As a Software Engineer at Trainline, you are at the heart of Europe’s leading independent rail and coach travel platform. Your work directly impacts how millions of people plan, book, and experience their journeys. By building and scaling robust, high-performance systems, you enable seamless travel experiences across complex, multi-modal networks.
This role is both technically demanding and strategically significant. You will contribute to a platform that handles massive concurrency, integrates with numerous external carrier APIs, and delivers real-time data to a global user base. Whether you are working on.NET Core microservices, cloud infrastructure on AWS, or optimizing API resiliency, you are solving real-world distributed systems challenges that require a blend of architectural foresight and precise execution.
You can expect a fast-paced environment where collaboration is paramount. At Trainline, engineering is not just about writing code; it is about understanding the business, participating in the full software development lifecycle, and driving improvements in how the team delivers value. You will be challenged to think deeply about system design, maintainability, and the end-to-end user experience.
Recruiter Screening
reportedThe title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.
What to demonstrate
- Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
- Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
- Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year
How to prepare
- Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
- Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
- Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter
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
Behavioral Interviews
reportedWhat you say here is written down by each interviewer and compared afterwards, so the unit of evaluation is a claim someone else could check, not a well-told narrative. Two things make a story checkable: detail only a participant would hold, and a clean line around which part was yours. Vague ownership is the usual failure and it is usually accidental, because engineers say we about the team's work and we about their own, so the thing they personally built disappears into the plural. Name the part you wrote, and name who did the rest.
What to demonstrate
- Whether your details are ones a participant would hold and an observer would not: the constraint that ruled out the obvious approach, the first attempt that failed, the person who objected and on what grounds
- Whether ownership survives a direct question, since a follow-up to we decided is routinely who decided, and an answer that stays plural at that point is read as the work belonging to someone else
- Whether the numbers you quote are ones you would say identically to a former colleague with the dashboard open
How to prepare
- Go through each story replacing every we with either I or a named role (the on-call engineer, the reviewer, the other team) and check the story still holds together. Wherever it stops making sense you have found a part you cannot actually speak to
- Open the artefacts for two of your stories, the pull request, the design doc, the incident notes, and read them for dates and figures you have been rounding in the retelling. Correct your version to match
- For each story write the single sentence you would least want repeated to a former teammate, then either make it accurate or take it out
PracHub editorial advice for the preparation topics above.
Holding money in a floating-point type, or rounding it more than once
Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.
Paginating a growing table with limit and offset
Two unrelated defects share the idiom. Correctness: rows inserted or deleted between page requests shift the window, so a consumer walking an export skips rows and sees others twice, which for a customer-facing sync is silent data loss rather than an error anyone notices. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N times the page size and a deep page on a large table degrades from milliseconds to seconds. Keyset pagination over a stable, unique, indexed ordering -- where (created_at, id) < ($1, $2) order by created_at desc, id desc limit $3 -- is constant-cost per page and immune to shifting, on the precondition that the cursor columns never change value for a row, which disqualifies updated_at as a cursor.
Comparing floating-point values for equality, or holding money in them
Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.
Tests that assert on the implementation rather than the behaviour
Assert on what a caller can observe, not on the number of internal calls or the shape of a private field. A test that breaks on every refactor but still passes when the answer is wrong costs more than it protects.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Parse and verify a timestamped multi-signature webhook header
An inbound webhook carries a signature header of at most 1 KiB shaped t=<unix seconds>,v1=<64 hex chars>, with up to five v1 values during secret rotation and possibly unknown scheme keys. You hold the raw request body bytes and the currently active signing secrets. Write the parser and the verifier: accept when any active secret reproduces a signature and the timestamp is within a five-minute tolerance in either direction, reject otherwise. Single left-to-right pass over the header, no regular expression. State what is inside the MAC and why.
Approach
- Parse in one scan: split on
,, then on the first=only, since a value may itself contain=under a future scheme. Accepttexactly once and treat a secondtas a reject rather than last-wins. Push everyv1onto a short list and ignore any other key, so av2can be introduced later without breaking this verifier. - Say what is signed: HMAC-SHA256 over the exact byte string
<t>.<raw body bytes>, yielding 32 bytes or 64 hex characters. The timestamp sits inside the MAC because otherwise an attacker replays yesterday's body with its still-valid signature and only has to edit the header timestamp. - Hash the bytes as received. Verifying against a re-serialised JSON body is the usual defect: key order, whitespace and number formatting all change the bytes while the parsed objects compare equal, so signatures fail for honest senders and the popular 'fix' is to stop checking.
- Compare in constant time over fixed-length digests. Decode the hex to 32 bytes, accumulate
acc |= a[i] ^ b[i]across the whole length, and testacc == 0at the end. Evaluate every candidate without an early exit; at five candidates that is five HMACs over the body, linear in body size and negligible beside the network. - Apply the tolerance as a two-sided bound, rejecting when
|now - t| > 300seconds. A sender whose clock runs ahead of yours is an ordinary case, and an unbounded future timestamp is a free replay window. - Complexity: O(L) over the header producing k candidates, plus k HMACs at O(|body|) each. Space is O(k) beyond the body itself. Do the cheap rejections, including the tolerance check, before any cryptography runs.
Follow-up
- The body is 40 MB. What changes about where you verify, and what can you do before the whole body has arrived?
- A customer reports that signatures fail for exactly the requests whose body contains a non-ASCII character. What is your first hypothesis?
- How do you rotate the signing secret with no failed deliveries, and how long do both secrets stay live?
Find peak concurrent sandbox usage from run intervals
Given up to 5 million job_run rows for one tenant over one day, with run_id, started_at, finished_at, status and wall_clock_limit_seconds, report the maximum number of sandboxes running at once, the earliest instant that maximum is reached, and the first run_id that would breach a per-tenant cap of C. started_at is null while a run is queued; finished_at is null both for runs still executing and for runs in status lost. Treat a run as occupying [started_at, finished_at). Give the complexity and state how you handle each null.
Approach
- Turn each run into two sweep events,
(started_at, +1)and(end, -1), then sort the 2n events by timestamp with-1ordered before+1at equal timestamps. That tie-break is what makes the interval half-open, so a run finishing at 10:00:00 and one starting at 10:00:00 never overlap. - Decide each null out loud before sweeping, because each choice moves the answer. A null
started_atmeans queued and contributes nothing. A nullfinished_atwith statusrunningorleasedis clipped to the window end. Statuslosthas no observed end at all, so clip it atstarted_at + wall_clock_limit_secondson the grounds that the supervisor owns the timeout, and record that you did. The table'scheck (finished_at is null or started_at is not null)guarantees you never see an end without a start. - Sweep once, maintaining a running counter, the maximum, and the timestamp at which the maximum was first attained (update
peak_atonly on a strict increase, or you will report the last such instant instead of the earliest). Capture the firstrun_idwhose+1takes the counter to C+1 during the same sweep rather than in a second pass. - Complexity: O(n log n) dominated by the sort, O(n) space. If rows already arrive ordered by
started_at, a min-heap of end times gives O(n log k) time and O(k) space with k the peak concurrency, which is the better shape when the rows come from an index scan on(tenant_id, started_at). - If second resolution is acceptable, counting-sort the endpoints into an 86,400-slot delta array and prefix-sum it: O(n + T) time and O(T) space, which beats the comparison sort at 5 million rows. It answers only at second granularity, so state which resolution the cap is defined in.
Follow-up
- Now report peak concurrency per tenant for 10,000 tenants from one globally sorted stream. What changes about memory and about the sort?
- The cap has to be enforced at dispatch rather than reported afterwards. What does the admission check look like, and where does it race?
- How would you answer 'peak concurrency within any 5-minute window' without re-sorting?
Hold a tenant to a trailing sixty-second request limit
The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.
Approach
- Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while
front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request. - Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
- Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate,
prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact. - Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (
tokens,last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual. - Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
- Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.
Worked solution 25 min
- Implement the deque version and instrument the per-request pop count, then confirm total pops equal total pushes over a run.
- Generate a burst that places R requests in the last 100 ms of one minute and R more in the first 100 ms of the next.
- Run that burst through the exact deque, a fixed 60-second counter, and the weighted two-bucket estimate, recording admissions in the trailing 60 seconds at every instant.
- Size the memory as R x active tenants x 8 bytes at R = 1,000 and 200,000 tenants, and compare it against what a token bucket would need.
Follow-up
- One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
- Quotas rather than rate limits: the check is
select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes. - How do you return an accurate
Retry-Afterfrom the exact algorithm without a second scan?
Find the join that inflates every invoice total
invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.
Approach
- Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row:
sum(l.amount_minor)is the true total multiplied by the attempt count, andcount(p.attempt_id)is attempts times lines. Three lines and two attempts report double the money and six attempts. - Reject the reflex repair.
count(distinct p.attempt_id)does fix the count, because attempt_id is unique.sum(distinct l.amount_minor)does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match. - Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
- Keep invoices with no attempts by making the attempt branch a LEFT JOIN with
coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about. - Push each filter to its own grain:
where l.voided_at is nullbelongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants. - Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.
Worked solution 25 min
- Create one invoice with three lines of 1000, 1000 and 500 minor units and two payment attempts, then run the original query.
- Confirm it reports 5000 and 6 rather than 2500 and 2.
- Apply
sum(distinct l.amount_minor)and confirm the total becomes 1500, which is worse rather than better. - Write the two-CTE version with a LEFT JOIN and coalesce, and confirm 2500 and 2.
- Add a second invoice with lines and no attempts and confirm it still appears.
Follow-up
- Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with
filter (where ...)be better? - Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
- The same report is needed per tenant per month. What index makes the line CTE cheap?
Paginate a tenant's delivery export without skipping rows
A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.
Approach
- Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
- Replace the position with a value predicate over a stable, unique, indexed ordering:
where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window. - Index
(tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit. - Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
- Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.
Follow-up
- The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added
created_at <= $snapshotbound, and a materialised export table. - A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
- The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?
How do you make an API resilient?
How do you make an API resilient?
Approach
- Choose a partition key and say what query it makes expensive.
- State the consistency you need, and where you are willing to be stale.
- Fix the scope first: who calls this, how often, and what they do when it fails.
Follow-up
- What breaks first when traffic grows ten times?
- What would you drop to keep the system up under load?
Design a scalable booking system on AWS.
Design a scalable booking system on AWS.
Approach
- Name the failure you are designing for, then the recovery path.
- Fix the scope first: who calls this, how often, and what they do when it fails.
- State the consistency you need, and where you are willing to be stale.
Follow-up
- What would you drop to keep the system up under load?
- What breaks first when traffic grows ten times?
How would you scale a service on the cloud?
How would you scale a service on the cloud?
Approach
- Name the failure you are designing for, then the recovery path.
- Fix the scope first: who calls this, how often, and what they do when it fails.
- Choose a partition key and say what query it makes expensive.
Follow-up
- What would you drop to keep the system up under load?
- How does this behave when that dependency is down for an hour?
Describe the current architecture of the system you are working on.
Describe the current architecture of the system you are working on.
Approach
- Choose a partition key and say what query it makes expensive.
- Name the failure you are designing for, then the recovery path.
- Name the read and write paths separately; they rarely have the same bottleneck.
Follow-up
- How does this behave when that dependency is down for an hour?
- What would you drop to keep the system up under load?
Explain API middleware and talk about any custom ones you have impleme…
Explain API middleware and talk about any custom ones you have implemented.
Approach
- State your assumptions explicitly before working the problem.
- 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.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Explain SOLID principles and how you apply them.
Explain SOLID principles and how you apply them.
Approach
- State your assumptions explicitly before working the problem.
- Say what you would check first and why it is the highest-information step.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Design a machine-readable error contract for the gateway
The edge gateway serves roughly 30k requests/second to SDKs and CI pipelines that retry automatically. Today every failure returns 500 with a prose message that clients string-match on. Design the error contract: the response body fields, and the status code for a malformed body, a revoked credential, a scope the credential lacks, a row belonging to another tenant, a reused idempotency key sent with a different body, an exceeded rate limit, and an unreachable dependency. For each, state whether the client may retry and on what schedule. Deliverable: the envelope schema plus the status-to-retry table.
Approach
- Split the envelope by audience: a stable
codestring for programs, amessagedocumented as human-only and free to change, arequest_idthat joins to gateway logs, and adetailsarray for per-field problems. The code list is an enum that only ever grows. - Assign status by who has to change something: 400/422 for the caller's bytes, 401 for a credential that no longer authenticates, 403 for a scope or entitlement, 404 rather than 403 for a row in another tenant because 403 confirms the identifier exists, 409 for an idempotency conflict, 429 for a limit, 503 for a dependency.
- Derive retryability from the method and the idempotency key rather than from the status: a 5xx or a timeout is an unknown outcome, not a failure, so GET/PUT/DELETE may be retried under HTTP semantics and POST only when it carries an idempotency key.
- Put the schedule in the response: Retry-After on 429 and 503 overrides the client's own backoff; otherwise capped exponential backoff with full jitter, sleeping uniformly in [0, min(cap, base * 2^attempt)], bounded by a total attempt budget so retries expire before the caller's deadline.
- Write the negative rules into the published contract: clients must never parse
message, must tolerate unknowncodevalues by falling back to the status class, and a code's meaning is never redefined once shipped.
Worked solution 20 min
- Write the envelope as a JSON schema with four top-level fields and say which are guaranteed present on every error.
- Fill a seven-row table: condition, status, code string, retryable yes/no, and the schedule or the reason retrying cannot help.
- For each non-retryable row, write the one thing the caller must change (bytes, credential, plan, key) so nothing is marked non-retryable without a remedy.
- Add the unknown-outcome row for timeouts and 5xx separately from the other rows, and give it an action other than 'treat as failed'.
- Write two sentences of client guidance: honour Retry-After when present, apply full jitter otherwise, and stop at the attempt budget.
Follow-up
- A customer reports they retried a 500 from POST /v1/runs and ended up with two sandboxes billed. Whose bug is it, and what in your contract permits their reading?
- You need to add a new error code next quarter without a version bump. What did the v1 contract have to say for that to be non-breaking?
Regional error rate explodes after a dependency merely slows
A control-plane read replica in one region degrades from 4 ms to 120 ms. Within ninety seconds that region's gateway error rate rises from 0.01% to 40% and its p99 becomes bimodal, one mode near the old p99 and one at the client timeout. The other two regions are unaffected. The gateway retries control-plane reads three times with exponential backoff and no jitter. Give an ordered checklist that separates trigger from amplifier, the offered-load arithmetic, and the controls that break the loop.
Approach
- Split the incident into three questions before touching a control: what started it, what amplified it, and what would make recovery slow. Here they are the replica slowdown, the retry policy interacting with queueing, and a synchronised unjittered herd at recovery. They are different mechanisms and each needs its own control.
- Read the distribution rather than the mean. A bimodal p99 with one mode pinned at the client timeout is two populations, not one degraded path; split latency by cache hit and miss and confirm the fast mode is hits and the timeout mode is misses that reached the replica.
- Do the load arithmetic. Three retries turn one client request into up to four upstream requests, so offered load reaches roughly 4x on a dependency that is already slower, and it arrives at the worst moment. With utilisation approaching one, queueing delay grows superlinearly, which is why a 30x latency increase upstream does not produce a 30x increase downstream, it produces timeouts.
- Break the loop with controls that bound offered load rather than with more attempts: a concurrency limit on the control-plane client so at most N calls are in flight and the remainder fail fast, a circuit breaker scoped per dependency and region, and a retry budget capping retries at a small fraction of base traffic so amplification has a ceiling that does not depend on how many clients are retrying.
- Add full jitter to whatever retries survive, sleeping uniformly in [0, min(cap, base x 2^attempt)], so attempts de-correlate instead of arriving in waves aligned to the moment of failure.
- Decide the unreachable-dependency behaviour in advance, because it is the actual product decision underneath: serving from an expired credential cache keeps the product available while extending a revoked key's life past the stated bound, and failing closed converts a dependency degradation into a total outage. State the mode and the staleness number rather than letting the timeout choose.
Follow-up
- The replica recovers. Describe what happens in the first ten seconds with your controls in place versus without them.
- Which single metric would have paged before the error rate moved, and why is upstream latency by itself not it?
- Requests that fail fast under the concurrency limit still need an answer. What does the gateway return, and what does it do to the usage event it would otherwise have emitted?
For someone fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Measure before reasoning
- Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
- Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
- Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.
Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.
Practice prompt ↗Practice prompt ↗Practice prompt ↗Worked solution ↗02References, copies, and the bugs they produce
- Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
- Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
- Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.
Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.
Practice prompt ↗Practice prompt ↗Practice prompt ↗03Types, once, in a language that checks them
- Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
- Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
- Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.
Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.
Practice prompt ↗Practice prompt ↗04Concurrency, starting with what actually runs at the same time
- Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
- Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
- Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.
Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Debugging as a procedure rather than an instinct
- Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
- Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
- Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.
Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.
Practice prompt ↗Practice prompt ↗06Tests that catch the bug you are about to write
- Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
- Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
- Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.
Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.
Practice prompt ↗Practice prompt ↗07Debug something broken, out loud
- Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
- Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
- Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.
Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.
Practice prompt ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Nobody is scoring your stamina at three in the morning. What carries weight is which signal told you something was wrong, what you measured before touching anything, what you rolled back versus what you fixed forward, and why you picked one. 'We restarted it and it went away' is a story about not knowing.
How do you handle criticism?
How do you handle criticism?
Approach
- Pick a story where you made the decision, not one where you watched it.
- Name the disagreement and how you resolved it with evidence.
- Give the blast radius: what could have broken, and what you measured.
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 at Trainline?
Why do you want to work at Trainline?
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
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
Argue against failing open when the control plane is unreachable
The gateway caches credential-to-context decisions with a sixty-second TTL. A design proposal says that when control-plane reads fail, pods should keep serving from expired entries indefinitely so a control-plane outage never becomes a product outage. You believe that converts every revocation into an unbounded one. Describe a design you argued against while it was still a live proposal: what you measured or modelled to make the case, what you conceded, who decided, and what happened afterwards. Say what would have changed your mind before the decision, not after it.
Approach
- Reframe it from a values argument into a bounded-staleness argument. Both sides already accept the cache; the disagreement is only about the ceiling on how long a revoked credential keeps authorising. Put a number on the table — serve stale for up to fifteen minutes, then fail closed — and make the other side argue against a number rather than against a principle.
- Bring arithmetic rather than adjectives: the rate of revocations with revoked_reason in ('suspected_leak','auth_version_bump'), the observed distribution of control-plane unavailability, and the product of the two, which is expected requests served by revoked credentials per outage-hour. At 30k requests/second the unbounded version is not a subtle exposure and the number says so.
- Concede the strong half of the opposing case first, because that is what buys you the room: failing closed turns one service's outage into a total outage across three regions, and a control plane doing tens of writes per second is not engineered to the gateway's availability target. A proposal you have not steelmanned reads as reflex.
- Propose the asymmetry that usually resolves this: stale entitlements cost bounded money (a quota fifteen minutes out of date over-serves by a computable amount), while a stale revocation costs unbounded access. Split the cached decision by what it authorises, give the two halves different staleness ceilings, and let the entitlement half fail open while the revocation half fails closed.
- State the propagation dependency plainly, since it is the part that is missed: validity is also derived from the principal's auth_version, so password reset and sign-out-everywhere flow through this same cache. A design that bounds staleness for explicit revocation and not for auth_version bumps has only fixed half of it.
- Say who decided, and what you did afterwards in either outcome: write the decision down with its number and a review date, and instrument the exposure you were worried about so the next round of the argument is settled by data instead of by seniority.
Follow-up
- Publish-subscribe invalidation is lossy under a partition, and a TTL is the only hard bound. What TTL do you pick, and what does it cost you at 30k requests/second?
- The key was revoked because it was found in a public repository. Does your answer change, and where does that urgency live in the design?
- You lost the argument and six weeks later the failure you predicted happens. What do you say in the review, and what do you not say?
- 01
How do you handle criticism?
- 02
Why do you want to work at Trainline?
- 03
The gateway caches credential-to-context decisions with a sixty-second TTL. A design proposal says that when control-plane reads fail, pods should keep serving from expired entries indefinitely so a control-plane outage never becomes a product outage. You believe that converts every revocation into an unbounded one. Describe a design you argued against while it was still a live proposal: what you measured or modelled to make the case, what you conceded, who decided, and what happened afterwards. Say what would have changed your mind before the decision, not after it.
Is this an official Trainline interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Trainline. Rounds and questions reflect what candidates have reported, not a process Trainline has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long does the interview process typically take?
Most candidates experience a process lasting 2–4 weeks. It is generally well-structured, but you should be proactive in requesting updates if you haven't heard back within the promised timeframe.
PracHub interview research ↗What is the best way to prepare for the coding test?
Focus on writing clean, testable code rather than just trying to finish quickly. Practice refactoring existing snippets and be prepared to explain your design decisions regarding scalability and error handling.
PracHub interview research ↗Is the technical interview purely about language-specific knowledge?
No. While you will be asked about your primary language (like C#), the focus is often on your underlying engineering principles, how you handle ambiguity, and your ability to communicate complex technical ideas.
PracHub interview research ↗What should I know about Trainline’s culture?
They value collaboration, transparency, and continuous improvement. They look for engineers who are not only technically proficient but also eager to contribute to a team-first environment.
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
Explore related engineering environments and compare the preparation focus.
Vercel Software Engineer Interview Guide
Debugging practice for Vercel software engineer prep built on a reproduction you trust, narrowing to the smallest failing input, and changing one variable per run so the result stays attributable.
Together Ai Software Engineer Interview Guide
Failure-handling practice for Together Ai software engineer candidates: requests whose outcome a timeout leaves unknown, where a retry is safe, and deduplicating at the receiver instead of assuming exactly-once delivery.
Lambda Software Engineer Interview Guide
Failure-handling practice for Lambda software engineer candidates: requests whose outcome a timeout leaves unknown, where a retry is safe, and deduplicating at the receiver instead of assuming exactly-once delivery.
CoreWeave Software Engineer Interview Guide
Failure-handling practice for CoreWeave software engineer candidates: requests whose outcome a timeout leaves unknown, where a retry is safe, and deduplicating at the receiver instead of assuming exactly-once delivery.