endpoint reads resource filtered by tenant_id and status, ordered by updated_at DESC, and returns each row plus the owner's display name from app_user and the actor of that resource's latest resource_revision. p99 is 55 ms at 10 rows per page and 1.4 s at 200. Database telemetry shows 401 statements per request, each under 1 ms, and nothing in the slow-query log. Diagnose the cause and give the fix, stating the statement count per request and the p99 you expect afterwards.
Approach
- Read the counters before forming a theory. 401 statements for 200 rows is one driver query plus two per row, and sub-millisecond execution with an empty slow-query log rules out a bad plan. The time is round trips, which is why it is invisible in every per-query metric and scales with rows returned rather than with filter selectivity.
- Name the two per-row statements from their normalised text: a single-row app_user lookup by user_id, and a resource_revision lookup by resource_id ordered by version DESC LIMIT 1. Confirm by dropping those two response fields and watching the statement count fall to one. That locates the calls in the serialisation layer, not the repository.
- Check that the arithmetic accounts for the whole gap. Measure one round trip to the replica in isolation; 400 trips at roughly 3 ms of network plus 0.2 ms of execution is about 1.3 s on top of a 55 ms baseline, which matches. If the multiplication had fallen short, the N+1 would only be part of the story and you would keep looking.
- Batch both lookups. Collect owner_user_ids and resource_ids from the driver query, then issue WHERE tenant_id = $1 AND user_id = ANY($2) for the users, and PostgreSQL's SELECT DISTINCT ON (resource_id) ... WHERE resource_id = ANY($2) ORDER BY resource_id, version DESC for the latest revision, which the UNIQUE (resource_id, version) index serves directly. On an engine without DISTINCT ON, use a lateral join or a row_number window. Three statements per request at any page size.
- Keep the tenant predicate in the batched query. The per-row version was implicitly scoped because its ids came from tenant-scoped rows; a batched user_id = ANY(...) with no tenant_id is an unscoped read that behaves correctly only as long as the id list is trustworthy.
- Re-measure at 10, 50 and 200 rows and confirm the statement count is constant. Latency should now track bytes returned.
Follow-up
- The page size is capped at 200 today. What breaks first if it is raised to 2,000, and is it still this bug?
- How do you stop the next N+1 from reaching production, given that no individual query is slow and the endpoint's tests pass?
- The latest-revision actor is only used to render an avatar. Make the case for denormalising it onto resource, and name the write anomaly that introduces.
Suggested practice checklist
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.
Know where your numbers came from
Bring the two or three numbers the story rests on and know how they were collected. A p99 whose timer starts inside your handler excludes the time a request spent queued, so it can sit flat while users wait longer. Give the window, the percentile and what the measurement left out, or drop the number.
Parlez-moi d'une situation où vous avez rencontré un problème.
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?
Comment gérez-vous une situation de conflit?
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.
- 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?
Quels sont vos points forts et vos axes d'amélioration?
Medium · behavioural and engineering judgement
Approach
- Name the disagreement and how you resolved it with evidence.
- Give the blast radius: what could have broken, and what you measured.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
Can you describe your experience with virtual reality or specific simulation tools?
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.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?