As a Software Engineer at Pursuit Aerospace, you play a critical role in bridging the gap between high-level engineering objectives and the functional software systems that power the company’s aerospace operations. This position is not merely about writing code; it is about contributing to the structural integrity and efficiency of aerospace processes, requiring a blend of technical precision and a commitment to continuous learning.
You will be working in an environment where your contributions have a tangible impact on the business. Whether you are automating workflows, developing internal tools, or supporting complex engineering projects, your work directly influences how the team operates. Pursuit Aerospace values engineers who bring a proactive mindset, a willingness to tackle complex problems, and the ability to thrive in a collaborative, often tight-knit team environment.
Initial Screening
reportedThe person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.
What to demonstrate
- Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
- Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
- Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural
How to prepare
- Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
- Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
- Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one
In-Person Component
reportedBecause the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.
What to demonstrate
- Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
- Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
- Whether you can be redirected in the middle of an answer without restarting it from the beginning
- Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it
How to prepare
- Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
- Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
- Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing
PracHub editorial advice for the preparation topics above.
Retrying an irreversible external call without an owned idempotency key
Timeouts and connection resets are ambiguous by construction - the partner may have committed before the response was lost - so a generic retry policy around an HTTP client is, in this domain, a machine for buying two labels and dispatching two trucks. Relying on the partner's deduplication is not a substitute, because their window is usually short, their key is often derived from fields you may legitimately change on retry, and many older integrations have no such concept at all. The workable pattern is to generate the key yourself, persist it with an explicit unknown status before the call, and resolve ambiguity by querying the partner for that key rather than re-issuing; the sweeper that does this is the component that has to be correct, not the call site. Teams get this right for payments and then forget that a carrier tender, a warehouse work release and an EDI shipping notice have the same shape.
Treating the database as the truth about physical stock
Every system in this domain drifts from reality, through miscounts, unrecorded damage, mis-scans and theft, so a design with no adjustment path forces the drift somewhere worse. The usual shape is a CHECK constraint or application guard that clamps a position at zero: the negative that would have exposed a missing receipt is silently swallowed, and the first visible symptom is a picker standing at an empty location for an order the system promised. The correct posture is to let the invariant fail loudly - record the cycle-count adjustment as a first-class movement with its own reason code, surface the variance, and let planning see that this location is unreliable - because an adjustment that is indistinguishable from a normal pick destroys the only signal that would have found the root cause. A related version of the same mistake is deleting or updating a movement row to fix a mistake, which leaves the position correct and the history a lie.
Not asking what the system looks like if it dies halfway through
For any multi-step write, say what state remains if the process stops between step two and step three, and what brings it back: a single transaction, a saga with compensating actions, an outbox, or a reconciliation job. Partial failure is routine at any real call volume, so 'that shouldn't happen' is an answer with nothing behind it.
Optimising an axis nobody named
Ask which resource is actually scarce here: wall-clock latency, throughput, memory footprint, cost per request, or engineering time. Shaving a constant factor off an in-memory step is wasted effort when the same function makes a blocking remote call inside the loop.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Trace a suspect lot forward to every shipped order line
lot_edge(parent_lot_id, child_lot_id) has 400 million rows with btree indexes on both (parent_lot_id, child_lot_id) and (child_lot_id, parent_lot_id). lot_shipment(lot_id, order_line_id) maps lots to shipped lines. Given one suspect received lot, return every downstream order_line_id. The reachable subgraph can reach 2 million lots and 12 levels; the full edge table does not fit in memory. Give the traversal, its complexity in terms of the reachable subgraph, and what you do if the data contains a cycle.
Approach
- Genealogy is a directed acyclic graph, not a tree: a production run consumes several lots and one lot splits across many shipments, so a node is reached by several paths and a tree traversal revisits it combinatorially. Breadth-first from the suspect lot over the
(parent_lot_id, ...)index, with a hash set of visited lot ids, is O(V' + E') where V' and E' count only the reachable subgraph — 2 million 8-byte ids is 16 MB of payload and a few tens of MB with open-addressing overhead, which is why the visited set is affordable and the edge table's 400 million rows never enter memory. - Batch each level into one indexed read. Issue
WHERE parent_lot_id = ANY($1)with a few thousand parents per call instead of a query per node: at 12 levels and 2 million nodes, per-node round trips dominate the wall clock entirely, and the recall's deadline is a real-world response time, not a query-plan aesthetic. The batch size is the tuning knob between round-trip count and per-statement planning cost. - Keep the visited set even though the schema says DAG. Rework and repack loops — material returned into an earlier lot — do occur, and an unguarded traversal on a cyclic edge set does not terminate; a set membership test is O(1) and turns a hang into a correct answer plus a data-quality alert. Also cap the traversal by level and by node count, and fail loudly on the cap rather than returning a partial set silently.
- Join to
lot_shipmentonce at the end over every visited lot id, not over the leaves and not per node. A lot is routinely both shipped and consumed: half a lot goes out on an order line while the other half is repacked into a child lot, so it has alot_shipmentrow and outgoing edges at the same time and is not a leaf. Restricting the join to nodes with no outgoing edges drops exactly those order lines, which is under-recall — the one failure mode this trace is not allowed to have. One bulk join against an index onlot_idover the whole visited set is a single pass; per-node lookups multiply the round-trip problem by the number of visited lots. - State the semantics honestly: reachability answers which shipped lines could contain material from the suspect lot, and without propagating edge quantities it is a superset of physical containment. For a recall that is the correct direction of error — over-recall, never under-recall — and saying so is the point, because a trace that returns quickly and incompletely will be believed. The upstream question, which received lots are in a suspect shipped unit, is the same traversal over the reverse index, which is why both indexes exist.
Follow-up
- A recursive CTE would express this in ten lines. What is its cost here if the index on the traversal column is missing, and how would you notice?
- You need the quantity of suspect material in each downstream line, not just reachability. What has to be on the edge, and what does that do to the traversal?
- The traversal must run while a production run is writing new edges. What does a consistent answer mean here, and what do you cut it at?
Re-sum a movement ledger and report positions that disagree
inventory_movement holds 3 billion immutable rows: item_id, node_id, lot_id (NULL when the item is lot-untracked), state, delta_qty (signed int64 in the item's smallest transacting unit), uom_code, and reversed_by. inventory_position holds one row per (item_id, node_id, lot_id, state) with qty, version and last_movement_id, using lot_id = 0 as the sentinel for untracked. Up to 80 million distinct keys. Report every key whose summed movements disagree with the stored qty, while writes continue. State your time and space bounds and how you bound the comparison.
Approach
- Fix the key first.
inventory_movement.lot_idis NULL for untracked items andinventory_position.lot_idis 0, so the group key is(item_id, node_id, COALESCE(lot_id, 0), state). Getting this wrong does not error — it silently produces two groups that each look like a variance, and the report becomes noise nobody reads. - Cut the ledger at a watermark, but do not take
W = max(movement_id)at the start of the scan. A sequence hands out an id before the inserting transaction commits, so at the instant you read that maximum there are ids below it still in flight and invisible to your snapshot. Summingmovement_id <= Wmisses them on this run, and advancingreconciled_through_movement_idtoWmakes the miss permanent: every later incremental run starts above those ids and they are never summed again. That is a hole in the ledger's own re-derivation, not a transient skew. - Take a watermark that is provably settled instead. Bound write transactions with a statement or transaction timeout so 'the longest write' is a number T, record
(observed_at, max_movement_id)samples periodically, and use asW_safethe largest sampled id whoseobserved_atis older than T — every id at or below it has committed or rolled back. Sum rows withmovement_id <= W_safe, compare only against position rows withlast_movement_id <= W_safe, treat anything newer as a write that raced you rather than a variance, and advancereconciled_through_movement_idonly toW_safeso the next run starts there instead of re-reading 3 billion rows. If the ledger carries a commit timestamp, cutting on that is the same guarantee without the sampling table. - Sum
delta_qtyas int64, and include reversal rows.reversed_byis a back-pointer for audit, not an exclusion filter: an original of +10 and its reversal of -10 must both be summed to reach 0. Excluding the original while keeping the reversal produces -10 and a false variance on every corrected key. - Guard the denomination rather than trusting it.
uom_codeis stored per row because pack factors change over time, so reject — do not sum — any row whoseuom_codeis not the item's smallest transacting unit, and report those keys separately. A mixed-denomination sum is arithmetically meaningless and looks exactly like a real variance. - Hash aggregation is O(n) time and O(distinct keys) space: 80 million entries at roughly 40-56 bytes each in a typical runtime is 3-5 GB, so quote the number. The bounded-memory alternative is an external sort-merge on the group key — O(n log n) comparisons, resident memory bounded by the merge fan-in rather than by key count, and it streams — or hash-partition by
hash(item_id) % Pand run P independent passes for 1/P of the peak.
Worked solution 25 min
- Write the group key expression down explicitly, including the NULL-to-0 mapping, before any aggregation code.
- Derive
W_safefrom a sample older than the transaction-duration bound T, and state in one sentence which rows on each side are in scope — this is the part that makes the report trustworthy under live writes, and the part that decides whether an in-flight row is re-read next run or lost forever. - Implement the hash aggregation with an int64 accumulator and a
uom_codeassertion per row, emitting rejected rows to a second output rather than into the sum. - Build a fixture with four keys: one clean, one with a movement missing from the position, one with an original plus its reversal, one with a row in cases where the item transacts in eaches.
- Run it, then re-run with the fixture rows shuffled, with a concurrent writer appending past the watermark, and with a writer that reserved an id below
max(movement_id)and commits only after the scan finishes — then run the incremental pass and confirm that row is summed there.
Follow-up
- A key shows a variance of exactly one pick, repeatedly, at one node. What do you look at first, and what would distinguish a duplicate movement from a missed one?
- The job takes six hours and the variance report is stale by the time anyone reads it. How would you make it incremental without losing the guarantee that it re-derives from the ledger?
- Who writes the correcting movement, and what reason code does it carry?
Find the hottest write keys under a fixed memory budget
One day of writes to inventory_movement is 200 million rows; each carries (item_id, node_id). Distinct pairs reach 60 million. Return the 200 pairs with the most writes so they can be moved behind a single writer before the next peak. First give the exact answer assuming you may hold 60 million counters, then give an answer under a 200 MB budget, and state the error guarantee that second answer carries. Both must be one pass over the day's rows unless you argue for a second.
Approach
- Exact version: one hash pass to build
(item_id, node_id) -> int64, then a single scan of the map maintaining a min-heap of size K = 200, pushing and popping when the incoming count exceeds the heap root. O(n + d log K) with n = 2e8 and d = 6e7, versus O(d log d) if you sort all 60 million entries — a factor of roughly log d / log K, about 3.4x, for no benefit. The heap must be a min-heap: the root is the weakest survivor, so eviction is O(log K). - Cost the map honestly. 60 million entries at a 16-byte key, an int64 count and per-entry overhead is 2-3 GB in most runtimes, which is the whole reason for the second version rather than a theoretical aside.
- Bounded version: Misra-Gries with m counters. Keep at most m keys with counts; on a key already present increment it; on a new key when fewer than m are held, insert at 1; otherwise decrement every held counter and drop those reaching zero. With m = 1,000,000 at roughly 40 bytes per counter, that is about 40 MB, comfortably inside 200 MB alongside the heap.
- State the guarantee as a number, not as 'approximate'. Misra-Gries returns an estimate satisfying f - n/(m+1) <= f_hat <= f, so with n = 2e8 and m = 1e6 every count is an undercount by at most 200. If the 200th hottest pair has hundreds of thousands of writes, a 200-count error cannot reorder the top of the list — and if the counts near rank 200 are within 200 of each other, the ranking there was never meaningful for the decision being made. The sketch also merges: run one per shard, combine, prune back to m, and the bound holds with n the combined length.
- If exactness is genuinely required, argue for two passes rather than more memory: the Misra-Gries pass yields a candidate superset of size m, and a second pass counts only those candidates exactly in O(m) memory. That is the standard way to get an exact top-K under a memory bound, and the cost is one extra read of the day's rows.
Follow-up
- The decision downstream is which keys get a single writer. Does a rank error at position 190 change that decision, and what does the answer tell you about how much accuracy to buy?
- Write volume on one key is bursty — a promotion concentrates it into twenty minutes. Does a daily top-K find it, and what window would?
- How do you keep the sketch across a worker restart without replaying the day?
Add a required unit column and uniqueness to a live ledger
inventory_movement holds 4.2 billion rows on PostgreSQL 15 and takes about 2,000 inserts per second around the clock. Two changes are required with no downtime: add uom_code TEXT NOT NULL, whose truthful value must come from the conversion factors in force at each row's occurred_at rather than from today's item master; and make idempotency_key UNIQUE, although duplicates predating the retry fix already exist. Give the ordered steps, the lock each one takes, its expected duration and its rollback.
Approach
- ADD COLUMN with a non-volatile DEFAULT has been metadata-only since PostgreSQL 11, so no rewrite, but it still takes ACCESS EXCLUSIVE for an instant. That lock queues behind any long-running reader while blocking every statement behind it, so one 10-minute analytics scan turns a millisecond change into a 10-minute outage. Set lock_timeout to 2s and retry in a loop rather than waiting.
- Reach NOT NULL without a scan under an exclusive lock: ADD CONSTRAINT ... CHECK (uom_code IS NOT NULL) NOT VALID, which is instant, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE and runs concurrently with inserts, then SET NOT NULL, which PostgreSQL 12 and later proves from the validated check and so skips its own scan. Drop the redundant check afterwards.
- The truthful backfill is the long pole and must read the versioned conversion factor as of each row's occurred_at, not the current item master, or a supplier moving a case from 12 units to 10 retroactively reinterprets last quarter. Batch by primary-key range, a few thousand rows per transaction, with a sleep between batches and a resumable cursor in a control table. At 5,000 rows a second, 4.2 billion rows is roughly ten days of throttled work: that is a schedule, not a failure, and it must not block the deploy.
- The UNIQUE index cannot be built until the duplicates are gone. CREATE UNIQUE INDEX CONCURRENTLY aborts on the first violation and leaves an INVALID index behind that is ignored by queries but still carries update overhead on every insert, and it can only be removed with DROP INDEX CONCURRENTLY. So: find the duplicates, resolve each one as a compensating reversal rather than a DELETE, since every duplicate key is a double-applied movement and deleting it fixes the constraint while leaving the position wrong.
- Then CREATE UNIQUE INDEX CONCURRENTLY, which makes two passes over the table and waits out concurrent transactions, so budget hours here, and note it cannot run inside a transaction block: a migration framework that wraps every step in BEGIN will fail on this one. Promote it with ALTER TABLE ... ADD CONSTRAINT ... UNIQUE USING INDEX, which is a brief ACCESS EXCLUSIVE.
- Question whether the constraint should cover all history. At roughly 40 bytes per key plus overhead, a btree over 4.2 billion rows is in the hundreds of gigabytes, while a retry never arrives a year later. A partial unique index bounded by a fixed occurred_at literal, rotated forward on a schedule, gives the same protection for a fraction of the size, and the predicate must be a literal because now() is not immutable.
Worked solution 45 min
- Write the migration as numbered steps with the lock mode, expected duration and rollback beside each, including the lock_timeout retry wrapper on every ACCESS EXCLUSIVE step.
- On a copy at reduced scale, run ADD COLUMN while a long SELECT is open and observe writers queueing behind the lock request, then repeat with lock_timeout set and confirm the migration backs off instead.
- Run the batched backfill against a versioned conversion table, kill it mid-run and restart it, confirming the control-table cursor resumes without redoing or skipping a batch.
- Seed duplicate idempotency keys, attempt CREATE UNIQUE INDEX CONCURRENTLY, and confirm the INVALID index left behind, then drop it concurrently, resolve the duplicates as reversals and rebuild.
- Compare the full unique index size against the partial one bounded by a fixed occurred_at literal.
Follow-up
- Sequence the application deploy against these steps. When exactly may the code start writing uom_code, and when may it start requiring it?
- The backfill is on day six and the conversion-factor table is itself corrected. What happens to the rows already written?
- How do you roll back after SET NOT NULL is in place and half the traffic depends on the column?
Decide whether available-to-promise is derived or stored
Available to promise at one node equals inventory_position.qty in state on_hand minus the sum of allocation.qty in status held or committed for that key. The read sits inside checkout: 4,000 reads per second at peak, p99 budget 20 ms. inventory_position holds 40 million rows; during a promotion a single hot item-node pair carries about 2,000 open allocations. Decide whether to compute this per read or to maintain available_qty on the position row. Give both paths, the failure mode of each, and what the browse read may do that the allocate read may not.
Approach
- Cost the derived read in tuples, not in plans. One index probe into the position is trivial; the aggregate over allocation touches one tuple per open claim on that key. Even if only a tenth of peak lands on the hot pair, 400 reads per second times 2,000 rows is 800,000 index tuples per second for one item at one node, and it gets worse exactly as the key gets hotter. A partial index on status IN ('held','committed') does not help, because the row count is the cost, not the selectivity.
- Maintaining available_qty on the position row is nearly free on the contention axis, which is the point most candidates miss. The allocator already takes a row lock or a compare-and-set on that same position row, so the extra column costs no lock that was not already held, and the read collapses to a single primary-key probe well inside 20 ms.
- It is not free on the correctness axis. available_qty becomes a second derived value that can drift, so it joins qty under the same reconciliation job with its own re-sum against the allocation table, and a drift alert on that comparison is the operational price of the read speed.
- Split the two reads rather than arguing about one. The browse read may serve from a replica or a short-TTL cache and must err low, showing less than is there. The allocate read must be on the primary and inside the same transaction as the conditional update, because replication lag peaks during the write burst that made the key hot. Conflating them is what makes a replica read look safe in a benchmark.
- Pick the TTL against the drain rate, with a number. If a promotion burns 2,000 units in a minute, a 30-second cache is a thousand units stale, so the displayed figure should be a banded signal such as in stock or low stock rather than an exact count that the allocate step will then contradict.
- Do not denormalise availability across nodes into a single network number. A yes-somewhere answer cannot be allocated against, so sourcing will disagree with the page, and the customer sees a promise retracted after checkout rather than an honest out-of-stock before it.
Follow-up
- Roll this out on a live system. How do both paths run side by side, what do you compare, and what result gates the cut-over?
- The reconciliation job finds available_qty drifted by 3 on one key. What do you do, and what must you not do?
- Lot-tracked items multiply the position rows per item-node. Does your answer change, and where does the read now spend its time?
What makes you better than the other candidates?
What makes you better than the other candidates?
Approach
- Clarify what is being asked and what a complete answer contains.
- Work from the requirement backwards to the design.
- 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?
What do you know about our company?
What do you know about our company?
Approach
- Say what you would check first and why it is the highest-information step.
- State your assumptions explicitly before working the problem.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Why do you want to work here?
Why do you want to work here?
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
- What assumption would you test first?
- How would you know your answer was wrong?
Fan shipment status out to five thousand customer webhooks
Customers subscribe to shipment status by webhook: 5,000 endpoints, some with an 8 second p99, some down for a day. The upstream is folded leg status, not raw observations. Requirements are per-shipment ordering at each subscriber, at-least-once delivery with a stable dedupe id, and no subscriber able to slow another. Assume 3 million legs a day, about six status transitions each, and a mean of three subscribers per shipment. Design the fan-out and quantify what folding first saves you.
Approach
- Fan out transitions, not observations, and do the arithmetic before designing delivery. Raw ingest averages 9,000 observations a second, about 780 million a day. Three million legs at six transitions each is 18 million notification sources a day, 208 per second, and three subscribers apiece is 54 million deliveries a day, 625 per second. Folding first is worth a factor of about 43 - 208 sources per second instead of 9,000, and 625 deliveries per second instead of roughly 27,000 - and that factor dominates every subsequent optimisation, so settle it before tuning delivery.
- Keep delivery state as a durable per-subscriber cursor over a per-subscriber queue rather than a row per attempt. At 54 million deliveries a day a row per attempt is roughly 19.7 billion rows a year before a single retry is counted, and retries only add to it; a cursor plus a bounded in-flight window gives retry and progress without any of them. Per-event rows earn their cost only where you must later prove a specific delivery in a dispute, so scope them to the subscribers who contract for that rather than to all 5,000.
- Preserve ordering deliberately: a bounded in-flight window greater than one can reorder. Give each subscriber's queue a small number of lanes keyed by hash(shipment_id), so order is strict within a lane and parallelism comes from lanes. Ordering is then a property of the key, not of a hopeful single-threaded consumer.
- Isolate subscribers with a per-subscriber concurrency cap drawn against a shared worker pool. By Little's law an 8 second endpoint receiving 20 notifications per second wants 160 workers in flight; without a cap it takes them from the pool and starves the other 4,999. The cap converts a slow subscriber into its own growing backlog instead of a shared outage.
- Handle the dead endpoint as a bounded liability: circuit-break after consecutive failures, stop attempting, let the cursor lag, and retain the queue to a fixed horizon such as 72 hours. At the horizon, advance the cursor, record a gap, and tell the subscriber to re-sync from the pull API. Unbounded retention for one dead customer is how the fan-out store becomes the next incident.
- Make at-least-once an explicit contract rather than an accident: every payload carries a stable event id and the leg's status_rank, so a customer can discard a duplicate or a stale retry on their side, and that requirement is documented rather than discovered.
Worked solution 30 min
- Derive the notification rate from legs per day, transitions per leg and subscribers per shipment, set it beside the raw observation rate, and state the reduction factor that justifies folding first.
- Choose between per-event delivery rows and a per-subscriber cursor, and show how the chosen shape still preserves per-shipment ordering under retry.
- Set a per-subscriber concurrency cap and compute an 8 second endpoint's in-flight occupancy at its share of traffic.
- Write the circuit-breaker states, the retention horizon and the re-sync contract for a subscriber that returns after a day offline.
Follow-up
- A customer demands exactly-once. What can you actually offer, and what must they implement to get the effect they want?
- One subscriber covers 40 percent of all shipments. What breaks first in a per-subscriber queue design, and what do you shard on instead?
- How do you replay a day of notifications for one subscriber after a bug on your side, without re-sending to the other 4,999?
Two committed allocations against a position holding one unit
Six oversells in a week, all on the twenty hottest item-node pairs, all inside promotion minutes. In one case inventory_position.qty was 1 and two allocation rows exist with qty 1 and status 'committed', created 4 ms apart; both transactions committed cleanly. The code reads availability, then issues UPDATE inventory_position SET qty = qty - :n, version = version + 1 WHERE item_id = :i AND node_id = :d AND lot_id = :l AND state = 'on_hand' AND version = :v, then inserts the allocation. A 200-thread load test across 50,000 SKUs never reproduces it. Find the defect and prove it.
Approach
- Reconstruct from the data before reading the code. Pull inventory_movement and allocation for that key ordered by identifier with the version values involved. Two commits that both proceeded from version v prove the compare-and-set did not gate anything; a single row correctly stepping v then v+1 with an oversell anyway would instead send you to the read path.
- Name the defect: the affected-row count of the conditional UPDATE is never tested. When version has moved the UPDATE matches nothing and affects zero rows, but control flow continues into the INSERT and the transaction commits, so the WHERE version = :v clause is decoration. A compare-and-set only protects you if the zero-row outcome changes what happens next.
- Explain why the load test is silent. With 200 threads spread over 50,000 SKUs, the chance that two allocators land on one row inside the UPDATE's execution window is negligible, while production traffic concentrates on twenty rows. Pin the same test to a single SKU with a position of 1 and N threads, assert that exactly one allocation survives, and it fails on the first run. That test is the proof, and it belongs in the suite.
- Rule out the decoy rather than ignoring it. If availability is read from a replica, lag peaks exactly when write volume does, so the read is often stale. But a stale read only mis-sizes the attempt: with a working compare-and-set the write still fails and retries, so the replica is a latency and retry-rate problem, not the oversell. Say that explicitly instead of fixing two things and learning nothing.
- Fix and then choose the mechanism deliberately. Test the rowcount; on zero rows re-read the position from the primary and retry with bounded attempts and jittered backoff, and never reuse the version captured before the failed attempt. Compare-and-set is lock-free but burns retries under contention on a single hot row; SELECT ... FOR UPDATE inside a short transaction queues writers predictably and is usually calmer for twenty hot keys; above that, funnel allocations for one item-node key through a single writer and accept the added latency.
- State what the isolation level does and does not do, since identical code is not equally safe everywhere. On PostgreSQL READ COMMITTED, which is the default and what this deployment runs, the second UPDATE blocks on the row lock, re-evaluates its predicate against the newly committed row, matches nothing and affects zero rows; control flow falls into the INSERT and the transaction commits, and that is the oversell. InnoDB's REPEATABLE READ ends the same way for this statement, because an UPDATE takes a current read rather than the snapshot read: it blocks, applies against the latest row, affects zero rows and does not abort. PostgreSQL REPEATABLE READ is the one that does not oversell, and not because the code is right: it raises SQLSTATE 40001 on a row a concurrent transaction updated after the snapshot, which aborts the whole transaction, so the unchecked INSERT can never commit. That is failing closed on an exception no handler anticipated, on one of the three configurations, at the cost of an error the caller has to catch and retry. Two of three commit the oversell, so the rowcount check is not optional, and the invariant belongs in a continuous monitor as well as in a test: assert per key that the sum of allocation.qty in status 'held' or 'committed' does not exceed the position quantity, and let the reconciliation job re-sum movements into positions and raise a variance when they disagree.
Follow-up
- An allocation must span two positions atomically. What changes in your mechanism, and what if the two live in different shards?
- Under the retry design, what stops a hot key from livelocking when fifty allocators all retry together?
- The six oversells are already promised to customers. What is the compensating path, and what does it write to inventory_movement?
For a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Numbers before diagrams
- Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
- Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
- Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.
Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.
Practice prompt ↗Practice prompt ↗Worked solution ↗02One system, from requirements to schema
- Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
- Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
- Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.
Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.
Practice prompt ↗Practice prompt ↗03The consistency you are actually buying
- Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
- Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
- Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.
Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.
Practice prompt ↗Practice prompt ↗04Failure is the design
- For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
- Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
- Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
- Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.
Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Scaling the hot path
- Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
- Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
- Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
- Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.
Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.
Practice prompt ↗Practice prompt ↗06Keep the coding hand in, at the bar that applies to you
- Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
- Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
- Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.
Deliverable: One problem hardened to review standard, with its test list and one written review comment.
Practice prompt ↗Practice prompt ↗07Defend it while being interrupted
- Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
- Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
- Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.
Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.
Describe your background in the field.
Describe your background in the field.
Approach
- Name the disagreement and how you resolved it with evidence.
- 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?
Can you describe (x) project you mentioned on your resume?
Can you describe (x) project you mentioned on your resume?
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.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
What tool should every engineer carry with them at all times?
What tool should every engineer carry with them at all times?
Approach
- Name the disagreement and how you resolved it with evidence.
- Close with what you would do differently, concretely.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
- 01
Describe your background in the field.
- 02
Can you describe (x) project you mentioned on your resume?
- 03
What tool should every engineer carry with them at all times?
Is this an official Pursuit Aerospace interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Pursuit Aerospace. Rounds and questions reflect what candidates have reported, not a process Pursuit Aerospace 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 process is generally quick, often moving from a screen to a final decision within a few weeks. However, timelines can vary, so maintain clear communication with your recruiter.
PracHub interview research ↗Should I prepare for a whiteboard coding test?
While some technical evaluation is expected, the focus at Pursuit Aerospace often leans more toward your past experience and problem-solving process rather than abstract algorithm memorization.
PracHub interview research ↗What is the best way to stand out?
Be authentic and show enthusiasm for the company’s mission. The team values candidates who are genuinely interested in what they do and who communicate clearly and humbly.
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