This guide covers what a Software Engineer at Washington University in St. Louis is expected to do and how to prepare for the interview.
Preparation focus
editorialNo round sequence has been reported for this company, so work the categories below and confirm the format with your recruiter.
What to demonstrate
- Breadth across SQL, experimentation and product reasoning
- Ability to state assumptions before choosing a method
How to prepare
- Drill the practice exercises below and time yourself
- Prepare three quantified stories about decisions you drove
PracHub editorial advice for the preparation topics above.
Running a schema change as though the lock lasts as long as the statement
In PostgreSQL an ALTER TABLE that needs an ACCESS EXCLUSIVE lock must first wait for every open transaction touching that table, and while it waits, later queries needing a conflicting lock queue behind it rather than overtaking it. A DDL statement that would execute in milliseconds, issued while a thirty-second analytics query is open, therefore stalls all traffic on that table for thirty seconds: the outage length is set by the longest open transaction, not by the change. The defences are specific and worth knowing by name - set lock_timeout low and retry rather than queue, add columns without a volatile default so no table rewrite occurs (from version 11 a non-volatile default is a metadata-only change), build indexes with CREATE INDEX CONCURRENTLY while accepting that it cannot run inside a transaction block and leaves an invalid index behind if it fails, and add constraints as NOT VALID followed by a separate VALIDATE CONSTRAINT, which takes a weaker lock.
Assuming an isolation level prevents the anomaly you actually have
Isolation levels are named by the SQL standard but implemented differently, so any claim about one is only true of a named engine. PostgreSQL defaults to READ COMMITTED, where every statement takes a fresh snapshot, so two statements inside one transaction can legitimately disagree about the same row. Its REPEATABLE READ is snapshot isolation: it removes non-repeatable and phantom reads but permits write skew, where two transactions each read a set, each conclude their own write is safe, both commit, and the combined result violates a constraint that no single row expresses. Only SERIALIZABLE closes that, and it closes it by aborting a transaction with a serialization failure (SQLSTATE 40001), which means the guarantee is theoretical unless the application has a retry loop. InnoDB's REPEATABLE READ is a different mechanism again - plain SELECTs read a consistent snapshot while locking reads and writes see the latest committed row - so a read-modify-write inside one transaction can act on a value that the transaction's own earlier SELECT never returned.
Assuming the input fits in memory
Ask how large the input is in bytes before committing to an in-memory algorithm; beyond that point the options are a single streaming pass, an external sort with bounded buffers, or a sketch that trades exactness for constant memory. An algorithm that assumes random access to the whole input is a different algorithm from one that sees each element once.
Trusting input because it came from your own front end
Anything crossing a trust boundary is hostile: parameterise queries instead of building SQL by concatenation, validate against an allow-list rather than a deny-list, and bound the size of anything you allocate from a request. Raising this unprompted in an API or design question is a cheap and unusually strong signal.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Collapse a redelivered event batch into per-aggregate high-water marks
You drain a batch of up to 5,000,000 events, each (aggregate_id BIGINT, aggregate_version INT, event_type, payload). The log guarantees order within one aggregate only; the batch merges 64 partitions, and a relay failover has redelivered a range, so an older version for an aggregate can appear after a newer one. Given a map of last_applied_version per aggregate, produce the events worth applying, at most one per (aggregate_id, version), plus the count discarded. Target O(n) time. State the memory for 2,000,000 distinct aggregates and what you do when it does not fit.
Approach
- One pass, one hash map from aggregate_id to the highest version kept, and a discard counter. An event whose version is at or below last_applied_version for its aggregate is dropped without further work, which is the whole reason the event carries its version rather than a delta. O(n) expected time, O(d) space in distinct aggregates.
- Keep the maximum, never the last occurrence. The redelivered range means the final appearance of an aggregate in the batch can be an older version than one seen earlier in the same batch, so last-wins applies stale state over newer state and the projection regresses with no error anywhere.
- Cost the memory instead of calling it large: an 8-byte key plus a 4-byte version is 12 bytes of payload, and an open-addressed table held at a 0.7 load factor costs roughly 17 bytes per entry before per-slot metadata, so 2,000,000 aggregates is tens of megabytes in a native layout and several times that in a runtime that boxes both key and value.
- If the distinct set exceeds memory, partition on hash(aggregate_id) mod P and reduce each partition independently. Every event for one aggregate hashes to the same partition, so the per-partition result is exact and the merge is concatenation rather than a second reduction.
- Reject sorting the batch by (aggregate_id, version) as the default. It is O(n log n) and buys nothing, because max is associative and commutative and needs no ordering; sorting earns its cost only when the downstream consumer must receive the events in order rather than a per-aggregate winner.
- Separate the two mechanisms out loud: in-batch deduplication does not make the consumer idempotent, because the same event redelivered tomorrow arrives in a different batch entirely. The projection write itself still has to be keyed on (aggregate_id, version).
Follow-up
- The payload is a patch rather than a snapshot, so applying only the highest version loses the intermediate changes. What changes in your reduction?
- How do you detect that version 7 arrived while version 6 was never delivered, and what should the consumer do about the gap?
- Two events for one aggregate carry the same version with different payloads. Which one is wrong, and how would you find out?
Find overlapping job attempts and peak concurrency from lease records
A day of job_run history yields about 50,000,000 attempt records: (job_run_id, job_type, attempt, started_at, finished_at which is NULL when the worker died, lease_expires_at). Leases expire on a clock, so a job that outran its lease ran twice. Produce (a) every job_run_id whose attempts overlapped in wall-clock time and (b) the peak number of simultaneously running attempts per job_type with the minute it occurred. Target O(n log n). State how you treat a NULL finished_at and what clock skew does to your answer.
Approach
- Define the interval before sorting anything: an attempt occupies [started_at, COALESCE(finished_at, lease_expires_at)). finished_at is observed and lease_expires_at is only a promise, so every attempt without a finish contributes an estimate and the whole result is a lower bound on overlap rather than an exact count.
- For peak concurrency, sweep: emit 2n endpoints, sort by (timestamp, kind) with ends ordered before starts at equal timestamps, then walk the sequence maintaining a counter per job_type and record each type's maximum with its timestamp. O(n log n) dominated by the sort, O(n) space, or O(1) extra if the sort is external and the walk streams.
- For overlap detection, do not compare attempts pairwise. A single global sort by (job_run_id, started_at) gives both the grouping and the order; within a group, keep the maximum end seen so far and report an overlap exactly when the next start is less than that running maximum, which is one linear pass after the sort.
- Half-open intervals matter and are easy to get wrong: with closed intervals an attempt ending at the same millisecond another begins reads as concurrency two, and across 50,000,000 records that artefact swamps the real signal.
- State the clock caveat: started_at and finished_at are written by different workers, so under skew of a few hundred milliseconds an apparent overlap shorter than that bound is not evidence. Filter reported overlaps by a minimum duration, or prefer timestamps written by whichever component heartbeats the lease.
- Scale the sort rather than assuming it fits: the sweep emits two endpoints per attempt, so 50,000,000 records become 100,000,000 endpoints, and at roughly 24 bytes each, an 8-byte timestamp plus a 4-byte job_type plus a kind flag padded to alignment, that is about 2.4 GB of sort keys before any scratch space. Either push the ordering into the database behind an index on (job_type, started_at) or run an external merge sort in chunks; the overlap pass sorts n records rather than 2n, so it is the cheaper of the two.
Worked solution 30 min
- Write the interval derivation with the COALESCE and state in one line which of the two end sources is observed and which is assumed.
- Write the concurrency sweep: the endpoint tuples, the sort key including the end-before-start tie-break, and the per-job_type counter.
- Hand-trace four attempts of one job, two disjoint and two overlapping by three seconds, and confirm the overlap detector fires exactly once.
- Add the skew filter as a minimum overlap duration, state the value you chose, and justify it from how the timestamps are written.
Follow-up
- A handler is not idempotent and you have found 400 overlapping jobs. Which of them actually caused damage, and what would you query to find out?
- Peak concurrency for one job_type is 4 against a configured cap of 4. Is the cap working, or is the data hiding attempts that never started?
- How would you compute both answers incrementally as records arrive rather than in a daily batch?
Archive a resource graph without breaking live references or recursing
Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.
Approach
- Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
- Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
- Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
- Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
- Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
- Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.
Follow-up
- The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
- The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
- An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?
Replace offset paging on the resource feed with keyset
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.
- State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
- Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.
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?
- What does the cursor do when the row it points at has since been deleted?
Explain why the owner filter ignores the listing index
The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.
Approach
- Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
- For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
- Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
- Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.
- Answer the two negatives explicitly. owner_user_id alone is not a left prefix of the new index, so it degrades to a full scan of the index at best. Ordered by updated_at, the query still seeks on the (tenant, owner) pair but must sort, because only created_at is ordered within that pair.
- Measure both sides with EXPLAIN (ANALYZE, BUFFERS) and compare estimated against actual rows at the lowest node — a 2M-versus-200 misestimate there is usually what chose the plan, and adding an index will not fix a statistics problem.
Worked solution 25 min
- Load 2M resource rows across 5k owners in one tenant, run the query under EXPLAIN (ANALYZE, BUFFERS), and record the node reading the most rows plus the Sort Method line.
- Create (tenant_id, owner_user_id, created_at DESC) concurrently and re-run, confirming the Sort node is gone and actual rows fall to about 20.
- Run the two negative cases and capture the plan for each.
- Re-run the original tenant listing query to confirm the new index has not displaced the index that query depends on.
Follow-up
- 90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
- A dashboard runs this for 40 owners in one page load. What changes about the design?
- How do you roll this index out on a table taking 1.2k writes/second, and what does it cost on every insert from then on?
Keep one tenant's bulk export from starving projection updates
The worker fleet runs about 600 jobs/second out of job_run, claimed under a lease and heartbeated while running. Most handlers finish under 200 ms; exports run for minutes. One tenant enqueues 50,000 exports. Design the scheduling so that tenant's exports slow down and nothing else does: the queue split, the concurrency caps and where they are enforced, the claim query, and the backpressure signal that stops accepting new work. State the in-flight worker count each class needs at its arrival rate, and what the cap costs the tenant that hits it.
Approach
- Do the concurrency arithmetic first, because it determines the pool shapes. Required in-flight work is arrival rate times service time: 600/second of 200 ms handlers needs about 120 workers, while one export per second at 180 seconds needs 180 on its own. The long tail dominates any pool it shares, so a single pool sized from the mean is consumed by exports while 50 ms projections queue behind them. That is head-of-line blocking, and adding capacity does not fix it because the ratio is what is wrong.
- Split by job class rather than by priority. Priority inside one pool still lets a running export hold its worker for the next four minutes - there is no preemption for a handler already executing. Separate queues and separate worker processes give each class a floor the other cannot take.
- Enforce the per-tenant cap inside the claim, not at enqueue. Claim with SELECT ... WHERE status='queued' AND run_after <= now() AND job_type=$1 ORDER BY run_after, job_run_id FOR UPDATE SKIP LOCKED LIMIT 1, restricted to tenants currently under their cap. Derive the running count from rows in status 'running' with a live lease rather than from a counter incremented at claim and decremented at completion: the derived count self-heals when a worker dies, and the counter leaks a slot every time one does.
- Make the ordering fair over tenants instead of over jobs. Strict FIFO with a cap still forces the claim to scan past 50,000 rows belonging to a capped tenant before it finds anyone else's work, so cost grows with backlog depth. Pick the tenant first from a small per-tenant queue-depth summary, then claim within that tenant, which makes the scan proportional to the number of active tenants instead.
- Close the loop with backpressure and deduplication. Reject or defer enqueues once a class's depth passes a bound, and rely on UNIQUE (job_type, dedupe_key) WHERE status IN ('queued','running') so a retrying producer collapses into one row instead of multiplying the backlog. Watch oldest-queued-age per class rather than depth: depth means nothing without a service rate, while age is the user-visible latency directly.
Worked solution 25 min
- Compute required in-flight workers per class from arrival rate times service time, and show what one shared pool does when both classes contend.
- Write the claim statement with the per-tenant restriction, and state how the running count is derived so it survives a worker being killed.
- Trace what the claim scans when one tenant holds 50,000 queued rows, under FIFO and under tenant-first selection.
- Choose the backpressure threshold and the signal it watches, then state what the producer receives when it trips.
Follow-up
- An export outruns its lease, so two copies run. What must the handler do to make the second copy harmless, and what does it write to guarantee that?
- With a cap of 4, the tenant's 50,000 exports now take hours. What do you tell them, and does the answer change if they are the largest tenant on the platform?
- Queue depth is flat but oldest-queued-age is climbing. Name the mechanisms that produce that exact pair of signals.
Make the resource write endpoint safe for retrying clients
The edge API accepts POST /v1/resources at 1.2k writes/second peak under a 400 ms p99 budget, and clients retry on timeout. idempotency_key has PRIMARY KEY (tenant_id, idempotency_key) plus request_fingerprint CHAR(64), state in ('in_flight','succeeded','failed'), response_status, response_body, resource_id, locked_until and expires_at. Specify the exact sequence the handler runs: what the second request does when it arrives while the first is still executing, what a retry carrying a different body receives, what happens when a process dies mid-request, and how the table is kept from growing without bound.
Approach
- Start from the race rather than the happy path. Insert the key row first, in its own short transaction, with state 'in_flight' and locked_until set to now() plus the request deadline. The unique constraint on (tenant_id, idempotency_key) is the arbitration mechanism: exactly one caller commits the insert, and every other caller takes a unique violation (SQLSTATE 23505) and becomes a follower. Reading the table and then inserting is a check-then-act race that both callers pass, and they pass it most often under the load that generates the retries.
- Give the follower a branch for every state it can observe, with no fall-through to doing the work: fingerprint mismatch means the key is being reused for a different request, so reject with 422 and never replay the stored response; 'succeeded' replays response_status and response_body verbatim; 'failed' allows a fresh attempt; 'in_flight' with locked_until in the future is answered with 409 and a retry hint, or a short bounded wait, because the only alternative is executing the effect twice.
- Commit the effect and the record of the effect together. The resource insert, its resource_revision row, its outbox_event row and the transition of the key row to 'succeeded' all happen in one transaction, so no crash can leave the work done and the key still 'in_flight'. The earlier insert is deliberately a separate transaction: it has to be visible to a concurrent caller before the work starts, which an uncommitted row is not.
- Close the wedged-key path. A process that dies after the insert leaves a row that would otherwise block that key forever, which is what locked_until exists for. Reclaim with a single conditional statement, UPDATE ... SET state='in_flight', locked_until=now()+interval WHERE ... AND state='in_flight' AND locked_until < now(), and treat a zero rowcount as losing the reclaim. Two statements that read and then update reintroduce the original race one layer down.
- Size the retention before it becomes an incident. At the 1.2k/second peak a 24-hour window is on the order of 100 million rows, and the table grows with traffic rather than with data, so the sweep is load-bearing. Batched deletes driven by an index on expires_at keep it bounded. Range-partitioning by day and dropping whole partitions is cheaper, but PostgreSQL requires every partition-key column to appear in the primary key, so the key becomes (tenant_id, idempotency_key, created_at) and a retry that straddles the boundary no longer collides with its original. State that cost rather than meeting it later.
Follow-up
- The stored response body averages 200 KB and this table is now the largest in the database. What do you store instead, and what does a replay return once the body has been pruned?
- A client library generates a fresh idempotency key on every attempt. What breaks, which layer should have caught it, and does the server have any defence?
- The first attempt succeeded but its response was lost, and the client retries 30 hours later, after the key expired. What does the second attempt do, and is that acceptable?
Listing latency scales with page size, not with filters
The tenant listing endpoint reads resource filtered by tenant_id and status, ordered by updated_at DESC, and returns each row plus the owner's display name from app_user and the actor of that resource's latest resource_revision. p99 is 55 ms at 10 rows per page and 1.4 s at 200. Database telemetry shows 401 statements per request, each under 1 ms, and nothing in the slow-query log. Diagnose the cause and give the fix, stating the statement count per request and the p99 you expect afterwards.
Approach
- Read the counters before forming a theory. 401 statements for 200 rows is one driver query plus two per row, and sub-millisecond execution with an empty slow-query log rules out a bad plan. The time is round trips, which is why it is invisible in every per-query metric and scales with rows returned rather than with filter selectivity.
- Name the two per-row statements from their normalised text: a single-row app_user lookup by user_id, and a resource_revision lookup by resource_id ordered by version DESC LIMIT 1. Confirm by dropping those two response fields and watching the statement count fall to one. That locates the calls in the serialisation layer, not the repository.
- Check that the arithmetic accounts for the whole gap. Measure one round trip to the replica in isolation; 400 trips at roughly 3 ms of network plus 0.2 ms of execution is about 1.3 s on top of a 55 ms baseline, which matches. If the multiplication had fallen short, the N+1 would only be part of the story and you would keep looking.
- Batch both lookups. Collect owner_user_ids and resource_ids from the driver query, then issue WHERE tenant_id = $1 AND user_id = ANY($2) for the users, and PostgreSQL's SELECT DISTINCT ON (resource_id) ... WHERE resource_id = ANY($2) ORDER BY resource_id, version DESC for the latest revision, which the UNIQUE (resource_id, version) index serves directly. On an engine without DISTINCT ON, use a lateral join or a row_number window. Three statements per request at any page size.
- Keep the tenant predicate in the batched query. The per-row version was implicitly scoped because its ids came from tenant-scoped rows; a batched user_id = ANY(...) with no tenant_id is an unscoped read that behaves correctly only as long as the id list is trustworthy.
- Re-measure at 10, 50 and 200 rows and confirm the statement count is constant. Latency should now track bytes returned.
Follow-up
- The page size is capped at 200 today. What breaks first if it is raised to 2,000, and is it still this bug?
- How do you stop the next N+1 from reaching production, given that no individual query is slow and the endpoint's tests pass?
- The latest-revision actor is only used to render an avatar. Make the case for denormalising it onto resource, and name the write anomaly that introduces.
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 ↗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 ↗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.
Narrate an outage you owned from page to postmortem
Pick an incident you personally drove, ideally one where writes were affected rather than reads. In six to eight minutes: state the symptom as it first appeared on a dashboard, the blast radius you established before you knew the cause, the mitigation you applied and when, the mechanism you eventually proved, and the follow-up that would prevent a repeat. Bring numbers: error rate, tenants affected, minutes to mitigate, minutes to resolve. If you cannot name what you measured, choose a different incident.
Approach
- Open on the signal rather than the cause: which metric at which percentile moved, on which service, at what time, so the listener follows the same evidence you had rather than a conclusion you already reached.
- Separate mitigation from diagnosis out loud. State what you did to stop the bleeding (flag off, shed traffic, drain a lease, roll back a deploy) and say plainly that you did it before the mechanism was known, because those are two jobs with different deadlines.
- Establish blast radius in countable terms: how many tenants, how many writes, and crucially whether the effect was loss or only delay. An append-only revision table or a pending outbox row means the change survived and the projection was merely behind, which is a repair rather than a data-loss incident.
- Prove the mechanism instead of asserting it. Name the trace span that grew, the plan that flipped to a sequential scan, the lease that expired, plus one alternative you ruled out and the signal that stayed flat while you ruled it out.
- Close on the durable fix and its cost, distinguishing what landed that week from what needed an expand-and-contract migration across several deploys, and say which of the two you actually finished.
Follow-up
- What would you do differently in the first five minutes, given the same dashboard and no more information?
- Which follow-up action did you deliberately not take, and why was dropping it the right call?
- How did you convince yourself the mitigation was safe to apply while the cause was still unknown?
Turn a code review disagreement into a decision
A colleague's change updates a row with UPDATE resource SET version = version + 1 WHERE resource_id = $1 AND version = $2 and treats an affected-row count of zero as a successful no-op. You read that as a silently lost update; they think returning 200 is friendlier to clients than returning a conflict. Describe how you have handled a review disagreement of this shape: what goes in the comment, when you leave the thread, and who decides. Then write the comment you would leave here, in under 80 words.
Approach
- Sort the disagreement before writing anything. A silently discarded write is a correctness claim about data; the choice between 409 and 412 is taste. Only the first justifies blocking a merge, and saying which one you are doing is most of the value of the comment.
- Make the claim reproducible in the comment itself with an interleaving rather than a principle: A reads version 7, B reads version 7, B commits version 8, A's predicate matches zero rows, A is told it succeeded and A's edit is gone.
- Offer the alternative with its cost attached: return 409 carrying the current version and the revision that won, so the client can re-read and re-apply. Note that automatic retry is not the fix, because a retry re-reads the winner's state and reapplies an intent formed against data that no longer exists.
- Apply an escalation rule you can state: two round trips on the thread, then a call, and the service's owner decides rather than the reviewer. A reviewer who cannot be overruled is a bottleneck with extra steps.
- Close in writing wherever the decision lands, so the next reader finds the reasoning in the code or the ticket instead of in a collapsed review thread.
Follow-up
- Where would you put the test that fails if someone reintroduces the swallowed zero rowcount?
- The author says clients cannot handle a 409. How do you check whether that is true?
- How do you handle the same review comment when the author is more senior than you and in a hurry?
Unblock an engineer without taking the keyboard
A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.
Approach
- Ask before diagnosing, and ask for things answerable from data they already have: the attempt count on the job rows that produced duplicates, the handler's observed duration against its lease expiry, and whether the duplicate rows share a natural key that a unique constraint could have caught.
- Teach the shape rather than the answer. A lease cannot distinguish a dead worker from a slow one, so a handler that outruns its lease is running twice by design, and deploys deliver the other half by killing handlers mid-run on every rollout. Both of their candidate theories produce identical duplicate rows, which is why the evidence has to come from timings rather than from argument.
- Hand over a checklist they execute: a natural key on every write the handler performs so the second copy collides rather than appends, the record of intent written before any external effect, a lease heartbeat while running, and the metric that shows it working.
- Keep ownership with them deliberately. Pair on the first write, then step back; if you finish it yourself you have closed one ticket and left the same person stuck on the next redelivery.
- Close on the systemic gap that let two days pass, which is usually a missing dashboard for attempt counts or an undocumented at-least-once contract, and fix that rather than only the bug.
Follow-up
- How would you distinguish a genuine double-delivery from a lease expiry using only the data already stored?
- Their handler calls an external endpoint before recording that it did. What do you tell them to change first?
- What do you do the third time the same person brings you the same class of bug?
- 01
Pick an incident you personally drove, ideally one where writes were affected rather than reads. In six to eight minutes: state the symptom as it first appeared on a dashboard, the blast radius you established before you knew the cause, the mitigation you applied and when, the mechanism you eventually proved, and the follow-up that would prevent a repeat. Bring numbers: error rate, tenants affected, minutes to mitigate, minutes to resolve. If you cannot name what you measured, choose a different incident.
- 02
A colleague's change updates a row with UPDATE resource SET version = version + 1 WHERE resource_id = $1 AND version = $2 and treats an affected-row count of zero as a successful no-op. You read that as a silently lost update; they think returning 200 is friendlier to clients than returning a conflict. Describe how you have handled a review disagreement of this shape: what goes in the comment, when you leave the thread, and who decides. Then write the comment you would leave here, in under 80 words.
- 03
A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.
Is this an official Washington University in St. Louis interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Washington University in St. Louis. Rounds and questions reflect what candidates have reported, not a process Washington University in St. Louis has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗What topics does Washington University in St. Louis test in interviews?
Washington University in St. Louis interviews most often cover Statistical Analysis, Stakeholder Management, Technical Interviewing, Scientific research (thesis/dissertation work), and Requirements Elicitation. The exact emphasis depends on the specific role you apply for.
PracHub interview research ↗Sources & methodology 3 sources ↗
Official role evidence, timestamped platform data and clearly labeled preparation advice.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
platform · Accessed 2026-09-30 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
platform · Accessed 2026-09-30 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
platform · Accessed 2026-09-30