Auria Software Engineer Interview Questions & Guide 2026

The 60-second brief

As a Software Engineer at Auria, you are at the forefront of engineering solutions that support critical infrastructure, defense, and space-related ground systems. This role is not merely about writing code; it is about architecting and maintaining the sophisticated software frameworks that power high-stakes projects, such as the Sentinel program and various Ground C2 Systems. Your work directly impacts the reliability and performance of complex hardware and algorithmic systems.

Ask whether any round happens inside an existing repository instead of a blank file. Reading unfamiliar code, isolating a fault and making the smallest correct change is a different skill from writing a function from scratch, and it needs its own practice.

Focus areas

• Choose indexes from the query's access path
• Bound every outbound call with a timeout
• Paginate large result sets with keyset cursors

Overview

The environment at Auria is characterized by its focus on mission-critical applications where precision and scalability are paramount. You will collaborate with cross-functional teams of systems engineers, signal processing experts, and domain specialists to solve challenging problems in fields like Digital Signal Processing and RF Simulation. This position is ideal for engineers who thrive in environments that value deep technical expertise, architectural integrity, and the ability to bridge the gap between abstract requirements and tangible, operational software.

The Auria interview process

Auria candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Recruiter Screen

The person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.

What to demonstrate

• Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
• Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
• Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural

How to prepare

• Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
• Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
• Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one

Interview with Hiring Manager

Part of what is being decided is what your manager's week looks like once you are on the team: how they find out that something you shipped is broken, and whether they hear it from you or from a customer. The material that moves this round is therefore not the launch, it is the week after. Describe how you knew the change was working, which number you watched and for how long, and what you would have reverted to. A project that ends at the word shipped leaves the manager guessing about the part they care about most.

What to demonstrate

• How a change of yours was verified in production, and whether that check existed before the deploy or was assembled afterwards once something looked wrong
• Whether the rollback story is specific, including the case where a revert is not enough: once a migration has dropped a column or rewritten data, going back is its own change with its own risk
• Whether you can describe an incident you caused, how it was found, and how much time passed between it starting and anyone noticing
• Whether bad news in your account travels early and from you, or consistently arrives via somebody else

How to prepare

• For your last significant change, write down what you watched after the deploy, for how long, and the number that would have made you revert. If nothing was watched, say that plainly instead of inventing a dashboard.
• Work out the rollback answer for a change that touched stored data, and be able to say what made it reversible or what you would have had to do instead. That distinction is a level signal on its own.
• Write the two-minute version of an incident you owned: what broke, how it surfaced, what you did in the first ten minutes, and the change that stopped it recurring. Get the timeline straight, because this is the story most likely to be interrupted with questions.

Where candidates lose points

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.

Sorting when the problem never required a total order

Match the algorithm to the guarantee actually needed: the top k comes from a size-k heap in O(n log k) time and O(k) space, distinctness needs a set rather than an ordering, and a small bounded integer key range admits a linear counting pass. A full O(n log n) sort is the right default only when you genuinely need everything in order.

Abandoning working code to chase the optimal solution

Get the straightforward version correct, state its complexity, and only then optimise, keeping the working version until the faster one passes the same cases. A correct quadratic solution with a stated path to linear beats a half-written optimal one that never ran.

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

1. 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.
2. 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.
3. 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.
4. 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.
5. Choose the late-event policy from what the projection is keyed on. The projection upserts on (aggregate_id, aggregate_version) and discards a version it has already applied, so a late event is safe to apply out of order and correctness never depended on the merge at all. Apply it, recompute the affected feed page, and count lateness so the 30-second budget can be re-derived from data rather than folklore.
6. Say what the merge does not buy: ordering is guaranteed within one aggregate by the log's partitioning, and no watermark makes the cross-aggregate order authoritative. Two events from different aggregates in the same millisecond have no true order, so the feed's order is a presentation choice that must be stable rather than correct.

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?
• One partition is ten minutes behind because its producer is slow. Do you stall the feed or emit without it?

Diff a projection against the primary without per-row point reads

Hard · reconciliation · range hashing · throttling

The listing projection has drifted and some rows show a stale version. The primary holds 40,000,000 resource rows across 12,000 tenants while serving 1,200 writes and 14,000 reads per second. The obvious repair, reading each resource row and comparing its version against the projection, is correct and would eventually finish. Explain precisely why it is unacceptable here, then give a diff that finds the differing rows, state its complexity, and make it safe to run against a live primary. Replication lag is usually under 100 ms and is not bounded.

Approach

1. Quantify the naive cost rather than calling it slow: 40,000,000 point reads at even 0.5 ms each is over five hours serialised, and the only lever is concurrency, which is exactly what you cannot spend. The primary's pool is sized for the write path, and 40,000,000 random reads evict the buffer cache that sustains the 85 percent cache hit rate, so the audit degrades the system it is auditing.
2. Replace random access with one ordered pass per side. Both sides can be read in (tenant_id, resource_id) order, which is a sequential scan on each and a merge join in O(n) time and O(1) memory. For a dense diff that is the whole answer, and it reads the primary once instead of 40,000,000 times.
3. For the expected sparse case, compare range hashes instead of rows: partition the key space, compute per range an order-independent aggregate over hash(resource_id, version), compare aggregates, and descend only into ranges that differ. With d differing rows and branching factor B, at most d ranges mismatch per level, so the drill-down examines O(d log_B(n/d)) ranges and reads full rows only in mismatching leaves.
4. Aggregate with a sum modulo 2^64 or a multiset hash, never XOR. XOR is order-independent but self-cancelling, so two rows wrong in the same way, or a row duplicated on one side, leave the range aggregate matching and the range is declared clean.
5. Pin the comparison to a point in time or it reports lag as drift: consider only rows whose updated_at is older than now minus a lag margin, and re-check each candidate mismatch individually before repairing. At 1,200 writes per second a diff without this reports thousands of false positives, and an unattended repairer would then overwrite live rows with stale values.
6. Make the run resumable and throttled: batch by range key, persist the last completed range, and watch a signal such as replica lag or primary CPU, pausing rather than pressing on. A reconciliation that cannot be stopped and resumed gets killed halfway and restarted from zero, which is how a repair becomes an incident.

Worked solution (35 min)

1. Compute the naive cost explicitly at 40,000,000 reads and 0.5 ms each, then at 100 concurrent, and state what those connections do to a pool already carrying 1,200 writes per second.
2. Write the merge-join version over (tenant_id, resource_id) and state its memory.
3. Define the range aggregate: the range key, the per-row hash input, and the combining function, with one sentence excluding XOR.
4. Work an example with 40,000,000 rows, branching factor 256 and 5 differing rows, and count the ranges examined.
5. Add the watermark filter and the resume point, and name the throttle signal the loop watches.

Expected result: A rejection of per-row point reads backed by the time cost and the cache-eviction argument, a single ordered merge join as the dense-case answer at O(n) time and O(1) memory, a range-hash drill-down examining O(d log_B(n/d)) ranges using a sum or multiset hash rather than XOR, a watermark excluding recently written rows, and a resumable throttled run loop.

Verify it yourself

• Injecting one known stale row makes exactly one leaf range mismatch and the drill-down reaches it.
• Duplicating a row on one side is detected, which is precisely the case XOR would miss.
• Running the diff twice against an idle system reports zero mismatches both times.

Follow-up

• The diff reports 900 stale rows. How do you decide between patching those rows and rebuilding the projection from resource_revision?
• Same job, but the projection lives in a search index that cannot be scanned in key order. What changes?
• How would you run this continuously at low cost instead of only as incident response?

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

1. 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.
2. 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.
3. 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.
4. 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.
5. Frame the hash preimage so concatenation cannot collide: delimit or length-prefix the method, path and body, otherwise one request's fields can be rearranged into another request with the same byte stream and the same fingerprint.
6. Name the refusals and their consequence: no case folding, no dropping of null-valued keys, no Unicode normalisation. Each makes two different requests fingerprint alike, and the resulting failure is the worst one this table has, since the second request is answered with the first one's stored response and its effect never happens.

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?
• The endpoint takes 1,000 requests per second with 256 KB bodies. What does hashing cost, and does it belong at the edge or in the core service?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

resource carries version INT NOT NULL DEFAULT 1. resource_revision holds revision_id, resource_id, version, actor_user_id, change_kind, patch JSONB, request_id, created_at with UNIQUE (resource_id, version). outbox_event holds aggregate_type, aggregate_id, aggregate_version, event_type, payload, status. A PUT carries the version the client read. Write the exact statements for the single transaction that applies the edit, records the revision and enqueues 'resource.updated', and give the handler's branch on zero affected rows. Then say what PostgreSQL 16 does under READ COMMITTED when two of these updates hit one row at once.

Approach

1. One transaction, three writes, no network call inside it: UPDATE resource SET title = $3, version = version + 1, updated_at = now() WHERE resource_id = $1 AND tenant_id = $4 AND version = $2; then INSERT the resource_revision row at version $2 + 1; then INSERT the outbox_event row at the same aggregate_version. The event goes to a table rather than a broker because no transaction spans both.
2. Branch on the affected-row count before doing anything else. Zero has three causes — stale version, wrong tenant, row gone — so re-read once and map to 409 carrying the current version, or 404 for an id outside the caller's tenant, which also stops the endpoint confirming that another tenant's id exists.
3. State the engine behaviour instead of assuming it. Under READ COMMITTED the second UPDATE blocks on the row lock, and when the first commits PostgreSQL re-evaluates the WHERE clause against the newly committed row, so the version predicate now fails and the statement reports zero rows. Under REPEATABLE READ the identical collision raises SQLSTATE 40001 instead, so the handler must fold both shapes into one conflict response.
4. Keep UNIQUE (resource_id, version) even though the predicate already serialises writers. It is what makes a lost update unwritable if any other path ever reaches the revision table, and it converts a logic bug into 23505 rather than into a silently missing history row.
5. Refuse to auto-retry the whole PUT. A retry re-reads the winner's state and reapplies an intent formed against data that no longer exists — the silent overwrite the version token was added to detect. Return the conflict; merge field-wise only if the patches are provably disjoint.
6. Note that now() is the transaction timestamp in PostgreSQL, so resource.updated_at, the revision's created_at and the outbox row share one instant, which is what later makes reconciliation between the three tables unambiguous.

Follow-up

• A client sends the version it read ten minutes ago and the resource has moved three versions. What is in your 409 so it can resolve the conflict without a full re-fetch?
• Two editors, two disjoint fields, no overlap. Does your answer still refuse the second write, and should it?
• Every write now touches a second hot table. How do you keep the outbox insert and its partial index from becoming the write bottleneck at 1.2k writes/second?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

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

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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)

1. 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.
2. 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.
3. Run the two negative cases and capture the plan for each.
4. Re-run the original tenant listing query to confirm the new index has not displaced the index that query depends on.

Expected result: With (tenant_id, owner_user_id, created_at DESC) the plan is an Index Scan with no Sort, roughly 20 rows and a handful of buffers, sub-millisecond. owner_user_id alone still scans; ORDER BY updated_at seeks on the pair and then sorts that owner's rows.

Verify it yourself

• Actual rows at the leaf node is on the order of the LIMIT, not the tenant's row count.
• No Sort node appears above the index scan.
• Buffers read stays constant as one owner's row count grows; if it grows with it, the ORDER BY is not being satisfied by the index.
• pg_stat_user_indexes shows idx_scan climbing on the new index and not on the listing index for this endpoint.

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?

How do you approach the integration of software with hardware or signal processing components?

Medium

Approach

1. Clarify what is being asked and what a complete answer contains.
2. Say what you would check first and why it is the highest-information step.
3. Work from the requirement backwards to the design.

Follow-up

• How would you know your answer was wrong?
• What assumption would you test first?

Can you walk me through a complex project you led or contributed significantly to?

Medium

Approach

1. Work from the requirement backwards to the design.
2. State your assumptions explicitly before working the problem.
3. Say what you would check first and why it is the highest-information step.

Follow-up

• What assumption would you test first?
• How would you know your answer was wrong?

Keep one tenant's bulk export from starving projection updates

Medium · backpressure · fair scheduling · head-of-line blocking · leases

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

1. 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.
2. 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.
3. 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.
4. 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.
5. 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)

1. Compute required in-flight workers per class from arrival rate times service time, and show what one shared pool does when both classes contend.
2. Write the claim statement with the per-tenant restriction, and state how the running count is derived so it survives a worker being killed.
3. Trace what the claim scans when one tenant holds 50,000 queued rows, under FIFO and under tenant-first selection.
4. Choose the backpressure threshold and the signal it watches, then state what the producer receives when it trips.

Expected result: Separate queues and worker pools per job class, sized from rate times service time (about 120 for 200 ms work, about 180 for exports). A per-(job_type, tenant) cap enforced inside the claim with FOR UPDATE SKIP LOCKED, with the running count derived from live leases so it self-heals. Tenant-first selection keeps claim cost proportional to active tenants rather than backlog depth. Enqueue is rejected on class depth, duplicates collapse on dedupe_key, and the alarm is oldest-queued-age per class.

Verify it yourself

• A tenant with 50,000 queued exports never occupies more than its cap's worth of workers.
• Claim cost does not grow as that tenant's backlog grows.
• Killing a worker mid-job returns its cap slot within the lease interval with no manual step.
• Projection p99 is unchanged while the export backlog drains.

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.

p99 jumped on one listing filter while p50 stayed flat

Medium · query planning · composite index · percentiles · tail latency

After a release that added an owner_user_id filter to the resource listing, p99 rose from 90 ms to 1.9 s while p50 stayed at 40 ms. Traffic and row counts are unchanged. resource carries the index (tenant_id, status, updated_at DESC, resource_id DESC). The new query filters tenant_id and owner_user_id, orders by updated_at DESC, resource_id DESC, and takes 20 rows. On PostgreSQL, explain the shape of the regression, prove it from a query plan, and give the index you would add.

Approach

1. Start from the shape. A flat p50 with a moved p99 means a subset of requests changed cost, not all of them, so the first job is naming the subset. Bucket the endpoint's latency by the tenant's row count; the natural hypothesis is that large tenants are a small share of requests and all of the tail.
2. Get the plan for the new query on a large tenant with EXPLAIN (ANALYZE, BUFFERS). Expect an index scan over the tenant's range, a filter discarding most of it, then a Sort feeding the Limit, possibly reporting Sort Method: external merge Disk. Read actual rows on the scan node, not estimated.
3. Explain why the existing index cannot serve it. A composite B-tree is seekable only as a left prefix, and with no equality predicate on status the scan cannot treat updated_at as an ordering, because rows in the tenant's range are ordered by status first. Everything matching must be read and sorted before LIMIT 20 can apply, so a tenant with 400,000 rows pays 400,000 rows to return 20.
4. Add (tenant_id, owner_user_id, updated_at DESC, resource_id DESC). Equality on the first two columns leaves the index ordered by updated_at within that pair, so the plan becomes an index scan that stops after 20 rows with no Sort node. PostgreSQL can scan a B-tree backwards, so the DESC markers matter only if the two sort columns ever disagree in direction; keeping them explicit documents the order the keyset cursor depends on.
5. Price the fix. This is a fourth index on a table taking about 1.2k writes/second, and every insert and version bump maintains it. Justify it against the query it serves, and check whether it makes an existing index redundant, which here it does not, since the original still serves the status-filtered default listing.
6. Re-measure per tenant-size bucket rather than in aggregate. A fleet-wide p99 can improve while the largest tenant is still on the old plan.

Follow-up

• The endpoint paginates with OFFSET. What does page 500 cost with your index, and what does the keyset version cost?
• How would you have caught this before release, given that a 10,000-row seed database produces the same plan shape at an unnoticeable cost?
• If a fourth index were unacceptable on write grounds, what else could serve this query?

Suggested practice checklist

For a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.

Day 1 — Numbers before diagrams

• Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
• Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
• Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.

Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.

Day 2 — One system, from requirements to schema

• Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
• Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
• Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.

Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.

Day 3 — The consistency you are actually buying

• Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
• Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
• Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.

Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.

Day 4 — Failure is the design

• For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
• Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
• Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
• Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.

Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.

Day 5 — Scaling the hot path

• Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
• Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
• Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
• Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.

Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.

Day 6 — Keep the coding hand in, at the bar that applies to you

• Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
• Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
• Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.

Deliverable: One problem hardened to review standard, with its test list and one written review comment.

Day 7 — Defend it while being interrupted

• Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
• Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
• Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.

Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.

What the behavioural round is actually measuring

Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.

What is your experience with the specific tech stack requested for this program?

Medium · behavioural and engineering judgement

Approach

1. Pick a story where you made the decision, not one where you watched it.
2. Close with what you would do differently, concretely.
3. Give the blast radius: what could have broken, and what you measured.

Follow-up

• How did you know your change caused the improvement?
• What would you do differently if you ran that again?

Can you describe a time you had to optimize a system for better performance or reliability?

Medium · behavioural and engineering judgement

Approach

1. Pick a story where you made the decision, not one where you watched it.
2. State the situation in two sentences and spend the rest on the reasoning.
3. 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?

Tell me about a conflict you had with a coworker and how you resolved it.

Medium · behavioural and engineering judgement

Approach

1. Name the disagreement and how you resolved it with evidence.
2. State the situation in two sentences and spend the rest on the reasoning.
3. Close with what you would do differently, concretely.

Follow-up

• What would you do differently if you ran that again?
• What did you decide not to do, and why?

Can you describe your process for working within a multi-disciplinary team?

Medium · behavioural and engineering judgement

Approach

1. Close with what you would do differently, concretely.
2. Name the disagreement and how you resolved it with evidence.
3. 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?