AGC Biologics Software Engineer Interview Questions & Guide 2026

The 60-second brief

As a Software Engineer at AGC Biologics, you are a vital contributor to the digital infrastructure that supports complex biopharmaceutical manufacturing. This role is not merely about writing code; it is about creating robust solutions that facilitate life-changing medical advancements. You will work at the intersection of technology and biotechnology, ensuring that systems are scalable, reliable, and compliant with the rigorous standards of the pharmaceutical industry.

The title spans product, platform and infrastructure work, and which of those the seat actually is decides whether design or algorithms carries more weight in your preparation. The posting rarely settles it; what the team is on call for usually does.

Focus areas

• Model bitemporal coverage and retroactive eligibility changes
• Authorise every identified read against current consent
• Hold accumulator limits under concurrent claim adjudication

Overview

The position offers a unique opportunity to influence how data and software integrate into global manufacturing processes. You will collaborate with cross-functional teams, including validation, engineering, and operations, to solve real-world problems. Whether you are working on technical transfers or process improvements, your work directly impacts the efficiency and quality of the products delivered to patients worldwide. It is a role for those who value precision, technical excellence, and the collaborative spirit required to drive innovation in a high-stakes environment.

The AGC Biologics interview process

AGC Biologics candidates report 2 rounds · ≈ 2-4 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 Phase

The same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.

What to demonstrate

• In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
• In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
• Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes

How to prepare

• Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
• Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
• For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first

Where candidates lose points

Assuming admission, discharge and transfer messages arrive in the order the events happened.

Interface engines route by message type across separate queues and retry independently, so a discharge can land before the admission it closes and an update can land before the registration it modifies. Ordering has to come from the sender's event timestamp plus a per-encounter sequence, and the consumer has to apply out-of-order and late-arriving events correctly rather than rejecting them, because rejection turns a recoverable ordering issue into permanent data loss that nobody notices until a report is short.

Treating a medical record number or a member ID as a globally unique key and joining on it directly.

These identifiers are unique only within the authority that issued them. Two facilities in one network routinely have the same medical record number for different people, and member IDs get reissued when someone changes plans. Joining on the bare value merges two patients' records, which is the most damaging failure available in this domain, and it passes every test written against a single-facility fixture because the collision only appears once a second source is connected.

Assuming fixed-width integer arithmetic cannot overflow

In languages with fixed-width integers, including C, C++, Java, Go and Rust, computing a midpoint as (lo + hi) / 2 overflows once the sum passes the type's maximum, so write lo + (hi - lo) / 2 instead. Say which language you are in: arbitrary-precision integers, as in Python or Ruby, remove this specific hazard and none of the others.

Saying 'eventually consistent' without naming the anomaly a user would see

Describe the concrete symptom you are choosing to accept: the author reloads and their own comment is missing for two seconds, or two devices show different balances for a minute. The class of consistency model is a technical label; the tolerable anomaly is the actual product decision.

Questions & practice

Reconstruct what the chart displayed at five million past instants

Hard · external sort · as-of join · io bound

observation_result holds 200 million versions: observation_id, enterprise_person_id, filler_order_id, loinc_code, value_numeric, result_status, collected_ts, issued_ts, version, supersedes_observation_id. An incident review hands you 5 million queries of (enterprise_person_id, filler_order_id, loinc_code, as_of_instant). For each, return the version that was visible at that instant, meaning the one with the greatest issued_ts at or before it. Neither side fits in memory. The per-query scan is correct. Explain precisely why it is too slow, then give a plan with its complexity.

Approach

1. Name the clock before naming an algorithm. Visibility is issued_ts, the release time. collected_ts is when the specimen was drawn and can precede release by hours, so ordering on it reports a correction as visible long before anyone could have seen it. No amount of index work rescues the wrong column.
2. Be exact about why the naive plan fails, because the interviewer is testing whether you can tell arithmetic cost from I/O cost. Per query the chain scan is O(V_k) and the arithmetic is trivial, but 5 million independent lookups into a 200-million-row structure that does not fit in RAM is 5 million random reads. The job is bounded by seeks per query, not by comparisons, and buying a faster comparison changes nothing.
3. Convert random access into sequential access. Hash-partition both sides on the chain key (enterprise_person_id, filler_order_id, loinc_code) into P shards sized to fit memory, sort each shard's versions by (chain key, issued_ts) and its queries by (chain key, as_of_instant), and sweep the pair in lockstep. Total O((V + Q) log(V + Q)) with external sort, replacing Q seeks with two sequential passes.
4. Inside a chain the sweep is linear, not logarithmic, because queries are visited in ascending as_of order and the version pointer only moves forward: O(V_k + Q_k) per chain. Binary search per query is the better shape only when Q is small relative to V and the versions are already indexed and resident.
5. Return the empty answer as a distinct outcome. A query whose as_of precedes the first issued_ts means nothing was displayed, which is not the same as the earliest value, and is frequently the exact fact the review is chasing.
6. Return a version later marked entered_in_error if it was live at the instant asked about. Reconstructing the past means reporting what was on the screen, including what was wrong, and quietly substituting today's truth defeats the purpose of the exercise.

Follow-up

• Read replicas lagged 40 seconds at the time. Does your answer describe what the clinician actually saw, and how would you bound the difference?
• Serve the same question online for a single chart at a p99 under 50ms. What changes?
• One partner changed its filler_order_id format mid-year, so the chain key is not stable. How does that appear in your output, and how do you detect it rather than returning empty answers?

Group transfer-chained encounters into episodes with out-of-order arrival

Medium · graph traversal · cycle detection · memoisation

You have up to 5 million encounter rows for a reporting month, in arbitrary order: encounter_id, enterprise_person_id, facility_id, encounter_class, admit_ts, discharge_ts (nullable), status, prior_encounter_id (nullable, set on transfer), source_event_ts. prior_encounter_id may reference a row that arrives later or never arrives. Group rows into episodes, one per maximal transfer chain, returning episode root, person, earliest admit_ts, latest discharge_ts (null if any member is still open) and member count. Report cycles instead of following them. Target O(N) expected time.

Approach

1. Recognise the shape: every row has at most one prior_encounter_id, so this is a forest of parent pointers, not a general graph. A hash map from encounter_id to row is the whole index you need, and the work is O(N) expected rather than anything sort-shaped.
2. Resolve roots by iterative pointer-following with write-back memoisation: from each node walk prior_encounter_id until you reach a node with no parent, a parent absent from the input, or a node whose root is already known, then stamp the discovered root onto every node on that path. Each node is finalised once, so total work stays O(N) even when one chain is very long.
3. Do not recurse. A malformed feed can produce a chain tens of thousands deep, and a recursive resolver dies on stack depth taking the whole batch with it. An explicit loop with write-back is both faster and bounded.
4. Detect cycles with three-state marking: unvisited, on the current path, finalised. On reaching a node that is on the current path, emit a data-quality record naming every encounter_id in the cycle and exclude that component from the episode output. Choosing an arbitrary root instead would produce an episode that looks plausible and will be believed.
5. Keep a dangling prior_encounter_id distinct from a genuine root. It means the chain starts outside the window or has not arrived, so flag the episode left-truncated and carry that flag into the output: an episode with an unknown start cannot be used for length of stay or readmission counting without biasing both downward.
6. Compute aggregates during the same pass that assigns roots: minimum admit_ts, maximum discharge_ts with null propagation so any open member leaves the episode open, and exclude cancelled and entered_in_error members from the aggregates while keeping them retrievable.

Follow-up

• Two encounters name each other as prior_encounter_id. What does your detector emit, and who needs to see it?
• One episode spans two facilities whose clocks differ by four seconds. Do you order on source_event_ts or admit_ts, and what does the choice cost?
• A chain is cut in half by the month boundary. What does this month's report say, and how do you make next month's agree with it?

Flatten overlapping coverage spans into a primary-payer timeline

Medium · sweep line · intervals · bitemporal

For one enterprise_person_id you hold up to 10,000 coverage_span rows: coverage_id, payer_id, plan_id, effective_date, termination_date (null means open-ended), coverage_order (1 is primary), status, valid_from and valid_to (system time). Given a system instant T and a date range D0 to D1, return disjoint business-date intervals covering that range, each labelled with the active coverage_id of lowest coverage_order, and with uncovered stretches returned explicitly. termination_date is the last covered day. Only status active counts. Target O(N log N) time, O(N) space.

Approach

1. Apply the system-time slice before any interval logic: keep rows where valid_from is at or before T and valid_to is after T. That single predicate is what makes the answer 'what we believed at T' rather than 'what we believe now', and a retroactive termination loaded after T leaks straight into the output if you skip it.
2. Convert each surviving row to a half-open day interval from effective_date up to termination_date plus one day, using positive infinity for a null termination. Half-open removes every off-by-one at the join between a termination and the next plan's effective date, which is where same-day switches get turned into false gaps.
3. Sweep: emit 2N boundary events, sort by date in O(N log N), and hold the active set in a min-heap keyed by (coverage_order, coverage_id) with lazy deletion, popping the top while it has already expired at the current boundary. Each coverage is pushed once and popped once, so the sweep is O(N log N) total and O(N) space.
4. Emit an output interval whenever the heap top changes between consecutive boundaries, and emit an explicit uncovered interval when the heap empties. A gap reported as a gap is a different answer from a gap omitted, and downstream the difference is a denial versus a silent assumption of coverage.
5. Fix the tie-break and state it: lowest coverage_order, then lowest coverage_id. Two rows at order 1 is a data error, and without a deterministic rule the coordination-of-benefits answer changes between two runs over identical data.
6. Clip to D0 and D1 last, so a coverage that starts before D0 still contributes its correct order inside the window.

Worked solution (30 min)

1. Filter to rows live at T, then map each to a half-open interval and a (coverage_order, coverage_id) priority.
2. Build the boundary list, sort it, and sweep with a lazily-deleted min-heap.
3. At each boundary, drop expired heap tops and compare the new top to the previous one, emitting an interval on change.
4. Emit an uncovered interval whenever the heap is empty between two boundaries.
5. Clip the emitted list to D0 through D1 and assert the intervals are disjoint and contiguous.

Expected result: Window 2026-01-01 to 2026-12-31 with three live rows: coverage 1 at order 1 effective 01-01 terminating 06-30, coverage 2 at order 2 effective 03-01 open-ended, coverage 3 at order 1 effective 09-01 open-ended. Output is three intervals: 01-01 to 06-30 labelled coverage 1, 07-01 to 08-31 labelled coverage 2, 09-01 to 12-31 labelled coverage 3, with no gaps. Delete coverage 2 and the middle interval becomes an explicit uncovered stretch of 07-01 to 08-31.

Verify it yourself

• The emitted intervals are sorted, pairwise disjoint, and their union is exactly D0 through D1 once uncovered stretches are included.
• Every boundary date in the output is D0, the day after D1, some effective_date, or some termination_date plus one day. Any other date means a boundary was synthesised.
• Removing the lowest-order coverage promotes exactly the next order and leaves every boundary it did not touch unchanged.
• Choosing T before the load time of a retroactive termination returns the pre-correction timeline, and T after it returns the corrected one, from the same table.

Follow-up

• Run the same query at two system instants and diff the timelines. What did the correction change, and what does each extra instant cost you?
• A termination arrives with an effective date earlier than the service date of a claim that already adjudicated and paid. What does the timeline now say, and what has to happen to that claim?
• Two enrolment files disagree on coverage_order for a dependent. What does your tie-break do, and what should the system do instead of tie-breaking?

Return the displayable version of each analyte with a window function

Medium · window functions · versioned rows · indexing

observation_result is append-only: observation_id, enterprise_person_id, encounter_id, accession_id, placer_order_id, filler_order_id, loinc_code, value_numeric, unit_ucum, result_status in ('registered','preliminary','final','amended','corrected','entered_in_error'), collected_ts, issued_ts, version, supersedes_observation_id. A chart view needs the currently displayable value for each (accession_id, loinc_code) for one person over the last 90 days, excluding analytes whose newest version is entered_in_error but keeping analytes whose older versions were. Write the query, give the index that serves it, and say honestly which part of the plan the index cannot remove.

Approach

1. Recognise the shape: this is a top-1-per-group over a version chain, not a filter. Pick the newest version first, then apply the status predicate to the winner — the order is the whole exercise.
2. Use DISTINCT ON (accession_id, loinc_code) with ORDER BY accession_id, loinc_code, version DESC, observation_id DESC on PostgreSQL; the ORDER BY must start with the DISTINCT ON expressions or it is a syntax error. Use ROW_NUMBER() OVER (PARTITION BY ... ORDER BY version DESC, observation_id DESC) with an outer WHERE rn = 1 if you need portability, since a window function cannot be referenced in the WHERE of its own query level.
3. Push only person and time into the inner WHERE. Pushing result_status <> 'entered_in_error' inside promotes a retracted result's predecessor back onto the chart, which is the bug the prompt is testing for.
4. Index (enterprise_person_id, collected_ts DESC) so the 90-day slice is an index range scan rather than a scan of the person's whole history.
5. Be straight about the limit: because collected_ts carries a range predicate, no b-tree index can also deliver rows pre-ordered by (accession_id, loinc_code, version DESC), so the sort stays in the plan. That is acceptable because one person's 90-day slice is hundreds to low thousands of rows, and it is a quicksort in work_mem rather than an external merge — verify that rather than assume it.

Follow-up

• A correction arrives carrying the same filler_order_id and the same version number as the row it corrects. What breaks, and what do you add to the sort key?
• The same query for a cohort of 50,000 people instead of one person: does DISTINCT ON still hold up, and at what point do you materialise a latest-version table with a partial index?
• How would you show the clinician that the displayed value was amended, given the chart query returns only the winner?

Diagnose why an eligibility lookup ignores its index

Medium · indexing · query planning · partial indexes

A 40-million-row coverage_span serves eligibility lookups with: SELECT coverage_id, plan_id, coverage_order FROM coverage_span WHERE enterprise_person_id = $1 AND valid_to = 'infinity' AND effective_date <= $2 AND (termination_date IS NULL OR termination_date >= $2). An index exists on (effective_date). EXPLAIN (ANALYZE, BUFFERS) shows a sequential scan reading 1.8 million buffers at 2.4 seconds. Explain why that index cannot help, propose the index plus any schema change that makes the predicate index-friendly, and state what you expect the new plan and buffer count to be.

Approach

1. Read the predicate for selectivity before touching the index. effective_date <= $2 matches most of a 40-million-row history, so a leading range scan on that column returns a large fraction of the table and the planner correctly prefers a sequential scan over millions of random heap fetches. The index is not being ignored; it is being rejected on cost.
2. Put the equality column first. enterprise_person_id = $1 selects a handful of rows out of 40 million, so it must lead the composite index; a b-tree can only use columns after the first range predicate as filters, not as search bounds.
3. Turn the open-ended termination into something indexable. The OR ... IS NULL branch forces either a BitmapOr or a post-index filter. Add coverage_end date GENERATED ALWAYS AS (COALESCE(termination_date, DATE '9999-12-31')) STORED and rewrite the predicate to coverage_end >= $2; both COALESCE over a column and a date literal are immutable, so the generated column is legal.
4. Make the index partial on WHERE valid_to = 'infinity'. Most rows in a bitemporal table are superseded beliefs, so the partial index is a fraction of the full one, stays in cache, and encodes the predicate for free rather than re-checking it per row.
5. Consider the range alternative honestly: a daterange(effective_date, termination_date + 1, '[)') column with a GiST index and the && operator handles overlap queries and supports an EXCLUDE constraint against overlapping active coverage, but GiST lookups are slower than b-tree for this point-in-time pattern. Pick b-tree for the read path and keep GiST only if you also need the overlap constraint.
6. Verify rather than assert: re-run EXPLAIN (ANALYZE, BUFFERS) and read Rows Removed by Filter and the shared hit/read split, not just the total time.

Worked solution (30 min)

1. Run EXPLAIN (ANALYZE, BUFFERS) on the original and write down estimated versus actual rows for each predicate, which shows effective_date <= $2 is not selective.
2. Add the generated column and rewrite the predicate to use coverage_end.
3. Create the partial composite index and ANALYZE the table so the planner has statistics for the new column.
4. Re-run EXPLAIN (ANALYZE, BUFFERS) and confirm an index scan with Rows Removed by Filter at or near zero.
5. Measure the partial index size against the full-table equivalent with pg_relation_size to confirm the cache argument is real and not assumed.

Expected result: The (effective_date) index is rejected because effective_date <= $2 is a low-selectivity leading range predicate over 40M rows; the planner's estimated cost of random heap access exceeds a sequential scan. Fix: ALTER TABLE coverage_span ADD COLUMN coverage_end date GENERATED ALWAYS AS (COALESCE(termination_date, DATE '9999-12-31')) STORED; CREATE INDEX CONCURRENTLY idx_coverage_active ON coverage_span (enterprise_person_id, effective_date, coverage_end) WHERE valid_to = 'infinity'; and rewrite the last predicate as coverage_end >= $2. The new plan is an Index Scan on idx_coverage_active, Rows Removed by Filter near zero, buffers in the low tens rather than 1.8 million, and sub-millisecond execution.

Verify it yourself

• Buffers drop from roughly 1.8 million to double digits — an order-of-magnitude change in buffers, not just in wall-clock time, is what proves the index is doing the work.
• Rows Removed by Filter is 0 or near it; a large value means a predicate is still being re-checked per row and belongs in the index or the partial clause.
• The rewritten and original predicates return identical rows for a member with an open-ended coverage and for one terminated exactly on $2.
• pg_relation_size on the partial index is a small fraction of the same index built without the WHERE clause.

Follow-up

• After the change the plan is still a sequential scan for one particular member. What single query tells you whether that is stale statistics, a genuinely huge row count for that person, or parameter-dependent plan caching?
• Your generated column needs backfilling on a live table. Does adding a STORED generated column rewrite it, and what does that mean for your maintenance window?
• How would you serve the same lookup when the answer must be as of a past system-time instant rather than the current belief, given your index is partial on valid_to = 'infinity'?

What specific programming languages and frameworks do you use to ensure software reliability?

Medium

Approach

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

Follow-up

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

How have you applied your software skills to support cell culturing or similar laboratory processes?

Medium

Approach

1. Say what you would check first and why it is the highest-information step.
2. Work from the requirement backwards to the design.
3. 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?

Idempotent admission feed ingestion under replay and reordering

Easy · idempotency · at-least-once · event-ordering · queues

The gateway consumes HL7v2 admit, discharge, transfer and update messages over MLLP: a few million a day with fourfold nightly bursts, across twelve parallel workers. Delivery is at-least-once — a lost acknowledgement makes the sender resend, and an interface restart replays a window. Message types are routed through separate queues and retried independently, so a discharge can arrive before the admission it closes. Define the idempotency key and the assumption it rests on, the mechanism that makes deduplication atomic under concurrency, and how the encounter row reaches the same final state under any arrival order.

Approach

1. Reject the message control identifier on its own: it is unique only per sending application, and senders roll the counter over. Use the tuple of sending facility, sending application, control identifier and the message timestamp, or a SHA-256 of the raw message body when the header cannot be trusted to be stable. Write the assumption down — no control identifier reuse within the timestamp's resolution — and count near-duplicate bodies with distinct keys so you learn in production when it stops holding.
2. Make deduplication a write-side decision: insert the key into a ledger under a UNIQUE constraint in the same transaction that applies the effect. A select-then-insert in application code is not sufficient, because two workers holding the same replayed message both pass the existence check before either inserts; the unique violation is the only thing that serialises them. Deduplicating on read costs you the effect already having been applied.
3. Order per encounter rather than per queue. Carry the sender's event timestamp and apply with UPDATE ... WHERE source_event_ts < :incoming plus a row_version bump, so an older message loses deterministically regardless of which worker wins the race.
4. Make out-of-order arrival constructive instead of a rejection: a discharge for an unknown encounter inserts the encounter in status discharged with the discharge fields set, and the later admission fills only what is missing. Rejecting the orphan turns a recoverable ordering problem into permanent loss that surfaces months later as a census report that is short.
5. Prove the property rather than assert it: replay a fixture stream containing a duplicate, an early discharge and a late correction under several random permutations and assert the final encounter rows are identical. If the final state depends on the permutation, the merge rule is wrong and the fixture is fine.

Worked solution (25 min)

1. Write the idempotency key as a concrete tuple, then list each field's uniqueness scope and which sender behaviour would break it.
2. Write the ledger insert and the effect apply as one transaction, and state what the handler returns on unique violation.
3. Write the conditional UPDATE that enforces per-encounter ordering, and show what happens when a late update loses.
4. Define the merge rule for a discharge preceding its admission, naming which fields each message may set and which it must not overwrite.
5. Permute a ten-message fixture five ways and compare the final rows field by field.

Expected result: A composite idempotency key with its assumption stated and monitored, deduplication enforced by a UNIQUE constraint inside the effect transaction rather than by an application-level existence check, per-encounter ordering driven by the sender's event timestamp, and an upsert-and-fill merge rule that makes the final state independent of arrival order.

Verify it yourself

• Replaying the same message ten times produces one encounter row and one ledger entry.
• Five random permutations of the same fixture stream yield identical final rows.
• Two workers processing the identical replayed message simultaneously produce exactly one effect, with the loser observing a unique violation.
• A discharge arriving before its admission leaves the encounter discharged rather than rejected or re-opened.

Follow-up

• A sender re-emits an entire day after a restart with the same control identifiers but fresh message timestamps. What does your key do, and what would you change?
• Two update messages for one encounter carry identical sender event timestamps. How do you break the tie deterministically across twelve workers?

Gateway instances OOM every five days with flat traffic

Medium · memory leak · heap profiling · idempotency

Interoperability gateway instances are OOM-killed with exit 137 every five to six days. Resident memory grows about 400MB a day per instance and never falls; restarting resets it. Message throughput, connection count and partner mix are unchanged over that period. The gateway keeps an in-process set of recently seen idempotency keys — (sending facility, sending application, message control ID, event timestamp) — in front of the durable ledger. Give the ordered investigation and say what evidence would rule the idempotency cache out.

Approach

1. Establish that this is a retained-object leak before profiling. Compare resident memory with live heap measured immediately after a forced collection, sampled daily. Resident memory rising while post-collection live heap stays flat points at allocator fragmentation, off-heap or native buffers, or thread stacks, and a heap dump will implicate the wrong object. Rising post-collection live heap is a genuine leak.
2. Find what the growth is linear in. Plot daily growth against messages processed, connections accepted and wall-clock hours across instances carrying different traffic. Proportional to messages means something is retained per message; proportional to connections means per-connection state is not released on abnormal close, such as a half-open socket whose FIN never arrives.
3. Diff two heap dumps taken six hours apart on the same instance, sorted by retained size, and identify the dominator. An unbounded set or map shows up as one root holding millions of small entries. This is a five-minute answer once you have the dumps, which is precisely why the first two steps exist: they tell you whether the dump can answer the question at all.
4. Check the idempotency cache directly against arithmetic rather than suspicion. At single-digit millions of messages a day and a key of roughly 80 bytes plus per-entry overhead, an unevicted set grows by hundreds of megabytes a day, which matches the observed 400MB. A live gauge of its entry count that stays flat is the evidence that rules it out; a gauge tracking messages processed convicts it.
5. Fix and restore the invariant: bound the cache by size or by a time window, and keep the durable unique constraint as the authority. An in-process set is a latency optimisation and cannot be the correctness mechanism, because it does not survive a restart and is not shared across instances — so a replay right after a deploy would bypass it entirely.
6. Verify over 48 hours that post-collection live heap is flat, and alert on the cache's entry-count gauge rather than on resident memory, which only tells you after the growth has already happened.

Follow-up

• How long must the time window be? Tie the number to the sender's replay behaviour after an interface restart and to how far back the broker redelivers.
• If growth had tracked connections instead of messages, what would you look at first, and how would a half-open MLLP connection show up in the socket table?

Suggested practice checklist

Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.

Day 1 — Coding, one pass at shallow depth

• Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
• For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
• Fix nothing today. The value of the pass is the unfixed record.

Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.

Day 2 — Design, one pass at shallow depth

• Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
• After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
• Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.

Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.

Day 3 — Fundamentals and the practical rounds

• Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
• Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
• Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.

Deliverable: Eight scored short answers and one written reading of unfamiliar code.

Day 4 — The rounds that are about you, and the map

• Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
• Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
• Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.

Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.

Day 5 — First chosen area, to the depth you skipped

• Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
• After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
• Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.

Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.

Day 6 — Second chosen area, where the gap is coverage rather than speed

• Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
• Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
• Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.

Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.

Day 7 — Reassemble the loop

• Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
• Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
• Reduce the week to one page holding only the rules you can state without reading them.

Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.

Know where your numbers came from

Bring the two or three numbers the story rests on and know how they were collected. A p99 whose timer starts inside your handler excludes the time a request spent queued, so it can sit flat while users wait longer. Give the window, the percentile and what the measurement left out, or drop the number.

What is your approach to giving and receiving constructive feedback within a team?

Medium · behavioural and engineering judgement

Approach

1. Close with what you would do differently, concretely.
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 did you decide not to do, and why?

Can you describe a time you had to adapt to a change in project requirements?

Medium · behavioural and engineering judgement

Approach

1. Close with what you would do differently, concretely.
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

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

Tell me about a time you had to work with a difficult coworker; how did you handle the situation?

Medium · behavioural and engineering judgement

Approach

1. Give the blast radius: what could have broken, and what you measured.
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 would you do differently if you ran that again?
• How did you know your change caused the improvement?

What motivates you to perform at your best in a fast-paced environment?

Medium · behavioural and engineering judgement

Approach

1. Name the disagreement and how you resolved it with evidence.
2. Give the blast radius: what could have broken, and what you measured.
3. Pick a story where you made the decision, not one where you watched it.

Follow-up

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