Interview guides · Beast Code

Beast Code Software Engineer Interview Questions & Guide 2026

Code reading practice for Beast Code software engineer prep: work out what unfamiliar code does before judging it.

The 60-second brief

A Software Engineer at Beast Code is a critical contributor to a mission-driven team dedicated to modernizing defense and commercial technology. You are not just writing code; you are building intuitive, high-impact solutions that directly enhance the capabilities of the warfighter and solve complex, real-world problems. Whether you are working on game-based learning programs or advanced 3D modeling solutions, your work sits at the intersection of cutting-edge innovation and practical, mission-critical application.

Find out what you will be typing into. A shared plain-text editor with no autocomplete, compiler or test runner changes what you have to hold in your head, and practising inside your own configured environment hides exactly that gap.

Focus areas

Build at-least-once pipelines with explicit deduplication horizons
Bound blast radius with per-tenant concurrency limits
Keep money in integer minor units

Overview

Working at Beast Code requires a unique blend of technical autonomy and deep collaboration. You will operate within a matrix organization, partnering closely with designers, product owners, and fellow developers to shepherd projects through the entire software development lifecycle. The environment is fast-paced and encourages "outside the box" thinking, making it an ideal space for engineers who are passionate about architecture, scalability, and adopting new technologies to maintain a technological edge.

The Beast Code interview process

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

Initial Screening

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

What to demonstrate

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

How to prepare

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

Technical 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.

Cultural Fit 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

Multiple Team Meetings

When a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.

What to demonstrate

Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened

How to prepare

Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure

Where candidates lose points

Holding money in a floating-point type, or rounding it more than once

Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.

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.

Trusting input because it came from your own front end

Anything crossing a trust boundary is hostile: parameterise queries instead of building SQL by concatenation, validate against an allow-list rather than a deny-list, and bound the size of anything you allocate from a request. Raising this unprompted in an API or design question is a cheap and unusually strong signal.

Hardcoding to the sample inputs

Solve the stated problem rather than the two examples; special-casing a literal to make a sample pass is obvious immediately and reads as either a misunderstanding or an attempt to fake progress. If you genuinely cannot generalise yet, say which part is a stub and what would replace it.

Questions & practice

Solve a logic-based brain teaser.

Medium · data structures and algorithms

Approach

Restate the input: its shape, its size, and what is guaranteed about it.
Walk one small example through your approach before writing the whole thing.
Choose the data structure from the access pattern, not from familiarity.

Follow-up

How does this change if the input no longer fits in memory?
Which test case would catch an off-by-one here?

Implement a Fibonacci sequence.

Medium · data structures and algorithms

Approach

Restate the input: its shape, its size, and what is guaranteed about it.
Choose the data structure from the access pattern, not from familiarity.
Name the brute-force solution and its complexity before improving on it.

Follow-up

What is the worst case, and how likely is it on real data?
How does this change if the input no longer fits in memory?

Write a function to solve factorials.

Medium · data structures and algorithms

Approach

Choose the data structure from the access pattern, not from familiarity.
State the target complexity and say which constraint rules the naive version out.
Walk one small example through your approach before writing the whole thing.

Follow-up

How does this change if the input no longer fits in memory?
What is the worst case, and how likely is it on real data?

Solve a classic coding challenge, such as FizzBuzz, with follow-up constraints.

Medium · data structures and algorithms

Approach

Name the brute-force solution and its complexity before improving on it.
State the target complexity and say which constraint rules the naive version out.
Restate the input: its shape, its size, and what is guaranteed about it.

Follow-up

How does this change if the input no longer fits in memory?
Which test case would catch an off-by-one here?

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

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.
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'.
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).
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].
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)).
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.

Worked solution (30 min)

Build four fixtures. A: 12 jobs, two branches of 100 s and 95 s that share no job. B: fixture A plus one back edge. C: two disjoint paths tied at 100 s. D: the shared-prefix case, one job of 10 s feeding a 5 s job and a 4 s job, so the longest path is 15 s and the runner-up is 14 s.
Run Kahn; on fixture B confirm it emits fewer than V nodes, then run the residual-subgraph DFS and print the actual cycle.
Compute earliest_finish forward and latest_finish backward, and list the zero-slack set for each fixture.
For each zero-slack job v, recompute the makespan with duration[v] := 0 to get T0(v), and record both the correct bound T - T0(v) and the wrong one, T - second_longest_path, side by side.
Apply the shortening for real (20 s off the critical branch of A, 10 s off the shared prefix of D) and diff the recomputed makespan against each prediction.

Expected result: Fixture A: makespan 100 s, and shortening by 20 s leaves 95 s, a gain of 5 s. Both formulas agree here, because the 95 s branch avoids the shortened job. Fixture D: makespan 15 s, and shortening by 10 s leaves 5 s, a gain of the full 10 s, which T - T0(v) = 15 - 5 = 10 predicts and T - second_longest = 1 does not. Fixture C: the zero-slack set covers both tied paths, and shortening a job on one of them alone gains nothing, since the other path still runs 100 s.

Verify it yourself

The cycle printed by the DFS is a real cycle: every consecutive pair is an edge present in the input.
Summing durations along any zero-slack path equals the makespan exactly.
For every candidate, the predicted gain equals the makespan recomputed from scratch with the shortened duration. Assert the equality; do not eyeball it.
Fixture D is the regression test for the bound, and a fixture whose branches are disjoint is not: it passes under either formula and proves nothing.

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?

Explain why the metering dashboard scans every daily partition

Medium · indexing · partition pruning · sargability · tenant skew

usage_event is range-partitioned daily on ingested_at and holds tenant_id, workspace_id, environment, sku, quantity numeric(20,6), occurred_at and ingested_at. The only relevant index is on (occurred_at). A dashboard runs select sku, sum(quantity) from usage_event where tenant_id = $1 and date_trunc('hour', occurred_at) >= $2 and environment = 'production' group by sku, and EXPLAIN shows a sequential scan of every partition. Give each distinct reason, rewrite the predicate so an index can serve it, propose the index, and state the write cost its column order adds.

Approach

Separate the three causes rather than blaming one. First, date_trunc('hour', occurred_at) wraps the column, so the predicate is not sargable against a btree on the bare column. Second, pruning keys off ingested_at while the query constrains occurred_at, so no partition can be excluded. Third, even made sargable, (occurred_at) is not tenant-leading, so for one tenant among thousands the scan reads the whole time range and discards nearly all of it.
Rewrite the bound carefully, because the obvious rewrite is only conditionally equivalent. date_trunc('hour', x) >= $2 equals x >= $2 only when $2 is already hour-aligned; for an arbitrary $2 it means x >= date_trunc('hour', $2) + interval '1 hour'. Normalise the parameter in the caller and leave the column bare.
Restore pruning with a second, redundant predicate on the partition key: ingested_at >= $2 - interval '<late-data horizon>'. State both sides of it. It prunes to a handful of partitions, and it silently omits any event whose ingest lagged past that horizon, which is precisely what a producer replay produces. Either document the horizon as a stated bound, or partition on occurred_at and move the problem into the dedup window instead.
Propose (tenant_id, occurred_at) include (sku, quantity) per partition. A partial index where environment = 'production' mostly saves size rather than selectivity, since production dominates the three environments; take it if non-production is a meaningful share and skip it otherwise.
Price the write path honestly. At roughly 250M rows/day each extra index is another insert plus WAL per row, and a tenant-leading key scatters inserts across one hot leaf per active tenant instead of appending to a single rightmost leaf, so page dirtying and random I/O both rise. An INCLUDE payload widens every leaf entry and enlarges the index accordingly.
Add the index-only-scan caveat before someone reports it as a regression: on a freshly appended table the visibility map is not yet set for recent pages, so the INCLUDE columns still cost heap fetches until autovacuum has been through, and the newest hour is exactly the data the dashboard reads.

Worked solution (30 min)

Build 30 daily partitions with skewed tenants, one holding about 40% of the rows, then ANALYZE.
Run explain (analyze, buffers) on the original query and record how many partitions were scanned and the rows removed by filter.
Apply the rewritten predicate and the index, re-run, and confirm the plan lists only the partitions inside the ingested_at bound.
Re-run with $2 set to a non-hour-aligned timestamp and confirm the rewritten and original predicates return identical rows.
Insert an event with ingested_at six hours past occurred_at and check whether the pruning predicate excludes it.

Expected result: Three causes named separately: a non-sargable date_trunc, pruning on ingested_at against a predicate on occurred_at, and a non-tenant-leading index. A rewrite to occurred_at >= <hour-normalised $2> plus a redundant ingested_at >= $2 - horizon, an index on (tenant_id, occurred_at) including sku and quantity, a per-row write cost stated, and the late-data omission the pruning predicate introduces stated as a known bound.

Verify it yourself

The plan lists only partitions inside the ingested_at bound.
Rows removed by filter on the index scan is small relative to rows returned.
The rewritten and original predicates agree for both hour-aligned and non-aligned bounds.
A deliberately late-ingested event is visible or omitted exactly as the stated horizon predicts.

Follow-up

CREATE INDEX CONCURRENTLY is not supported on a partitioned parent. Give the sequence that gets this index onto 400 existing partitions without blocking ingest.
One tenant holds 200 times the median row count and the dashboard still times out for them with the index in place. What changes?
Should this read hit usage_rollup_hourly instead? State what that costs in freshness and what the watermark lets you promise.

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

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.
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.
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.
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.
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.
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?

Explain hoisting and its potential consequences in code execution.

Medium

Approach

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

Follow-up

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

Explain variable scoping in your primary language.

Medium

Approach

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

Follow-up

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

Leasing sandboxed job runs without double execution

Medium · leases · fencing · state machines · metering

job-runner leases work from a queue and executes customer workloads in sandboxes with CPU, memory, wall-clock and egress limits. Run durations span 200 ms to 30 minutes, thousands are concurrent, and job_run.status must reach exactly one terminal state among succeeded, failed, timed_out, cancelled and lost. A worker can pause 45 seconds for garbage collection, or be briefly partitioned, after its sandbox has exited but before it writes the outcome. Design the lease duration, the fencing, and timeout ownership, and state how billable_seconds is computed for a run whose worker never returns.

Approach

Resolve the lease-duration conflict rather than picking a compromise. A lease shorter than the longest legitimate run re-dispatches work that is still executing; a lease long enough for a 30-minute run leaves a crashed 200 ms run undetected for half an hour. Use a short lease - about 30 seconds - renewed by a heartbeat every 10 seconds while the supervisor is alive, each renewal pushing leased_until to now() + 30 s. Detection latency is then one lease, not one heartbeat: a dead worker's last successful heartbeat landed at most 10 s before it died, so leased_until expires 20 to 30 s after the death, plus whatever interval the reaper scans on. What the 3:1 ratio of lease to heartbeat buys is headroom - one or two missed heartbeats change nothing, and a stall only becomes a re-dispatch once it outlives the lease remaining when it began, between 20 and 30 s here. Both costs follow from those two numbers: heartbeat write load proportional to concurrent runs (3,000 runs at a 10-second interval is about 300 updates/second), and any pause longer than the lease - the 45-second GC - re-dispatching a run that is perfectly healthy.
Fence the outcome write so the store, not the worker's memory, arbitrates: UPDATE job_run SET status = $1, finished_at = $2, exit_code = $3 WHERE run_id = $4 AND status = 'running' AND lease_token = $5, and require exactly one row affected. Zero rows means this worker was fenced while it was paused, and its correct behaviour is to discard the result, not to retry - the retry is how a resumed worker overwrites a newer attempt.
Make the status machine monotone with a check constraint or a trigger so no terminal row can move, and model a retry as a new row with parent_run_id set and attempt incremented rather than a reset of the old one. That is what keeps how many times did this actually execute answerable afterwards and keeps each attempt's resource usage attributable to itself.
Put the wall-clock timeout in the supervisor, never in the workload: a customer-supplied program asked to time itself out will not. Distinguish timed_out, which the supervisor observed and caused, from lost, which nobody observed at all, and leave exit_code null unless the supervisor actually saw the process exit. Recording lost as failed asserts an outcome no one witnessed and then bills and retries on that assertion.
Decide billable_seconds for the unobserved case explicitly, because it cannot be derived. It is not now() - started_at, which bills queue and pause time the customer never consumed. The two defensible policies are leaving it null while status = 'lost' and billing nothing, or flooring it at the last heartbeat's observed running time as a stated lower bound; pick one and write it down. Add a per-tenant concurrency cap on sandbox slots so one tenant cannot occupy the whole fleet while these questions are being answered.

Worked solution (30 min)

Draw the timeline: last successful heartbeat at t, sandbox exits at t, worker pauses from t to t+45 s, so leased_until = t+30 s and the lease expires there. Mark where the second worker starts and where the first worker's write arrives, then say how short the pause would have had to be to change nothing.
Write the fenced terminal update and state the expected rows-affected in both the healthy case and the fenced case.
Size the heartbeat: 3,000 concurrent runs at a 10-second interval is about 300 row updates/second on job_run - decide whether that lands on the same table and what it contends with.
Write the billable_seconds rule for each terminal status and say which status leaves it null.

Expected result: A short lease of roughly 30 seconds renewed by a 10-second heartbeat, so detection latency is one lease - 20 to 30 seconds after a worker dies, plus the reaper's scan interval - independent of the 200 ms to 30 minute run spread, with one or two missed heartbeats tolerated rather than fatal; a terminal write fenced by AND status = 'running' AND lease_token = $x where zero rows affected means discard; retries as new rows with parent_run_id; a supervisor-owned wall-clock kill that separates timed_out from lost; billable_seconds left null while status = 'lost', or floored at the last heartbeat as a stated policy, never inferred from wall-clock.

Verify it yourself

In your timeline the fenced write must affect zero rows. If it affects one, the lease alone was doing the work and the token adds nothing.
At 3,000 concurrent runs and a 10-second heartbeat, check 300 updates/second against the other write load on that table before accepting the interval; if it is the dominant writer, the heartbeat belongs somewhere else. Check the ratio too: at lease = 3 x interval a lost heartbeat is survivable, whereas at lease = interval every scheduling hiccup is a double execution.
State whether a duplicate execution is billed, then confirm that summing billable_seconds across a parent and its retry matches that stated policy rather than matching it by accident.

Follow-up

A worker returns from a 45-second GC pause and writes succeeded. Walk through what the database does statement by statement.
Heartbeats are now a few hundred writes a second against job_run. How do you keep that off the path that matters, and what do you lose by moving it?
The re-dispatched workload has an external side effect the first attempt already performed. What does your design owe the customer, and what can it not fix?

Invoice detail latency triples after an ORM relationship refactor

Easy · n+1 query · orm · latency regression

An invoice detail endpoint returned in 40 ms at p99 last week. After a refactor replaced a hand-written join with ORM relationship access it returns in 1.4 s, and the regression grows with the number of invoice_line_item rows on the invoice. Database CPU rose, but no statement in the slow-query log exceeds 3 ms. You have request traces with per-span SQL, the ORM statement log, and a staging copy of the data. Produce an ordered diagnostic checklist, the measurement that confirms the cause before any code change, and the fix.

Approach

Count statements per request before reading any statement duration. A slow-query log hides this class by construction, because every individual query is fast and only their number is wrong; take one trace and count SQL spans.
Establish proportionality rather than asserting it: sample invoices with 5, 20, 60 and 200 line items and plot statements per request against line count. A straight line of slope 1 through an intercept of one or two identifies a lazy relationship load, and no index or cache would move that line.
Locate the emitting attribute access in the refactored code and check whether the same shape repeats one level deeper, for instance a tax or adjustment collection hanging off each line, which turns the cost quadratic.
Fix with a bounded statement count: either one join that fetches invoice and lines together, or two statements where the second is WHERE invoice_id = $1 AND tenant_id = $2. Keep tenant_id in the predicate so the read stays tenant-scoped even though invoice_id already implies it.
Choose between the two deliberately: the join duplicates the wide parent row across N children on the wire, the two-statement form avoids that for one extra round trip. Prefer the join for narrow parents and the split for wide ones.
Pin it with a per-request statement-count assertion in a test that varies line count, because a latency assertion passes on a small fixture and would not have caught this.

Follow-up

The endpoint now also needs per-line tax rows. Show the shape that keeps statement count constant instead of reintroducing the same defect one level down.
How does this change if a transaction-pooling proxy sits between the service and the database, so each statement may land on a different backend session?
The same page paginates invoices with LIMIT and OFFSET. Why is that a second, independent defect, and what replaces it?

Suggested practice checklist

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

Day 1 — Numbers before diagrams

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

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

Day 2 — One system, from requirements to schema

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

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

Day 3 — The consistency you are actually buying

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

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

Day 4 — Failure is the design

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

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

Day 5 — Scaling the hot path

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

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

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

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

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

Day 7 — Defend it while being interrupted

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

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

What the behavioural round is actually measuring

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

Describe a time you had to collaborate with a cross-functional team to meet a deadline.

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Give the blast radius: what could have broken, and what you measured.
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?
What did you decide not to do, and why?

Tell us about your professional background and your journey in technology.

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
State the situation in two sentences and spend the rest on the reasoning.
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?

Reverse a webhook ordering decision after measuring its cost

Medium · reversing decisions · head-of-line blocking · at-least-once · api contracts

You argued for strict per-subscription ordering in webhook-delivery, which means one in-flight attempt per subscription. It shipped. Three months later a single unresponsive endpoint holds one subscription's queue at a six-hour backlog, and two customers report events arriving out of order anyway once their own retries are counted. Describe a decision you reversed: what you originally optimised for, the measurement that changed your mind, what the reversal cost in engineering time and customer change, and how you told the people who had already built on the original guarantee.

Approach

State the original decision as a trade you made knowingly. Ordering across a network requires a single in-flight attempt per subscription, and its price is head-of-line blocking whenever one endpoint is slow. 'We priced it wrong' is a much stronger opening than 'we did not realise', and it is usually the true one.
Bring the measurement that flipped it, not the anecdote: backlog age at the ninety-ninth percentile per subscription, the share of subscriptions where one slow endpoint gated an otherwise healthy queue, and the delivery throughput lost to serialisation. A reversal justified by complaints is indistinguishable from a reversal justified by fatigue.
Name what you learned about the guarantee itself, which is the engineering content of this story. At-least-once delivery means a retried event already arrives after newer ones and the consumer already must be idempotent, so a guarantee the customer has to defend against anyway was never worth what it cost to provide.
Describe the migration, because reversing a published contract is the hard half and the part candidates skip. Parallel attempts behind a per-subscription flag, a monotonically increasing sequence number added to the envelope so order-sensitive consumers can sort or discard, documentation that states at-least-once and unordered in those words, and a deprecation measured in quarters because the client is a pinned SDK inside a build pipeline you cannot see or redeploy.
Give the cost in the two currencies that matter: engineer-weeks, and how many customers had to change code. Then say who you told before it shipped rather than in a changelog afterwards, and which large customer you left on the old behaviour and for how long.
Close with the signal you now weight differently, stated as something you would do earlier next time: measuring the blocking cost on the slowest decile of endpoints before committing to the guarantee, rather than after a customer noticed.

Follow-up

A customer insists they need ordering. What do you offer them that is not global serialisation?
How did you choose the deprecation window given that you cannot see or redeploy the clients?
What would have to be true for you to reverse back?