Interview guides · DeZURIK

DeZURIK Software Engineer Interview Questions & Guide 2026

Software engineer interview prep for DeZURIK: original coding, system design and debugging practice, each with a worked solution.

The 60-second brief

The Software Engineer role at DeZURIK is a critical function that bridges the gap between sophisticated industrial engineering and modern software solutions. As a company renowned for its high-performance valves and flow control products, DeZURIK relies on software talent to drive innovation, optimize production, and maintain the complex systems that support its global operations.

State every complexity claim with the assumption sitting under it. Hash lookup is O(1) on average and only for a hash that spreads your actual keys; comparison-based sorting cannot beat n log n, though counting or radix sort can when the keys are bounded integers.

Focus areas

Make every write idempotent under retry
Choose indexes from the query's access path
Trace a symptom to a mechanism under load

Overview

In this position, you will be responsible for building, maintaining, and refining the tools and applications that define the company’s digital footprint. You will work within an environment that balances traditional industrial excellence with the need for agile, robust software. Success in this role requires not just technical proficiency, but the ability to translate complex business requirements into high-quality, reliable code that directly impacts the efficiency of DeZURIK operations.

The DeZURIK interview process

DeZURIK candidates report 4 rounds · ≈ 3-5 weeks. The stages below are what candidates describe, not a published process.

Initial Screening

Half of this call is the part candidates treat as small talk: start date, notice period, work authorisation and its timing, location and time zone, on-call, and the number. Those are what kill offers late, after several engineers have each spent a day. Surfacing a hard constraint now costs you nothing and occasionally buys you something, since a loop compressed to fit a competing deadline can usually only be arranged if it is asked for early. The common failure is deflecting the compensation question twice, then discovering at offer stage that the band never reached your number.

What to demonstrate

Whether your hard constraints are compatible with the role before a loop gets booked: earliest start, notice period, what authorisation you hold and when it needs action, days on site, willingness to carry a pager
Whether you give a compensation range with something behind it, such as current total compensation or a competing timeline, rather than leaving the band untested
Whether your stated timeline is real, since a competing deadline raised now is something scheduling can sometimes work around and the same deadline raised at offer stage usually is not

How to prepare

Write each constraint down in one line before the call and state them as facts rather than negotiating them live under a question you were not expecting
Set your range from two or three current data points for that level and location, and name the structure you are quoting in, so the number is comparable to the one they are holding
If another process is running, say where it stands and by when, and ask directly whether this loop can be scheduled inside that window

Technical Skills Assessment

Most of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.

What to demonstrate

Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
Whether a failing case is isolated and explained before any edit is made to the code

How to prepare

From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.

Interpersonal Alignment

An unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.

What to demonstrate

Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating

How to prepare

Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes

Final Evaluation

Coding rounds mostly set a floor. They decide whether you clear the bar, not where you land on the ladder. Level tends to come out of the design discussion and the ownership stories, so the question worth auditing beforehand is whether the scope you describe matches the scope of the job. Work that stops at your own service, or a story whose hard part was writing the code rather than getting several people to agree on an interface, reads a level below where you think you are interviewing, and that gap is usually resolved downwards.

What to demonstrate

Whether the largest thing you describe owning ran end to end — the decision, the migration path, the rollout, and what you did when it went wrong — or stopped at the change you merged
Whether design answers include what you would not build, what you would defer, and what you would measure before committing, rather than only what the boxes are
Whether a disagreement in a story was settled with something checkable — a benchmark, a prototype, a written proposal — instead of by seniority or by waiting it out
Whether you can say which calls you made alone and which you escalated, and why the line sat where it did

How to prepare

Write your largest piece of owned work as a timeline of decisions — who decided what, when, and what you did when the plan broke — then delete every sentence whose subject is "we" and see how much survives
Take one system you know well and drill the migration answer: how old and new paths run side by side under live traffic, how you compare their outputs, what the rollback is once writes are going to both, and which step you would not automate
Map each line of the ladder in the job posting to a specific thing you have done, find the line you cannot support, and prepare the closest evidence you have plus an honest account of the gap

Where candidates lose points

Choosing an index from the columns a query mentions rather than from how it filters and orders

A composite B-tree index on (a, b, c) can be seeked only as a left prefix: equality on a, then equality on b, then a range or an ordering on c. A query that filters on b alone cannot seek into it at all and at best gets a full scan of the index; a query that filters a and ranges on b gets no benefit from c, because the index is only sorted by c within a fixed (a, b) pair. The practical consequence is that one index per column is close to useless for multi-predicate queries while a single correctly ordered composite index turns a scan into a lookup. The ordering half is what gets missed: if the index cannot satisfy the ORDER BY, the database must read every matching row and sort before the limit can apply, so a LIMIT 20 over a million matching rows still reads a million rows.

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.

A queue or buffer with no bound

Every producer-consumer boundary needs a capacity and a policy for reaching it: block the producer, shed load, or drop the oldest entry. Unbounded buffering converts a temporary slowdown into memory exhaustion and hides the backpressure signal that would have revealed the consumer was falling behind.

Check-then-act on shared state

Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.

Questions & practice

Collapse a redelivered event batch into per-aggregate high-water marks

Easy · hashing · at-least-once · aggregation

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).

Worked solution (20 min)

Write the pass: look up last_applied_version, skip if the event's version is not greater, otherwise upsert into the keep-map only when the incoming version exceeds the version already held, incrementing the discard counter on every skip.
Hand-trace one aggregate whose events arrive as v5, v3, v4, v5 with last_applied_version = 2, and confirm the output holds v5 once while the counter reads 3.
Compute the table footprint for 2,000,000 entries at 12 bytes of payload and a 0.7 load factor, then state the multiplier for a runtime that boxes keys and values.
Add the hash-partitioning fallback and say in one sentence why the per-partition results need no cross-partition merge logic.

Expected result: A single-pass O(n) reduction keyed on aggregate_id that keeps the maximum version rather than the last occurrence, O(d) space with the byte cost stated for 2,000,000 aggregates, a hash-partitioning fallback keyed on aggregate_id, and an explicit statement that batch-local deduplication does not replace a projection write keyed on (aggregate_id, version).

Verify it yourself

Shuffling the batch order changes nothing in the output, which is the test that separates max-wins from last-wins.
Feeding the same batch twice produces an identical keep-set and only doubles the discard counter.
Every emitted event has a version strictly greater than the supplied last_applied_version for its aggregate.

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?

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.
Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.

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?
How would you make the window survive a process restart, and is it worth the cost?

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

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?

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

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.
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.
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.
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.
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.
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

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.

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?

Is this job going to be challenging enough for you?

Medium

Approach

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

Follow-up

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

What classes or professional development are you currently pursuing?

Medium

Approach

State your assumptions explicitly before working the problem.
Work from the requirement backwards to the design.
Clarify what is being asked and what a complete answer contains.

Follow-up

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

Choose what to break when replication lag reaches forty seconds

Hard · replication lag · read-after-write · consistency vs availability · load shedding

Reads are served from two replicas: 14k requests/second, about 85% absorbed by cache, so roughly 2.1k reads/second reach the database. Writes go to the primary at 1.2k/second. A tenant's backfill drives replication lag from under 100 ms to 40 seconds and it is still climbing. Sessions that have just written are pinned to the primary. Decide, endpoint class by endpoint class, whether to serve stale, fail, or route to the primary, and justify each choice with the load it adds to the primary. Then state what you would have built beforehand.

Approach

Establish blast radius before cause, because mitigation and diagnosis have different deadlines. The decisive arithmetic is what happens if the database reads move to the primary: 2.1k reads/second on top of 1.2k writes/second roughly triples its operation count, on the node already absorbing the backfill that caused this. Reads and writes are not equal in cost, so treat that as an argument against a blanket move rather than as a capacity model - but it is enough to rule out routing everything to the primary.
Classify endpoints by what staleness costs, not by how important they feel. Reads whose staleness is invisible - listings, search, counters - stay on the replica and return the watermark so the client can tell. Reads that immediately follow that same session's write keep their primary pin, which is a small bounded slice of traffic rather than the whole 2.1k/second. Reads that feed a decision with a side effect - authorisation, quota, the read half of a read-modify-write - must not be stale at all, because a 40-second-old permission row is the stale-permission failure wearing a different costume; those go to the primary or fail.
Shed instead of queueing. If the must-be-fresh class alone exceeds the primary's headroom, refuse its lowest-value slice with 503 and a retry-after. A request queued behind a saturated primary holds a connection for a client that has already given up, and the retry storm that follows is what turns degradation into an outage. Bound the connection pool per role so the read fallback cannot consume the write path's connections - that bulkhead is the single decision that determines whether writes survive the next ten minutes.
Attack the cause in parallel, since it is the one thing that can be stopped. The backfill is the load generator. A backfill that reads replication lag as its throttle signal and pauses above a threshold would have made this a non-event, with batch sizes small enough that each batch's write volume is a fraction of what a replica can apply per second. That is most of the answer to what should have existed beforehand.
Name the mechanism you would prefer over session pinning. Capture the write position at commit and require the read path to be at or past it: compare the primary's pg_current_wal_lsn() at commit time against the replica's pg_last_wal_replay_lsn(), and fall back to the primary only for the specific request that is ahead of the replica. Session pinning is the cheap approximation and it over-pins - every read in the window goes to the primary whether or not it needed to, which is a share of the cost being paid right now.

Worked solution (35 min)

List the endpoints in three buckets - staleness invisible, staleness visible to the writer only, staleness unsafe - and attach the share of the 2.1k reads/second each bucket carries.
Compute the primary's operation count under each routing option and mark which options are arithmetically available.
Write the pin rule and its window, then the shed rule: which slice, what status code, what retry-after.
Write the backfill's throttle predicate against a measured lag value, including its pause threshold and resume condition.

Expected result: Staleness-invisible reads stay on the replica and return a watermark. Reads following the session's own write keep a bounded primary pin. Reads that gate a side effect go to the primary or fail, with the lowest-value slice shed as 503 plus retry-after if headroom runs out. Connection pools are bounded per role so the read fallback cannot starve writes. The backfill is throttled on measured replication lag. The durable fix is a commit-position token compared against the replica's replay position, replacing blanket session pinning.

Verify it yourself

The traffic routed to the primary under the plan is stated as a rate, not as a category.
No element of the plan increases the write path's connection usage.
The backfill throttle references a measured lag value and has an explicit resume condition.
Every endpoint left on a replica returns something with which a client can detect staleness.

Follow-up

Lag returns to normal in nine minutes. Which mitigation do you remove first, and which one stays permanently?
A user reports their change did not save, and the write committed. Trace the path that produces that report and name the signal that would have shown it before the report arrived.
The replica is 40 seconds behind but otherwise healthy. Do you take it out of rotation? What does that do to the other replica's lag?

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

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.
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.
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.
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.
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.
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 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.

Day 1 — Measure 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.

Day 2 — References, 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.

Day 3 — Types, 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.

Day 4 — Concurrency, 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.

Day 5 — Debugging 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.

Day 6 — Tests 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.

Day 7 — Debug 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.

Outage answers are graded on instruments

Nobody is scoring your stamina at three in the morning. What carries weight is which signal told you something was wrong, what you measured before touching anything, what you rolled back versus what you fixed forward, and why you picked one. 'We restarted it and it went away' is a story about not knowing.

Describe a challenging project you have worked on.

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Give the blast radius: what could have broken, and what you measured.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

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

Explain the technical problems you faced during a past project.

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
Name the disagreement and how you resolved it with evidence.
Close with what you would do differently, concretely.

Follow-up

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

Please provide a brief overview of your resume and professional background.

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
State the situation in two sentences and spend the rest on the reasoning.
Give the blast radius: what could have broken, and what you measured.

Follow-up

How did you know your change caused the improvement?
What did you decide not to do, and why?

How did you approach solving those technical problems while collaborating with others?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
State the situation in two sentences and spend the rest on the reasoning.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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