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Swish Analytics Software Engineer Interview Guide

Software engineer interview prep for Swish Analytics: original coding, system design and debugging practice, each with a worked solution.

The 60-second brief

As a Software Engineer at Swish Analytics, you will play a critical role in building and scaling the next generation of sports betting technology, predictive modeling platforms, and real-time data integration systems. Swish Analytics operates at the intersection of high-volume data science, sports analytics, and consumer-facing software. Your work will directly impact how massive quantities of live sports data are ingested, processed, and visualized for sportsbooks, media companies, and professional bettors.

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

Focus areas

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

Overview

The engineering team is responsible for developing high-throughput APIs, designing robust database schemas, and crafting intuitive, data-rich user interfaces. This role requires a unique blend of backend efficiency and frontend precision, as you will be tasked with transforming complex statistical outputs into clean, actionable insights. Whether you are optimizing database queries to handle millions of real-time updates or building highly interactive visualization dashboards, your contributions will directly drive the core products of the company.

Because is a highly focused, fast-moving organization, engineers are expected to take immense ownership of their projects. You will not just write code; you will architect solutions, manage deployments, and collaborate closely with data scientists to ensure predictive models are seamlessly integrated into production systems. This is an environment that rewards self-reliance, technical curiosity, and a genuine passion for sports and data.

The Swish Analytics interview process

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

Initial Conversation

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

Hands-on Technical Challenge

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.

Architectural Discussion

You cannot drill a format you do not know, so put the preparation into material that travels. Three pieces of your own work, each rehearsed until you can take a follow-up you did not anticipate, will carry a conversation or a code walkthrough equally well. Specificity is what separates that from filler. A number needs its definition before it means anything: a p99 is over some window and measured at some hop, and a server-side figure excludes the queueing and network time a client would see. The number you cannot qualify is the one to leave out.

What to demonstrate

  • Whether your examples carry detail only someone who did the work would hold, such as what the binding constraint actually was, which alternative you rejected and why it was worse, and what you measured on each side of the change
  • Whether a number survives one follow-up, meaning you can say what it was measured over and whether it moved because of your change or merely alongside it
  • Whether a failure is described with the specific change that followed it, rather than a lesson stated in general terms
  • Whether your part in a team effort is stated accurately, including what other people did

How to prepare

  • Write a page on each of three projects covering the constraint, the option you rejected, the measurement before and after, and what went wrong. Cut any line you cannot take a follow-up on, since you are writing the parts you will be pressed on rather than a summary.
  • Recover the real figures while you still have access: request volume, data size, latency with its percentile and window, team size, timeline. Note where each came from, whether a dashboard, a design document or memory, and mark the estimates so you can say which they are out loud.
  • Take your weakest project story to someone who works in a different area and have them ask why four times in succession. The point where you run out of answer is the part to go and re-read before the round.

Team-fit Discussion

An unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.

What to demonstrate

  • Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
  • Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
  • Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating

How to prepare

  • Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
  • Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
  • Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes

Where candidates lose points

One shared connection pool for every tenant and every query class

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

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.

Treating a network call as though it were a local function call

A remote call can be slow, fail, or return after you stopped waiting, so name the timeout, the retry policy, and what the caller sees while the dependency is down. A call with no timeout turns one slow dependency into an exhausted thread or connection pool in every service upstream of it.

Assuming fixed-width integer arithmetic cannot overflow

In languages with fixed-width integers, including C, C++, Java, Go and Rust, computing a midpoint as (lo + hi) / 2 overflows once the sum passes the type's maximum, so write lo + (hi - lo) / 2 instead. Say which language you are in: arbitrary-precision integers, as in Python or Ruby, remove this specific hazard and none of the others.

Questions & practice

Hold a tenant to a trailing sixty-second request limit

Medium · sliding-window · two-pointer · rate-limiting · tenant-skew

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

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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-After from the exact algorithm without a second scan?

Schedule ordered webhook retries with a heap of subscription queues

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

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

Approach

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

Follow-up

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

Fold a deduplicated usage stream into hourly rollups

Easy · aggregation · deduplication · watermarks · exact-arithmetic

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

Approach

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

Worked solution (25 min)

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

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

Verify it yourself

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

Follow-up

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

Rebuild an hourly rollup with deduplication and late-arrival accounting

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

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

Approach

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

Follow-up

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

Paginate a tenant's delivery export without skipping rows

Easy · keyset pagination · indexing · multi-tenancy

A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.

Approach

  1. Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
  2. Replace the position with a value predicate over a stable, unique, indexed ordering: where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window.
  3. Index (tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit.
  4. Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
  5. Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.

Worked solution (20 min)

  1. Load about 50k deliveries for one tenant, then walk them with the OFFSET query while a writer inserts 10 rows/second, collecting every returned delivery_id.
  2. Compare the distinct ids collected against the set of ids that existed when the walk started, and record the shortfall.
  3. Repeat the walk with the keyset query and confirm every pre-existing id is returned exactly once.
  4. Run explain (analyze, buffers) on page 1 and page 500 of each form and compare shared buffer hits.

Expected result: A keyset query using the row comparison (created_at, delivery_id) < ($2, $3) with order by created_at desc, delivery_id desc limit 100, backed by (tenant_id, created_at, delivery_id) scanned backwards; plus the statement that OFFSET drops rows under concurrent insert and costs time proportional to the offset, and that updated_at is disqualified because it mutates on retry.

Verify it yourself

  • Buffer counts for page 500 are the same order of magnitude as page 1.
  • Every id present before the walk is returned exactly once when only inserts run concurrently.
  • The query plan shows no Sort node above the index scan.
  • Two rows sharing a created_at value to the microsecond are both returned across the page boundary.

Follow-up

  • The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added created_at <= $snapshot bound, and a materialised export table.
  • A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
  • The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?

Describe a secure AWS architecture for hosting a high-availability sports analytics API.

Medium · system design and architecture

Approach

  1. Choose a partition key and say what query it makes expensive.
  2. State the consistency you need, and where you are willing to be stale.
  3. Name the failure you are designing for, then the recovery path.

Follow-up

  • How does this behave when that dependency is down for an hour?
  • What breaks first when traffic grows ten times?

How would you containerize a multi-service Node.js and Python application using Docker for a production deployment?

Medium · system design and architecture

Approach

  1. Name the read and write paths separately; they rarely have the same bottleneck.
  2. Name the failure you are designing for, then the recovery path.
  3. Choose a partition key and say what query it makes expensive.

Follow-up

  • What breaks first when traffic grows ten times?
  • What would you drop to keep the system up under load?

What strategies do you use to troubleshoot and debug performance bottlenecks in a containerized microservices environment?

Medium · system design and architecture

Approach

  1. Name the failure you are designing for, then the recovery path.
  2. Fix the scope first: who calls this, how often, and what they do when it fails.
  3. Name the read and write paths separately; they rarely have the same bottleneck.

Follow-up

  • What breaks first when traffic grows ten times?
  • How does this behave when that dependency is down for an hour?

Explain how you would utilize D3.js alongside React to visualize complex, multi-dimensional sports datasets.

Medium

Approach

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

Follow-up

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

How do you manage application state when dealing with high-frequency data updates from a live sports feed?

Medium

Approach

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

Follow-up

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

An idempotent create endpoint that returns a one-time secret

Easy · idempotency · race conditions · api design

POST /v1/api-keys inserts a tenant_api_key row (tenant_id, workspace_id, name, key_prefix, secret_hash, scopes, status, auth_version, created_at, expires_at) and returns the plaintext secret exactly once, since only its SHA-256 is stored. Write volume is tens per second and clients retry on timeout. Design the idempotency mechanism: what the key is scoped by, where the record lives, its retention, what happens when the same key arrives with a different body, what happens when a retry arrives while the first request is still in flight, and what a replay returns for the secret.

Approach

  1. Scope the key by tenant, not globally: uniqueness is on (tenant_id, idempotency_key), or one tenant's key collides with another's and the second caller receives a stored response for a request it never sent. Store a fingerprint of the request alongside it - method, path and a hash of the canonicalised body - so a mismatch is detectable.
  2. Let the unique constraint decide the race instead of application logic. In the same transaction as the credential insert, run INSERT INTO idempotency_record (tenant_id, key, request_fingerprint, status) VALUES (...) ON CONFLICT DO NOTHING RETURNING id; no returned row means this is a replay, and the existing record is then read. A select-then-insert here loses to itself under concurrency in exactly the way this endpoint is meant to prevent.
  3. Write the three replay outcomes as a decision table rather than as prose: same fingerprint and completed returns the stored response; same fingerprint and still in flight returns 409 with Retry-After, without blocking and without executing; different fingerprint returns 422, because replaying a stored response for a mutated body would tell the caller a credential was created for parameters it never sent.
  4. Handle the secret as the part that makes this endpoint different from an ordinary idempotent create. The plaintext cannot be regenerated from secret_hash, so either the stored response holds it - making the idempotency record a secret at rest whose retention is now the secret's exposure window - or a replay returns the key metadata without the secret and the documentation says a lost response is resolved by listing keys and revoking the orphan. The second is the safer default precisely because credentials are listable and revocable.
  5. Set retention from the client's retry budget, not from a round number: the record must outlive the SDK's maximum total retry duration, so twenty-four hours is defensible if that budget is minutes. After expiry the key is reusable and a very late retry creates a second credential, which is acceptable here only because the object is listable and revocable, and would not be for an unlistable side effect. Expire with a scheduled delete on an index over created_at.

Worked solution (20 min)

  1. Write the DDL for the idempotency record, including the unique constraint that makes the concurrent case impossible rather than unlikely.
  2. Write the four outcomes - fresh, replay-completed, replay-in-flight, fingerprint-mismatch - as a decision table with the HTTP status for each.
  3. Decide what a replay returns for the plaintext secret and write the exact sentence the API reference has to carry about it.
  4. Pick a retention and justify it from the client library's own maximum retry duration rather than from a round number.

Expected result: Unique (tenant_id, idempotency_key); the record inserted in the same transaction as the credential with ON CONFLICT DO NOTHING deciding the winner; 200 with the stored response, 409 for in-flight, 422 for a fingerprint mismatch; a documented rule that the plaintext secret is not replayed and a lost response is resolved by listing and revoking; retention exceeding the SDK retry budget, for example twenty-four hours.

Verify it yourself

  • Two concurrent requests with the same key must leave exactly one row in tenant_api_key. If the logic is select-then-insert, they will not.
  • A retry with a mutated body must not return 200. If it does, the fingerprint is not being compared or is not covering the fields that matter.
  • The retention you chose must be strictly longer than the client library's maximum total retry duration. Look that number up rather than assuming it.

Follow-up

  • Two requests with the same key arrive concurrently. Show the exact statements and say which one loses, and how it finds out.
  • The client receives a timeout, retries, and gets 409 in-flight. What should the client library do next, and for how long?
  • How does the design change if the created object is not listable - a one-off payout, say - so an orphan cannot be found afterwards?

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

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

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.

Day 1 — Fix 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.

Day 2 — One 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.

Day 3 — One 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.

Day 4 — Only 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.

Day 5 — Your 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.

Day 6 — The 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.

Day 7 — Taper

  • 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.

The failure that was yours

Keep one story where the bad call was yours rather than a dependency's or a manager's. Name the check that would have caught it, whether you added that check afterwards, and whether it has fired since. Answers that route blame outward end the conversation early; answers that end in a guardrail someone still relies on tend to open it up.

Tell me about a time when you encountered ambiguous requirements or incomplete documentation on a project. How did you proceed?

Medium · behavioural and engineering judgement

Approach

  1. Pick a story where you made the decision, not one where you watched it.
  2. Name the disagreement and how you resolved it with evidence.
  3. Give the blast radius: what could have broken, and what you measured.

Follow-up

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

Own the incident where invoices undercounted metered usage

Medium · incident response · at-least-once · billing correction · postmortem

A metering consumer acknowledged each batch before committing the fold into usage_rollup_hourly. A rolling deploy restarted consumers mid-batch for two hours; roughly 1.4M usage_event rows were acknowledged and never folded, and 61 invoices sealed against the resulting rollups before anyone noticed. Take the owner's role. Describe an incident of comparable blast radius you owned: how it surfaced, the query that sized the loss, what you stopped first, and how the money was corrected. Give a wall-clock timeline and one thing you got wrong while it was still live.

Approach

  1. Open with the invariant that broke and the direction of the error, because they determine everything else: acknowledging before committing makes the consumer at-most-once, so this loses events rather than duplicating them, and loss raises no error anywhere. A listener who hears 'we lost revenue silently' knows immediately why detection took two hours.
  2. Size it with a stated reconciliation rather than an adjective: sum(quantity) from usage_event grouped by (tenant_id, sku, hour of occurred_at) over the window, against usage_rollup_hourly.quantity_sum on the same keys, filtered to environment='production' because staging and sandbox are metered but not billed. Then bisect by hour and tenant until single cells explain the gap. Say how long that ran and whether a replica could serve it while the incident was live.
  3. Separate mitigation from fix and say which came first. Mitigation is holding the sealing job, because a sealed row is frozen by design and every minute of sealing converts a recoverable rollup into an invoice correction. The fix is moving the acknowledgement after the commit, which re-introduces duplicates that the dedup check on (tenant_id, idempotency_key) must now absorb.
  4. State the correction path in the domain's own terms: sealed periods are never edited, so each affected tenant gets an adjustment line on the next invoice with kind='adjustment' and voided_by_line_id pointing at the line it reverses, priced against the same rate tier and carrying the watermark it priced against. That is four separate numbers — tenants affected, minor units, the cycle the adjustment lands in, and when customers were told.
  5. Close on one prevention control with its cost, not five: a per-hour reconciliation comparing raw sum to rollup sum that pages above a threshold. Name the threshold and the false-page rate you accepted, because a detector nobody will keep staffed is not prevention.
  6. Name a mistake you made inside the response window — the wrong first hypothesis, a mitigation that made it worse — rather than a design mistake from six months earlier. That is the part candidates rehearse away and interviewers weight heavily.

Follow-up

  • Your fix moves the acknowledgement after the commit. What breaks now, and what absorbs it?
  • One undercharged tenant has since churned. Do you bill them, and who decides?
  • How would you have caught this in ten minutes instead of two hours, and what would that detector cost you in pages per week?

Argue against failing open when the control plane is unreachable

Hard · revocation · cache staleness · fail-open vs fail-closed · influence

The gateway caches credential-to-context decisions with a sixty-second TTL. A design proposal says that when control-plane reads fail, pods should keep serving from expired entries indefinitely so a control-plane outage never becomes a product outage. You believe that converts every revocation into an unbounded one. Describe a design you argued against while it was still a live proposal: what you measured or modelled to make the case, what you conceded, who decided, and what happened afterwards. Say what would have changed your mind before the decision, not after it.

Approach

  1. Reframe it from a values argument into a bounded-staleness argument. Both sides already accept the cache; the disagreement is only about the ceiling on how long a revoked credential keeps authorising. Put a number on the table — serve stale for up to fifteen minutes, then fail closed — and make the other side argue against a number rather than against a principle.
  2. Bring arithmetic rather than adjectives: the rate of revocations with revoked_reason in ('suspected_leak','auth_version_bump'), the observed distribution of control-plane unavailability, and the product of the two, which is expected requests served by revoked credentials per outage-hour. At 30k requests/second the unbounded version is not a subtle exposure and the number says so.
  3. Concede the strong half of the opposing case first, because that is what buys you the room: failing closed turns one service's outage into a total outage across three regions, and a control plane doing tens of writes per second is not engineered to the gateway's availability target. A proposal you have not steelmanned reads as reflex.
  4. Propose the asymmetry that usually resolves this: stale entitlements cost bounded money (a quota fifteen minutes out of date over-serves by a computable amount), while a stale revocation costs unbounded access. Split the cached decision by what it authorises, give the two halves different staleness ceilings, and let the entitlement half fail open while the revocation half fails closed.
  5. State the propagation dependency plainly, since it is the part that is missed: validity is also derived from the principal's auth_version, so password reset and sign-out-everywhere flow through this same cache. A design that bounds staleness for explicit revocation and not for auth_version bumps has only fixed half of it.
  6. Say who decided, and what you did afterwards in either outcome: write the decision down with its number and a review date, and instrument the exposure you were worried about so the next round of the argument is settled by data instead of by seniority.

Follow-up

  • Publish-subscribe invalidation is lossy under a partition, and a TTL is the only hard bound. What TTL do you pick, and what does it cost you at 30k requests/second?
  • The key was revoked because it was found in a public repository. Does your answer change, and where does that urgency live in the design?
  • You lost the argument and six weeks later the failure you predicted happens. What do you say in the review, and what do you not say?