At Systems Engineering and Assessment Ltd, the Software Engineer role is pivotal to delivering robust, mission-critical solutions. You will be responsible for designing, developing, and maintaining complex software systems that support high-stakes engineering projects. Your work directly influences the reliability and performance of systems that the company's clients depend on for safety and operational success.
This role requires a blend of technical precision and strategic thinking. You will work within multidisciplinary teams, bridging the gap between hardware capabilities and software functionality. Whether you are working on Product Lifecycle Management (PLM) tools or advanced systems integration, your contributions will be central to the engineering lifecycle.
Initial 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 Deep-Dives
reportedInput bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.
What to demonstrate
- Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
- Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
- Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
- Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply
How to prepare
- For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
- For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
- Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last
Behavioral Assessments
reportedYour first answer is not really what is scored. It buys the follow-up questions, and those decide the round. An interviewer with fifteen minutes takes one thread and pushes on it four or five times, so a story you can only tell at a single level of detail collapses under the third why. That is an argument for fewer stories known deeply rather than one prepared per prompt. Four or five pieces of work you can still explain down to the code you changed and the argument you had about it will cover nearly anything asked in this round.
What to demonstrate
- Whether a story holds as the questioning moves from what you did to why that instead of the alternative, and then to what you would change knowing what you know now
- Whether you can re-cut a project to answer the question actually asked rather than delivering a rehearsed block that answers an adjacent one
- Whether your level of detail is chosen rather than habitual: going down to the schema when the question is about the data model, staying out of it when the question is about the person who disagreed with you
How to prepare
- Pick four projects and write the chain out four levels deep for each: what you did, why that, why not the alternative, and what would have to be true for the alternative to have won. Where you cannot reach the fourth level, you have a placeholder rather than a story
- Have someone ask why three times in a row on a single thread with nothing else added, and mark the point where you start repeating a sentence you already said. That point is where the interviewer stops learning anything
- Build a one-page index instead of an answer bank: the common prompts in this round (disagreement, a failure that was yours, thin requirements, a deadline you missed, work you inherited) mapped to which of your four projects you would use for each, so the choosing is done now rather than while an interviewer waits
PracHub editorial advice for the preparation topics above.
Serialising a tenant's writes through select ... for update on a single counter row
It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.
Treating a timed-out write as a failed write
A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.
Writing code before the input contract is pinned down
Before the first line, state the types, the size bounds, whether duplicates, negatives or an empty input are possible, whether the input is sorted, whether you may mutate it, and what the function returns when nothing matches. Every one of those answers changes the code, and discovering one at minute twenty costs a rewrite you no longer have time for.
Trusting input because it came from your own front end
Anything crossing a trust boundary is hostile: parameterise queries instead of building SQL by concatenation, validate against an allow-list rather than a deny-list, and bound the size of anything you allocate from a request. Raising this unprompted in an API or design question is a cheap and unusually strong signal.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Seal an hour under late data with bounded memory
Metering ingest reads 256 partitions at 10,000 to 40,000 events/second. Events carry occurred_at and ingested_at, and during a producer replay the gap between them is hours. Seal each UTC hour once no more than 50 parts per million of that hour's eventual quantity can still arrive, using memory that does not grow with the size of the replay. Define the watermark, the lateness parameter and how you measure it, the structure holding open hours, and the write that performs the seal. State what an idle partition does to your watermark.
Approach
- Two clocks, two jobs. Bucket by
occurred_at, because that is the hour the customer is billed for, and advance the watermark oningested_at, because that is what the fold has consumed and whatsource_max_ingested_atrecords. Conflating them is what makes late data invisible. - The global watermark is the min over partitions of each partition's committed
ingested_at, not the max: the fold is trustworthy only as far as the slowest partition. The consequence is that one idle partition pins the watermark forever and nothing seals, so an idle partition must promote its watermark to wall clock after a stated idle timeout, and that timeout becomes a correctness parameter, because a partition that is slow rather than idle gets sealed past. - Choose the lateness L from the measured distribution of
ingested_at - occurred_at, weighted by quantity rather than by event count. The target is 50 ppm of the hour's quantity, and a replay is rare in events while carrying disproportionate mass, so an event-weighted quantile picks an L that is comfortably wrong at exactly the moment it matters. - Measure that quantile in bounded memory. A Greenwald-Khanna summary gives epsilon-approximate quantiles in O((1/epsilon) log(epsilon n)) space; a t-digest costs more per merge but has relative error that tightens at the tails, which is the half of the distribution you are reading at p99.99. Keep a separate summary per tenant class, because one tenant's batch importer is not the population.
- Hold open hours in a min-heap keyed by
hour_start. When the watermark advances, pop every hour withhour_end + L < Wand seal it: O(log H_open) per advance and O(1) amortised per event to touch its bucket. Memory is open hours multiplied by distinct(tenant, workspace, sku)keys, so cap the number of simultaneously open hours and spill the oldest intousage_rollup_hourlyasstatus='open'with arevisionbump. While an hour is open the row is upsertable, so the store is your overflow. - The seal itself is a conditional write:
update ... set status='sealed', sealed_at=now() where status='open' returning .... Two sealers race on every restart, and the loser must see zero rows and stop rather than write a second value. After the seal, an event for that hour is not an upsert but an adjustment, andsource_max_ingested_atis what proves it arrived afterwards.
Follow-up
- A replay starts during the sealing window for a period you are about to close. What do you do, and what is the customer-visible consequence of each option?
- Your measured quantity-weighted p99.99 lateness is six hours and the invoice must be issued at 02:00 UTC on the first. How do you reconcile those two numbers?
- How would you detect that L has drifted before it costs you an hour's quantity?
Order a job dependency graph and find its critical path
A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.
Approach
- Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
- Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
- Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order:
earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E). - Second pass in reverse topological order for
latest_finish, thenslack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain.slack[v] = 0is exactly the statement that some longest path runs through v; equivalently, the longest path through v has lengthT - slack[v]. - The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by
min(d, T - L_avoid(v)), whereL_avoid(v)is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, whilemin(10, 15 - 14)predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan ismax(T - d, L_avoid(v)). - Compute
L_avoid(v)the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, sinceL_avoid(v)only ever matters through that max: setduration[v] := 0, recompute the makespan asT0(v) = max(T - duration[v], L_avoid(v)), and the gain ismin(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.
Worked solution 30 min
- Build four fixtures. A: 12 jobs, two branches of 100 s and 95 s that share no job. B: fixture A plus one back edge. C: two disjoint paths tied at 100 s. D: the shared-prefix case, one job of 10 s feeding a 5 s job and a 4 s job, so the longest path is 15 s and the runner-up is 14 s.
- Run Kahn; on fixture B confirm it emits fewer than V nodes, then run the residual-subgraph DFS and print the actual cycle.
- Compute
earliest_finishforward andlatest_finishbackward, and list the zero-slack set for each fixture. - For each zero-slack job v, recompute the makespan with
duration[v] := 0to getT0(v), and record both the correct boundT - T0(v)and the wrong one,T - second_longest_path, side by side. - Apply the shortening for real (20 s off the critical branch of A, 10 s off the shared prefix of D) and diff the recomputed makespan against each prediction.
Follow-up
- Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
- Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
- Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?
Schedule ordered webhook retries with a heap of subscription queues
Design the in-memory scheduler for webhook delivery. Up to 20 million rows sit in status pending or failed_retryable across 200,000 subscriptions, each row carrying next_attempt_at and attempt_count, and each endpoint having a circuit breaker. Deliveries for one subscription must be attempted in order, so at most one attempt per subscription may be in flight. Support due(now), complete(delivery, outcome) and insert(delivery) in O(log S), where S is the subscription count rather than the delivery count. Give the backoff formula you schedule retries with.
Approach
- Key the global heap by subscription, not by delivery. Each subscription owns a FIFO of its due deliveries in event order; the heap holds one entry per eligible subscription, keyed by its head's
next_attempt_at. That is 200,000 heap entries instead of 20 million, and it makes the one-in-flight rule structural rather than a check somebody can forget. due(now): peek the minimum. If its key is in the future, sleep until then instead of spinning. Otherwise pop it, move the subscription into an in-flight set, and do not re-push it. A subscription absent from the heap cannot be dispatched twice, which is precisely how ordering is preserved.complete: on success, drop the head and re-push the subscription keyed by its new head, or leave it out when the queue empties. On a retryable failure, incrementattempt_countand setnext_attempt_at = now + uniform(0, min(cap, base * 2^attempt)), sampled uniformly across the whole interval. That is full jitter; deterministic backoff re-synchronises the herd you just created.- Circuit breaker: park the subscription in a second heap keyed by its half-open time, so an endpoint dead for six hours costs one heap entry and zero attempts rather than consuming worker slots. Admit exactly one probe at half-open and close the breaker only on its success.
- Say the price of the ordering guarantee out loud. One in-flight attempt per subscription means an endpoint answering in 10 seconds drains at 0.1 deliveries/second however many workers you run, and its backlog grows until it recovers. If the customer does not need order, allow k in flight and document delivery as unordered; that is the trade, and it is a product decision.
- All three operations are O(log S) with O(S) resident heap memory and the queues themselves backed by the store. The database-backed equivalent is a partial index on
(subscription_id, next_attempt_at) where status in ('pending','failed_retryable')claimed withFOR UPDATE SKIP LOCKED, and the write-back must be fenced onlease_tokenso a worker that stalled and resumed cannot overwrite a newer attempt.
Follow-up
- One subscription has 4 million queued deliveries. What stops it from starving the other 199,999, and what does your heap look like under that load?
- A customer requests redelivery of last Tuesday's events. Where do those rows enter your structure, and what keeps them from reordering live traffic?
- The process restarts. How much state do you rebuild, and what stops every subscription from being attempted in the same second?
Explain why the metering dashboard scans every daily partition
usage_event is range-partitioned daily on ingested_at and holds tenant_id, workspace_id, environment, sku, quantity numeric(20,6), occurred_at and ingested_at. The only relevant index is on (occurred_at). A dashboard runs select sku, sum(quantity) from usage_event where tenant_id = $1 and date_trunc('hour', occurred_at) >= $2 and environment = 'production' group by sku, and EXPLAIN shows a sequential scan of every partition. Give each distinct reason, rewrite the predicate so an index can serve it, propose the index, and state the write cost its column order adds.
Approach
- Separate the three causes rather than blaming one. First,
date_trunc('hour', occurred_at)wraps the column, so the predicate is not sargable against a btree on the bare column. Second, pruning keys off ingested_at while the query constrains occurred_at, so no partition can be excluded. Third, even made sargable, (occurred_at) is not tenant-leading, so for one tenant among thousands the scan reads the whole time range and discards nearly all of it. - Rewrite the bound carefully, because the obvious rewrite is only conditionally equivalent.
date_trunc('hour', x) >= $2equalsx >= $2only when $2 is already hour-aligned; for an arbitrary $2 it meansx >= date_trunc('hour', $2) + interval '1 hour'. Normalise the parameter in the caller and leave the column bare. - Restore pruning with a second, redundant predicate on the partition key:
ingested_at >= $2 - interval '<late-data horizon>'. State both sides of it. It prunes to a handful of partitions, and it silently omits any event whose ingest lagged past that horizon, which is precisely what a producer replay produces. Either document the horizon as a stated bound, or partition on occurred_at and move the problem into the dedup window instead. - Propose
(tenant_id, occurred_at) include (sku, quantity)per partition. A partial indexwhere environment = 'production'mostly saves size rather than selectivity, since production dominates the three environments; take it if non-production is a meaningful share and skip it otherwise. - Price the write path honestly. At roughly 250M rows/day each extra index is another insert plus WAL per row, and a tenant-leading key scatters inserts across one hot leaf per active tenant instead of appending to a single rightmost leaf, so page dirtying and random I/O both rise. An INCLUDE payload widens every leaf entry and enlarges the index accordingly.
- Add the index-only-scan caveat before someone reports it as a regression: on a freshly appended table the visibility map is not yet set for recent pages, so the INCLUDE columns still cost heap fetches until autovacuum has been through, and the newest hour is exactly the data the dashboard reads.
Worked solution 30 min
- Build 30 daily partitions with skewed tenants, one holding about 40% of the rows, then ANALYZE.
- Run
explain (analyze, buffers)on the original query and record how many partitions were scanned and the rows removed by filter. - Apply the rewritten predicate and the index, re-run, and confirm the plan lists only the partitions inside the ingested_at bound.
- Re-run with $2 set to a non-hour-aligned timestamp and confirm the rewritten and original predicates return identical rows.
- Insert an event with ingested_at six hours past occurred_at and check whether the pruning predicate excludes it.
Follow-up
- CREATE INDEX CONCURRENTLY is not supported on a partitioned parent. Give the sequence that gets this index onto 400 existing partitions without blocking ingest.
- One tenant holds 200 times the median row count and the dashboard still times out for them with the index in place. What changes?
- Should this read hit
usage_rollup_hourlyinstead? State what that costs in freshness and what the watermark lets you promise.
Rebuild an hourly rollup with deduplication and late-arrival accounting
From usage_event (event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at), produce the values usage_rollup_hourly should hold for one tenant over one day: per (workspace_id, sku, hour_start) the deduplicated quantity_sum, event_count and source_max_ingested_at, bucketed by occurred_at. Duplicates share (tenant_id, idempotency_key). Also report, per hour, the running total across the day and the share of quantity that arrived more than two hours after the hour began. Write the query, and state which duplicates a daily unique index cannot catch.
Approach
- Deduplicate in its own CTE before any aggregation, because a SUM cannot be un-summed:
row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id) = 1. Include the tiebreaker. Without it the surviving row is non-deterministic when two duplicates share an ingested_at, and a rollup described as deterministically recomputable then disagrees with itself between runs. - Bucket on occurred_at and nothing else, and pin the timezone explicitly.
date_trunc('hour', timestamptz)truncates in the session's TimeZone setting, so the same query run by a session set to a non-UTC zone buckets differently; use the three-argumentdate_trunc('hour', occurred_at, 'UTC')on PostgreSQL 16 or later, ordate_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC'before that. Filterenvironment = 'production'explicitly, since metering covers three environments and billing covers one. - Aggregate to the grain with
sum(quantity),count(*)andmax(ingested_at). The last is not decoration: it is the watermark the row consumed up to, and without it there is no way to prove afterwards what a number did and did not include. - Compute the late share inside the dedup-and-aggregate step as a conditional aggregate,
sum(quantity) filter (where ingested_at > hour_start + interval '2 hours'), then divide by the hour's total. Compute the running total as a window over the already aggregated rows:sum(quantity_sum) over (partition by workspace_id, sku order by hour_start rows between unbounded preceding and current row). Running either over raw rows puts the duplicates back. - Answer the index question exactly. The unique constraint is on (ingested_day, tenant_id, idempotency_key), because a unique index on a partitioned table must contain the partition key. It therefore deduplicates only within one ingest day and admits a duplicate whose retry crosses midnight or whose replay runs a week later. That is why this CTE dedups across the whole window being recomputed, and why the dedup horizon is a correctness parameter rather than a retention cost.
- Keep the numeric type all the way through. quantity is numeric so the sums are exact; a cast to double precision anywhere in this pipeline reintroduces drift that surfaces only as a few unreconcilable cents per tenant per month, long after the query is out of anyone's mind.
Follow-up
- A dispute forces the same recompute over 40 days for one tenant. What changes about the dedup CTE's memory use and the chosen plan, and what would you do about it?
- Two runs a minute apart return different quantity_sum values for an hour that is already closed. Give two mechanisms that produce that, and the single query that distinguishes them.
- Express the same rollup incrementally so it does not re-scan the day each time the watermark advances. What does the incremental version stop being able to answer?
How do you approach debugging complex legacy codebases?
How do you approach debugging complex legacy codebases?
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?
When faced with an ambiguous technical requirement, what steps do you …
When faced with an ambiguous technical requirement, what steps do you take to clarify and proceed?
Approach
- Work from the requirement backwards to the design.
- State your assumptions explicitly before working the problem.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
How do you ensure software reliability when working on mission-critica…
How do you ensure software reliability when working on mission-critical systems?
Approach
- Work from the requirement backwards to the design.
- Clarify what is being asked and what a complete answer contains.
- State your assumptions explicitly before working the problem.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Design the batch ingest endpoint metering agents retry into
A customer-run agent posts usage events in batches of up to 1,000 to metering-ingest with a 30-second timeout and at-least-once retry of the whole batch. Each event carries event_id, idempotency_key, sku, quantity and occurred_at; the server adds ingested_at, and usage_event is partitioned daily on ingested_at with unique (ingested_day, tenant_id, idempotency_key). Design the endpoint: the request shape, what the response says when 900 events are new, 90 are duplicates and 10 are malformed, the status code, and the agent's algorithm on timeout. Then state the deduplication horizon and justify it against that unique constraint.
Approach
- Fix the per-item outcome taxonomy first, because the status code follows from it: accepted, duplicate, and rejected with a permanent code. A duplicate is a success; reporting it as an error makes the agent either re-send revenue it already delivered or drop it.
- Allow only permanent failures per item. A transient per-item failure inside a 200 invites the agent to discard that event, so anything transient escalates to a 5xx for the entire batch. A 200 is then a promise that every event not marked rejected is committed and durable.
- Return 200 with a results array aligned by index and carrying the event id, so the agent can retry precisely the subset that needs it and quarantine the ten malformed events instead of hard-looping a poison batch forever. Cap the batch at 1,000 items and a byte size, with 413 beyond it and 429 with Retry-After for backpressure.
- Deduplicate per event, never per batch: the agent may split, merge or reorder a retried batch, so a batch-level key matches nothing on the second attempt. The key is (tenant_id, idempotency_key), and the tenant comes from the resolved credential; a tenant id present in the body is compared against it, never trusted.
- Size the horizon as a correctness parameter. The unique index includes the partition key, so it deduplicates only within one day: a retry that crosses midnight, or a replay run a week later, passes straight through it. A separate dedup store keyed (tenant_id, idempotency_key) with a TTL exceeding the agent's maximum retry window plus the longest replay you intend to support is what actually enforces the invariant, which makes its retention a correctness setting rather than a cost knob.
- Order the commit against the response and the acknowledgement: commit then respond at the endpoint, and downstream commit the fold then acknowledge the message. Acknowledging first turns a crash into silently lost revenue with no error raised anywhere.
Worked solution 40 min
- Write the request body schema with the batch envelope and one event, and state which fields the server assigns rather than accepts.
- Write the 200 response for the 900/90/10 case, showing three result entries, one of each outcome, with the rejected one carrying a permanent code.
- Write the rule separating per-item rejection from whole-batch failure, and list which conditions fall on each side.
- Compute the dedup horizon from the agent's retry window plus the replay window you support, and say where the dedup state lives and how it ages out.
- Write the agent's pseudocode for timeout, 5xx, 429 and 200-with-rejections, four branches, and mark which branch may drop an event.
- Trace the crash between commit and response, and between fold and acknowledgement, and say what each produces.
Follow-up
- The agent times out at 30 seconds having received nothing. What exactly does it do next, and what in your design makes that safe?
- Ten events are rejected every hour for a week and nobody notices. What does the endpoint owe the customer beyond a per-item 4xx code?
- A replay pushes 40 million events through this endpoint in an hour. Which part of your design degrades first?
Metering partition crash-loops and the sealing watermark freezes
One metering-ingest partition has stopped advancing. Lag grows linearly, the consumer restarts about every 40 seconds, and the same offset appears in every startup log while other partitions stay healthy. Events are committed in batches of a few thousand and the acknowledgement follows the commit. Sealing is six hours away and source_max_ingested_at for that partition's tenants is frozen. Give an ordered checklist, a containment action available within minutes, and the durable fix, saying what each does to exactly-once accounting.
Approach
- Distinguish a poison record from a capacity problem in one measurement: compare the offset and the exception across restarts. An identical pair every time is deterministic failure on one record, whereas a throughput problem still advances the offset between crashes.
- Read the record from a separate consumer group so the bytes can be inspected without perturbing the stuck consumer, then classify the defect: schema violation, a quantity failing the non-negative check, a null workspace, an unmappable SKU enum, or a payload past a size limit. That classification decides whether this is a producer bug or a missing consumer guard.
- Account for batch granularity before acting. With commits of a few thousand, one bad record fails thousands of good ones, so the blast radius is the batch. Halve the batch around the offset to isolate the record, or move to per-record error isolation so the radius becomes the record.
- Contain by diverting that record to a dead-letter store with its raw bytes and offset, then resume. This is safe here precisely because the acknowledgement follows the commit: the good records from the failed batch are re-consumed and absorbed by the uniqueness check on (tenant_id, idempotency_key) rather than counted twice.
- Make the fix durable with per-record error isolation, a bounded poison counter, and an alert on dead-letter rate rather than on lag alone, since lag only reveals this after the sealing margin has already been eaten.
- Check the horizon before replaying anything. The unique index lives on a daily-partitioned table and therefore includes the partition key, so it deduplicates within a day only; a replay landing on a later ingest day needs the separate dedup store or it double-counts into a tenant's bill.
Follow-up
- Move the acknowledgement before the commit and describe exactly what is lost and what is duplicated in each of the two crash windows.
- Sealing is in six hours and the partition will not drain in time. What do you seal on, and what does the invoice have to record so the difference is explainable later?
- A producer replays two weeks of events next month. Which part of your fix stops holding, and what is the dedup horizon you would actually configure?
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.
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.
Describe a time you had to solve a complex technical issue under a tig…
Describe a time you had to solve a complex technical issue under a tight deadline.
Approach
- Name the disagreement and how you resolved it with evidence.
- Close with what you would do differently, concretely.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
How do you handle conflicting requirements from different engineering …
How do you handle conflicting requirements from different engineering stakeholders?
Approach
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
- Close with what you would do differently, concretely.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
Describe your experience with PLM systems and how you integrate softwa…
Describe your experience with PLM systems and how you integrate software solutions into mechanical or systems engineering workflows.
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Give the blast radius: what could have broken, and what you measured.
- Close with what you would do differently, concretely.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
- 01
Describe a time you had to solve a complex technical issue under a tight deadline.
- 02
How do you handle conflicting requirements from different engineering stakeholders?
- 03
Describe your experience with PLM systems and how you integrate software solutions into mechanical or systems engineering workflows.
Is this an official Systems Engineering and Assessment Ltd interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Systems Engineering and Assessment Ltd. Rounds and questions reflect what candidates have reported, not a process Systems Engineering and Assessment Ltd 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?
The timeline can vary, and candidates report a thorough assessment. Expect the process to move at a pace that lets interviewers evaluate your skills against the company's project needs.
PracHub interview research ↗How should I prepare for the technical portion of the interview?
Focus on your past projects. Be prepared to discuss the specific technical challenges you faced, the decisions you made, and the outcomes of your work.
PracHub interview research ↗Is there a focus on specific technologies?
Candidates report that, alongside general engineering principles, interviewers look for adaptability. Be prepared to discuss how you have applied your skills across different tools and environments.
PracHub interview research ↗What differentiates successful candidates?
Successful candidates demonstrate both deep technical competence and a genuine interest in the broader engineering context of the company's work.
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
