A Software Engineer at Service Corporation International (SCI) plays a vital role in maintaining and evolving the technology platforms that support the company’s expansive operations. As the largest provider of funeral, cremation, and cemetery services in North America, Service Corporation International relies on robust software to manage complex logistics, customer service records, and operational data. You will be responsible for building, maintaining, and optimizing the applications that keep this critical infrastructure running efficiently.
This role is not just about writing code; it is about contributing to a stable, high-scale environment where your work directly impacts the company’s ability to serve families during difficult times. You will work within an IT organization that values technical proficiency in the.NET ecosystem and web-based frameworks. Success in this position requires a balance of technical competence and the ability to articulate how your software solutions solve specific business challenges.
Initial Technical Conversation
reportedWhat this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.
What to demonstrate
- Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
- Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
- Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly
How to prepare
- Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
- Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
- Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live
Onsite or Video Interview
reportedWhere the day includes a partner from product, design or data, that conversation is weighted like the technical ones and prepared for least. They are deciding one thing: whether having you in the room makes their decisions cheaper. That means options with costs attached, not implementation detail and not "it depends". An estimate someone can plan against — a range, the assumption that would push it to the high end, and what you would drop to hit the low one — is worth more than a confident single number, which everyone present already knows is wrong.
What to demonstrate
- Whether an estimate comes as a range with the assumption most likely to break it, and states what a specific scope cut would actually buy
- Whether a technical constraint is handed over as a choice with consequences on their side, rather than as a verdict they have no standing to argue with
- Whether you establish what decision is on the table before proposing anything
- Whether risk is raised while it can still change the plan, with the trigger that would confirm it, instead of reported afterwards as a slip
How to prepare
- Take a project that shipped late and write the two-sentence warning you could have given three weeks earlier, naming what you would have needed decided at that point
- Rehearse one estimate out loud until it arrives in three parts: the range, the single assumption that would blow it, and the smallest thing you would cut to protect the date
- Rewrite an objection you have actually made — the "we can't do that" version — as two options with their costs, so the choice ends up with the person who owns it
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.
Finishing a solution without stating its complexity
Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.
Choosing a schema before the access patterns are known
Write the queries first, with their filters, sort orders, cardinalities and which ones sit on the latency-critical path, then design tables and indexes to serve them. An index nothing queries still costs write throughput and storage, and a hot query with no supporting index becomes a full scan that only hurts once the table is big.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Order a job dependency graph and find its critical path
A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.
Approach
- Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
- Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
- Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order:
earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E). - Second pass in reverse topological order for
latest_finish, thenslack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain.slack[v] = 0is exactly the statement that some longest path runs through v; equivalently, the longest path through v has lengthT - slack[v]. - The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by
min(d, T - L_avoid(v)), whereL_avoid(v)is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, whilemin(10, 15 - 14)predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan ismax(T - d, L_avoid(v)). - Compute
L_avoid(v)the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, sinceL_avoid(v)only ever matters through that max: setduration[v] := 0, recompute the makespan asT0(v) = max(T - duration[v], L_avoid(v)), and the gain ismin(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.
Follow-up
- Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
- Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
- Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?
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.
Worked solution 25 min
- Write both key tuples down before any code: dedup key
(tenant_id, idempotency_key), cell key(tenant_id, workspace_id, sku, hour_start), withhour_startderived fromoccurred_atin UTC. - Build a 10,000-row fixture containing one event duplicated three times under the same
idempotency_key, two events sharing anidempotency_keyacross differenttenant_idvalues, one event whoseoccurred_atis two hours before itsingested_at, and onestagingevent inside an otherwise production cell. - Fold it and assert each of those four expectations separately rather than eyeballing a grand total.
- Re-run with the input shuffled and diff the output files.
- Size the dedup set for 250M keys using the load-factor arithmetic and write the number down next to the fixture.
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?
Rebuild an hourly rollup with deduplication and late-arrival accounting
From usage_event (event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at), produce the values usage_rollup_hourly should hold for one tenant over one day: per (workspace_id, sku, hour_start) the deduplicated quantity_sum, event_count and source_max_ingested_at, bucketed by occurred_at. Duplicates share (tenant_id, idempotency_key). Also report, per hour, the running total across the day and the share of quantity that arrived more than two hours after the hour began. Write the query, and state which duplicates a daily unique index cannot catch.
Approach
- Deduplicate in its own CTE before any aggregation, because a SUM cannot be un-summed:
row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id) = 1. Include the tiebreaker. Without it the surviving row is non-deterministic when two duplicates share an ingested_at, and a rollup described as deterministically recomputable then disagrees with itself between runs. - Bucket on occurred_at and nothing else, and pin the timezone explicitly.
date_trunc('hour', timestamptz)truncates in the session's TimeZone setting, so the same query run by a session set to a non-UTC zone buckets differently; use the three-argumentdate_trunc('hour', occurred_at, 'UTC')on PostgreSQL 16 or later, ordate_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC'before that. Filterenvironment = 'production'explicitly, since metering covers three environments and billing covers one. - Aggregate to the grain with
sum(quantity),count(*)andmax(ingested_at). The last is not decoration: it is the watermark the row consumed up to, and without it there is no way to prove afterwards what a number did and did not include. - Compute the late share inside the dedup-and-aggregate step as a conditional aggregate,
sum(quantity) filter (where ingested_at > hour_start + interval '2 hours'), then divide by the hour's total. Compute the running total as a window over the already aggregated rows:sum(quantity_sum) over (partition by workspace_id, sku order by hour_start rows between unbounded preceding and current row). Running either over raw rows puts the duplicates back. - Answer the index question exactly. The unique constraint is on (ingested_day, tenant_id, idempotency_key), because a unique index on a partitioned table must contain the partition key. It therefore deduplicates only within one ingest day and admits a duplicate whose retry crosses midnight or whose replay runs a week later. That is why this CTE dedups across the whole window being recomputed, and why the dedup horizon is a correctness parameter rather than a retention cost.
- Keep the numeric type all the way through. quantity is numeric so the sums are exact; a cast to double precision anywhere in this pipeline reintroduces drift that surfaces only as a few unreconcilable cents per tenant per month, long after the query is out of anyone's mind.
Worked solution 40 min
- Generate a day of events for one tenant across three workspaces and two SKUs, inject 2% duplicates sharing an idempotency_key with a later ingested_at, and inject 1% whose ingested_at is six hours past occurred_at.
- Write the dedup CTE alone and confirm its row count drops by exactly the injected duplicate count.
- Aggregate and compare quantity_sum against a ground truth computed outside SQL from the generator's own records.
- Add the two window and filtered aggregates, and confirm the final hour's running total equals the ungrouped day total for each (workspace, sku).
- Run the whole query twice over static data and diff the two outputs.
- Re-run with the session TimeZone set to a non-UTC zone and confirm hour_start does not move.
Follow-up
- A dispute forces the same recompute over 40 days for one tenant. What changes about the dedup CTE's memory use and the chosen plan, and what would you do about it?
- Two runs a minute apart return different quantity_sum values for an hour that is already closed. Give two mechanisms that produce that, and the single query that distinguishes them.
- Express the same rollup incrementally so it does not re-scan the day each time the watermark advances. What does the incremental version stop being able to answer?
Paginate a tenant's delivery export without skipping rows
A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.
Approach
- Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
- Replace the position with a value predicate over a stable, unique, indexed ordering:
where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window. - Index
(tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit. - Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
- Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.
Follow-up
- The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added
created_at <= $snapshotbound, and a materialised export table. - A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
- The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?
How have you utilized WCF (Windows Communication Foundation) in your p…
How have you utilized WCF (Windows Communication Foundation) in your previous projects?
Approach
- Clarify what is being asked and what a complete answer contains.
- 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?
What is your level of experience working with Angular for front-end de…
What is your level of experience working with Angular for front-end development?
Approach
- Work from the requirement backwards to the design.
- Say what you would check first and why it is the highest-information step.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
A resumable usage export that never skips a row
Customers pull their own rows from usage_event through GET /v1/usage to reconcile against their own systems. The table is append-only and partitioned daily on ingested_at; a large tenant adds millions of rows a day while the export is being walked, and a client may pause for hours and resume. Specify the cursor, the index it requires, the tenant scoping, the ordering guarantee you can honestly offer, the per-page cost as the walk deepens, and what the client must do to avoid missing rows.
Approach
- Rule out LIMIT/OFFSET on two independent grounds and say both, because fixing only one leaves the other. Correctness: rows inserted between page requests shift the window, so a walking client skips rows and repeats others, which for a reconciliation consumer is silent data loss rather than an error anyone sees. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N x page_size and a deep page degrades from milliseconds to seconds.
- Use keyset pagination over a stable, unique, indexed ordering: WHERE tenant_id = $1 AND (ingested_at, event_id) > ($2, $3) ORDER BY ingested_at, event_id LIMIT $4, carrying the last row's pair as the cursor. The row-value comparison navigates a composite btree directly, so each page is O(log n + page_size) and stays constant as the walk deepens. The precondition is that the cursor columns never change value for a row, which ingested_at satisfies and updated_at would not.
- Lead the index with tenant_id - (tenant_id, ingested_at, event_id) - which is simultaneously the correctness guard and the plan choice. An index on (ingested_at) alone forces a filter across every tenant's rows, and on a table where one tenant holds most of them that is fine only for that tenant and terrible for everyone else. Because ingested_at is also the partition key, a resumed cursor prunes to the partitions from the cursor forward.
- Name the visibility hazard rather than assuming it away: in PostgreSQL now() is transaction start time, so a transaction that starts at T, inserts, and commits at T+8 s writes a row whose ingested_at is T but which becomes visible only at T+8 s. A walker that has already passed T never returns it. The gap equals the writer's longest transaction, so the mitigation is either a safety lag - serve only rows older than now() minus the longest permitted transaction - or a client that re-walks a trailing overlap window and deduplicates on event_id, which is stable and unique.
- Offer the guarantee you can actually keep: ordering by (ingested_at, event_id) with no claim whatsoever about occurred_at order, and completeness only behind the safety lag or with the documented overlap-and-dedup obligation on the client. Saying this in the API reference is part of the design, because the client's reconciliation logic is what has to absorb it.
Worked solution 20 min
- Write the keyset query and the exact index it needs, then read the query plan and confirm there is no Sort node.
- Insert rows concurrently while walking with a LIMIT/OFFSET pager and count the distinct rows returned against the rows that exist; repeat with the keyset pager and compare.
- Construct the out-of-order commit case by hand: open a transaction, insert, hold it open while the walker passes that timestamp, then commit, and check whether the walker ever returns that row.
- Choose the mitigation - safety lag or client overlap plus event_id dedup - and write down the number it depends on.
Follow-up
- The customer wants to reconcile by occurred_at instead of ingested_at. What breaks, and what would you offer them in its place?
- One tenant starts a full-history export. How do you keep it from occupying every connection in the pool?
- A customer reports a missing row. What do you check first, and what would each answer tell you?
Webhook workers leak until OOM and drop in-flight deliveries
webhook-delivery workers grow from 400 MB to a 2 GB limit over about 36 hours, are OOM-killed, restart, and repeat. Each restart abandons in-flight attempts, so webhook_delivery rows sit in in_flight until their leases expire and the backlog spikes. The live set measured after a forced full collection also grows. The fleet serves tens of thousands of subscriptions, several thousand of which have been failing for weeks. Give an ordered checklist, the measurement separating retention from fragmentation, and the fix.
Approach
- Separate the two failure shapes with one measurement: track resident set size against the live set after a forced full collection. A live set that climbs monotonically is retention; a flat live set under a rising RSS is fragmentation, off-heap or native allocation, or an allocator that never returns pages. The stated symptom puts this in the first category, which rules out allocator tuning as a fix.
- Characterise the curve rather than the total. Growth linear in uptime implies an unbounded structure keyed by something that keeps arriving; step growth implies buffering a large object. Correlate the slope against event rate and separately against the count of distinct subscriptions seen, because those two diverge and only one of them will fit.
- Diff two heap snapshots an hour apart by retained size grouped by dominant root, not by allocation count, which is dominated by short-lived objects and will point at the wrong thing.
- Expect a per-subscription map with no eviction: circuit-breaker or backoff state created on first failure and never removed, so the retained set grows with endpoints that have ever failed, and the several thousand permanently dead endpoints hold theirs forever.
- Fix in two places. Bound the in-memory structure with a size-capped LRU or a TTL keyed on last use, and move state that must survive a restart onto the subscription or webhook_delivery row, since the worker holding it in memory is exactly why a restart loses it.
- Repair the second-order damage separately, because it will outlive the leak: workers claim by compare-and-set with leased_until, so a bounded lease returns in_flight rows to pending on a known schedule, and a graceful shutdown releases leases instead of waiting them out.
Follow-up
- The backlog spike after a restart is itself a thundering herd against customer endpoints. What stops the recovery from becoming a second incident?
- Suppose the live set had been flat while RSS still climbed. Name two causes and the measurement that separates them.
- How would you size the LRU, and what does a miss on an evicted circuit-breaker entry cost a customer whose endpoint is down?
For someone fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Measure before reasoning
- Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
- Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
- Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.
Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.
Practice prompt ↗Practice prompt ↗Worked solution ↗02References, copies, and the bugs they produce
- Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
- Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
- Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.
Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.
Practice prompt ↗Practice prompt ↗03Types, once, in a language that checks them
- Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
- Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
- Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.
Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.
Practice prompt ↗Practice prompt ↗04Concurrency, starting with what actually runs at the same time
- Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
- Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
- Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.
Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Debugging as a procedure rather than an instinct
- Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
- Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
- Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.
Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.
Practice prompt ↗Practice prompt ↗06Tests that catch the bug you are about to write
- Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
- Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
- Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.
Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.
Practice prompt ↗Practice prompt ↗07Debug something broken, out loud
- Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
- Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
- Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.
Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Team size, service count and tickets closed say very little. Seniority shows in the decision you owned: what you chose not to build, which constraint you traded away, whose objection you had to resolve before anything could move. A large project where you executed someone else's plan is a small story.
Can you explain a complex technical challenge you faced and how you re…
Can you explain a complex technical challenge you faced and how you resolved it?
Approach
- Pick a story where you made the decision, not one where you watched it.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
Describe your experience working within an engineering team and your p…
Describe your experience working within an engineering team and your process for collaborating on code.
Approach
- Close with what you would do differently, concretely.
- Give the blast radius: what could have broken, and what you measured.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
Tell us about your last several projects and the specific role you pla…
Tell us about your last several projects and the specific role you played in them.
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
Explain your experience with MVC (Model-View-Controller) architecture.
Explain your experience with MVC (Model-View-Controller) architecture.
Approach
- Name the disagreement and how you resolved it with evidence.
- Give the blast radius: what could have broken, and what you measured.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
- 01
Can you explain a complex technical challenge you faced and how you resolved it?
- 02
Describe your experience working within an engineering team and your process for collaborating on code.
- 03
Tell us about your last several projects and the specific role you played in them.
- 04
Explain your experience with MVC (Model-View-Controller) architecture.
Is this an official Service Corporation International interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Service Corporation International. Rounds and questions reflect what candidates have reported, not a process Service Corporation International 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 take?
The process is generally fast-paced, often concluding within two weeks from the initial contact.
PracHub interview research ↗Who will I be interviewing with?
You will typically meet with an Architect during the initial phase, followed by a panel that includes an IT Director and senior developers.
PracHub interview research ↗What is the best way to stand out?
Be prepared to provide detailed, project-specific examples of your technical work. Candidates who can clearly articulate their contributions and handle technical questions with precision stand out the most.
PracHub interview research ↗Is the work environment highly collaborative?
Yes, you will be expected to work closely with other developers and leadership to ensure project success. Being able to communicate your ideas effectively is a key component of the team culture.
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
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