Interview guides · Scadea

Scadea Software Engineer Interview Questions & Guide 2026

Testing practice for Scadea software engineer candidates: the boundaries worth covering, and why flakiness is a real defect.

The 60-second brief

A Software Engineer at Scadea plays a pivotal role in delivering robust technical solutions for a diverse range of clients. You are expected to operate as a high-impact contributor who can navigate complex technical landscapes, from core Java development to sophisticated framework architecture. Your work directly influences the reliability and scalability of the software products provided to Scadea’s client base.

Browser-facing seats are not covered by algorithm practice. Scope in state ownership, what the page does on a slow or failed request, and how you would diagnose something that renders correctly but feels slow.

Focus areas

Bound blast radius with per-tenant concurrency limits
Build at-least-once pipelines with explicit deduplication horizons
Make every write idempotent under client retries

Overview

The role requires a blend of deep language expertise and the ability to solve architectural challenges in real-time. Because Scadea operates as a services-oriented firm, you will often find yourself working on varied projects, requiring you to adapt quickly to different team dynamics and technical requirements. This position is ideal for engineers who thrive on technical variety and are motivated by the challenge of solving critical problems in environments where code efficiency and design patterns are paramount.

The Scadea interview process

Scadea candidates report 6 rounds · ≈ 4-6 weeks. The stages below are what candidates describe, not a published process.

Recruiter Screening

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

What to demonstrate

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

How to prepare

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

Aptitude Testing

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.

Live Coding Sessions

Input bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.

What to demonstrate

Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply

How to prepare

For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last

Technical Discussions

What this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.

What to demonstrate

Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly

How to prepare

Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live

Behavioral Fit Evaluation

Your 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 Offer

Coding 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

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.

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

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

Optimising an axis nobody named

Ask which resource is actually scarce here: wall-clock latency, throughput, memory footprint, cost per request, or engineering time. Shaving a constant factor off an in-memory step is wasted effort when the same function makes a blocking remote call inside the loop.

Writing code before the input contract is pinned down

Before the first line, state the types, the size bounds, whether duplicates, negatives or an empty input are possible, whether the input is sorted, whether you may mutate it, and what the function returns when nothing matches. Every one of those answers changes the code, and discovering one at minute twenty costs a rewrite you no longer have time for.

Questions & practice

How would you structure a new project from the ground up?

Medium · data structures and algorithms

Approach

Choose the data structure from the access pattern, not from familiarity.
State the target complexity and say which constraint rules the naive version out.
Name the brute-force solution and its complexity before improving on it.

Follow-up

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

String manipulation techniques using Java 8 features.

Medium · data structures and algorithms

Approach

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

Follow-up

What is the worst case, and how likely is it on real data?
Which test case would catch an off-by-one here?

Explain how you handle concurrency and multithreading in your applications.

Medium · data structures and algorithms

Approach

Walk one small example through your approach before writing the whole thing.
Name the brute-force solution and its complexity before improving on it.
Restate the input: its shape, its size, and what is guaranteed about it.

Follow-up

What is the worst case, and how likely is it on real data?
Which test case would catch an off-by-one here?

Seal an hour under late data with bounded memory

Hard · watermarks · late-data · quantile-sketch · conditional-write

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 on ingested_at, because that is what the fold has consumed and what source_max_ingested_at records. 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 with hour_end + L < W and 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 into usage_rollup_hourly as status='open' with a revision bump. 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, and source_max_ingested_at is 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.

Expected result: The quantity-weighted p99.99 is hours larger than the event-weighted one. Choosing L from the event-weighted number lets roughly the tail's 3% of quantity land after the seal, 600 times the 50 ppm target. With the min watermark and no idle timeout, the stalled partition blocks all sealing; with a 60-second timeout, the stall is sealed past and its events arrive late.

Verify it yourself

Post-seal quantity as a fraction of the hour's total is at or below 50 ppm at the chosen L, measured rather than assumed.
The second sealer's update reports zero rows affected and writes nothing.
Heap size equals the number of open hours and does not grow with the replay's event count.

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?

Model credential revocation so history survives the delete

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

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

Approach

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

Follow-up

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

Explain why the metering dashboard scans every daily partition

Medium · indexing · partition pruning · sargability · tenant skew

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

Approach

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

Worked solution (30 min)

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

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

Verify it yourself

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

Follow-up

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

Write a program to identify the employee with the highest and lowest salary from a list.

Medium

Approach

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

Follow-up

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

How do you resolve circular dependencies in Spring?

Medium

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?

Sealing a billing period against late-arriving usage

Medium · idempotency · immutable ledger · rounding · conditional writes

billing seals a tenant's period once the metering watermark passes the period end, prices the sealed rollups against the plan (included allowance, tier boundaries, negotiated discount) and writes invoice_line_item rows: quantity numeric(28,6), unit_price_micros bigint, amount_minor bigint, currency char(3), source_rollup_watermark timestamptz. The job is re-run after failures and two workers may attempt the same tenant. Events legitimately arrive with occurred_at inside the period and ingested_at after it. Specify the seal transition, the idempotency key for every write including the payment-processor call, the rounding position and mode, and the fate of a post-seal event.

Approach

Make sealing a conditional write rather than a read followed by a write: UPDATE usage_rollup_hourly SET status = 'sealed', sealed_at = now() WHERE tenant_id = $1 AND hour_start >= $2 AND hour_start < $3 AND status = 'open' RETURNING rollup_id. Two concurrent sealers cannot both win because the loser's predicate no longer matches, and the loser learns it lost from an empty result rather than from a lock timeout.
Gate the seal on the watermark, not on the clock - the period closes when the fold is trustworthy past period_end - but cap the wait, because an unsealed period blocks the whole billing run. Seal anyway after a stated maximum (six hours past period end is a reasonable default) and let whatever arrives afterwards become an adjustment. The trade-off is named in one line: adjustments are cheap, a missed billing cycle is not.
Make every write converge under re-run. Line items are idempotent on (invoice_id, sku, rate_tier) with ON CONFLICT DO UPDATE permitted only while the invoice is draft; the payment-processor call carries (tenant_id, billing_period_start) as its idempotency key, because a timeout there is an unknown outcome, not a failed one, and a plain retry is how a customer gets charged twice.
Keep the arithmetic exact and round exactly once. quantity stays numeric, the rate is an integer unit_price_micros in millionths of a minor unit because per-request prices are genuinely below a cent, and amount_minor = round_half_even(quantity x unit_price_micros / 1,000,000) is computed once per line and stored. Tiering walks the boundaries in order emitting one line per (sku, rate_tier) with tier 0 as the included allowance at price zero; the invoice total is the sum of stored amount_minor values and is never recomputed downstream from quantity and rate.
Handle the post-seal event as a forward-only correction. The sealed rollup is frozen, so the difference between the sealed value and the restated value becomes a new invoice_line_item with kind = 'adjustment' and voided_by_line_id pointing at the line it reverses, carrying its own source_rollup_watermark. Nothing is edited in place, because the original line is the only evidence of what the customer was actually charged and is precisely what a dispute, a refund or an audit asks to see.

Worked solution (35 min)

Write the seal as a single conditional UPDATE ... WHERE status = 'open' RETURNING and state exactly what the losing worker sees.
Price one line by hand: quantity 4,318,904.250000 at unit_price_micros 1200 gives 5,182.6851 minor units and 5,183 after a single half-even rounding. Then do quantity 2,500,000.000000 at unit_price_micros 1, which lands exactly on 2.5 and rounds to 2 under half-even but 3 under half-up.
Take four hundred synthetic lines with fractional minor units and compute the total two ways - sum of per-line rounded amounts, and a single rounding of the summed exact amounts - then record the gap.
Trace one event with occurred_at inside the period and ingested_at two days after the seal all the way to the row that eventually reflects it.

Expected result: Seal as a conditional UPDATE from status = 'open' gated on the metering watermark with a bounded maximum wait; line writes idempotent on (invoice_id, sku, rate_tier) and the processor call idempotent on (tenant_id, billing_period_start); amount_minor rounded once per line with half-even; late data becoming an adjustment line that points at the line it reverses and records its own watermark, with no in-place edit anywhere.

Verify it yourself

Re-running the entire billing job on an already-invoiced period must produce identical line rows and must not call the payment processor a second time. If it charges again, the idempotency key was on the retry rather than on the business fact.
Over four hundred lines with uniformly distributed fractional parts the two totals should differ by a handful of minor units - standard deviation around sqrt(400/12), about 5.8, with a worst case of 200. If your two totals match exactly, the test data has no fractional parts and has not exercised the trap.
Every line carries a source_rollup_watermark, and two lines for the same SKU with different watermarks should explain their difference without opening any other table.

Follow-up

Two workers start the same tenant's seal a millisecond apart and the winner crashes after sealing but before writing any line. What does the second worker observe, and is the resulting invoice correct?
Show the divergence between rounding each line and rounding the total once over four hundred lines, and say which direction it goes.
A customer disputes a charge from two quarters ago. Which rows answer it, and which single design decision made that answer possible?

One tenant's counter writes stall the whole connection pool

Medium · lock contention · connection pool · multi-tenancy

A change that made a per-tenant usage counter correct now produces site-wide latency whenever one large tenant writes: unrelated endpoints time out waiting for a connection while database CPU stays low and no statement is slow. The change wraps the counter update in a transaction that takes SELECT ... FOR UPDATE on one row, calls an external pricing service, then updates and commits. Give an ordered checklist, the arithmetic that bounds that tenant's write rate, and three repairs with the cost each one accepts.

Approach

Separate waiting from working. Low database CPU alongside high application latency points at a queue, so instrument connection-acquisition wait separately from query execution time; that queue forms in the application and is invisible in database metrics, which is why the database looks healthy throughout.
Confirm the lock rather than assuming it: sample waiting sessions and group by wait event, relation and tuple. Contention concentrated on one tuple belonging to one tenant is the signature; a deadlock would instead show the database aborting transactions after its detection timeout, which is not happening here.
Do the arithmetic out loud. Throughput on a serialised row is one divided by the lock hold time, and the hold spans the external call, so a 20 ms pricing call caps that tenant near 50 writes per second no matter how many pods run. Every waiter also holds a pooled connection while it queues, so the shared pool drains and unrelated tenants fail at acquisition.
Repair one: shrink the critical section to a single statement with the price resolved before the transaction opens. Cost is a stale price for the duration of one request and a second round trip; benefit is a hold time measured in the database's own execution time.
Repairs two and three change where the contention lives rather than how long it is held. Sharding the counter into per-(tenant, bucket) rows and summing on read multiplies write throughput by the shard count, at the cost of an aggregate on every read and a shard count you must size against the largest tenant rather than the median. Accumulating in memory and flushing periodically removes the per-write round trip entirely, paid for with a bounded loss window on crash, which is acceptable for a rate limiter and not for a billing counter.
Contain independently of which repair wins: a separate pool or per-tenant concurrency cap for this write class, a statement timeout low enough that a pathological query dies before it accumulates waiters, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks.

Follow-up

What would a genuine deadlock look like here, which two code paths would produce one, and how does the database's response differ from what you observed?
If a transaction-pooling proxy sits in front of the database, which of your three repairs changes behaviour, and what stops working that would have worked on a direct connection?
The counter also enforces a quota. Why is SELECT the count and then INSERT still wrong after you have fixed the contention?

Suggested practice checklist

Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.

Day 1 — Coding, one pass at shallow depth

Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
Fix nothing today. The value of the pass is the unfixed record.

Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.

Day 2 — Design, one pass at shallow depth

Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.

Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.

Day 3 — Fundamentals and the practical rounds

Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.

Deliverable: Eight scored short answers and one written reading of unfamiliar code.

Day 4 — The rounds that are about you, and the map

Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.

Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.

Day 5 — First chosen area, to the depth you skipped

Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.

Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.

Day 6 — Second chosen area, where the gap is coverage rather than speed

Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.

Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.

Day 7 — Reassemble the loop

Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
Reduce the week to one page holding only the rules you can state without reading them.

Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.

'We rewrote it in a better framework' is not a story

A migration is a cost you chose to pay, not an achievement. The story is what the old system made expensive, what you measured before committing, what kept serving traffic during the cutover, and what you would have done if the numbers had come back flat. Without those, a rewrite reads as taste.

Describe your experience with migrating projects from Ant to Maven and the common issues encountered.

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
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 did you decide not to do, and why?
What would you do differently if you ran that again?

Unblock an engineer on a job run that finished twice

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

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

Approach

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

Follow-up

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

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

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.
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.
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.
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.
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.
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?