Interview guides · Apollo GraphQL

Apollo GraphQL Software Engineer Interview Questions & Guide 2026

Apollo GraphQL software engineer prep on databases: indexes, query plans, and the anomalies an isolation level still permits.

The 60-second brief

As a Software Engineer at Apollo GraphQL, you are at the forefront of the GraphQL ecosystem, building the tools and infrastructure that enable organizations to transition toward a graph-based architecture. This role is not just about writing code; it is about solving complex distributed systems problems and improving the developer experience for thousands of teams globally. You will work on products that allow engineers to manage data graphs, improve performance, and scale their API development workflows.

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

Focus areas

Keep money in integer minor units
Evolve APIs without breaking pinned SDK clients
Scope every query and cache key by tenant

Overview

The work is intellectually demanding and highly impactful. You will likely contribute to core Apollo products, such as Apollo Router, Apollo Studio, or the client-side libraries that power modern frontend applications. You will operate in an environment that values technical depth, architectural rigor, and a deep understanding of how data flows through modern applications. If you enjoy building developer-centric tools and navigating the complexities of high-scale distributed systems, this role offers a unique opportunity to influence the future of how software communicates.

Apollo GraphQL values engineers who can balance high-level architectural thinking with the practical, hands-on implementation details required to keep a mission-critical platform stable.

The Apollo GraphQL interview process

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

Recruiter Screening

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

Technical Conversations

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.

Real-time Problem Solving

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

Behavioral Assessment

Many of these questions are about something that went wrong, and the grading sits mostly in the hours after you knew. Who found out first, whether that was you or an alert or a user, how long it took you to say it out loud, and whether the people who needed the news got it while they could still act on it. Engineers under-tell this part because it feels like confessing. The pattern it is looking for is the opposite: the quiet fix, an incident absorbed without telling anyone, after which nothing changed and the same failure is still available.

What to demonstrate

How the problem was found, and whether that route was one you had built or one that happened to you, since a user reporting it first means your instrumentation did not cover that failure
Whether time-to-detect and time-to-tell are separate numbers in your account and whether you know both, because a fast fix that nobody heard about until the retro is a different answer from a slow one that was announced immediately
Whether the resolution left something durable behind, a check that fires or a default that changed, rather than depending on people remembering to be careful
Whether you can say what the failure cost without either inflating it or waving it away

How to prepare

Reconstruct one incident you were part of as a timeline with clock times: first bad request, first signal, first person who knew, first message outside the team, mitigation, permanent fix. The gaps between those entries are what gets asked about
Look up the configuration of the signal that caught it, including its evaluation window and threshold. An alert defined on a five-minute aggregate cannot fire until the condition holds across that window, which puts a floor under time-to-detect that has nothing to do with how severe the failure was. Be able to say what that floor was and whether anyone had chosen it deliberately
Prepare one story where you escalated early and the severity turned out to be smaller than you thought, including what it cost the people you pulled in. Without it, every answer you give about raising alarms is unfalsifiable

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.

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.

Listing technologies instead of trade-offs

Name the property the design needs first, such as ordered range scans, multi-entity transactions, cheap appends, or a predictable p99, then pick something that provides it and say what it gives up in exchange. Almost any component is defensible once you state the requirement it satisfies and the one it sacrifices.

Finishing a solution without stating its complexity

Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.

Questions & practice

Parse and verify a timestamped multi-signature webhook header

Easy · parsing · hmac · constant-time-compare · replay-protection

An inbound webhook carries a signature header of at most 1 KiB shaped t=<unix seconds>,v1=<64 hex chars>, with up to five v1 values during secret rotation and possibly unknown scheme keys. You hold the raw request body bytes and the currently active signing secrets. Write the parser and the verifier: accept when any active secret reproduces a signature and the timestamp is within a five-minute tolerance in either direction, reject otherwise. Single left-to-right pass over the header, no regular expression. State what is inside the MAC and why.

Approach

Parse in one scan: split on ,, then on the first = only, since a value may itself contain = under a future scheme. Accept t exactly once and treat a second t as a reject rather than last-wins. Push every v1 onto a short list and ignore any other key, so a v2 can be introduced later without breaking this verifier.
Say what is signed: HMAC-SHA256 over the exact byte string <t>.<raw body bytes>, yielding 32 bytes or 64 hex characters. The timestamp sits inside the MAC because otherwise an attacker replays yesterday's body with its still-valid signature and only has to edit the header timestamp.
Hash the bytes as received. Verifying against a re-serialised JSON body is the usual defect: key order, whitespace and number formatting all change the bytes while the parsed objects compare equal, so signatures fail for honest senders and the popular 'fix' is to stop checking.
Compare in constant time over fixed-length digests. Decode the hex to 32 bytes, accumulate acc |= a[i] ^ b[i] across the whole length, and test acc == 0 at the end. Evaluate every candidate without an early exit; at five candidates that is five HMACs over the body, linear in body size and negligible beside the network.
Apply the tolerance as a two-sided bound, rejecting when |now - t| > 300 seconds. A sender whose clock runs ahead of yours is an ordinary case, and an unbounded future timestamp is a free replay window.
Complexity: O(L) over the header producing k candidates, plus k HMACs at O(|body|) each. Space is O(k) beyond the body itself. Do the cheap rejections, including the tolerance check, before any cryptography runs.

Follow-up

The body is 40 MB. What changes about where you verify, and what can you do before the whole body has arrived?
A customer reports that signatures fail for exactly the requests whose body contains a non-ASCII character. What is your first hypothesis?
How do you rotate the signing secret with no failed deliveries, and how long do both secrets stay live?

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

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.
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.
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.
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.
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.
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)

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.
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.
Fold it and assert each of those four expectations separately rather than eyeballing a grand total.
Re-run with the input shuffled and diff the output files.
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?

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.

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?

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.

Worked solution (40 min)

Generate a day of events for one tenant across three workspaces and two SKUs, inject 2% duplicates sharing an idempotency_key with a later ingested_at, and inject 1% whose ingested_at is six hours past occurred_at.
Write the dedup CTE alone and confirm its row count drops by exactly the injected duplicate count.
Aggregate and compare quantity_sum against a ground truth computed outside SQL from the generator's own records.
Add the two window and filtered aggregates, and confirm the final hour's running total equals the ungrouped day total for each (workspace, sku).
Run the whole query twice over static data and diff the two outputs.
Re-run with the session TimeZone set to a non-UTC zone and confirm hour_start does not move.

Expected result: One query: a dedup CTE using row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id); aggregation by (workspace_id, sku, hour_start) producing quantity_sum, event_count and max(ingested_at) plus a filtered late-quantity aggregate; a running total as a window over the aggregated rows; and an explicit statement that the partition-scoped daily unique index cannot catch a duplicate that crosses the ingest-day boundary or arrives in a later replay.

Verify it yourself

Row count after dedup equals raw count minus injected duplicates, exactly.
The running total at the final hour equals the ungrouped day total for that (workspace, sku).
Two consecutive runs over static data return byte-identical rows.
Every hour_start is hour-aligned in UTC and unchanged by the session's TimeZone setting.

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?

Model credential revocation so history survives the delete

Easy · soft delete · partial indexes · check constraints · audit history

tenant_api_key stores key_id, tenant_id, workspace_id, name, key_prefix, secret_hash, scopes text[], status (active, revoked, expired, compromised), auth_version, created_at, expires_at, last_used_at, revoked_at, revoked_reason. Rotation inserts a new row and revocation never deletes, because an incident review asks which credential served a request last quarter. Write the constraints that enforce: a label is unique only among a tenant's live keys, revoked_at and status can never disagree, and scopes is never empty. Then write the authentication lookup predicate, and name one column in this table that must stay out of it.

Approach

Reach for a partial unique index rather than a plain UNIQUE: create unique index on tenant_api_key (tenant_id, name) where revoked_at is null. Any number of revoked rows may share a label, the live namespace stays unique per tenant, and the revoked majority is not in the index at all, so it stays small on a table that only grows.
Tie the nullable timestamp to the enum so the two cannot drift: check ((revoked_at is not null) = (status in ('revoked','compromised'))) and check ((revoked_at is null) = (revoked_reason is null)). A revocation that records no reason is the one an incident review cannot use.
Write the emptiness check as check (cardinality(scopes) > 0), not array_length(scopes, 1) > 0. array_length returns NULL for an empty array, a CHECK constraint passes when its expression is NULL, so the array_length version accepts exactly the value it was written to reject.
Make the lookup a single index probe with every liveness condition inside it: where secret_hash = $1 and revoked_at is null and (expires_at is null or expires_at > now()) and auth_version = $2, backed by a unique index on secret_hash. Nothing is filtered in application code, so there is no path that forgets a clause.
Keep last_used_at out of that predicate. It is written asynchronously and is allowed to lag by a minute, so it is a usage signal; feeding it into an authorisation decision makes the decision depend on a write that may be late, batched away or lost.
Flag the modelling smell while you are here: expired is derivable from expires_at < now(), so storing it as a status obliges a job to keep it true and guarantees the column is wrong between the expiry instant and that job's next run. Derive it in the predicate; keep the stored status for states that are decisions rather than clock readings.

Follow-up

Rotation issues a replacement while the old key stays live for a 30-day overlap. What does the uniqueness rule become, and what does the UI show to tell two same-named keys apart?
A password reset bumps the principal's auth_version. No row in this table changed. How does the next request fail, and what query counts how many keys that bump just killed?
A key turns up in a public repository. Which columns let you find it, and what do you write to the row?

Given an existing system, what are the flaws, and how would you redesign them?

Medium

Approach

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

What considerations guide your decisions when implementing new technologies?

Medium

Approach

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.
Work from the requirement backwards to the design.

Follow-up

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

Can you walk me through a challenging software project you worked on, including your technical decisions, the challenges faced, and the outcome?

Medium

Approach

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

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?

A rare job-run overwrite that logging makes disappear

Hard · fencing tokens · leases · concurrency

About one job run in fifty thousand records billable_seconds matching no observed sandbox lifetime, and a few rows show worker_id changing after finished_at was already set. It does not reproduce: debug logging around the terminal write made it vanish for two weeks before it returned. Runs last from 200 ms to 30 minutes, the lease is 60 seconds and is renewed while a run executes. Give an ordered checklist, the mechanism, and a fix that makes the illegal write impossible rather than merely rarer.

Approach

Mine the evidence instead of chasing a repro: select rows where updated_at is later than finished_at, or where a terminal status was written twice, and join them to attempt history to recover both writer identities. The defect has already happened tens of times and the rows are the recording.
State the signature before measuring it. If the cause is a lease that expired while the original worker was stalled, affected runs should cluster where the gap between the last renewal and the terminal write exceeds the lease, and should correlate with worker pause metrics rather than with workload shape.
Read the disappearance honestly. Logging inside the window changed the timing and lowered the probability; it is evidence about how narrow the window is, not a fix. Reproduce by widening the window on purpose, shortening the lease and injecting a pause between sandbox exit and the terminal write, rather than by adding more instrumentation.
Name the mechanism precisely: the lease expires during a stall such as a long garbage-collection pause or a brief partition, the run is re-dispatched, and the original worker then wakes and writes its terminal state over the new attempt's row. A lease alone cannot stop this, because the check and the write are separated by the stall.
Fix by fencing the write itself: UPDATE job_run SET status = $2, finished_at = $3, billable_seconds = $4 WHERE run_id = $1 AND status = 'running' AND lease_token = $5, with zero rows affected interpreted as having been fenced rather than as success. The token lives on the row so the store arbitrates, not the worker's memory.
Keep the state machine honest: a retry inserts a new row pointing at parent_run_id rather than resetting the old one, and a run whose worker vanished terminates as lost with billable_seconds null, because recording failure asserts an outcome nobody observed and then bills and retries on that assertion.

Follow-up

The supervisor also emits a usage event on completion. What does the fenced worker do about the event it already emitted, and how does metering absorb it?
Lease renewal is itself a network call. What happens when a renewal times out, and how does the worker decide whether it still holds the lease?
Why is lengthening the lease past the longest legitimate run the wrong lever, and what breaks if you do it anyway?

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.

Bounding a vague ticket before writing code

When the requirements were thin, the interesting part is how you fenced the problem off: the assumption you wrote down, who you got to confirm it, the narrow version you shipped first so the rest stayed cheap to change. Guessing and being right is luck. Guessing in writing, where someone could correct you, is method.

What would your previous managers say your biggest strengths and weaknesses were?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

Unblock an engineer on a job run that finished twice

Easy · mentoring · fencing tokens · lease expiry · debugging method

An engineer two weeks into the team brings you a job_run row showing status succeeded with an exit_code written by a worker declared dead ten minutes earlier; the retry attempt also shows succeeded. They have spent a day adding logging and are no closer. You have twenty minutes and you do not want to take the keyboard. Describe how you unblock someone: the question you ask first, what you let them find themselves, the concept you name and when, and how you check the next day that they own the fix rather than having watched you produce it.

Approach

Ask what they expect rather than what they see: which statement set status to succeeded, and what did it check before writing? That question points directly at the update's WHERE clause, which is where the answer lives, and it costs them nothing to answer, so it does not read as a test.
Let them build the timeline themselves from the row: queued_at, started_at, leased_until, finished_at and worker_id, on both the original run and the retry. Two different worker_ids with a lease expiry between them tells the whole story, and they will see it before you say it.
Name the concept once the evidence has earned it. A lease bounds time; it does not prevent a write. The store has to reject a stale writer, which means the update carries a fencing token the row compares — update job_run set status = 'succeeded' where run_id = $1 and lease_token = $2 and status = 'running' — and a long garbage-collection pause or a brief partition is enough to produce what they are looking at.
Point at the second, less obvious half and let them decide it: 'lost' exists in the status enum precisely so a run whose worker vanished is not recorded as failed, because failed asserts an outcome nobody observed and the system then bills and retries on that assertion. Ask them what these two rows should have said.
Leave them with the next step rather than the patch — a test that kills the first worker after the sandbox exits and before the row is written — and say when you are available again, so the offer is real rather than polite.
Check ownership the next day by what they produced, not by asking if it went well: a test that reproduces the window proves they understood it; a test that only asserts the new WHERE clause proves they copied it. Ask them to explain it to a third person and listen for whether the explanation is theirs.

Follow-up

They propose a longer lease instead of a token. What do you say, and what breaks when legitimate runs last thirty minutes?
How can you tell whether your explanation landed or they simply deferred to you?
The same engineer hits a variant of this next month. What did you fail to teach the first time?

Ship metered billing with a named deduplication horizon

Medium · technical debt · deduplication · deadline pressure · detectors

Metered billing must be on in three weeks. usage_event is partitioned daily, so its unique index must include the partition key and deduplicates only within a day: a producer retry that crosses midnight, or a replay run a week later, gets through. A cross-partition dedup store is two weeks you do not have. Describe shipping with debt you named in advance: what you shipped, what you wrote down, the detector you added, the trigger and date for paying it off, and what you would have refused to ship under the same pressure.

Approach

Show you can separate the two kinds of debt, because that distinction is what the question actually probes. Debt that costs engineering time later is shippable on a deadline. Debt that silently corrupts a number a customer gets charged for is not shippable unless the corruption is detectable, and detectability is the whole negotiation.
Make the exposure narrow and measured rather than gestural. The hole is duplicates whose occurrences straddle a UTC day boundary, plus any replay older than partition retention. Measure it before arguing about it: how often an idempotency_key recurs at all, and the distribution of the gap between first and last occurrence. If the ninety-ninth percentile of that gap is four minutes, the residual risk is a small band around midnight and you can say so numerically.
Add the detector before the feature, not after. A nightly job counting keys that appear in more than one partition is one grouped scan over recent partitions, and it converts a silent overcount into a page. State what it costs to run and what it fires on.
Buy the cheap half of the real fix immediately: extend partition retention so the dedup horizon exceeds the producer's maximum retry window plus the longest replay you intend to support. That reframes retention as a correctness parameter rather than a storage cost, which is the sentence you need on record before someone optimises the bill.
Make repayment mechanical instead of aspirational: a dated entry with a named owner, plus a threshold that pulls the date forward — first detector hit above N events, or first customer dispute. Debt with a trigger gets paid; debt with only a date does not.
Answer the second half honestly by naming what you would refuse under identical pressure: the sealing path, because a sealed row is frozen and a wrong number there stops being a bug and becomes an adjustment line, a dispute and an audit question.

Follow-up

The detector fires on forty duplicate events for one tenant, and two of their invoices have already sealed. What happens next?
Whom did you tell that the billing numbers had a known hole, and in what words?
Finance asks you to cut storage by shortening partition retention. What do you say, and to whom?