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CPB Contractors Software Engineer Interview Questions & Guide 2026

Software engineer prep for CPB Contractors: design exercises weighted toward the constraints of B2B software and infrastructure work.

The 60-second brief

As a Software Engineer at CPB Contractors, you are stepping into a critical role that bridges the gap between high-level engineering strategy and on-site operational execution. Unlike traditional software roles focused solely on digital products, this position requires you to apply technical rigor, data-driven problem solving, and robust engineering principles to complex infrastructure and construction projects. You will act as the technical backbone for large-scale initiatives, ensuring that systems, processes, and technical workflows are executed with precision.

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

  • Evolve APIs without breaking pinned SDK clients
  • Bound blast radius with per-tenant concurrency limits
  • Make every write idempotent under client retries

Overview

Your impact is measured by your ability to optimize project delivery, maintain uncompromising safety standards, and leverage technical systems to solve the unique challenges inherent in heavy civil, utilities, and building construction. You will collaborate with cross-functional teams, including project managers, site supervisors, and external stakeholders, to translate technical requirements into tangible, successful project outcomes. This role is ideal for engineers who thrive in high-stakes environments where technical proficiency directly influences the success of major infrastructure developments.

The CPB Contractors interview process

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

Initial Screening

The title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.

What to demonstrate

  • Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
  • Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
  • Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year

How to prepare

  • Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
  • Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
  • Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter

Technical Interviews

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

Panel Interviews

Nobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.

What to demonstrate

  • Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
  • Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
  • Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
  • Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience

How to prepare

  • Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
  • For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
  • Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed

Final Selection

Where the day includes a partner from product, design or data, that conversation is weighted like the technical ones and prepared for least. They are deciding one thing: whether having you in the room makes their decisions cheaper. That means options with costs attached, not implementation detail and not "it depends". An estimate someone can plan against — a range, the assumption that would push it to the high end, and what you would drop to hit the low one — is worth more than a confident single number, which everyone present already knows is wrong.

What to demonstrate

  • Whether an estimate comes as a range with the assumption most likely to break it, and states what a specific scope cut would actually buy
  • Whether a technical constraint is handed over as a choice with consequences on their side, rather than as a verdict they have no standing to argue with
  • Whether you establish what decision is on the table before proposing anything
  • Whether risk is raised while it can still change the plan, with the trigger that would confirm it, instead of reported afterwards as a slip

How to prepare

  • Take a project that shipped late and write the two-sentence warning you could have given three weeks earlier, naming what you would have needed decided at that point
  • Rehearse one estimate out loud until it arrives in three parts: the range, the single assumption that would blow it, and the smallest thing you would cut to protect the date
  • Rewrite an objection you have actually made — the "we can't do that" version — as two options with their costs, so the choice ends up with the person who owns it

Where candidates lose points

Treating a timed-out write as a failed write

A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.

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.

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.

Comparing floating-point values for equality, or holding money in them

Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.

Questions & practice

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.

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?

Locate a billing reconciliation gap without rescanning ninety million events

Hard · reconciliation · dimensional-bisection · watermarks · hypothesis-testing

A tenant's sealed invoice total is 0.4% below the sum of its raw usage_event rows for the period. That tenant has 90 million events over 30 days in a table partitioned daily on ingested_at, and its rollups carry source_max_ingested_at, revision and sealed_at. Recomputing all 30 days from raw is correct, and you are not going to do it. Give the procedure that locates the divergent (workspace, sku, hour) cell, the cost of each probe, and the one query you run before any of it.

Approach

  1. Run the free query first. Sum raw quantity for the period restricted to ingested_at <= source_max_ingested_at of the sealed rollups, and compare that against the unrestricted sum. The rollup stores the watermark precisely so this can be answered without a scan. If the whole 0.4% sits above the watermark, nothing is broken: it is late data, it becomes an adjustment line, and the investigation ends in one query.
  2. Only if the gap survives that test do you bisect, and you bisect by dimension rather than by rows. Compare 30 per-day totals, then inside the offending day compare the 6 SKUs, then the workspaces, then the 24 hours. That is roughly 30 + 6 + W + 24 grouped probes, each an indexed range scan over one daily partition for one tenant, against O(N) per attempt for the naive re-fold.
  3. Quantify why naive is not merely slow but unusable mid-incident: at a generous 200,000 rows/second sequential, 90 million rows is about 7.5 minutes per attempt, you will want ten attempts, and every one competes for I/O on the same partitions live ingest is writing. The diagnostic worsens the backlog it is diagnosing.
  4. Before fetching each comparison, state what it would look like under each hypothesis. Two adjacent hours off by equal and opposite amounts is occurred_at versus ingested_at bucketing. A whole day offset by exactly N hours is a timezone applied at the wrong layer. A gap confined to one SKU in one workspace is an environment filter. The same (tenant_id, idempotency_key) present in two ingested_day partitions is the dedup horizon losing a retry that crossed midnight.
  5. Make the next bisection cheap by storing the aggregate you keep recomputing. A per-(tenant_id, ingested_day) count and quantity checksum turns step two from thirty probes into one read, and it is the same number the reconciliation job already produces.
  6. Whatever you find, the sealed period does not change value. The correction is an adjustment line pointing at the line it reverses, carrying its own source_rollup_watermark, because the original invoice is the evidence of what the customer was charged.

Follow-up

  • The gap is 0.4% in one direction on one day and 0.4% the other way the next day. What does that shape rule in, and what does it rule out?
  • How do you distinguish a duplicate from a restatement, given revision and recomputed_at on the rollup?
  • Ingest is still running while you investigate. What makes your two numbers comparable at all?

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

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

  1. 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.
  2. Compute the p99.99 lateness two ways, event-weighted and quantity-weighted, and put the two numbers side by side.
  3. 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.
  4. Set the idle-partition timeout to 60 seconds, repeat the stall, and measure how much quantity arrives after the seal.
  5. 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?

Find the join that inflates every invoice total

Medium · join fan-out · aggregation grain · sql correctness

invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.

Approach

  1. Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row: sum(l.amount_minor) is the true total multiplied by the attempt count, and count(p.attempt_id) is attempts times lines. Three lines and two attempts report double the money and six attempts.
  2. Reject the reflex repair. count(distinct p.attempt_id) does fix the count, because attempt_id is unique. sum(distinct l.amount_minor) does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match.
  3. Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
  4. Keep invoices with no attempts by making the attempt branch a LEFT JOIN with coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about.
  5. Push each filter to its own grain: where l.voided_at is null belongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants.
  6. Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.

Worked solution (25 min)

  1. Create one invoice with three lines of 1000, 1000 and 500 minor units and two payment attempts, then run the original query.
  2. Confirm it reports 5000 and 6 rather than 2500 and 2.
  3. Apply sum(distinct l.amount_minor) and confirm the total becomes 1500, which is worse rather than better.
  4. Write the two-CTE version with a LEFT JOIN and coalesce, and confirm 2500 and 2.
  5. Add a second invoice with lines and no attempts and confirm it still appears.

Expected result: The original sum equals the true line total multiplied by the number of payment attempts and the count equals attempts times lines; the corrected query pre-aggregates each branch to invoice grain in separate CTEs and LEFT JOINs them with coalesce, with voided lines excluded inside the line CTE and the tenant predicate applied to both branches.

Verify it yourself

  • The money total is unchanged by adding or removing payment attempts.
  • An invoice with zero attempts appears with its total and a zero count.
  • Two lines with identical amounts both contribute to the total.
  • Voiding a line changes the money figure and leaves the attempt count alone.

Follow-up

  • Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with filter (where ...) be better?
  • Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
  • The same report is needed per tenant per month. What index makes the line CTE cheap?

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.

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?

How do you approach the integration of new technical systems within an existing project framework?

Medium

Approach

  1. State your assumptions explicitly before working the problem.
  2. Clarify what is being asked and what a complete answer contains.
  3. Work from the requirement backwards to the design.

Follow-up

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

Give an example of how you have used data or technical analysis to improve a project process.

Medium

Approach

  1. Work from the requirement backwards to the design.
  2. State your assumptions explicitly before working the problem.
  3. Clarify what is being asked and what a complete answer contains.

Follow-up

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

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?

Hourly rollups merge one hour and lose another

Easy · timezone · rollups · reconciliation

Reconciliation flags one tenant on one day. Summing usage_event.quantity by hour of occurred_at gives 24 non-empty hours, but usage_rollup_hourly holds 23 rows for that tenant, workspace and SKU, one of which carries roughly the sum of two adjacent hours. Other days reconcile exactly, and the affected date matches a civil-time transition. hour_start is documented as truncated to the hour in UTC. You have both tables, the rollup job source, and its runtime environment. Give an ordered checklist, the mechanism, and the correction path for a day that may already be sealed.

Approach

  1. Bisect by dimension until one cell explains the whole difference: tenant, then day, then SKU, then hour. A defect confined to a single transition date already rules out deduplication and late arrival, both of which are indifferent to which hour an event lands in.
  2. Read the truncation with its precondition stated: date_trunc on a timestamptz value is evaluated in the session TimeZone, not in UTC. If the job connects without pinning that setting, it inherits the server or container default.
  3. Follow that to the collision: in a zone that observes daylight saving, two distinct UTC hours map to the same local wall-clock label at the autumn transition, so both fold into one key under the unique constraint on (tenant_id, workspace_id, sku, hour_start) and their quantities sum into one row. At the spring transition a label never occurs and the row is simply absent.
  4. Confirm from data rather than from reading code: run the same aggregate twice, once with the session pinned to UTC and once with the job host zone, and check that the second reproduces the stored rollup exactly.
  5. Fix at the source by pinning the connection to UTC explicitly, or by truncating on occurred_at AT TIME ZONE 'UTC', rather than relying on a default that differs between a developer machine, CI and production.
  6. Correct according to status, not convenience: an open hour is recomputed with revision incremented, a sealed hour is frozen and the difference becomes an adjustment line on the next invoice with voided_by_line_id pointing at the line it reverses.

Follow-up

  • The same job also emits a daily figure for a dashboard. Why can a correct hourly rollup still produce a wrong day, and what does the tenant's billing timezone have to do with it?
  • How would you detect this class automatically rather than waiting for reconciliation, given that it only manifests twice a year per zone?

Suggested practice checklist

Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.

Day 1 — Diagnostic, scored before you study anything

  • Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
  • Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
  • Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.

Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.

Day 2 — Largest gap: find the boundary rather than the subject

  • Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
  • Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
  • Re-attempt one of them from blank four hours later with nothing open.

Deliverable: A sub-skill map with the two blocking sub-skills circled.

Day 3 — Drill the blocking sub-skill by repeating the shape

  • Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
  • State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
  • Have someone else read your one-sentence rule and find the precondition you left out.

Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.

Day 4 — Second gap, plus maintenance on the strongest area

  • Run the same sub-skill decomposition on the second-largest gap in half the time.
  • Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
  • Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.

Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.

Day 5 — The gap that is not a skill

  • Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
  • Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
  • Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.

Deliverable: Two recordings with a counted reduction in time-to-first-question.

Day 6 — Retest under day-one conditions

  • Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
  • For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
  • Write down which single block you would still lose the offer on.

Deliverable: A second scored rubric placed beside the first, with one named remaining risk.

Day 7 — Full loop under interview conditions

  • Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
  • Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
  • Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.

Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.

What the behavioural round is actually measuring

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

Describe a situation where you had to influence a decision based on your technical assessment.

Medium · behavioural and engineering judgement

Approach

  1. Pick a story where you made the decision, not one where you watched it.
  2. State the situation in two sentences and spend the rest on the reasoning.
  3. Close with what you would do differently, concretely.

Follow-up

  • What did you decide not to do, and why?
  • How did you know your change caused the improvement?

How do you manage relationships with subcontractors and suppliers to ensure technical standards are met?

Medium · behavioural and engineering judgement

Approach

  1. Close with what you would do differently, concretely.
  2. State the situation in two sentences and spend the rest on the reasoning.
  3. Give the blast radius: what could have broken, and what you measured.

Follow-up

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

How do you communicate complex technical information to non-technical stakeholders or site teams?

Medium · behavioural and engineering judgement

Approach

  1. State the situation in two sentences and spend the rest on the reasoning.
  2. Give the blast radius: what could have broken, and what you measured.
  3. Close with what you would do differently, concretely.

Follow-up

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

How do you handle a situation where a design or technical requirement is not feasible on-site?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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