As a Software Engineer at SBB (Swiss Federal Railways), you are at the heart of one of the most critical infrastructures in Switzerland. Your work directly influences how millions of passengers travel daily, ensuring that complex logistics, ticketing, and scheduling systems run with precision and reliability. Whether you are working on backend services, customer-facing applications, or internal operational tools, your code has a tangible, real-world impact on the efficiency of the Swiss transport network.
This role is both challenging and rewarding, requiring a balance of technical rigor and a deep understanding of the user experience. You will collaborate with diverse, cross-functional teams, including Product Owners, infrastructure specialists, and other engineers, to solve problems that are unique to the scale of national transport. Joining SBB means contributing to a mission-driven organization where stability, innovation, and public service intersect.
Initial Contact
reportedBecause the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.
What to demonstrate
- Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
- Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
- Whether you can be redirected in the middle of an answer without restarting it from the beginning
- Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it
How to prepare
- Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
- Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
- Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing
Technical Assessments
reportedWhat 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
Team Interviews
reportedWhen 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
PracHub editorial advice for the preparation topics above.
Shipping a migration and the code that depends on it as a single change
During any rolling deploy, and for as long as a rollback remains possible, old and new code execute against the same schema at the same time. A migration that drops or renames a column breaks every instance that has not restarted yet, and code that requires a column the migration has not applied breaks every instance that restarted early. The discipline is expand then contract: add the new column nullable, write both shapes, backfill in batches, move reads across once the backfill is verified, and only then stop writing the old shape and drop it - four deploys, usually spread over days. It feels disproportionate until the first rollback, at which point it is the only reason the previous version still runs.
Running a schema change as though the lock lasts as long as the statement
In PostgreSQL an ALTER TABLE that needs an ACCESS EXCLUSIVE lock must first wait for every open transaction touching that table, and while it waits, later queries needing a conflicting lock queue behind it rather than overtaking it. A DDL statement that would execute in milliseconds, issued while a thirty-second analytics query is open, therefore stalls all traffic on that table for thirty seconds: the outage length is set by the longest open transaction, not by the change. The defences are specific and worth knowing by name - set lock_timeout low and retry rather than queue, add columns without a volatile default so no table rewrite occurs (from version 11 a non-volatile default is a metadata-only change), build indexes with CREATE INDEX CONCURRENTLY while accepting that it cannot run inside a transaction block and leaves an invalid index behind if it fails, and add constraints as NOT VALID followed by a separate VALIDATE CONSTRAINT, which takes a weaker lock.
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.
Assuming the bug is in the framework
Suspect your own code first: read the stack trace top to bottom, check which versions are actually installed rather than which ones you believe are, and reproduce in isolation before blaming a library that thousands of people run daily. When the fault really is upstream, you need that minimal reproduction to say so credibly anyway.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Identify the heaviest tenants in a five-minute window under memory pressure
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.
Canonicalise a request body into a stable idempotency fingerprint
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.
Worked solution 25 min
- Write the serialiser: recursive emit with a depth counter, objects sorted by UTF-8 key bytes, arrays in order, strings escaped by one fixed rule, numbers emitted as their original token.
- Run it over three bodies: the same object with keys reordered, the same object with \u0041 written as A, and one with a nested array reversed. The first two must produce identical bytes and the third must not.
- Take the id 9007199254740993, round-trip it through a double, show it returns as 9007199254740992, then state the rule that prevents this.
- Define the hash preimage explicitly with its delimiters, and construct a pair of (path, body) inputs that would collide without them.
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?
Track a rolling failure rate per destination for circuit decisions
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.
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?
Denormalise tenant onto revisions and backfill it live
resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.
Approach
- Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
- Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
- Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
- Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.
- Step three, backfill: batch by primary key rather than by created_at so the cursor is dense and resumable — UPDATE resource_revision rr SET tenant_id = r.tenant_id FROM resource r WHERE r.resource_id = rr.resource_id AND rr.revision_id > $1 AND rr.revision_id <= $1 + 5000 AND rr.tenant_id IS NULL — committing per batch and persisting the cursor. Throttle on replica replay lag and on dead-tuple count, since each batch writes 5,000 new row versions. Run the backfill before the index exists so those updates can stay HOT.
- Step four, index then enforce then contract: CREATE INDEX CONCURRENTLY (cannot run inside a transaction block, scans the table twice, waits on open transactions, and leaves an INVALID index to drop concurrently if it fails); ADD CONSTRAINT ... CHECK (tenant_id IS NOT NULL) NOT VALID, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE, after which SET NOT NULL uses the validated check instead of re-scanning on PostgreSQL 12 and later. Only then move the audit reads onto the column and, in a later deploy, delete the join path.
Worked solution 40 min
- Write the five steps as separate scripts and state, for each, the lock mode it acquires and the deploy it pairs with.
- On a 20M-row copy, run the ADD COLUMN while a 30-second transaction holds a lock on the table, and record how long unrelated queries queue behind it.
- Run the batched backfill at 5,000 rows, kill it mid-run, restart from the persisted cursor, and confirm no row is processed twice and none is skipped.
- Build the index concurrently under concurrent write load, then add the CHECK ... NOT VALID, VALIDATE it and SET NOT NULL, timing each.
- Compare the audit-feed plan before and after: join-and-filter versus an index seek with no Sort.
Follow-up
- The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
- How do you verify the backfill actually finished, given rows are still being inserted while it runs?
- A resource must now be movable between tenants. What does that do to the composite foreign key and to the revisions already written?
Find version gaps and relay lag with window functions
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.
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?
Questions regarding specific technologies mentioned in your CV.
Questions regarding specific technologies mentioned in your CV.
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?
In your own words, describe the role you are applying for and what you…
In your own words, describe the role you are applying for and what you would do in this position.
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
- What assumption would you test first?
- How would you know your answer was wrong?
Make the resource write endpoint safe for retrying clients
The edge API accepts POST /v1/resources at 1.2k writes/second peak under a 400 ms p99 budget, and clients retry on timeout. idempotency_key has PRIMARY KEY (tenant_id, idempotency_key) plus request_fingerprint CHAR(64), state in ('in_flight','succeeded','failed'), response_status, response_body, resource_id, locked_until and expires_at. Specify the exact sequence the handler runs: what the second request does when it arrives while the first is still executing, what a retry carrying a different body receives, what happens when a process dies mid-request, and how the table is kept from growing without bound.
Approach
- Start from the race rather than the happy path. Insert the key row first, in its own short transaction, with state 'in_flight' and locked_until set to now() plus the request deadline. The unique constraint on (tenant_id, idempotency_key) is the arbitration mechanism: exactly one caller commits the insert, and every other caller takes a unique violation (SQLSTATE 23505) and becomes a follower. Reading the table and then inserting is a check-then-act race that both callers pass, and they pass it most often under the load that generates the retries.
- Give the follower a branch for every state it can observe, with no fall-through to doing the work: fingerprint mismatch means the key is being reused for a different request, so reject with 422 and never replay the stored response; 'succeeded' replays response_status and response_body verbatim; 'failed' allows a fresh attempt; 'in_flight' with locked_until in the future is answered with 409 and a retry hint, or a short bounded wait, because the only alternative is executing the effect twice.
- Commit the effect and the record of the effect together. The resource insert, its resource_revision row, its outbox_event row and the transition of the key row to 'succeeded' all happen in one transaction, so no crash can leave the work done and the key still 'in_flight'. The earlier insert is deliberately a separate transaction: it has to be visible to a concurrent caller before the work starts, which an uncommitted row is not.
- Close the wedged-key path. A process that dies after the insert leaves a row that would otherwise block that key forever, which is what locked_until exists for. Reclaim with a single conditional statement, UPDATE ... SET state='in_flight', locked_until=now()+interval WHERE ... AND state='in_flight' AND locked_until < now(), and treat a zero rowcount as losing the reclaim. Two statements that read and then update reintroduce the original race one layer down.
- Size the retention before it becomes an incident. At the 1.2k/second peak a 24-hour window is on the order of 100 million rows, and the table grows with traffic rather than with data, so the sweep is load-bearing. Batched deletes driven by an index on expires_at keep it bounded. Range-partitioning by day and dropping whole partitions is cheaper, but PostgreSQL requires every partition-key column to appear in the primary key, so the key becomes (tenant_id, idempotency_key, created_at) and a retry that straddles the boundary no longer collides with its original. State that cost rather than meeting it later.
Worked solution 20 min
- Draw a timeline with the retry arriving 150 ms into a 900 ms first attempt, and mark the instant each row becomes visible to the other transaction.
- Write the insert-first statement, then a five-row branch table for the follower: fingerprint mismatch, 'succeeded', 'failed', live 'in_flight', expired 'in_flight' - one action each.
- Write the single conditional UPDATE that reclaims an expired in_flight row, and argue why two concurrent reclaimers cannot both see a non-zero rowcount.
- Compute rows per day at peak, name the index the sweep uses, and check the sweep rate against the insert rate.
Follow-up
- The stored response body averages 200 KB and this table is now the largest in the database. What do you store instead, and what does a replay return once the body has been pruned?
- A client library generates a fresh idempotency key on every attempt. What breaks, which layer should have caught it, and does the server have any defence?
- The first attempt succeeded but its response was lost, and the client retries 30 hours later, after the key expired. What does the second attempt do, and is that acceptable?
Every query on one table stalls for forty seconds mid-deploy
During a release on PostgreSQL, every query touching resource times out for about 40 seconds and then recovers with no intervention. The release ran one migration, ALTER TABLE resource ADD COLUMN archived_reason TEXT, and the migration log shows it completing in 6 ms. Unrelated tables showed no change in error rate. Explain how a 6 ms statement caused a 40-second stall, give the ordered checks you would run on a live system to confirm it, and give the migration procedure that prevents a repeat.
Approach
- Separate the statement's duration from the lock's duration. ADD COLUMN with no default is a catalogue-only change and genuinely runs in milliseconds, but it requires ACCESS EXCLUSIVE, and it cannot acquire that until every transaction already touching the table has finished.
- Account for the queueing, which is the part that surprises people. A lock request that is waiting blocks later requests for conflicting modes behind it rather than letting them overtake, so one long-open transaction holds the DDL and the DDL holds all the traffic. The stall length is set by the longest open transaction, not by the size of the change.
- Confirm on a live system in this order: pg_stat_activity for that table ordered by xact_start, looking for the oldest transaction and specifically for state = idle in transaction; then pg_locks where granted = false to find the waiter; then join them on pid to name blocker and blocked. pg_blocking_pids() does that join for you and is the fastest single call.
- Prevent rather than merely time it better. Set lock_timeout to a second or two on the migration session so the DDL abandons the queue after a bounded wait and is retried, instead of holding it for as long as the oldest transaction lives. Be exact about what that buys: queries arriving during the wait still queue behind the pending ACCESS EXCLUSIVE request, so each attempt costs them up to one lock_timeout of added latency. The outage goes from 40 seconds to about one second per attempt, not to zero. Also run migrations away from deploy-time peaks, and put a statement timeout and an idle-in-transaction timeout on the analytics role that opens the long transactions.
- Know the lock each change takes, since the mitigation differs by change. A column with a non-volatile default is a metadata-only change from PostgreSQL 11 and still needs the brief ACCESS EXCLUSIVE; an index needs CREATE INDEX CONCURRENTLY, which cannot run inside a transaction block and leaves an INVALID index to drop if it fails; a check or foreign key is added NOT VALID and then VALIDATE CONSTRAINT as a separate statement under a weaker lock.
Follow-up
- The same release also wants NOT NULL on that column. What is the sequence that gets there without a long lock?
- Your lock_timeout retry fails ten times in a row because the analytics transaction is always open. What do you change?
- How does this differ on MySQL with InnoDB online DDL, and what is the equivalent of the waiting-lock queue there?
For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Rebuild the primitives by implementing them
- Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
- Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
- For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.
Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Arrays under an invariant: two pointers, sliding window, binary search
- Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
- Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
- Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.
Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.
Practice prompt ↗Practice prompt ↗03Sorting, heaps, and the greedy argument that has to be proved
- Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
- Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
- Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.
Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.
Practice prompt ↗Practice prompt ↗04Recursion, memoisation, and the step to a table
- Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
- Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
- Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.
Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Graphs, where most of the work is choosing the traversal
- Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
- Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
- Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.
Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.
Practice prompt ↗Practice prompt ↗06One day for everything that is not an algorithm
- Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
- Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
- Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.
Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.
Practice prompt ↗Practice prompt ↗07Solve out loud, under time
- Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
- Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
- Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.
Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Team size, service count and tickets closed say very little. Seniority shows in the decision you owned: what you chose not to build, which constraint you traded away, whose objection you had to resolve before anything could move. A large project where you executed someone else's plan is a small story.
Parlez-nous de votre expérience. (Tell us about your experience.)
Parlez-nous de votre expérience. (Tell us about your experience.)
Approach
- 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.
- Close with what you would do differently, concretely.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
Wie gehen Sie mit neuen Herausforderungen bzw. Themen um, welche Sie n…
Wie gehen Sie mit neuen Herausforderungen bzw. Themen um, welche Sie noch nicht kennen oder beherrschen? (How do you handle new challenges or topics that you do not yet know or master?)
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 would you do differently if you ran that again?
- How did you know your change caused the improvement?
Why do you want to work in Switzerland?
Why do you want to work in Switzerland?
Approach
- Pick a story where you made the decision, not one where you watched it.
- Close with what you would do differently, concretely.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
How will you work in a group?
How will you work in a group?
Approach
- 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.
- 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?
- 01
Parlez-nous de votre expérience. (Tell us about your experience.)
- 02
Wie gehen Sie mit neuen Herausforderungen bzw. Themen um, welche Sie noch nicht kennen oder beherrschen? (How do you handle new challenges or topics that you do not yet know or master?)
- 03
Why do you want to work in Switzerland?
- 04
How will you work in a group?
Is this an official SBB interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at SBB. Rounds and questions reflect what candidates have reported, not a process SBB has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How difficult are the technical interviews?
The technical interviews are considered fair and focused on your actual experience. You should be prepared to discuss the details of your CV rather than solving abstract, disconnected puzzles.
PracHub interview research ↗What is the typical timeline for the hiring process?
The process is generally efficient, with many candidates receiving feedback within a week of their interviews.
PracHub interview research ↗What is the most important thing to prepare for?
Be ready to explain your motivation for joining SBB and show a genuine interest in the challenges of the Swiss rail network.
PracHub interview research ↗Is the work environment collaborative?
Yes, SBB places a high value on teamwork. You will likely meet the entire team during the final stages of the process to ensure a mutual fit.
PracHub interview research ↗Sources & methodology 3 sources ↗
No official company page is cited. Rounds and questions come from candidate reports and PracHub editorial material; each source shows the date it was read.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
PracHub page · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-22