IPCA Group Software Engineer Interview Questions & Guide 2026

The 60-second brief

A Software Engineer at IPCA Group plays a vital role in supporting the technological infrastructure that underpins pharmaceutical manufacturing. You are not just writing code; you are building the digital backbone that enables quality control, equipment monitoring, and process optimization across various production lines, including tablet, liquid, and injection manufacturing.

For a design discussion, a catalogue of architectures is worth less than the ability to turn a vague requirement into a data model and an API contract. A box diagram with no schema under it collapses at the first follow-up question.

Focus areas

• Detect concurrent edits instead of losing writes
• Trace a symptom to a mechanism under load
• Bound every outbound call with a timeout

Overview

Your impact is felt directly on the factory floor and in the laboratory. By developing and maintaining systems that track chemical testing, chromatography data, and preventive maintenance, you ensure that IPCA Group maintains its high standards of operational excellence. This role is ideal for engineers who thrive at the intersection of software development and industrial manufacturing, where your work directly contributes to the precision and reliability of life-saving pharmaceutical products.

The IPCA Group interview process

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

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.

Professional Background Evaluation

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

Cultural Alignment Assessment

What this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.

What to demonstrate

• Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
• Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
• Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly

How to prepare

• Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
• Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
• Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live

Final HR Review

Nobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.

What to demonstrate

• Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
• Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
• Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
• Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience

How to prepare

• Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
• For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
• Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed

Where candidates lose points

Paginating with LIMIT/OFFSET over a set that changes while the client is reading it

OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.

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.

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.

Comparing floating-point values for equality, or holding money in them

Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.

Questions & practice

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

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

Worked solution (30 min)

1. Write the edge-loading query with the tenant predicate on both endpoints and state what it does with a cross-tenant edge.
2. Implement iterative Tarjan with an explicit stack and confirm on a three-node cycle that it emits one component of size three.
3. Build the transpose restricted to the visited set and mark every node with an in-edge from outside it as refused, carrying the referrer id.
4. Run Kahn over the condensation and verify the emitted order against the referrer-before-referenced rule.
5. Size the CSR arrays for 2,000,000 nodes and 8,000,000 edges and compare against a boxed adjacency map.

Expected result: An iterative O(V+E) traversal over a tenant-scoped CSR subgraph, SCC condensation so cycles archive atomically as one component, a transpose-based refusal list naming the external referrer for each blocked resource, and a Kahn topological order over the condensation, with recursion replaced by an explicit stack because of graph depth rather than style.

Verify it yourself

• A two-node cycle with no external referrers archives as one unit, and adding a single external referrer moves both nodes to the refusal list.
• No resource appears in the archive order before something that references it, except within its own component.
• A counter on edge visits matches the loaded edge count exactly, confirming each edge is traversed once.

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?

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?

Find overlapping job attempts and peak concurrency from lease records

Medium · sweep line · intervals · leases

A day of job_run history yields about 50,000,000 attempt records: (job_run_id, job_type, attempt, started_at, finished_at which is NULL when the worker died, lease_expires_at). Leases expire on a clock, so a job that outran its lease ran twice. Produce (a) every job_run_id whose attempts overlapped in wall-clock time and (b) the peak number of simultaneously running attempts per job_type with the minute it occurred. Target O(n log n). State how you treat a NULL finished_at and what clock skew does to your answer.

Approach

1. Define the interval before sorting anything: an attempt occupies [started_at, COALESCE(finished_at, lease_expires_at)). finished_at is observed and lease_expires_at is only a promise, so every attempt without a finish contributes an estimate and the whole result is a lower bound on overlap rather than an exact count.
2. For peak concurrency, sweep: emit 2n endpoints, sort by (timestamp, kind) with ends ordered before starts at equal timestamps, then walk the sequence maintaining a counter per job_type and record each type's maximum with its timestamp. O(n log n) dominated by the sort, O(n) space, or O(1) extra if the sort is external and the walk streams.
3. For overlap detection, do not compare attempts pairwise. A single global sort by (job_run_id, started_at) gives both the grouping and the order; within a group, keep the maximum end seen so far and report an overlap exactly when the next start is less than that running maximum, which is one linear pass after the sort.
4. Half-open intervals matter and are easy to get wrong: with closed intervals an attempt ending at the same millisecond another begins reads as concurrency two, and across 50,000,000 records that artefact swamps the real signal.
5. State the clock caveat: started_at and finished_at are written by different workers, so under skew of a few hundred milliseconds an apparent overlap shorter than that bound is not evidence. Filter reported overlaps by a minimum duration, or prefer timestamps written by whichever component heartbeats the lease.
6. Scale the sort rather than assuming it fits: the sweep emits two endpoints per attempt, so 50,000,000 records become 100,000,000 endpoints, and at roughly 24 bytes each, an 8-byte timestamp plus a 4-byte job_type plus a kind flag padded to alignment, that is about 2.4 GB of sort keys before any scratch space. Either push the ordering into the database behind an index on (job_type, started_at) or run an external merge sort in chunks; the overlap pass sorts n records rather than 2n, so it is the cheaper of the two.

Follow-up

• A handler is not idempotent and you have found 400 overlapping jobs. Which of them actually caused damage, and what would you query to find out?
• Peak concurrency for one job_type is 4 against a configured cap of 4. Is the cap working, or is the data hiding attempts that never started?
• How would you compute both answers incrementally as records arrive rather than in a daily batch?

Denormalise tenant onto revisions and backfill it live

Hard · denormalisation · online migration · backfill

resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.

Approach

1. Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
2. Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
3. Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
4. Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.
5. Step three, backfill: batch by primary key rather than by created_at so the cursor is dense and resumable — UPDATE resource_revision rr SET tenant_id = r.tenant_id FROM resource r WHERE r.resource_id = rr.resource_id AND rr.revision_id > $1 AND rr.revision_id <= $1 + 5000 AND rr.tenant_id IS NULL — committing per batch and persisting the cursor. Throttle on replica replay lag and on dead-tuple count, since each batch writes 5,000 new row versions. Run the backfill before the index exists so those updates can stay HOT.
6. Step four, index then enforce then contract: CREATE INDEX CONCURRENTLY (cannot run inside a transaction block, scans the table twice, waits on open transactions, and leaves an INVALID index to drop concurrently if it fails); ADD CONSTRAINT ... CHECK (tenant_id IS NOT NULL) NOT VALID, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE, after which SET NOT NULL uses the validated check instead of re-scanning on PostgreSQL 12 and later. Only then move the audit reads onto the column and, in a later deploy, delete the join path.

Follow-up

• The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
• How do you verify the backfill actually finished, given rows are still being inserted while it runs?
• A resource must now be movable between tenants. What does that do to the composite foreign key and to the revisions already written?

Find version gaps and relay lag with window functions

Medium · window functions · event tables · frame semantics

outbox_event holds event_id, aggregate_type, aggregate_id, aggregate_version, event_type, payload, status ('pending','published','dead'), attempts, created_at, published_at. A projection is missing rows and you must decide whether the relay skipped events or the consumer dropped them. Write three queries over the last seven days: one listing every aggregate_id whose published aggregate_version sequence has a hole, one giving per-day counts with a running total, and one returning the newest published event per aggregate. For each, say where the window function is evaluated relative to WHERE and LIMIT. PostgreSQL 16.

Approach

1. Gaps: compute lead(aggregate_version) OVER (PARTITION BY aggregate_id ORDER BY aggregate_version) in a subquery, then filter next_version <> aggregate_version + 1 in the outer query. Window functions are evaluated after WHERE, GROUP BY and HAVING and before the outer ORDER BY and LIMIT, so the predicate cannot sit in the same WHERE clause and PostgreSQL 16 has no QUALIFY.
2. Say what the seven-day filter does to the answer: it truncates every partition, so the first row per aggregate has no predecessor inside the window and a hole spanning the boundary is invisible. Widen the window, or join to resource.version as the authority for the true maximum.
3. Running total: SELECT date_trunc('day', created_at) AS d, count() AS n, sum(count()) OVER (ORDER BY date_trunc('day', created_at) ROWS UNBOUNDED PRECEDING). An aggregate inside a window call is legal because grouping runs before windowing. The grouping key is unique per row here so ROWS and RANGE agree, but write the frame anyway — over ungrouped rows with tied timestamps the default RANGE frame pulls in every peer row and the total jumps.
4. Newest per aggregate: DISTINCT ON (aggregate_id) ... ORDER BY aggregate_id, aggregate_version DESC is the cheap PostgreSQL-only form when an index matches that order; row_number() OVER (PARTITION BY aggregate_id ORDER BY aggregate_version DESC) = 1 is the portable form and needs a subquery for the same evaluation-order reason as the gap query.
5. Interpret rather than report: no gaps plus a normal p95 of published_at - created_at points at the consumer; gaps or a fat lag tail point at the relay; rows still 'pending' with attempts > 0 point at neither, because they never left the database.
6. Be explicit that the partial index on (created_at, event_id) WHERE status = 'pending' does not serve any of these — they read published rows. Name the index a recurring monitor would need, and say why a query run twice a year may not deserve one.

Worked solution (30 min)

1. Write the three queries against seven days of data and confirm each returns without error.
2. In a scratch copy, delete one middle event for a single aggregate and confirm the gap query names that aggregate and the versions either side.
3. Run a running total over ungrouped rows ordered by date_trunc('second', created_at), once with the default frame and once with ROWS, and record where the two series diverge.
4. Compare the DISTINCT ON and row_number() plans on the same data and record rows-read for each.

Expected result: The gap query returns one row per hole with the version on each side; the daily query returns seven rows whose running total ends at the seven-day count; the newest-per-aggregate query returns exactly one row per aggregate_id present in the window. The RANGE running total is higher than the ROWS one at every tied second.

Verify it yourself

• The per-day counts sum to COUNT(*) over the same predicate.
• Re-running the gap query after reinserting the deleted event returns zero rows.
• count(*) of the newest-per-aggregate result equals count(DISTINCT aggregate_id) over the same filter.

Follow-up

• Relay failover redelivers events. Does a duplicate break the gap query, and how would you detect one from this table alone?
• Turn the gap check into a continuous monitor rather than a query someone runs after an incident. What does it watch?
• The consumer claims it never received event 4,812,006. What do you look at, in what order?

What are the core features of ASP.NET?

Medium

Approach

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

Follow-up

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

Explain the advantages and disadvantages of using.NET.

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?

What alternative programming languages are you proficient in?

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. 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?

Rebuild the search projection while it serves nine thousand queries

Medium · projections · eventual consistency · rebuild · watermark

The read-model service answers about 9k queries/second at a 120 ms p99 from a projection built off the event log, normally under 2 seconds behind. A mapping change forces a full rebuild from resource_revision, during which apply lag rises to minutes. Writes continue at 1.2k/second and events at 4k/second. Design the rebuild: how the new index is populated and cut over, how position is tracked per log partition, what the API returns alongside results so a client can tell a stale answer from a current one, and the criterion for cutting over.

Approach

1. Build into a second index and swap an alias rather than mutating the live one. The rebuild is then reversible by pointing the alias back, so a bad mapping costs a wasted rebuild instead of an outage. The price is peak storage for two full copies and double apply load during catch-up, and both numbers should be stated up front rather than discovered when disk fills.
2. Track position per log partition, not globally. Order is guaranteed only within an aggregate's partition, so progress is a vector, and the only number safe to publish is taken from the least advanced partition - the slowest one is what bounds completeness. Publishing the most advanced partition's position declares the projection current while another partition sits twenty minutes behind.
3. Make apply idempotent so the backfill and the live tail can overlap without a freeze. Each document records the aggregate_version it reflects, and any event at or below that version is discarded. This is why the event carries the full fact and not a delta: a delta cannot be discarded safely, and a consumer that calls back to read current state applies a state newer than the event it is processing, which is how a projection ends up with changes applied out of order.
4. Turn lag into a contract instead of a surprise. Return the watermark with every result set, and return the version produced by a write so the client can compare the two. A client that wrote version 7 and receives results at a watermark older than its own commit can show that its change is still landing, rather than rendering the previous value as current. Blocking the read until the projection catches up would convert a staleness problem into an availability problem at 9k queries/second, and choosing not to do that is the trade.
5. Define the cutover numerically. Writes are untouched by the rebuild - they commit to the primary and land in the outbox - so the only coupling is apply throughput. If catch-up applies slower than the 4k events/second arriving, it never converges. The cutover criterion is that lag is measurably decreasing and below a stated threshold, not that the backfill loop reached the end of its range.

Worked solution (30 min)

1. Write the rebuild sequence: snapshot source, populate the second index, attach the live tail, verify, swap alias, retire the old index - naming what is reversible at each step.
2. Define the per-partition position record and the single watermark derived from it, and show what each would report when one partition stalls.
3. Write the idempotent apply rule using the document's recorded aggregate_version, and replay one event twice against it.
4. State the cutover threshold in lag terms and the measurement that shows convergence rather than completion.

Expected result: A parallel index populated from resource_revision snapshots plus the live tail, cut over by an alias swap that is reversible. Position is stored per log partition and the published watermark comes from the least advanced one. Apply is idempotent on the document's recorded aggregate_version, so backfill and tail can overlap without a freeze. Results carry the watermark and writes return their version so clients can detect staleness. Cutover happens when lag is decreasing and under the stated threshold.

Verify it yourself

• Applying the same event twice leaves the document unchanged.
• The published watermark never advances past the least advanced partition.
• Pointing the alias back restores previous behaviour with no reindex.
• Catch-up rate exceeds the 4k events/second arrival rate before cutover, shown by the lag trend rather than by the loop finishing.

Follow-up

• The rebuilt index disagrees with the primary tables for 300 documents. Which is authoritative, and how do you decide without freezing writes?
• The rebuild doubles load on the log and pushes the projection p99 from 120 ms to 400 ms. What do you throttle, and which signal sets how much?
• Clients start polling until the watermark passes their write. What does that do at 9k queries/second, and what do you offer instead?

One customer endpoint stalls deliveries to every other destination

Medium · bulkhead · timeouts · saturation · retry amplification

The egress service delivers about 1.5k webhooks/second across 40,000 destinations, with a per-destination concurrency cap of 4 and a 10-second connect-plus-read timeout. Throughput falls to 300/second, queue depth climbs, and p99 delivery latency for unaffected destinations goes from 200 ms to minutes, while the error rate barely moves. One tenant holds 900 destination rows whose URLs share a hostname that now answers in 9.5 seconds. Explain the mechanism with the arithmetic, then give the containment in the order you would apply it.

Approach

1. Look at saturation before errors. A flat error rate with collapsing throughput says nothing is failing, things are waiting, so the first signal to pull is in-flight request count or pool wait time rather than the error counter. This is the distinction that decides the whole investigation.
2. Group in-flight work by resolved host, not by destination id. The cap is keyed per destination row, so 900 rows sharing one hostname buy 3,600 concurrent slots against a single host, each held for 9.5 seconds. The bulkhead was never a bulkhead for that host, and grouping by the wrong dimension is why the dashboard looked healthy.
3. Do the arithmetic in both directions. Required concurrency is arrival rate times latency, so 1.5k/second at 200 ms needs about 300 in flight, which is entirely consumed by 3,600 slow slots; conversely whatever concurrency is left sustains rate equals concurrency divided by 9.5 seconds, which is the 300/second you are seeing. Matching both numbers is what promotes this from a plausible story to the mechanism.
4. Explain why the circuit breaker never helped. It opens on consecutive failures, and a 9.5-second response inside a 10-second timeout is a success. Slow is not failing, so an error-rate breaker cannot see this; you need a slow-call ratio, a deadline propagated from the caller's remaining budget, or a concurrency limiter.
5. Contain in order: park the offending host so the shared pool drains, add a per-resolved-host concurrency cap alongside the per-destination one, give slow hosts their own queue so they cannot occupy the general pool, derive the timeout from the delivery deadline rather than a round number, and check the retry policy is not tripling load on a host that is already slow. Use backoff with full jitter so retries do not resynchronise on recovery.
6. State the invariant you are restoring: one tenant's endpoints degrade only that tenant's deliveries. That is a property to load-test for, not to assume from a config value.

Follow-up

• The host recovers to 80 ms. How long does the queue take to drain, and what does the drain do to the recovered host?
• Where should the 10-second timeout number actually come from?
• If that tenant had one destination row instead of 900, would the cap of 4 have saved you? What would you measure to be sure?

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.

The bug you decided not to fix

Engineers over-index on what they repaired. A stronger answer covers something you knowingly left broken: the alert you tuned down, the data inconsistency you documented instead of chasing, the cleanup you deferred past two quarters. Give the reasoning and the condition that would have reopened it, so it reads as a decision and not as neglect.

Tell me about yourself and your professional background.

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. 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?

Estimate work you have never done and defend the range

Hard · estimation · backfills · expand-contract

You are asked to estimate a change you have never attempted: add a column to a 100-million-row table, populate it, move reads across, and drop the old shape. Give a range with the assumptions that generate it, including batch size, the signal your backfill throttles on, and wall-clock hours, and name the three unknowns that would move the number most. Then describe a real estimate you gave under comparable ignorance: how you expressed its uncertainty, what you committed to, and how wrong you turned out to be.

Approach

1. Decompose into independently deployable steps before estimating anything: add the column nullable, write both shapes, backfill in batches, verify, move reads, stop writing the old shape, drop it. That is four deploys spread over days, and the calendar estimate is dominated by them rather than by the loop's runtime.
2. Do the arithmetic aloud for the part that has arithmetic in it: batch size times number of batches times per-batch duration, at a write rate the primary can absorb alongside roughly 1.2k writes per second of production traffic. The loop is throttled by replication lag and lock waits, not by how fast it can issue statements.
3. Price the schema step by its lock rather than its statement duration. In PostgreSQL an ALTER TABLE taking ACCESS EXCLUSIVE waits for every open transaction on that table while later queries queue behind it, so a millisecond change issued during a thirty-second analytics query stalls that table for thirty seconds. Adding a nullable column with a non-volatile default avoids a rewrite from version 11; a new index wants CREATE INDEX CONCURRENTLY, which cannot run inside a transaction block and leaves an invalid index behind if it fails.
4. Express the answer as a range whose endpoints each trace to a stated assumption, then name the cheapest experiment that collapses it, which is almost always running one real batch against the real table and multiplying.
5. Commit to a checkpoint rather than a completion date: the day you report a measured number from that first batch. That is a promise you can keep under uncertainty, and it is what the asker actually needs in order to plan.

Follow-up

• How do you verify the backfill genuinely finished, given rows written by production traffic while it ran?
• Where does the backfill resume from after a worker is killed mid-batch, and what makes that resume point trustworthy?
• Your first batch comes back ten times slower than assumed. What do you tell the person waiting on the estimate, and when?

Unblock an engineer without taking the keyboard

Easy · mentoring · leases · at-least-once

A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.

Approach

1. Ask before diagnosing, and ask for things answerable from data they already have: the attempt count on the job rows that produced duplicates, the handler's observed duration against its lease expiry, and whether the duplicate rows share a natural key that a unique constraint could have caught.
2. Teach the shape rather than the answer. A lease cannot distinguish a dead worker from a slow one, so a handler that outruns its lease is running twice by design, and deploys deliver the other half by killing handlers mid-run on every rollout. Both of their candidate theories produce identical duplicate rows, which is why the evidence has to come from timings rather than from argument.
3. Hand over a checklist they execute: a natural key on every write the handler performs so the second copy collides rather than appends, the record of intent written before any external effect, a lease heartbeat while running, and the metric that shows it working.
4. Keep ownership with them deliberately. Pair on the first write, then step back; if you finish it yourself you have closed one ticket and left the same person stuck on the next redelivery.
5. Close on the systemic gap that let two days pass, which is usually a missing dashboard for attempt counts or an undocumented at-least-once contract, and fix that rather than only the bug.

Follow-up

• How would you distinguish a genuine double-delivery from a lease expiry using only the data already stored?
• Their handler calls an external endpoint before recording that it did. What do you tell them to change first?
• What do you do the third time the same person brings you the same class of bug?