Interview guides · BS&A Software

BS&A Software Software Engineer Interview Guide

BS&A Software Software Engineer interview guide: reported rounds, questions and how candidates describe the process.

The 60-second brief

As a Software Engineer at BS&A Software, you are integral to building and maintaining the specialized software solutions that power local government operations. Your work directly impacts how municipalities manage critical data, from tax assessments to financial records, making efficiency and reliability the cornerstones of your output. You will be operating within a company that values stability, clear communication, and a deep understanding of the end-user’s needs. This role requires a blend of technical precision and a genuine interest in the business impact of your code. You will likely collaborate across departments, including support and operations, to ensure that the software not only meets functional requirements but also provides a seamless experience for government staff.

This guide is scoped to a Software Engineer candidate at BS&A Software.

Focus areas

Cloud Infrastructure
Infrastructure Engineering
Software Engineering Fundamentals

Where to focus your preparation

No round sequence has been reported for BS&A Software. Confirm the format with your recruiter.

Preparation focus

No round sequence has been reported for this company, so confirm the format with your recruiter and work the reported questions below.

What to demonstrate

Breadth across the topics this company reports testing
Whether you confirm the format before preparing for it

How to prepare

Ask the recruiter for the sequence, the duration of each stage and whether you will be writing code
Work the reported questions below and time yourself

Where candidates lose points

Structure your answers

Use the STAR method (Situation, Task, Action, Result) to keep your behavioral answers focused and impactful.

Ask questions

Use the time at the end of the interview to ask about the team’s current challenges or the company’s vision—it shows genuine interest.

Be authentic

The interviewers are looking for a long-term fit; being yourself helps them determine if you will be happy and successful in the role.

Always follow up with a thank-you note to your interviewers

It is a simple gesture that reinforces your professional courtesy and interest in the position.

Questions & practice

Merge partitioned event streams into one ordered feed with bounded lateness

Hard · k-way merge · watermarks · out-of-order streams

The read-model service consumes 64 log partitions carrying about 4,000 events per second in total. Each partition is ordered within itself, but partitions drift by up to 30 seconds, and the activity feed must present a tenant's events in occurred_at order. Produce the merge. State its complexity, the buffer it requires in events and in bytes, what happens when one partition is idle, and what you do with an event that arrives after you have already emitted its position. Payloads average 1 KB.

Approach

Merge with a min-heap over the 64 partition heads keyed on (occurred_at, event_id): O(log P) per event and O(n log P) overall. The tie-break on event_id is what makes the output deterministic when two partitions carry the same millisecond, which matters because the feed is paginated and a non-deterministic order reorders pages under the reader.
Emitting the heap head is only correct once every partition has produced everything up to that timestamp, so the emit condition is a watermark: the minimum across partitions of the highest occurred_at seen, less the allowed lateness. Events are held until the watermark passes them, which is what turns individually ordered streams into a jointly ordered one.
Size the buffer from the lateness rather than guessing: 4,000 events per second times 30 seconds is 120,000 buffered events, and at 1 KB each about 120 MB of heap. That number is the real price of the ordering guarantee and belongs in front of whoever asked for it.
Handle the idle partition explicitly, because it fails the feed rather than corrupting it: a partition with no traffic never advances its own maximum, so the watermark freezes and output stops entirely. Either every partition emits a periodic idle marker carrying the broker's current time, or the watermark falls back to wall clock for a partition silent beyond a threshold.

Follow-up

The lateness budget is raised to five minutes. What is the new buffer, and what besides memory changes?
The consumer restarts. Where does it resume from, and what does the feed look like for the first 30 seconds?

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.

Follow-up

The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?

Canonicalise a request body into a stable idempotency fingerprint

Medium · parsing · canonicalisation · hashing

idempotency_key.request_fingerprint is a SHA-256 over the method, path and canonicalised body, and a retry whose fingerprint differs must be rejected with 422 rather than served the stored response. Write the canonicaliser. Bodies are JSON up to 256 KB nested at most 32 levels; clients vary key order, whitespace and unicode escaping, and some send 64-bit ids as JSON numbers. Produce a deterministic byte string such that semantically identical bodies match and any semantic difference does not. State your complexity and name two normalisations you refuse to perform.

Approach

Parse once into a tree, then re-serialise under fixed rules: object keys sorted, array order preserved, one escaping convention, no insignificant whitespace. Parsing is O(n) and sorting keys is O(k log k) per object, so O(n log n) overall with O(depth) stack, and the 32-level cap is enforced during parsing because hostile nesting is how a canonicaliser becomes a stack overflow.
Sort keys by their UTF-8 bytes and say why the obvious implementation is wrong in some runtimes: a default string comparison that orders by UTF-16 code units places surrogate pairs, meaning code points from U+10000 up, below U+E000 to U+FFFF, which is not UTF-8 byte order, so two services written in different languages disagree on the same document.
Do not re-encode numbers through a double. IEEE-754 binary64 represents integers exactly only up to 2^53, so normalising a 19-digit id through a float changes it, and 1 against 1.0 cannot be reconciled without deciding whether they are the same value. Preserve the literal token, and require ids as strings at the API boundary if you want them comparable.
Reject duplicate keys rather than picking one. JSON permits them and parsers disagree, most keeping the last, so any choice you make ties the fingerprint to a parser detail that the code handling the request does not necessarily share.

Follow-up

A client sends the same logical request with an extra field your API ignores. Same key, different fingerprint, so you return 422. Is that the right answer?
Where does the fingerprint get computed relative to request decompression and the body-size limit?

Stop tag and share joins from fanning out a page

Medium · join fan-out · semi-joins · many-to-many

resource_tag is (resource_id, tag_id) with PK (resource_id, tag_id); resource_share is (resource_id, shared_with_user_id, permission). The tagged-and-shared listing inner-joins resource to both, filters tenant_id, tag_id = ANY($2) and shared_with_user_id = $3, orders by updated_at DESC and takes 50. Pages come back with fewer than 50 distinct resources and the total in the header is far too high. Explain the row multiplication, rewrite both the page query and the count query so each is correct, and name the index each one needs. PostgreSQL 16.

Approach

Do the arithmetic against the predicates that are actually there. An inner join emits one row per matching child row, and both joins are filtered: tag_id = ANY($2) admits only the requested tags, shared_with_user_id = $3 admits one user's share rows. So a resource holding three of the requested tags and shared with $3 once yields three rows, not one — the multiplier is its count of matching tags times its share rows for that single user, and that second factor is 1 unless the table admits duplicate (resource_id, shared_with_user_id) pairs. LIMIT 50 then limits rows rather than resources, and COUNT(*) counts pairs — the header is the product, not the population.
Reject DISTINCT as the fix. It deduplicates after the product has been built, so the planner must materialise and sort the fanned-out set before the LIMIT can apply, and it leaves any SUM or AVG in the same select list wrong.
Rewrite both filters as semi-joins, keeping resource as the only row source: AND EXISTS (SELECT 1 FROM resource_tag rt WHERE rt.resource_id = r.resource_id AND rt.tag_id = ANY($2)) and the same shape against resource_share. A semi-join stops at the first match per resource and preserves the driving index order, so ORDER BY updated_at DESC, resource_id DESC LIMIT 50 still stops after 50 rows.
Count with the same predicates and no join at all: SELECT count(*) FROM resource r WHERE r.tenant_id = $1 AND r.status = 'active' AND EXISTS (...) AND EXISTS (...). Nothing multiplies a resource, so the number is the population.

Follow-up

The filter changes from 'any of these tags' to 'all of these tags'. Rewrite it and state what it costs relative to the ANY form.
A resource can be shared with the same user twice under different permissions. Does your count change, and should it?

Replace offset paging on the resource feed with keyset

Easy · keyset pagination · composite index · sort order

resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.

Approach

Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.

Follow-up

The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?

Relay committed events to the log without gaps or reordering

Medium · transactional outbox · at-least-once · ordering

outbox_event rows are written in the same transaction as the state change and carry aggregate_type, aggregate_id, aggregate_version, payload and status, with a partial index on (created_at, event_id) WHERE status = 'pending'. The relay publishes about 4k events/second to a partitioned append-only log keyed by aggregate_id, with one leader per partition range holding a lease. Consumers must never miss an event; they may see one twice. Design the claim-publish-mark loop, and state exactly what consumers observe when a leader's lease expires while it is mid-batch.

Approach

Claim with SELECT ... WHERE status='pending' ORDER BY created_at, event_id LIMIT $batch FOR UPDATE SKIP LOCKED inside a transaction. SKIP LOCKED lets several relay workers share a range without serialising on each other's rows, and the partial index keeps the claim proportional to the backlog rather than to a table that is overwhelmingly published rows. At 4k/second a batch of 500 is eight claims per second, each an index scan of 500 entries.
Publish before marking, never the reverse, and say why it is a choice. Marking first loses the event outright if the process dies in the gap, and the loss is silent - nothing remains to retry, and it surfaces later as a projection missing a row. Publishing first can repeat the event, and repetition is what every consumer is already built to survive. That single ordering is the whole at-least-once guarantee.
Preserve the only ordering on offer. Partition by aggregate_id and never publish two events for one aggregate concurrently: claim in (created_at, event_id) order and publish sequentially within an aggregate, or hash aggregate_id to a worker slot. Order across aggregates is not available at any price here, which is why the event carries aggregate_version and the full fact rather than a delta - a consumer can then discard what it has already applied without coordinating with anyone.
State the failover behaviour precisely, because it is the consistency-versus-availability decision in this design. A lease expires because the holder is slow, and no mechanism distinguishes that from dead, so for the length of the lease window two leaders can publish the same claimed batch. The system accepts duplicates to avoid stalling publication for every aggregate in the range whenever one worker pauses. Consumers deduplicate on (aggregate_id, aggregate_version) and drop anything at or below what they have applied.

Follow-up

The relay is down 40 minutes and 9.6 million rows are pending. What does catch-up do to the primary, and what changes in the claim loop to survive it?
A consumer insists it never received an event. Which single query settles whether the relay lost it, and what does each answer look like?

Design the async export contract a client can resume safely

Hard · async operations · idempotency · leases

A tenant asks for a CSV of every resource. The work runs for minutes on the worker fleet through a job_run row carrying a lease, an attempt count and a dedupe_key, far past the edge's 400 ms budget. Callers are a browser that polls and a script that walks away and checks later. Specify what the submit call returns, the operation resource and its states, how a duplicate submit is handled, how a client learns about completion, what cancellation means given that a lease can expire mid-run, and how the result is fetched and when it expires.

Approach

Split the API in two. Submit returns 202 with an operation id and a location to poll, and never blocks on the work. The operation is a real resource with its own lifecycle - queued, running, succeeded, failed, cancelled - plus attempt, a monotonic progress figure, and a terminal error drawn from the same code taxonomy the synchronous endpoints use, so a client needs one error vocabulary rather than two.
Deduplicate at submit using job_run.dedupe_key, unique over (job_type, dedupe_key) while status is 'queued' or 'running': a repeat submit of the same logical export returns 200 with the existing operation instead of 202 with a new one, and the partial index deliberately permits a legitimate re-run once the first has finished. Pair it with the request's idempotency key so an HTTP-level retry of the submit is exact rather than merely similar.
Tell the poller how to poll: Retry-After on the polling response, a minimum interval enforced at the edge, and a documented maximum lifetime after which an operation is reaped. Polling is the contract of record; the webhook is the fast path, and both must lead to the same terminal state, so a client that receives the completion event and then polls anyway sees no contradiction.
Be exact about cancellation. A cancel request records intent; it cannot stop work already executing. The handler reads the flag at checkpoints, and because a lease expires on a clock that cannot distinguish a dead worker from a slow one, a second copy may start after the cancel was recorded - so the handler re-reads the flag immediately after claiming the lease. 'cancelled' becomes terminal only when no lease is outstanding; reporting it earlier shows a client a stopped job while a worker is still writing output.

Follow-up

An operation has said 'running' for 40 minutes and the worker is gone. What does the client see, and which columns in job_run decide that?
Two tenants each submit 50 exports at once. What in this contract stops one of them delaying the other?

How do you approach debugging a complex, multi-layered issue?

Medium · Technical and Problem Solving

Approach

Establish what changed and when, before forming any theory.
Pick a bisection that eliminates candidates whichever way it turns out.
Check the instrumentation before believing the symptom.
Separate the trigger from the cause; the deploy is rarely the bug.

Follow-up

What would you look at first, and what would it rule out?
How would you tell a cause from a coincidence here?

Suggested practice checklist

Built from the topics and questions BS&A Software candidates report; no round sequence has been reported.

Day 1 — Establish the BS&A Software format

No round sequence has been reported, so ask your recruiter for the sequence, the duration of each stage and whether you will write code.

Deliverable: A written reply from your recruiter confirming the format.

Day 2 — Work Cloud Infrastructure

Spend the session on Cloud Infrastructure, which BS&A Software candidates report being tested on.
Write one worked example in Cloud Infrastructure and time yourself on it.

Deliverable: One timed worked example in Cloud Infrastructure.

Day 3 — Work Infrastructure Engineering

Spend the session on Infrastructure Engineering, which BS&A Software candidates report being tested on.
Write one worked example in Infrastructure Engineering and time yourself on it.

Deliverable: One timed worked example in Infrastructure Engineering.

Day 4 — Work Software Engineering Fundamentals

Spend the session on Software Engineering Fundamentals, which BS&A Software candidates report being tested on.
Write one worked example in Software Engineering Fundamentals and time yourself on it.

Deliverable: One timed worked example in Software Engineering Fundamentals.

Day 5 — Answer out loud: Behavioral and Situational

Answer aloud, timed: Tell me about a time you had to make a tough decision.
Answer aloud, timed: What are your greatest strengths and weaknesses?

Deliverable: Spoken answers to 2 reported Behavioral and Situational question(s), under time.

Day 6 — Answer out loud: Technical and Problem Solving

Answer aloud, timed: How do you approach debugging a complex, multi-layered issue?

Deliverable: Spoken answers to 1 reported Technical and Problem Solving question(s), under time.

Day 7 — Dry run for BS&A Software

Run one full mock under time, then write down the two questions you most want to ask your interviewers.

Deliverable: A completed timed mock and two questions to ask.