A Software Engineer at Surrey Satellite Technology (SSTL) operates at the cutting edge of space exploration and satellite communications. This role is not merely about writing code; it is about engineering robust, reliable solutions for environments where hardware access is impossible once a mission launches. You will contribute to the design, development, and testing of software that powers micro-satellites, ensuring they function autonomously in the extreme conditions of space.
Your work directly impacts the success of satellite missions, ranging from imager electronics to complex payload management. Because Surrey Satellite Technology is a pioneer in the small satellite industry, you will be expected to balance technical precision with the practical, fast-paced realities of space hardware development. The role is highly collaborative, requiring you to bridge the gap between software logic and physical engineering constraints, such as vibration, radiation, and power management.
Technical Assessment
reportedMost of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.
What to demonstrate
- Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
- Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
- Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
- Whether a failing case is isolated and explained before any edit is made to the code
How to prepare
- From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
- Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
- Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.
Project Presentation
reportedWhen a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.
What to demonstrate
- Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
- Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
- Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
- Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened
How to prepare
- Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
- Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
- Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure
Group Exercise
reportedYou cannot drill a format you do not know, so put the preparation into material that travels. Three pieces of your own work, each rehearsed until you can take a follow-up you did not anticipate, will carry a conversation or a code walkthrough equally well. Specificity is what separates that from filler. A number needs its definition before it means anything: a p99 is over some window and measured at some hop, and a server-side figure excludes the queueing and network time a client would see. The number you cannot qualify is the one to leave out.
What to demonstrate
- Whether your examples carry detail only someone who did the work would hold, such as what the binding constraint actually was, which alternative you rejected and why it was worse, and what you measured on each side of the change
- Whether a number survives one follow-up, meaning you can say what it was measured over and whether it moved because of your change or merely alongside it
- Whether a failure is described with the specific change that followed it, rather than a lesson stated in general terms
- Whether your part in a team effort is stated accurately, including what other people did
How to prepare
- Write a page on each of three projects covering the constraint, the option you rejected, the measurement before and after, and what went wrong. Cut any line you cannot take a follow-up on, since you are writing the parts you will be pressed on rather than a summary.
- Recover the real figures while you still have access: request volume, data size, latency with its percentile and window, team size, timeline. Note where each came from, whether a dashboard, a design document or memory, and mark the estimates so you can say which they are out loud.
- Take your weakest project story to someone who works in a different area and have them ask why four times in succession. The point where you run out of answer is the part to go and re-read before the round.
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.
Checking a quota with a select and then writing
Under read-committed isolation, two concurrent transactions both observe a count below the limit and both insert, so the limit is exceeded by exactly the concurrency. Repeatable read does not rescue it either: it provides a stable snapshot, and this is write skew, which snapshot isolation permits by design. The options are serialisable isolation, which detects the conflict and aborts one transaction with a serialisation failure and therefore obliges the caller to retry; a single statement with the predicate inside the write; or a constraint that makes the surplus insert fail outright. The reason this pattern survives review is that it is correct in every test that runs one request at a time.
A queue or buffer with no bound
Every producer-consumer boundary needs a capacity and a policy for reaching it: block the producer, shed load, or drop the oldest entry. Unbounded buffering converts a temporary slowdown into memory exhaustion and hides the backpressure signal that would have revealed the consumer was falling behind.
Hardcoding to the sample inputs
Solve the stated problem rather than the two examples; special-casing a literal to make a sample pass is obvious immediately and reads as either a misunderstanding or an attempt to fake progress. If you genuinely cannot generalise yet, say which part is a stub and what would replace it.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Fold a deduplicated usage stream into hourly rollups
You are given one day of usage_event rows, up to 250 million, each carrying event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at and ingested_at. Produce usage_rollup_hourly cells keyed (tenant_id, workspace_id, sku, hour_start) with quantity_sum, event_count and source_max_ingested_at. An event counts once per (tenant_id, idempotency_key). The rollup grain has no environment column, so state your filter. One pass. Give your time and space bounds, and say what the deduplication actually costs in memory.
Approach
- Bucket on
occurred_at, neveringested_at:hour_start = date_trunc('hour', occurred_at at time zone 'UTC'). The two columns answer different questions.occurred_atsays which hour the customer is billed for;ingested_atsays how current the fold is. Using the second for the first makes late data invisible instead of correctable. - The fold is trivial and the deduplication is the entire cost, so price it before designing anything clever. An exact set over
(tenant_id, idempotency_key)at 250M entries, stored as a 16-byte 128-bit hash in an open-addressed table at 0.7 load factor, needs about 357M slots at 16 bytes each, roughly 5.7 GB. The fix is partitioning byhash(tenant_id) % Pso each shard holds 1/P of the set and no tenant's keys straddle shards. - Rule out a Bloom filter as a replacement, in the right direction: a false positive reports 'already seen' for an event never seen, so you drop a real event and lose revenue with no error raised. It is usable only as a negative pre-filter in front of the exact set, where a miss is conclusive and a hit must fall through to the real lookup.
- Accumulate in scaled integers, not binary floating point.
numeric(20,6)admits values below 10^14, so one event scaled to micro-units can reach 10^20, past int64's 9.22 x 10^18; use a 128-bit or arbitrary-precision accumulator unless you first bound the per-event maximum. binary64 represents integers exactly only to 2^53, about 9.01 x 10^15, and cannot represent 0.1 at all, so two runs that sum in different orders disagree. - Carry
source_max_ingested_at = max(ingested_at)over the events folded into each cell, and countevent_countover accepted, post-dedup events. Without that watermark there is no way to prove later what a number did and did not include, which is the first question any reconciliation asks. - State the environment filter explicitly, because the rollup grain cannot record it. A fold that quietly includes
stagingbills non-production traffic; one that quietly excludes it loses a cost signal. Production-only is the billing answer, and either way it belongs in the job name and the output metadata. Complexity: O(n) time, O(distinct dedup keys) space, dominated by the dedup set rather than by the cells.
Follow-up
- A producer retries at 23:59:59 and the retry lands at 00:00:01. The unique index on the daily-partitioned table must include the partition key. What gets double-counted, and what is the smallest change that fixes it?
- The consumer acknowledges its batch before committing the fold. Which failure loses revenue now, and which arrangement duplicates instead?
- What makes a re-run over the same day produce byte-identical rollups?
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?
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.
Worked solution 15 min
- Write the grammar on one line before coding:
header := field (',' field)*,field := key '=' value, split on the first=only. - Implement the parser to return
{t: int, v1: [hex, ...]}, rejecting a missingt, a duplicatet, anyv1that is not 64 hex characters, and a header over 1 KiB, all before any cryptography runs. - Implement the verifier: for each active secret compute
HMAC-SHA256(secret, f'{t}.'.encode() + raw_body), compare it in constant time against each parsedv1, and OR the results with no early exit. - Test with a valid signature; the same body with
tmoved 400 seconds into the past; the same body witht400 seconds into the future; a header carrying an unknownv2=alongside a validv1; and a body re-serialised with different JSON key order.
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?
Enforce a concurrent-run quota that survives simultaneous requests
A plan allows at most 20 concurrently running rows in job_run per tenant. The table holds run_id, tenant_id, workspace_id, status (queued, leased, running, succeeded, failed, timed_out, cancelled, lost), lease_token, leased_until, started_at and finished_at. Today the service runs select count(*) from job_run where tenant_id = $1 and status = 'running', compares the result to 20, then inserts. Under load a tenant exceeds the cap by exactly the number of concurrent requests. Name the anomaly, say which isolation levels do and do not prevent it, and give a version that holds, as SQL.
Approach
- Name it: write skew. Each transaction reads a predicate (the count of running rows), neither modifies what the other read, and both then insert rows that jointly violate an invariant no single row expresses. Read committed permits it. So does repeatable read, because snapshot isolation's first-updater-wins check fires only on conflicting row updates, and these are inserts touching disjoint rows.
- Enumerate the fixes with their real costs. SERIALIZABLE works: PostgreSQL's SSI tracks the predicate read and aborts one transaction with SQLSTATE 40001, which obliges the caller to retry and makes the abort rate rise with contention on a hot tenant. Folding the predicate into the write as
insert ... select ... where (select count(*) ...) < 20narrows the race to the statement's snapshot but does not close it under read committed. - Give the version that holds at read committed: serialise on a row both transactions must touch.
update tenant_concurrency set running = running + 1 where tenant_id = $1 and running < 20 returning runningupdates zero rows when the cap is reached, and zero rows is the rejection. This works because at read committed a blocked UPDATE re-evaluates its WHERE clause against the newly committed row; at repeatable read the same statement raises a serialisation error instead, so the isolation level changes the calling contract. - State the cost you just bought. That row is now a per-tenant serialisation point, so admission throughput for the tenant is bounded by one divided by the lock hold time; at a 2 ms hold that is roughly 500 admissions/second. Keep the critical section to the single UPDATE, with no network call or scheduling decision inside the transaction, and decrement in the same transaction that writes the terminal status.
- Close the leak the status enum implies: a run can end as
lost, so a crashed worker otherwise consumes a slot forever. Reconcile on a schedule againststatus = 'running' and leased_until < now(), and treat the counter as a fast path overjob_run, which stays the system of record.
Follow-up
- Write the retry loop for the SERIALIZABLE version. What does the caller see when it keeps aborting, and what bounds the retries?
- Two regions each keep a counter. What is the effective cap, and what does admission do when the counter store is unreachable?
- The cap changes mid-flight on a plan upgrade. Do running jobs get killed, and what does the counter row look like during the change?
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.
Questions regarding basic engineering knowledge and optical engineerin…
Questions regarding basic engineering knowledge and optical engineering concepts.
Approach
- Say what you would check first and why it is the highest-information step.
- State your assumptions explicitly before working the problem.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Explain the overall network and software architecture in a micro-satel…
Explain the overall network and software architecture in a micro-satellite.
Approach
- Work from the requirement backwards to the design.
- 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?
Leasing sandboxed job runs without double execution
job-runner leases work from a queue and executes customer workloads in sandboxes with CPU, memory, wall-clock and egress limits. Run durations span 200 ms to 30 minutes, thousands are concurrent, and job_run.status must reach exactly one terminal state among succeeded, failed, timed_out, cancelled and lost. A worker can pause 45 seconds for garbage collection, or be briefly partitioned, after its sandbox has exited but before it writes the outcome. Design the lease duration, the fencing, and timeout ownership, and state how billable_seconds is computed for a run whose worker never returns.
Approach
- Resolve the lease-duration conflict rather than picking a compromise. A lease shorter than the longest legitimate run re-dispatches work that is still executing; a lease long enough for a 30-minute run leaves a crashed 200 ms run undetected for half an hour. Use a short lease - about 30 seconds - renewed by a heartbeat every 10 seconds while the supervisor is alive, each renewal pushing leased_until to now() + 30 s. Detection latency is then one lease, not one heartbeat: a dead worker's last successful heartbeat landed at most 10 s before it died, so leased_until expires 20 to 30 s after the death, plus whatever interval the reaper scans on. What the 3:1 ratio of lease to heartbeat buys is headroom - one or two missed heartbeats change nothing, and a stall only becomes a re-dispatch once it outlives the lease remaining when it began, between 20 and 30 s here. Both costs follow from those two numbers: heartbeat write load proportional to concurrent runs (3,000 runs at a 10-second interval is about 300 updates/second), and any pause longer than the lease - the 45-second GC - re-dispatching a run that is perfectly healthy.
- Fence the outcome write so the store, not the worker's memory, arbitrates: UPDATE job_run SET status = $1, finished_at = $2, exit_code = $3 WHERE run_id = $4 AND status = 'running' AND lease_token = $5, and require exactly one row affected. Zero rows means this worker was fenced while it was paused, and its correct behaviour is to discard the result, not to retry - the retry is how a resumed worker overwrites a newer attempt.
- Make the status machine monotone with a check constraint or a trigger so no terminal row can move, and model a retry as a new row with parent_run_id set and attempt incremented rather than a reset of the old one. That is what keeps how many times did this actually execute answerable afterwards and keeps each attempt's resource usage attributable to itself.
- Put the wall-clock timeout in the supervisor, never in the workload: a customer-supplied program asked to time itself out will not. Distinguish timed_out, which the supervisor observed and caused, from lost, which nobody observed at all, and leave exit_code null unless the supervisor actually saw the process exit. Recording lost as failed asserts an outcome no one witnessed and then bills and retries on that assertion.
- Decide billable_seconds for the unobserved case explicitly, because it cannot be derived. It is not now() - started_at, which bills queue and pause time the customer never consumed. The two defensible policies are leaving it null while status = 'lost' and billing nothing, or flooring it at the last heartbeat's observed running time as a stated lower bound; pick one and write it down. Add a per-tenant concurrency cap on sandbox slots so one tenant cannot occupy the whole fleet while these questions are being answered.
Worked solution 30 min
- Draw the timeline: last successful heartbeat at t, sandbox exits at t, worker pauses from t to t+45 s, so leased_until = t+30 s and the lease expires there. Mark where the second worker starts and where the first worker's write arrives, then say how short the pause would have had to be to change nothing.
- Write the fenced terminal update and state the expected rows-affected in both the healthy case and the fenced case.
- Size the heartbeat: 3,000 concurrent runs at a 10-second interval is about 300 row updates/second on job_run - decide whether that lands on the same table and what it contends with.
- Write the billable_seconds rule for each terminal status and say which status leaves it null.
Follow-up
- A worker returns from a 45-second GC pause and writes succeeded. Walk through what the database does statement by statement.
- Heartbeats are now a few hundred writes a second against job_run. How do you keep that off the path that matters, and what do you lose by moving it?
- The re-dispatched workload has an external side effect the first attempt already performed. What does your design owe the customer, and what can it not fix?
One tenant's counter writes stall the whole connection pool
A change that made a per-tenant usage counter correct now produces site-wide latency whenever one large tenant writes: unrelated endpoints time out waiting for a connection while database CPU stays low and no statement is slow. The change wraps the counter update in a transaction that takes SELECT ... FOR UPDATE on one row, calls an external pricing service, then updates and commits. Give an ordered checklist, the arithmetic that bounds that tenant's write rate, and three repairs with the cost each one accepts.
Approach
- Separate waiting from working. Low database CPU alongside high application latency points at a queue, so instrument connection-acquisition wait separately from query execution time; that queue forms in the application and is invisible in database metrics, which is why the database looks healthy throughout.
- Confirm the lock rather than assuming it: sample waiting sessions and group by wait event, relation and tuple. Contention concentrated on one tuple belonging to one tenant is the signature; a deadlock would instead show the database aborting transactions after its detection timeout, which is not happening here.
- Do the arithmetic out loud. Throughput on a serialised row is one divided by the lock hold time, and the hold spans the external call, so a 20 ms pricing call caps that tenant near 50 writes per second no matter how many pods run. Every waiter also holds a pooled connection while it queues, so the shared pool drains and unrelated tenants fail at acquisition.
- Repair one: shrink the critical section to a single statement with the price resolved before the transaction opens. Cost is a stale price for the duration of one request and a second round trip; benefit is a hold time measured in the database's own execution time.
- Repairs two and three change where the contention lives rather than how long it is held. Sharding the counter into per-(tenant, bucket) rows and summing on read multiplies write throughput by the shard count, at the cost of an aggregate on every read and a shard count you must size against the largest tenant rather than the median. Accumulating in memory and flushing periodically removes the per-write round trip entirely, paid for with a bounded loss window on crash, which is acceptable for a rate limiter and not for a billing counter.
- Contain independently of which repair wins: a separate pool or per-tenant concurrency cap for this write class, a statement timeout low enough that a pathological query dies before it accumulates waiters, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks.
Follow-up
- What would a genuine deadlock look like here, which two code paths would produce one, and how does the database's response differ from what you observed?
- If a transaction-pooling proxy sits in front of the database, which of your three repairs changes behaviour, and what stops working that would have worked on a direct connection?
- The counter also enforces a quota. Why is SELECT the count and then INSERT still wrong after you have fixed the contention?
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.
For anything that touched live traffic, be ready to say how you would have undone it: a flag, a staged rollout, dual writes with the old path still authoritative. Once the old column is dropped or the source rows are overwritten there is no reverse, so name what you kept a copy of and for how long.
Give an example of a time where you solved a problem using communicati…
Give an example of a time where you solved a problem using communication skills.
Approach
- Pick a story where you made the decision, not one where you watched it.
- Close with what you would do differently, concretely.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
What are your interests outside of work?
What are your interests outside of work?
Approach
- Close with what you would do differently, concretely.
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
Why are you looking to leave your current role for this one?
Why are you looking to leave your current role for this one?
Approach
- Give the blast radius: what could have broken, and what you measured.
- Pick a story where you made the decision, not one where you watched it.
- Close with what you would do differently, concretely.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
What is your experience with link budgets?
What is your experience with link budgets?
Approach
- Close with what you would do differently, concretely.
- 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.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
- 01
Give an example of a time where you solved a problem using communication skills.
- 02
What are your interests outside of work?
- 03
Why are you looking to leave your current role for this one?
- 04
What is your experience with link budgets?
Is this an official Surrey Satellite Technology interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Surrey Satellite Technology. Rounds and questions reflect what candidates have reported, not a process Surrey Satellite Technology has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗Is the interview process difficult?
The difficulty is generally perceived as average. The focus is more on your ability to apply engineering logic to practical problems rather than solving complex, abstract algorithmic puzzles.
PracHub interview research ↗How much time should I spend preparing for the presentation?
You should dedicate significant time to your project presentation, as it is a core part of the assessment. Ensure you can explain the "why" behind your technical decisions, not just the "how."
PracHub interview research ↗What is the culture like at Surrey Satellite Technology?
It is generally described as friendly and relaxed. The team values direct communication and a "get the job done" mentality, focusing more on results than corporate hierarchy.
PracHub interview research ↗How long does the hiring process take?
Timelines can vary significantly. While the process is designed to be efficient, expect potential delays in communication. Stay proactive and professional in your follow-ups.
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
