Arkham Technology Software Engineer Interview Questions & Guide 2026

The 60-second brief

As a Software Engineer at Arkham Technology, you are a critical contributor to the company’s technical vision and product reliability. This role is not merely about writing code; it is about solving complex, domain-specific problems that directly impact the efficiency and success of the organization. You will work closely with cross-functional teams to translate business requirements into robust, scalable software solutions.

Getting the code to run is the floor. What usually separates answers is the case checked without prompting: empty input, a single element, duplicate keys, or a value that overflows the integer type you chose.

Focus areas

• Make every write idempotent under client retries
• Build at-least-once pipelines with explicit deduplication horizons
• Evolve APIs without breaking pinned SDK clients

Overview

This position offers a high level of strategic influence. Because Arkham Technology often operates in niche technical spaces, the work you do directly shapes the tools and infrastructure the company relies on daily. You will be expected to balance immediate project needs with long-term architectural health, making this an ideal environment for engineers who enjoy both deep technical engagement and collaborative problem-solving.

The Arkham Technology interview process

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

Phone Screen

The title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.

What to demonstrate

• Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
• Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
• Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year

How to prepare

• Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
• Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
• Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter

Stakeholder Interviews

Because the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.

What to demonstrate

• Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
• Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
• Whether you can be redirected in the middle of an answer without restarting it from the beginning
• Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it

How to prepare

• Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
• Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
• Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing

Where candidates lose points

Serialising a tenant's writes through select ... for update on a single counter row

It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.

One shared connection pool for every tenant and every query class

A single tenant with a large table and a missing index can occupy every connection with slow queries, and every other tenant then waits in connection acquisition -- a queue invisible in database metrics, because the database itself looks healthy while the application starves. Containment is bulkheads: separate pools or per-tenant concurrency caps for interactive requests, background jobs and exports, a statement timeout low enough that a pathological query dies before it accumulates, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks indefinitely. One caveat worth knowing in advance: if a transaction-pooling proxy sits in front of the database, session-scoped behaviour changes, so session-level advisory locks and settings applied outside a transaction do not survive the way they do on a direct connection.

Sharing mutable state with no stated owner

Say which thread, request or task owns each mutable structure, and what protects it when the answer is more than one: a lock, a queue that hands ownership across, or an immutable copy per reader. A structure documented as safe for concurrent reads is usually not safe for a concurrent write alongside those reads.

Reading the constraints as preamble rather than as part of the problem

The bounds are usually there to eliminate the obvious approach: n up to 10^5 makes an O(n^2) scan roughly 10^10 operations, far outside any per-test time budget, and an input larger than memory rules out loading it at all. When a bound is not given, ask for it, then say out loud which approach it kills.

Questions & practice

Order a job dependency graph and find its critical path

Medium · topological-sort · dag-longest-path · cycle-detection · critical-path

A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.

Approach

1. Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
2. Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
3. Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order: earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E).
4. Second pass in reverse topological order for latest_finish, then slack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain. slack[v] = 0 is exactly the statement that some longest path runs through v; equivalently, the longest path through v has length T - slack[v].
5. The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by min(d, T - L_avoid(v)), where L_avoid(v) is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, while min(10, 15 - 14) predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan is max(T - d, L_avoid(v)).
6. Compute L_avoid(v) the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, since L_avoid(v) only ever matters through that max: set duration[v] := 0, recompute the makespan as T0(v) = max(T - duration[v], L_avoid(v)), and the gain is min(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.

Follow-up

• Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
• Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
• Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?

Schedule ordered webhook retries with a heap of subscription queues

Medium · heap · scheduling · backoff · head-of-line-blocking

Design the in-memory scheduler for webhook delivery. Up to 20 million rows sit in status pending or failed_retryable across 200,000 subscriptions, each row carrying next_attempt_at and attempt_count, and each endpoint having a circuit breaker. Deliveries for one subscription must be attempted in order, so at most one attempt per subscription may be in flight. Support due(now), complete(delivery, outcome) and insert(delivery) in O(log S), where S is the subscription count rather than the delivery count. Give the backoff formula you schedule retries with.

Approach

1. Key the global heap by subscription, not by delivery. Each subscription owns a FIFO of its due deliveries in event order; the heap holds one entry per eligible subscription, keyed by its head's next_attempt_at. That is 200,000 heap entries instead of 20 million, and it makes the one-in-flight rule structural rather than a check somebody can forget.
2. due(now): peek the minimum. If its key is in the future, sleep until then instead of spinning. Otherwise pop it, move the subscription into an in-flight set, and do not re-push it. A subscription absent from the heap cannot be dispatched twice, which is precisely how ordering is preserved.
3. complete: on success, drop the head and re-push the subscription keyed by its new head, or leave it out when the queue empties. On a retryable failure, increment attempt_count and set next_attempt_at = now + uniform(0, min(cap, base * 2^attempt)), sampled uniformly across the whole interval. That is full jitter; deterministic backoff re-synchronises the herd you just created.
4. Circuit breaker: park the subscription in a second heap keyed by its half-open time, so an endpoint dead for six hours costs one heap entry and zero attempts rather than consuming worker slots. Admit exactly one probe at half-open and close the breaker only on its success.
5. Say the price of the ordering guarantee out loud. One in-flight attempt per subscription means an endpoint answering in 10 seconds drains at 0.1 deliveries/second however many workers you run, and its backlog grows until it recovers. If the customer does not need order, allow k in flight and document delivery as unordered; that is the trade, and it is a product decision.
6. All three operations are O(log S) with O(S) resident heap memory and the queues themselves backed by the store. The database-backed equivalent is a partial index on (subscription_id, next_attempt_at) where status in ('pending','failed_retryable') claimed with FOR UPDATE SKIP LOCKED, and the write-back must be fenced on lease_token so a worker that stalled and resumed cannot overwrite a newer attempt.

Follow-up

• One subscription has 4 million queued deliveries. What stops it from starving the other 199,999, and what does your heap look like under that load?
• A customer requests redelivery of last Tuesday's events. Where do those rows enter your structure, and what keeps them from reordering live traffic?
• The process restarts. How much state do you rebuild, and what stops every subscription from being attempted in the same second?

Fold a deduplicated usage stream into hourly rollups

Easy · aggregation · deduplication · watermarks · exact-arithmetic

You are given one day of usage_event rows, up to 250 million, each carrying event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at and ingested_at. Produce usage_rollup_hourly cells keyed (tenant_id, workspace_id, sku, hour_start) with quantity_sum, event_count and source_max_ingested_at. An event counts once per (tenant_id, idempotency_key). The rollup grain has no environment column, so state your filter. One pass. Give your time and space bounds, and say what the deduplication actually costs in memory.

Approach

1. Bucket on occurred_at, never ingested_at: hour_start = date_trunc('hour', occurred_at at time zone 'UTC'). The two columns answer different questions. occurred_at says which hour the customer is billed for; ingested_at says how current the fold is. Using the second for the first makes late data invisible instead of correctable.
2. The fold is trivial and the deduplication is the entire cost, so price it before designing anything clever. An exact set over (tenant_id, idempotency_key) at 250M entries, stored as a 16-byte 128-bit hash in an open-addressed table at 0.7 load factor, needs about 357M slots at 16 bytes each, roughly 5.7 GB. The fix is partitioning by hash(tenant_id) % P so each shard holds 1/P of the set and no tenant's keys straddle shards.
3. Rule out a Bloom filter as a replacement, in the right direction: a false positive reports 'already seen' for an event never seen, so you drop a real event and lose revenue with no error raised. It is usable only as a negative pre-filter in front of the exact set, where a miss is conclusive and a hit must fall through to the real lookup.
4. Accumulate in scaled integers, not binary floating point. numeric(20,6) admits values below 10^14, so one event scaled to micro-units can reach 10^20, past int64's 9.22 x 10^18; use a 128-bit or arbitrary-precision accumulator unless you first bound the per-event maximum. binary64 represents integers exactly only to 2^53, about 9.01 x 10^15, and cannot represent 0.1 at all, so two runs that sum in different orders disagree.
5. Carry source_max_ingested_at = max(ingested_at) over the events folded into each cell, and count event_count over accepted, post-dedup events. Without that watermark there is no way to prove later what a number did and did not include, which is the first question any reconciliation asks.
6. State the environment filter explicitly, because the rollup grain cannot record it. A fold that quietly includes staging bills non-production traffic; one that quietly excludes it loses a cost signal. Production-only is the billing answer, and either way it belongs in the job name and the output metadata. Complexity: O(n) time, O(distinct dedup keys) space, dominated by the dedup set rather than by the cells.

Worked solution (25 min)

1. Write both key tuples down before any code: dedup key (tenant_id, idempotency_key), cell key (tenant_id, workspace_id, sku, hour_start), with hour_start derived from occurred_at in UTC.
2. Build a 10,000-row fixture containing one event duplicated three times under the same idempotency_key, two events sharing an idempotency_key across different tenant_id values, one event whose occurred_at is two hours before its ingested_at, and one staging event inside an otherwise production cell.
3. Fold it and assert each of those four expectations separately rather than eyeballing a grand total.
4. Re-run with the input shuffled and diff the output files.
5. Size the dedup set for 250M keys using the load-factor arithmetic and write the number down next to the fixture.

Expected result: The triplicate contributes one event and its quantity once. The two same-key, different-tenant events both count, because the dedup key is the pair. The late event lands in the hour of its occurred_at while that cell's source_max_ingested_at advances to the later timestamp. The staging event is included or excluded per the stated filter and never silently.

Verify it yourself

• Shuffled input produces byte-identical cells, including event_count and source_max_ingested_at.
• The sum of quantity_sum across all cells equals the sum of quantity over the distinct dedup keys, computed independently.
• Swapping the accumulator to binary64 makes a fixture of 0.1-valued quantities disagree with the exact sum. That is a failing test, not a rounding note.

Follow-up

• A producer retries at 23:59:59 and the retry lands at 00:00:01. The unique index on the daily-partitioned table must include the partition key. What gets double-counted, and what is the smallest change that fixes it?
• The consumer acknowledges its batch before committing the fold. Which failure loses revenue now, and which arrangement duplicates instead?
• What makes a re-run over the same day produce byte-identical rollups?

Rebuild an hourly rollup with deduplication and late-arrival accounting

Hard · window functions · deduplication · event time vs ingest time · watermarks

From usage_event (event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at), produce the values usage_rollup_hourly should hold for one tenant over one day: per (workspace_id, sku, hour_start) the deduplicated quantity_sum, event_count and source_max_ingested_at, bucketed by occurred_at. Duplicates share (tenant_id, idempotency_key). Also report, per hour, the running total across the day and the share of quantity that arrived more than two hours after the hour began. Write the query, and state which duplicates a daily unique index cannot catch.

Approach

1. Deduplicate in its own CTE before any aggregation, because a SUM cannot be un-summed: row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id) = 1. Include the tiebreaker. Without it the surviving row is non-deterministic when two duplicates share an ingested_at, and a rollup described as deterministically recomputable then disagrees with itself between runs.
2. Bucket on occurred_at and nothing else, and pin the timezone explicitly. date_trunc('hour', timestamptz) truncates in the session's TimeZone setting, so the same query run by a session set to a non-UTC zone buckets differently; use the three-argument date_trunc('hour', occurred_at, 'UTC') on PostgreSQL 16 or later, or date_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC' before that. Filter environment = 'production' explicitly, since metering covers three environments and billing covers one.
3. Aggregate to the grain with sum(quantity), count(*) and max(ingested_at). The last is not decoration: it is the watermark the row consumed up to, and without it there is no way to prove afterwards what a number did and did not include.
4. Compute the late share inside the dedup-and-aggregate step as a conditional aggregate, sum(quantity) filter (where ingested_at > hour_start + interval '2 hours'), then divide by the hour's total. Compute the running total as a window over the already aggregated rows: sum(quantity_sum) over (partition by workspace_id, sku order by hour_start rows between unbounded preceding and current row). Running either over raw rows puts the duplicates back.
5. Answer the index question exactly. The unique constraint is on (ingested_day, tenant_id, idempotency_key), because a unique index on a partitioned table must contain the partition key. It therefore deduplicates only within one ingest day and admits a duplicate whose retry crosses midnight or whose replay runs a week later. That is why this CTE dedups across the whole window being recomputed, and why the dedup horizon is a correctness parameter rather than a retention cost.
6. Keep the numeric type all the way through. quantity is numeric so the sums are exact; a cast to double precision anywhere in this pipeline reintroduces drift that surfaces only as a few unreconcilable cents per tenant per month, long after the query is out of anyone's mind.

Follow-up

• A dispute forces the same recompute over 40 days for one tenant. What changes about the dedup CTE's memory use and the chosen plan, and what would you do about it?
• Two runs a minute apart return different quantity_sum values for an hour that is already closed. Give two mechanisms that produce that, and the single query that distinguishes them.
• Express the same rollup incrementally so it does not re-scan the day each time the watermark advances. What does the incremental version stop being able to answer?

Migrate a live partitioned event table without blocking ingest

Hard · online migration · ddl locks · expand and contract · partitioning

usage_event is range-partitioned daily on ingested_at, holds roughly 250M rows per day across 400 live partitions, and is written at 10-40k rows/second. Two changes are required: quantity must move from double precision to numeric(20,6), and a new environment column must become NOT NULL with a default of 'production'. Ingest cannot stop. Give the ordered plan, naming for each step the lock it takes, what that lock blocks, and roughly how long it is held. Identify the one step that cannot be rolled back cleanly once traffic depends on it.

Approach

1. Classify the two changes before planning anything. Adding a column with a non-volatile default has been metadata-only since PostgreSQL 11, so it is cheap. Changing double precision to numeric is not binary-coercible, so alter column ... type rewrites every partition under ACCESS EXCLUSIVE and rebuilds its indexes; on this volume that is hours of blocked ingest and is simply not an option, which is why the plan is expand-and-contract rather than one statement.
2. Expand: add quantity_numeric numeric(20,6) and environment with its default on the parent. Both are catalogue-only but both take a brief ACCESS EXCLUSIVE that cascades to partitions, so run each with lock_timeout set to a second or two and retry on failure. A queued ACCESS EXCLUSIVE request blocks every reader behind it, which is how a metadata-only change turns into an outage.
3. Dual-write: deploy producer code that populates both columns on every insert, and leave it running before anything reads the new column. This is the step that cannot be reverted cleanly. Once readers depend on quantity_numeric, reverting the writer leaves rows with a null there, and the gap is only discoverable by re-reading the old column, which the readers have stopped doing.
4. Backfill older partitions in batches keyed on the primary key, oldest first, committing every few thousand rows with a pause between batches, and skipping the partition still receiving writes until it rotates. Each batch is an ordinary UPDATE taking row locks only. The cost is bloat and WAL rather than blocking, so watch dead tuples and let autovacuum keep pace instead of wrapping 400 partitions in one transaction.
5. Make NOT NULL cheap with the three-step form: add constraint ... check (environment is not null) not valid (brief ACCESS EXCLUSIVE, no scan), then validate constraint (SHARE UPDATE EXCLUSIVE, scans while reads and writes continue), then set not null, which from PostgreSQL 12 uses the validated check and skips its own full scan. Do this per partition, then on the parent.
6. Switch and contract: move reads to the new column behind a flag, verify over a full period that both columns agree on freshly written rows, drop the old column (metadata-only), and only then remove the dual-write. Any index on the new column goes on with CREATE INDEX CONCURRENTLY per partition, since CIC is not supported on a partitioned parent: create the parent index with ONLY, build each child concurrently, then ALTER INDEX ... ATTACH PARTITION until the parent index becomes valid.

Worked solution (45 min)

1. On a scratch cluster, build 10 partitions of 2M rows each and run a writer at a few thousand inserts/second.
2. Run the naive type change and measure how long writes stall and how far ingest lag grows before killing it.
3. Run the expand step with lock_timeout = '2s' while the writer runs, and observe a clean lock timeout and retry instead of a pile-up of blocked readers.
4. Backfill in 5k-row batches and chart dead tuples and WAL generated per batch.
5. Run the not-valid, validate, set-not-null sequence and confirm from pg_stat_activity and timings that nothing held an exclusive lock through a full scan.
6. Add an index with CIC per partition plus ATTACH PARTITION and confirm the parent index reports valid only after the last attach.

Expected result: An ordered expand, dual-write, backfill, switch, contract plan with the lock named per step: metadata-only ADD COLUMN under a brief ACCESS EXCLUSIVE taken with lock_timeout; batched UPDATE backfill under row locks; CHECK NOT VALID, then VALIDATE under SHARE UPDATE EXCLUSIVE, then SET NOT NULL; CREATE INDEX CONCURRENTLY per partition with ATTACH PARTITION; and the dual-write-to-read switch identified as the step that cannot be reverted cleanly.

Verify it yourself

• No step in the plan holds ACCESS EXCLUSIVE for longer than the configured lock_timeout.
• The writer's error rate stays at zero through the rehearsal apart from the deliberate lock timeout.
• After VALIDATE, SET NOT NULL completes in milliseconds on a table where the naive form scanned every row.
• Freshly written rows agree on both columns for a full verification period before the old column is dropped.

Follow-up

• A CREATE INDEX CONCURRENTLY fails halfway through the partition list. What state is the table in, how do you detect it, and what do you run?
• The producer computes quantity itself. What happens to a request already in flight when the dual-write deploy lands, and does it matter?
• Give two queries that prove the backfill is complete: one cheap enough to run every minute, one authoritative.

What motivates you to stay engaged when faced with complex, long-term projects?

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

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

How do you approach learning new, niche technical domains on the job?

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. State your assumptions explicitly before working the problem.

Follow-up

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

Ship three breaking-looking changes without breaking pinned SDKs

Hard · versioning · backward compatibility · deprecation

GET /v1/runs/{id} returns status from the set queued, leased, running, succeeded, failed, timed_out, cancelled. You must add the terminal state lost, add an optional billable_seconds field, and start rejecting name values over 64 characters that were previously accepted. Clients are generated SDKs pinned inside customer build pipelines you cannot redeploy or reliably contact, some with strict deserialisers. Classify each change as compatible or breaking and justify it mechanically, then give the rollout: the versioning scheme, the transform old clients receive, the telemetry, and the sunset process.

Approach

1. Classify against a deserialiser rather than against intuition. A new enum value breaks a generated client that validates the enum or switches exhaustively; a new optional field breaks one generated with additionalProperties false; tightening validation breaks every caller already sending an 80-character name even though no field moved. All three are breaking for some real client, and that is the answer, not a technicality.
2. Choose the versioning axis on maintenance cost. A path version forks the handler, and each subsequent change multiplies the fork. A date-pinned version header, defaulted per credential at first use and pinned thereafter, keeps one internal model and a chain of ordered response transforms, so the handler stays single and the transforms compose.
3. Write the downgrade honestly. Older versions have no lost, so the transform maps it to the least-wrong existing value and the changelog says which; mapping it to failed asserts an outcome nobody observed, so pair it with a reason field the old version already exposes rather than inventing a field, and keep the mapping in one documented table instead of a conditional inside the serialiser.
4. Sequence the validation tightening separately from the payload changes: first measure how many tenants send names over 64 characters, then warn for a stated window with a Deprecation header and a non-fatal warning code, then reject only for versions dated after the change, never retroactively for a pinned client.
5. Deprecate from evidence, not from announcement: per-version, per-tenant, per-credential request counts from the gateway, a Sunset header carrying the date, direct contact for the handful of accounts that make up most of the tail, and short scheduled brownouts before the cutoff so the clients nobody can reach fail on your calendar rather than during their release.

Worked solution (40 min)

1. Write the three-row classification table with the client-side mechanism that breaks in each case, not just a verdict.
2. Pick the versioning scheme and write the header, the default-and-pin rule for a new credential, and where the transform chain sits relative to the handler.
3. Write the lost mapping for pre-change versions as an explicit table entry plus the changelog sentence a customer reads.
4. Write the validation timeline with four dated steps, and mark which one is visible to clients that never change their version.
5. Write the deprecation plan: the telemetry query by version and tenant, the headers, the brownout schedule, and the fallback for the largest holdout.

Expected result: All three changes classified breaking with the mechanism named, a date-pinned version header with one internal model and ordered downgrade transforms, a documented lost mapping, validation tightened only for versions after the change, and a sunset driven by per-version traffic with brownouts.

Verify it yourself

• No change is called additive without naming the client behaviour that makes it safe.
• A client that never changes its version header sees no behaviour change from any of the three.
• The plan names a measurement that tells you when sunset is safe.
• The lost mapping is stated once, in a table, not implied by serialiser branches.

Follow-up

• Which step of your rollout cannot be rolled back cleanly once a client has depended on it?
• A tenant pinned to a sunset version breaks their build the morning after the cutoff. What in your plan already lets you resolve it within an hour?
• Your transform chain is now six versions deep. What stops it becoming the fork you were avoiding?

Hourly rollups merge one hour and lose another

Easy · timezone · rollups · reconciliation

Reconciliation flags one tenant on one day. Summing usage_event.quantity by hour of occurred_at gives 24 non-empty hours, but usage_rollup_hourly holds 23 rows for that tenant, workspace and SKU, one of which carries roughly the sum of two adjacent hours. Other days reconcile exactly, and the affected date matches a civil-time transition. hour_start is documented as truncated to the hour in UTC. You have both tables, the rollup job source, and its runtime environment. Give an ordered checklist, the mechanism, and the correction path for a day that may already be sealed.

Approach

1. Bisect by dimension until one cell explains the whole difference: tenant, then day, then SKU, then hour. A defect confined to a single transition date already rules out deduplication and late arrival, both of which are indifferent to which hour an event lands in.
2. Read the truncation with its precondition stated: date_trunc on a timestamptz value is evaluated in the session TimeZone, not in UTC. If the job connects without pinning that setting, it inherits the server or container default.
3. Follow that to the collision: in a zone that observes daylight saving, two distinct UTC hours map to the same local wall-clock label at the autumn transition, so both fold into one key under the unique constraint on (tenant_id, workspace_id, sku, hour_start) and their quantities sum into one row. At the spring transition a label never occurs and the row is simply absent.
4. Confirm from data rather than from reading code: run the same aggregate twice, once with the session pinned to UTC and once with the job host zone, and check that the second reproduces the stored rollup exactly.
5. Fix at the source by pinning the connection to UTC explicitly, or by truncating on occurred_at AT TIME ZONE 'UTC', rather than relying on a default that differs between a developer machine, CI and production.
6. Correct according to status, not convenience: an open hour is recomputed with revision incremented, a sealed hour is frozen and the difference becomes an adjustment line on the next invoice with voided_by_line_id pointing at the line it reverses.

Follow-up

• The same job also emits a daily figure for a dashboard. Why can a correct hourly rollup still produce a wrong day, and what does the tenant's billing timezone have to do with it?
• How would you detect this class automatically rather than waiting for reconciliation, given that it only manifests twice a year per zone?

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.

What the behavioural round is actually measuring

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

Tell me about the last project you were a part of and your contributions to the team.

Medium · behavioural and engineering judgement

Approach

1. Close with what you would do differently, concretely.
2. Name the disagreement and how you resolved it with evidence.
3. Give the blast radius: what could have broken, and what you measured.

Follow-up

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

Describe your past experience as a team lead, specifically regarding challenges you faced.

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

When have you been in a difficult situation and how did you come to a resolution?

Medium · behavioural and engineering judgement

Approach

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