Interview guides · Adzuna

Adzuna Software Engineer Interview Questions & Guide 2026

Take-home style practice for Adzuna software engineer candidates: a small service, a fixed budget, a written tradeoff.

The 60-second brief

A Software Engineer at Adzuna plays a pivotal role in maintaining and evolving a global job search engine that processes massive amounts of data. You will be responsible for building scalable solutions that connect job seekers with opportunities, ensuring that the platform remains performant and accurate. Your work directly impacts how millions of users interact with employment data, making your contributions highly visible and strategically important.

Ask how many rounds there are and what each one is before you plan your weeks, because the composition is what your hours should follow. A loop with three coding rounds and one short design conversation deserves a different split from the reverse, and whoever schedules it will usually say plainly which it is.

Focus areas

Build at-least-once pipelines with explicit deduplication horizons
Make every write idempotent under client retries
Evolve APIs without breaking pinned SDK clients

Overview

The environment at Adzuna is characterized by a focus on practical, efficient engineering. You will be expected to balance technical excellence with the realities of product development, working closely with product managers and other stakeholders to deliver features that drive business value. Whether you are optimizing scraping scripts or refining core search architecture, your engineering decisions will have a measurable impact on the company’s ability to aggregate and present job market insights effectively.

The Adzuna interview process

Adzuna candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Initial Discussions

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

What to demonstrate

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

How to prepare

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

Technical Assessment

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

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.

Paginating a growing table with limit and offset

Two unrelated defects share the idiom. Correctness: rows inserted or deleted between page requests shift the window, so a consumer walking an export skips rows and sees others twice, which for a customer-facing sync is silent data loss rather than an error anyone notices. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N times the page size and a deep page on a large table degrades from milliseconds to seconds. Keyset pagination over a stable, unique, indexed ordering -- where (created_at, id) < ($1, $2) order by created_at desc, id desc limit $3 -- is constant-cost per page and immune to shifting, on the precondition that the cursor columns never change value for a row, which disqualifies updated_at as a cursor.

Saying 'eventually consistent' without naming the anomaly a user would see

Describe the concrete symptom you are choosing to accept: the author reloads and their own comment is missing for two seconds, or two devices show different balances for a minute. The class of consistency model is a technical label; the tolerable anomaly is the actual product decision.

Assuming the input fits in memory

Ask how large the input is in bytes before committing to an in-memory algorithm; beyond that point the options are a single streaming pass, an external sort with bounded buffers, or a sketch that trades exactness for constant memory. An algorithm that assumes random access to the whole input is a different algorithm from one that sees each element once.

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

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

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?

Parse and verify a timestamped multi-signature webhook header

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

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

Approach

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

Worked solution (15 min)

Write the grammar on one line before coding: header := field (',' field)*, field := key '=' value, split on the first = only.
Implement the parser to return {t: int, v1: [hex, ...]}, rejecting a missing t, a duplicate t, any v1 that is not 64 hex characters, and a header over 1 KiB, all before any cryptography runs.
Implement the verifier: for each active secret compute HMAC-SHA256(secret, f'{t}.'.encode() + raw_body), compare it in constant time against each parsed v1, and OR the results with no early exit.
Test with a valid signature; the same body with t moved 400 seconds into the past; the same body with t 400 seconds into the future; a header carrying an unknown v2= alongside a valid v1; and a body re-serialised with different JSON key order.

Expected result: The valid case accepts. Both out-of-tolerance cases reject, including the future one. The unknown v2 field is ignored and the v1 still verifies. The re-serialised body fails, which is correct and is exactly why the raw bytes must be retained.

Verify it yourself

Timing the comparison over 10,000 mismatching signatures that differ at the first character versus the last shows no separation you can measure.
Removing t from the MAC input makes an old captured body with its original signature pass. Write that test, watch it pass, then put t back.
Every rejection except the signature mismatch itself returns before an HMAC is computed.

Follow-up

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

Hold a tenant to a trailing sixty-second request limit

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

The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.

Approach

Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request.
Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate, prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact.
Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (tokens, last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual.
Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.

Follow-up

One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
Quotas rather than rate limits: the check is select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes.
How do you return an accurate Retry-After from the exact algorithm without a second scan?

Decide which facts an invoice line copies instead of joining

Medium · denormalisation · immutable records · foreign keys · rounding

invoice_line_item already denormalises tenant_id, which is reachable through invoice_id, and stores amount_minor even though quantity times unit_price_micros would recompute it. A reviewer asks you to normalise both away, and separately asks whether the tenant's legal name and billing address should be copied onto the invoice header. Decide each case. For every field you keep denormalised, name the read pattern or the invariant that justifies it, the anomaly the copy can develop, and the mechanism that prevents that anomaly here.

Approach

Split the question into two kinds of copy, because they fail differently. A copy of a currently mutable fact is a cache: it drifts and needs invalidation. A copy of a fact frozen at write time is not a cache at all, it is the record of what happened, and normalising it away destroys information the source no longer holds.
Keep tenant_id on the line. It costs 8 bytes, it leads every index on the table so no read is ever accidentally cross-tenant, and it turns a wrong join into an empty result rather than another tenant's money. Prevent the drift structurally: a unique constraint on invoice (invoice_id, tenant_id) plus a composite foreign key from the line on (invoice_id, tenant_id) makes a mismatched pair impossible, so the database enforces agreement instead of a code review.
Keep amount_minor. Rounding must happen exactly once, at a named site, with a stated mode (half-even here). If readers recompute from quantity and unit_price_micros, every reader owns a rounding decision, and half-up and half-even diverge systematically across thousands of lines rather than cancelling out. A check constraint can bound the stored value but deliberately cannot re-derive it.
Copy the legal name and billing address onto the invoice header, written once and never updated. The statement must show what was true when it was sealed, and the tenant record will change afterwards. This is a snapshot for the same reason source_rollup_watermark is stored per line: without it, nobody can reconstruct what the customer was told.
Name the read pattern that pays for all of it. Rendering, dispute response and export are per-tenant, per-period reads over thousands of lines that would otherwise join back to slowly changing dimensions that no longer hold the historical value. The write side is a once-per-period batch, so the extra columns cost nothing that matters.
Concede the case where the reviewer is right: a mutable operational attribute such as the tenant's current plan name has no business on a line. If a report wants it, join. If a statement needs the plan as of the period, that is another snapshot and it belongs on the header with the rest.

Worked solution (25 min)

Write the DDL: unique (invoice_id, tenant_id) on invoice, the composite FK from the line, and a comment on each denormalised column saying whether it is a snapshot or a cache.
Attempt to insert a line whose tenant_id differs from its invoice's and confirm the foreign key rejects it.
Rename a tenant, re-render a sealed invoice, and confirm the rendered name is the one stored on the header.
Recompute amount_minor from quantity times unit_price_micros for a thousand synthetic lines rounding half-up, sum both ways, and record the divergence from the stored half-even values.

Expected result: tenant_id kept and enforced by a composite foreign key to (invoice_id, tenant_id); amount_minor kept because rounding happens once with a stated mode; legal name and billing address snapshotted onto the invoice header and never updated; mutable operational attributes left normalised. Each decision paired with its read pattern, its anomaly, and the mechanism that prevents the anomaly.

Verify it yourself

A line cannot be attached to an invoice belonging to another tenant, and the rejection comes from the database rather than the application.
Re-rendering a sealed invoice after editing the tenant record produces identical output.
Rounding per line and rounding once at the end differ measurably over a thousand lines, which is the argument for storing the value.
No column on the line is recomputed by any reader.

Follow-up

Write the composite foreign key and the unique constraint it requires on the parent. What does it cost on every line insert, and what does it do to a bulk load?
A tenant is renamed after being invoiced. Which rows change, and what does the customer see on last quarter's PDF?
Where does currency live, and what breaks if a tenant's billing currency changes between two periods?

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?

How would you approach scraping specific data from a provided URL?

Medium

Approach

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

Follow-up

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

What development approach did you use in your previous job (e.g., Agile, Waterfall)?

Medium

Approach

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

Follow-up

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

What programming languages and tools have you utilized in your past professional projects?

Medium

Approach

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

Follow-up

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

Design a machine-readable error contract for the gateway

Easy · error taxonomy · retries · api contract

The edge gateway serves roughly 30k requests/second to SDKs and CI pipelines that retry automatically. Today every failure returns 500 with a prose message that clients string-match on. Design the error contract: the response body fields, and the status code for a malformed body, a revoked credential, a scope the credential lacks, a row belonging to another tenant, a reused idempotency key sent with a different body, an exceeded rate limit, and an unreachable dependency. For each, state whether the client may retry and on what schedule. Deliverable: the envelope schema plus the status-to-retry table.

Approach

Split the envelope by audience: a stable code string for programs, a message documented as human-only and free to change, a request_id that joins to gateway logs, and a details array for per-field problems. The code list is an enum that only ever grows.
Assign status by who has to change something: 400/422 for the caller's bytes, 401 for a credential that no longer authenticates, 403 for a scope or entitlement, 404 rather than 403 for a row in another tenant because 403 confirms the identifier exists, 409 for an idempotency conflict, 429 for a limit, 503 for a dependency.
Derive retryability from the method and the idempotency key rather than from the status: a 5xx or a timeout is an unknown outcome, not a failure, so GET/PUT/DELETE may be retried under HTTP semantics and POST only when it carries an idempotency key.
Put the schedule in the response: Retry-After on 429 and 503 overrides the client's own backoff; otherwise capped exponential backoff with full jitter, sleeping uniformly in [0, min(cap, base * 2^attempt)], bounded by a total attempt budget so retries expire before the caller's deadline.
Write the negative rules into the published contract: clients must never parse message, must tolerate unknown code values by falling back to the status class, and a code's meaning is never redefined once shipped.

Worked solution (20 min)

Write the envelope as a JSON schema with four top-level fields and say which are guaranteed present on every error.
Fill a seven-row table: condition, status, code string, retryable yes/no, and the schedule or the reason retrying cannot help.
For each non-retryable row, write the one thing the caller must change (bytes, credential, plan, key) so nothing is marked non-retryable without a remedy.
Add the unknown-outcome row for timeouts and 5xx separately from the other rows, and give it an action other than 'treat as failed'.
Write two sentences of client guidance: honour Retry-After when present, apply full jitter otherwise, and stop at the attempt budget.

Expected result: An envelope of { code, message, request_id, details[] } plus a table in which only 429 and 503 are unconditionally retryable, 5xx and timeouts are retryable only for idempotent methods or keyed POSTs, and every other row names a caller-side change.

Verify it yourself

No row requires the client to read message to decide what to do.
The cross-tenant row returns 404 and the justification mentions existence disclosure.
Timeout appears as its own outcome, distinct from both success and failure.
Retry-After is stated as authoritative over any client-side backoff calculation.

Follow-up

A customer reports they retried a 500 from POST /v1/runs and ended up with two sandboxes billed. Whose bug is it, and what in your contract permits their reading?
You need to add a new error code next quarter without a version bump. What did the v1 contract have to say for that to be non-breaking?

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

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.
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.
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.
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.
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.
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

For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.

Day 1 — Rebuild the primitives by implementing them

Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.

Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.

Day 2 — Arrays under an invariant: two pointers, sliding window, binary search

Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.

Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.

Day 3 — Sorting, heaps, and the greedy argument that has to be proved

Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.

Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.

Day 4 — Recursion, memoisation, and the step to a table

Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.

Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.

Day 5 — Graphs, where most of the work is choosing the traversal

Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.

Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.

Day 6 — One day for everything that is not an algorithm

Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.

Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.

Day 7 — Solve out loud, under time

Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.

Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.

The estimate and the day you knew

A slipped date is only a bad story if you sat on it. What matters is what you believed when you gave the number, the signal that told you it was wrong, how many days passed before you said so, and what you cut rather than asking for more time. Scope you defended counts as much as scope you dropped.

What are your long-term career prospects and goals?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

How do you feel about performing repetitive tasks as part of your development workflow?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
State the situation in two sentences and spend the rest on the reasoning.
Name the disagreement and how you resolved it with evidence.

Follow-up

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

Can you walk us through your past employment history and key contributions?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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