At Larson Design Group, the Software Engineer role bridges complex engineering requirements and the technical solutions the firm depends on. The role builds the digital systems that support the firm's architecture, engineering, and consulting projects. The work shapes how Larson Design Group manages data, designs infrastructure, and delivers outcomes for its clients.
This role requires a balance of technical proficiency and a deep understanding of business needs. You will often collaborate with cross-functional teams, including project managers and senior engineers, to translate non-technical requirements into robust, scalable software. Because Larson Design Group is a 100% employee-owned company, you will find that your contributions have a tangible impact on the firm's growth and operational efficiency.
Initial Screening
reportedThe 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 Discussions
reportedMost 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.
Behavioral Discussions
reportedYour first answer is not really what is scored. It buys the follow-up questions, and those decide the round. An interviewer with fifteen minutes takes one thread and pushes on it four or five times, so a story you can only tell at a single level of detail collapses under the third why. That is an argument for fewer stories known deeply rather than one prepared per prompt. Four or five pieces of work you can still explain down to the code you changed and the argument you had about it will cover nearly anything asked in this round.
What to demonstrate
- Whether a story holds as the questioning moves from what you did to why that instead of the alternative, and then to what you would change knowing what you know now
- Whether you can re-cut a project to answer the question actually asked rather than delivering a rehearsed block that answers an adjacent one
- Whether your level of detail is chosen rather than habitual: going down to the schema when the question is about the data model, staying out of it when the question is about the person who disagreed with you
How to prepare
- Pick four projects and write the chain out four levels deep for each: what you did, why that, why not the alternative, and what would have to be true for the alternative to have won. Where you cannot reach the fourth level, you have a placeholder rather than a story
- Have someone ask why three times in a row on a single thread with nothing else added, and mark the point where you start repeating a sentence you already said. That point is where the interviewer stops learning anything
- Build a one-page index instead of an answer bank: the common prompts in this round (disagreement, a failure that was yours, thin requirements, a deadline you missed, work you inherited) mapped to which of your four projects you would use for each, so the choosing is done now rather than while an interviewer waits
Final Decision
reportedCoding rounds mostly set a floor. They decide whether you clear the bar, not where you land on the ladder. Level tends to come out of the design discussion and the ownership stories, so the question worth auditing beforehand is whether the scope you describe matches the scope of the job. Work that stops at your own service, or a story whose hard part was writing the code rather than getting several people to agree on an interface, reads a level below where you think you are interviewing, and that gap is usually resolved downwards.
What to demonstrate
- Whether the largest thing you describe owning ran end to end — the decision, the migration path, the rollout, and what you did when it went wrong — or stopped at the change you merged
- Whether design answers include what you would not build, what you would defer, and what you would measure before committing, rather than only what the boxes are
- Whether a disagreement in a story was settled with something checkable — a benchmark, a prototype, a written proposal — instead of by seniority or by waiting it out
- Whether you can say which calls you made alone and which you escalated, and why the line sat where it did
How to prepare
- Write your largest piece of owned work as a timeline of decisions — who decided what, when, and what you did when the plan broke — then delete every sentence whose subject is "we" and see how much survives
- Take one system you know well and drill the migration answer: how old and new paths run side by side under live traffic, how you compare their outputs, what the rollback is once writes are going to both, and which step you would not automate
- Map each line of the ladder in the job posting to a specific thing you have done, find the line you cannot support, and prepare the closest evidence you have plus an honest account of the gap
PracHub editorial advice for the preparation topics above.
Paginating a growing table with limit and offset
Two unrelated defects share the idiom. Correctness: rows inserted or deleted between page requests shift the window, so a consumer walking an export skips rows and sees others twice, which for a customer-facing sync is silent data loss rather than an error anyone notices. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N times the page size and a deep page on a large table degrades from milliseconds to seconds. Keyset pagination over a stable, unique, indexed ordering -- where (created_at, id) < ($1, $2) order by created_at desc, id desc limit $3 -- is constant-cost per page and immune to shifting, on the precondition that the cursor columns never change value for a row, which disqualifies updated_at as a cursor.
Treating a timed-out write as a failed write
A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.
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.
Designing for a scale nobody asked for
Ask for request rate, data size, read-to-write ratio and expected growth, then size the simplest option first; one relational instance on current hardware covers a large share of real workloads. Reaching for shards, queues and a cache tier before any number has been quoted reads as pattern-matching rather than judgement.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Seal an hour under late data with bounded memory
Metering ingest reads 256 partitions at 10,000 to 40,000 events/second. Events carry occurred_at and ingested_at, and during a producer replay the gap between them is hours. Seal each UTC hour once no more than 50 parts per million of that hour's eventual quantity can still arrive, using memory that does not grow with the size of the replay. Define the watermark, the lateness parameter and how you measure it, the structure holding open hours, and the write that performs the seal. State what an idle partition does to your watermark.
Approach
- Two clocks, two jobs. Bucket by
occurred_at, because that is the hour the customer is billed for, and advance the watermark oningested_at, because that is what the fold has consumed and whatsource_max_ingested_atrecords. Conflating them is what makes late data invisible. - The global watermark is the min over partitions of each partition's committed
ingested_at, not the max: the fold is trustworthy only as far as the slowest partition. The consequence is that one idle partition pins the watermark forever and nothing seals, so an idle partition must promote its watermark to wall clock after a stated idle timeout, and that timeout becomes a correctness parameter, because a partition that is slow rather than idle gets sealed past. - Choose the lateness L from the measured distribution of
ingested_at - occurred_at, weighted by quantity rather than by event count. The target is 50 ppm of the hour's quantity, and a replay is rare in events while carrying disproportionate mass, so an event-weighted quantile picks an L that is comfortably wrong at exactly the moment it matters. - Measure that quantile in bounded memory. A Greenwald-Khanna summary gives epsilon-approximate quantiles in O((1/epsilon) log(epsilon n)) space; a t-digest costs more per merge but has relative error that tightens at the tails, which is the half of the distribution you are reading at p99.99. Keep a separate summary per tenant class, because one tenant's batch importer is not the population.
- Hold open hours in a min-heap keyed by
hour_start. When the watermark advances, pop every hour withhour_end + L < Wand seal it: O(log H_open) per advance and O(1) amortised per event to touch its bucket. Memory is open hours multiplied by distinct(tenant, workspace, sku)keys, so cap the number of simultaneously open hours and spill the oldest intousage_rollup_hourlyasstatus='open'with arevisionbump. While an hour is open the row is upsertable, so the store is your overflow. - The seal itself is a conditional write:
update ... set status='sealed', sealed_at=now() where status='open' returning .... Two sealers race on every restart, and the loser must see zero rows and stop rather than write a second value. After the seal, an event for that hour is not an upsert but an adjustment, andsource_max_ingested_atis what proves it arrived afterwards.
Worked solution 40 min
- Replay a day of events with a synthetic lateness distribution: 99.9% under two minutes, plus a 0.05% tail at four to six hours that carries 3% of total quantity.
- Compute the p99.99 lateness two ways, event-weighted and quantity-weighted, and put the two numbers side by side.
- Implement the min-heap of open hours with the watermark as the min over 256 partitions, then stall one partition for 20 minutes and observe what seals.
- Set the idle-partition timeout to 60 seconds, repeat the stall, and measure how much quantity arrives after the seal.
- Attempt the seal from two workers at once and confirm the conditional update lets exactly one through.
Follow-up
- A replay starts during the sealing window for a period you are about to close. What do you do, and what is the customer-visible consequence of each option?
- Your measured quantity-weighted p99.99 lateness is six hours and the invoice must be issued at 02:00 UTC on the first. How do you reconcile those two numbers?
- How would you detect that L has drifted before it costs you an hour's quantity?
Hold a tenant to a trailing sixty-second request limit
The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.
Approach
- Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while
front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request. - Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
- Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate,
prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact. - Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (
tokens,last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual. - Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
- Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.
Follow-up
- One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
- Quotas rather than rate limits: the check is
select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes. - How do you return an accurate
Retry-Afterfrom the exact algorithm without a second scan?
Order a job dependency graph and find its 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, thenslack[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] = 0is exactly the statement that some longest path runs through v; equivalently, the longest path through v has lengthT - 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)), whereL_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, whilemin(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 ismax(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, sinceL_avoid(v)only ever matters through that max: setduration[v] := 0, recompute the makespan asT0(v) = max(T - duration[v], L_avoid(v)), and the gain ismin(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
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-argumentdate_trunc('hour', occurred_at, 'UTC')on PostgreSQL 16 or later, ordate_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC'before that. Filterenvironment = 'production'explicitly, since metering covers three environments and billing covers one. - Aggregate to the grain with
sum(quantity),count(*)andmax(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?
Model credential revocation so history survives the delete
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')))andcheck ((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), notarray_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:
expiredis derivable fromexpires_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.
Worked solution 20 min
- Create the table with all three constraints on a scratch database and insert two revoked rows sharing (tenant_id, name); the partial index should accept both.
- Insert a second live row with that same name and confirm the violation names the partial index.
- Run
update tenant_api_key set revoked_at = now()leaving status = 'active' and confirm the CHECK rejects it; then tryinsert ... scopes = '{}'against both the cardinality and the array_length forms and note that only one rejects it. - Run
explain (analyze, buffers)on the lookup predicate for a live key and confirm an index scan on secret_hash with rows removed by filter equal to zero.
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?
How do you ensure your code remains maintainable and scalable?
How do you ensure your code remains maintainable and scalable?
Approach
- Say what you would check first and why it is the highest-information step.
- Work from the requirement backwards to the design.
- State your assumptions explicitly before working the problem.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
What is your process for troubleshooting a critical system error?
What is your process for troubleshooting a critical system error?
Approach
- Clarify what is being asked and what a complete answer contains.
- Say what you would check first and why it is the highest-information step.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Authorisation cache with a bounded revocation window
The edge gateway serves about 30k requests/second from roughly 120 pods across three regions and may add no more than 10 ms at p99. Each request presents an API key that must resolve to an authorisation context: tenant, workspace, scopes, entitlements and credential version. The control plane that owns those rows takes tens of writes/second. A revoked credential must stop authorising within a bound you state as a number. Design the cache - what is keyed, what invalidates it, how many tiers - and specify what the gateway does for the duration of a control-plane outage.
Approach
- Fix the entry shape before the topology: key on SHA-256 of the presented secret, value is the resolved context plus the principal's auth_version and a fetched_at. Cache negative lookups too, with a much shorter TTL and a bounded-size structure, because otherwise every sprayed invalid key is a control-plane round trip, and unbounded negative entries let a sprayer evict live ones.
- Compute the control-plane read load before choosing a TTL, and notice the multiplier is pods, not regions, when the cache is in-process: distinct_active_keys x pods / TTL. At 50,000 active credentials, 120 pods and a 60 s TTL that is 100,000 reads/second against a single-writer primary with read replicas, which is not serviceable - so the design needs two tiers, a per-region shared cache in front of the control plane with the in-process cache held to a few seconds.
- State the bound as the sum of the tiers, not as a hope: with a 10 s in-process TTL over a 60 s regional TTL, worst-case staleness absent any invalidation message is 70 s, because an in-process entry can be filled from a regional entry that was itself about to expire. Publish-subscribe invalidation on every credential and entitlement mutation makes the typical case sub-second, but it is lossy under partition, so the TTL is the only enforced bound and both tiers must subscribe.
- Make auth_version propagate through the same path: a password reset or sign-out-everywhere bumps the principal and revokes its keys with no hook of its own, so the invalidation publisher has to expand principal -> credentials and publish per key, or the cache keeps serving keys whose auth_version no longer matches.
- Decide the partition behaviour in advance and write it as two rules: on a cache hit past TTL, serve from the stale entry up to a grace ceiling (say 10 minutes); on a cache miss, refuse, because authorising something never seen converts a control-plane outage into an authorisation bypass. Worst-case revoked-key lifetime during an outage is then TTL + grace, about 11 minutes, and that number is the price of not turning a control-plane outage into a total data-plane outage.
- Protect the refill path: per-key single-flight so a mass invalidation or a cold pod does not stampede the control plane, TTL jitter so entries created together do not expire together, and a small separately replicated deny-list for compromised keys that is consulted on the hot path and survives control-plane loss.
Worked solution 35 min
- Write the cache entry shape - key, value fields, and which of those fields a request actually reads on the hot path - and mark which field makes a password reset propagate.
- Compute control-plane reads/second for TTLs of 10 s, 60 s and 300 s using distinct_keys x cache_instances / TTL, once with cache_instances = 3 regions and once with cache_instances = 120 pods, and note which of the two the in-process design actually implies.
- Enumerate the four states a revocation can be in - published and received, published and dropped, control plane unreachable, pod started after the publish - and write which entry serves the next request in each.
- Write the outage policy as two rules (hit past TTL within grace: serve; miss: refuse) and compute worst-case revoked-key lifetime as the sum of both tier TTLs plus the grace.
Follow-up
- A key is found in a public repository and must stop working in seconds, not minutes. What changes, and what does it cost on the request path?
- One region is partitioned from the control plane while the control plane itself is healthy. What do that region's pods do, and how do you distinguish this from a control-plane outage?
- How would you measure the actual revocation bound in production rather than asserting it from the configuration?
Metering partition crash-loops and the sealing watermark freezes
One metering-ingest partition has stopped advancing. Lag grows linearly, the consumer restarts about every 40 seconds, and the same offset appears in every startup log while other partitions stay healthy. Events are committed in batches of a few thousand and the acknowledgement follows the commit. Sealing is six hours away and source_max_ingested_at for that partition's tenants is frozen. Give an ordered checklist, a containment action available within minutes, and the durable fix, saying what each does to exactly-once accounting.
Approach
- Distinguish a poison record from a capacity problem in one measurement: compare the offset and the exception across restarts. An identical pair every time is deterministic failure on one record, whereas a throughput problem still advances the offset between crashes.
- Read the record from a separate consumer group so the bytes can be inspected without perturbing the stuck consumer, then classify the defect: schema violation, a quantity failing the non-negative check, a null workspace, an unmappable SKU enum, or a payload past a size limit. That classification decides whether this is a producer bug or a missing consumer guard.
- Account for batch granularity before acting. With commits of a few thousand, one bad record fails thousands of good ones, so the blast radius is the batch. Halve the batch around the offset to isolate the record, or move to per-record error isolation so the radius becomes the record.
- Contain by diverting that record to a dead-letter store with its raw bytes and offset, then resume. This is safe here precisely because the acknowledgement follows the commit: the good records from the failed batch are re-consumed and absorbed by the uniqueness check on (tenant_id, idempotency_key) rather than counted twice.
- Make the fix durable with per-record error isolation, a bounded poison counter, and an alert on dead-letter rate rather than on lag alone, since lag only reveals this after the sealing margin has already been eaten.
- Check the horizon before replaying anything. The unique index lives on a daily-partitioned table and therefore includes the partition key, so it deduplicates within a day only; a replay landing on a later ingest day needs the separate dedup store or it double-counts into a tenant's bill.
Follow-up
- Move the acknowledgement before the commit and describe exactly what is lost and what is duplicated in each of the two crash windows.
- Sealing is in six hours and the partition will not drain in time. What do you seal on, and what does the invoice have to record so the difference is explainable later?
- A producer replays two weeks of events next month. Which part of your fix stops holding, and what is the dedup horizon you would actually configure?
Roughly ninety minutes on weeknights with one longer weekend block. The plan cuts scope rather than compressing everything, on the assumption that one thing finished per night beats four half-started.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Fix the scope and take a cold baseline
- Read the role description and write the three things the loop will almost certainly test, then write an explicit not-doing list and keep it visible all week.
- Take one twenty-five-minute coding problem and one fifteen-minute design prompt cold, and write the single sentence naming what blocked each, because those two sentences decide where the remaining evenings go.
- Set the week's rule: one thing finished every night, including the night you only have forty minutes.
Deliverable: A one-page scope with a not-doing list and two cold attempts, each carrying one sentence on what blocked it.
Practice prompt ↗Practice prompt ↗Worked solution ↗02One pattern, written three times from blank
- Choose the single pattern most likely to appear in your loop and write it three times from an empty file rather than editing the previous attempt.
- On the third pass, write the invariant as a comment before the loop body and the complexity before the first line of code.
- Stop at ninety minutes even if the third version is imperfect, and write the one thing you would fix given another hour.
Deliverable: Three independent implementations of the same pattern plus a note on what changed between them.
Practice prompt ↗Practice prompt ↗03One design, only to the depth you can defend
- Take one system shape and go only as far as requirements, interface and data model, refusing to draw a box you could not survive a follow-up about.
- Attach one number to each non-functional requirement, deriving it rather than asserting it, and write the assumption the number rests on.
- Write the one tradeoff you are choosing against and the observation that would make you reverse it.
Deliverable: One design at interface-and-schema depth with derived numbers and one written reversible tradeoff.
Practice prompt ↗Practice prompt ↗04Only the fundamentals you will have to defend
- Write, in under two hundred words each, the answers to the two questions that follow almost any implementation: why this structure and not the obvious alternative, and what happens to this code at a hundred times the input.
- Write what an index actually costs: faster lookups on the indexed columns against a write that now maintains a second structure, plus the cases where the planner declines to use it anyway, low selectivity, or a predicate wrapping the column in a function.
- Delete any answer you cannot deliver aloud in under a minute, since an answer that needs reading is not an answer you have.
Deliverable: Three written answers, each under two hundred words and each timed aloud.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Your own work, timed
- Write a ninety-second and a four-minute version of your main project and time both aloud rather than reading them.
- Prepare the two follow-ups that always come: what you would do differently, and how you knew it worked.
- Put one number in the first sentence and be ready to say exactly where it came from and what it excludes.
Deliverable: Two timed narratives with one defensible number in the opening line.
Practice prompt ↗Practice prompt ↗06The one full rehearsal, in the weekend block
- Run a sixty-minute mock covering a coding round and a design round in one sitting with no break, because sustained attention is the thing evenings have not trained.
- Immediately afterwards, and before hearing any feedback, write the three moments you lost the thread.
- Spend the rest of the block only on those three moments, and on nothing you merely feel shaky about.
Deliverable: Mock notes naming three failure moments with a specific fix written under each.
Practice prompt ↗Practice prompt ↗07Taper
- Write the twenty-minute warm-up you will actually do on the morning: one problem you can already solve from a blank file, one design you can narrate, and nothing you have never seen.
- Re-read only your own notes from this week and open no new material.
- Write the logistics down: the editor or shared document you will be working in, whether execution and lookups are permitted, and the sentence you will use when you do not know something.
Deliverable: A one-page card holding the design structure, the project numbers, and the logistics.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Conflict answers where you were right and everyone came round are the weakest ones. Stronger: the evidence you went and collected, what would have changed your mind, and what you did in the weeks after the call went against you. Implementing a design you argued against, properly, is a specific and checkable behaviour.
Describe a time you had to explain a complex technical problem to a no…
Describe a time you had to explain a complex technical problem to a non-technical stakeholder. How did you ensure they understood?
Approach
- Close with what you would do differently, concretely.
- State the situation in two sentences and spend the rest on the reasoning.
- 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?
What is your experience with C# and SQL?
What is your experience with C# and SQL?
Approach
- Name the disagreement and how you resolved it with evidence.
- 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.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
Tell us about a time you contributed to a team goal that required cros…
Tell us about a time you contributed to a team goal that required cross-departmental cooperation.
Approach
- Name the disagreement and how you resolved it with evidence.
- Give the blast radius: what could have broken, and what you measured.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
Have you had to work with upper management and other departments to ga…
Have you had to work with upper management and other departments to gather and refine project requirements?
Approach
- Pick a story where you made the decision, not one where you watched it.
- Close with what you would do differently, concretely.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
- 01
Describe a time you had to explain a complex technical problem to a non-technical stakeholder. How did you ensure they understood?
- 02
What is your experience with C# and SQL?
- 03
Tell us about a time you contributed to a team goal that required cross-departmental cooperation.
- 04
Have you had to work with upper management and other departments to gather and refine project requirements?
Is this an official Larson Design Group interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Larson Design Group. Rounds and questions reflect what candidates have reported, not a process Larson Design Group has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How much time should I spend preparing for the technical portion?
Since the interviews are often conversational rather than coding-heavy, prioritize reviewing your past projects. Be prepared to explain the "why" behind your technical decisions rather than just memorizing syntax.
PracHub interview research ↗What is the company culture like?
Larson Design Group is 100% employee-owned, which fosters a culture of collaboration, shared responsibility, and long-term thinking. Candidates report that interviewers favor people invested in the success of the entire firm.
PracHub interview research ↗Will I be working on a specific team?
Yes, you will typically be aligned with a specific development group. The interview process is designed to ensure you are a good fit for that specific team's workflow and technical challenges.
PracHub interview research ↗How long does the hiring process typically take?
Timelines can vary. You can generally expect the progression from initial screen to offer to take a few weeks.
PracHub interview research ↗Sources & methodology 3 sources ↗
No official company page is cited. Rounds and questions come from candidate reports and PracHub editorial material; each source shows the date it was read.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
PracHub page · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-22
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