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Arvato Software Engineer Interview Questions & Guide 2026

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

The 60-second brief

As a Software Engineer at Arvato, you are at the intersection of complex logistics, digital transformation, and large-scale enterprise solutions. Arvato operates as a global leader in supply chain management and IT services, meaning your code directly impacts the efficiency of global commerce, customer service infrastructure, and data-driven supply chain operations. You are not just writing software; you are building the digital backbone that keeps complex, international business operations running smoothly.

Allocate preparation against your weakest link rather than your favourite topic. A strong algorithm habit usually comes with weak out-loud explanation of tradeoffs, and years of shipping usually come with rusty from-scratch implementation under a clock.

Focus areas

  • Evolve APIs without breaking pinned SDK clients
  • Make every write idempotent under client retries
  • Scope every query and cache key by tenant

Overview

This role requires a balance of technical precision and pragmatic problem-solving. Whether you are working on Cloud Network Engineering, Logistics automation, or RPA (Robotic Process Automation), you will be expected to translate high-level business requirements into scalable, maintainable technical solutions. You will collaborate with cross-functional teams, including product managers and operations experts, to ensure that the systems you build provide real-world value.

Expect to work in an environment where technical rigor meets business agility. You will find the work particularly rewarding if you enjoy solving real-world constraints, such as optimizing data flow in logistics or ensuring high availability for critical enterprise platforms. Success in this role is defined by your ability to deliver reliable code while maintaining a clear understanding of the broader organizational goals.

The Arvato interview process

Arvato candidates report 5 rounds · ≈ 4-6 weeks. The stages below are what candidates describe, not a published process.

Initial 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

Technical Evaluation

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

HR Interaction

When a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.

What to demonstrate

  • Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
  • Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
  • Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
  • Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened

How to prepare

  • Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
  • Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
  • Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure

Team Interaction

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.

Final Decision

A day like this is several different games in a row, and the expensive mistake is carrying the previous one into the next room. Coding rewards narrow precision and finishing inside a timer. Design rewards breadth, stated assumptions and naming what you are deliberately not building. Behavioural rewards specificity about people and decisions. Candidates who over-engineer a coding problem they were supposed to finish, or who start sketching class hierarchies before anyone has agreed what the system has to do, are usually still playing the last round. Between rooms, name out loud which game the next one is.

What to demonstrate

  • Whether the coding round ends with something that runs and has been traced against a degenerate input, rather than an extensible design that was never finished
  • Whether a design discussion opens by agreeing on traffic shape, read-to-write ratio and what is allowed to be stale, instead of proceeding from an architecture you arrived with
  • Whether a behavioural answer names a person, a disagreement and what you did about it, rather than describing the system the story happened inside
  • Whether the opening habits still appear late in the day: restating the problem, asking for constraints, saying the plan before typing

How to prepare

  • Book three mocks of different types back to back on one afternoon and ask each interviewer afterwards which round you answered in the wrong mode
  • Write a three-line opening script per round type — coding: restate, name the approach and its cost, then type; design: ask for scale, read-write mix and what must not break; behavioural: name the person, the stakes and the decision — and run it off a card so the switch is mechanical rather than remembered
  • Practise coding with a timer you do not extend, stopping when it stops, so the trained reflex is to finish a correct solution rather than to keep improving one

Where candidates lose points

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.

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.

Assuming the bug is in the framework

Suspect your own code first: read the stack trace top to bottom, check which versions are actually installed rather than which ones you believe are, and reproduce in isolation before blaming a library that thousands of people run daily. When the fault really is upstream, you need that minimal reproduction to say so credibly anyway.

Trusting input because it came from your own front end

Anything crossing a trust boundary is hostile: parameterise queries instead of building SQL by concatenation, validate against an allow-list rather than a deny-list, and bound the size of anything you allocate from a request. Raising this unprompted in an API or design question is a cheap and unusually strong signal.

Questions & practice

Write a Java program to accomplish a specific task, including unit tests.

Medium · data structures and algorithms

Approach

  1. Name the brute-force solution and its complexity before improving on it.
  2. Restate the input: its shape, its size, and what is guaranteed about it.
  3. Choose the data structure from the access pattern, not from familiarity.

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?

Can you read through a snippet of third-party code and explain its functionality?

Medium · data structures and algorithms

Approach

  1. Choose the data structure from the access pattern, not from familiarity.
  2. Restate the input: its shape, its size, and what is guaranteed about it.
  3. Name the brute-force solution and its complexity before improving on it.

Follow-up

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

How would you approach building an address formatter that handles different country formats?

Medium · data structures and algorithms

Approach

  1. State the target complexity and say which constraint rules the naive version out.
  2. Choose the data structure from the access pattern, not from familiarity.
  3. Restate the input: its shape, its size, and what is guaranteed about it.

Follow-up

  • Which test case would catch an off-by-one here?
  • How does this change if the input no longer fits in memory?

Design a set of classes for a library application based on a provided set of requirements.

Medium · data structures and algorithms

Approach

  1. Choose the data structure from the access pattern, not from familiarity.
  2. Name the brute-force solution and its complexity before improving on it.
  3. 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?
  • How does this change if the input no longer fits in memory?

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.

Worked solution (35 min)

  1. Reproduce the shape locally: generate 2 million events for one synthetic tenant over 5 days, fold them, then inject three defects, namely 0.2% of events bucketed by ingested_at, a handful of duplicate idempotency_key values whose retries cross midnight, and a block of events ingested after the seal.
  2. Write the watermark-bounded query first and record how much of the gap it explains on its own.
  3. Bisect by day, then SKU, then hour, recording the probe count and the rows each probe touches.
  4. For each located cell, write down the predicted signature before querying it, then check whether the data matches the prediction.
  5. Time a full re-fold of the 2 million rows and extrapolate to 90 million.

Expected result: The watermark-bounded query accounts for the post-seal block entirely and removes it from the investigation. The ingested_at bucketing appears as adjacent hours off by equal and opposite amounts. The midnight-crossing duplicates appear as one (tenant_id, idempotency_key) present in two ingested_day partitions. Bisection reaches each cell in fewer than 70 probes.

Verify it yourself

  • The three injected defects sum to the original gap with no unexplained remainder.
  • Each probe's row count is bounded by one day-partition for one tenant, not by the period.
  • Removing the source_max_ingested_at bound from the first query makes a gap appear, and grow, while ingest runs, even after every defect is removed.

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?

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.

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?

Rebuild an hourly rollup with deduplication and late-arrival accounting

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

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

Approach

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

Worked solution (40 min)

  1. Generate a day of events for one tenant across three workspaces and two SKUs, inject 2% duplicates sharing an idempotency_key with a later ingested_at, and inject 1% whose ingested_at is six hours past occurred_at.
  2. Write the dedup CTE alone and confirm its row count drops by exactly the injected duplicate count.
  3. Aggregate and compare quantity_sum against a ground truth computed outside SQL from the generator's own records.
  4. Add the two window and filtered aggregates, and confirm the final hour's running total equals the ungrouped day total for each (workspace, sku).
  5. Run the whole query twice over static data and diff the two outputs.
  6. Re-run with the session TimeZone set to a non-UTC zone and confirm hour_start does not move.

Expected result: One query: a dedup CTE using row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id); aggregation by (workspace_id, sku, hour_start) producing quantity_sum, event_count and max(ingested_at) plus a filtered late-quantity aggregate; a running total as a window over the aggregated rows; and an explicit statement that the partition-scoped daily unique index cannot catch a duplicate that crosses the ingest-day boundary or arrives in a later replay.

Verify it yourself

  • Row count after dedup equals raw count minus injected duplicates, exactly.
  • The running total at the final hour equals the ungrouped day total for that (workspace, sku).
  • Two consecutive runs over static data return byte-identical rows.
  • Every hour_start is hour-aligned in UTC and unchanged by the session's TimeZone setting.

Follow-up

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

Can you explain specific software patterns you have implemented in past projects?

Medium

Approach

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

Follow-up

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

Can you explain the difference between Java EE and Spring Boot?

Medium

Approach

  1. Clarify what is being asked and what a complete answer contains.
  2. Say what you would check first and why it is the highest-information step.
  3. State your assumptions explicitly before working the problem.

Follow-up

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

Authorisation cache with a bounded revocation window

Hard · caching · invalidation · availability · authz

The edge gateway serves about 30k requests/second from roughly 120 pods across three regions and may add no more than 10 ms at p99. Each request presents an API key that must resolve to an authorisation context: tenant, workspace, scopes, entitlements and credential version. The control plane that owns those rows takes tens of writes/second. A revoked credential must stop authorising within a bound you state as a number. Design the cache - what is keyed, what invalidates it, how many tiers - and specify what the gateway does for the duration of a control-plane outage.

Approach

  1. Fix the entry shape before the topology: key on SHA-256 of the presented secret, value is the resolved context plus the principal's auth_version and a fetched_at. Cache negative lookups too, with a much shorter TTL and a bounded-size structure, because otherwise every sprayed invalid key is a control-plane round trip, and unbounded negative entries let a sprayer evict live ones.
  2. Compute the control-plane read load before choosing a TTL, and notice the multiplier is pods, not regions, when the cache is in-process: distinct_active_keys x pods / TTL. At 50,000 active credentials, 120 pods and a 60 s TTL that is 100,000 reads/second against a single-writer primary with read replicas, which is not serviceable - so the design needs two tiers, a per-region shared cache in front of the control plane with the in-process cache held to a few seconds.
  3. State the bound as the sum of the tiers, not as a hope: with a 10 s in-process TTL over a 60 s regional TTL, worst-case staleness absent any invalidation message is 70 s, because an in-process entry can be filled from a regional entry that was itself about to expire. Publish-subscribe invalidation on every credential and entitlement mutation makes the typical case sub-second, but it is lossy under partition, so the TTL is the only enforced bound and both tiers must subscribe.
  4. Make auth_version propagate through the same path: a password reset or sign-out-everywhere bumps the principal and revokes its keys with no hook of its own, so the invalidation publisher has to expand principal -> credentials and publish per key, or the cache keeps serving keys whose auth_version no longer matches.
  5. Decide the partition behaviour in advance and write it as two rules: on a cache hit past TTL, serve from the stale entry up to a grace ceiling (say 10 minutes); on a cache miss, refuse, because authorising something never seen converts a control-plane outage into an authorisation bypass. Worst-case revoked-key lifetime during an outage is then TTL + grace, about 11 minutes, and that number is the price of not turning a control-plane outage into a total data-plane outage.
  6. Protect the refill path: per-key single-flight so a mass invalidation or a cold pod does not stampede the control plane, TTL jitter so entries created together do not expire together, and a small separately replicated deny-list for compromised keys that is consulted on the hot path and survives control-plane loss.

Worked solution (35 min)

  1. Write the cache entry shape - key, value fields, and which of those fields a request actually reads on the hot path - and mark which field makes a password reset propagate.
  2. Compute control-plane reads/second for TTLs of 10 s, 60 s and 300 s using distinct_keys x cache_instances / TTL, once with cache_instances = 3 regions and once with cache_instances = 120 pods, and note which of the two the in-process design actually implies.
  3. Enumerate the four states a revocation can be in - published and received, published and dropped, control plane unreachable, pod started after the publish - and write which entry serves the next request in each.
  4. Write the outage policy as two rules (hit past TTL within grace: serve; miss: refuse) and compute worst-case revoked-key lifetime as the sum of both tier TTLs plus the grace.

Expected result: A two-tier design with a single stated number: roughly 10 s in-process over 60 s regional, so about 70 s worst-case staleness without invalidation and typically under a second with it, plus a 10-minute stale-serve grace giving about 11 minutes worst case during a control-plane outage; fail-open on a warm hit, fail-closed on a miss; a separately replicated deny-list for compromised keys; per-key single-flight and TTL jitter on the refill path.

Verify it yourself

  • Recompute the read load with pods rather than regions as the multiplier. If the number exceeds what read replicas can serve, a single in-process tier is not a viable design and you need a shared tier, a longer TTL, or routing that gives each pod a slice of the key space.
  • Trace a password reset end to end. If your invalidation publishes only on credential rows, the reset revokes nothing and the auth_version field in the cache entry is decorative.
  • State the outage behaviour as a decision already made rather than as an alert. If the answer is a human paging someone, the availability target for the control plane has not actually been set.

Follow-up

  • A key is found in a public repository and must stop working in seconds, not minutes. What changes, and what does it cost on the request path?
  • One region is partitioned from the control plane while the control plane itself is healthy. What do that region's pods do, and how do you distinguish this from a control-plane outage?
  • How would you measure the actual revocation bound in production rather than asserting it from the configuration?

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

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.

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 is the most challenging work you have done in the past?

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. 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?

Where do you see yourself in five years?

Medium · behavioural and engineering judgement

Approach

  1. Close with what you would do differently, concretely.
  2. Pick a story where you made the decision, not one where you watched it.
  3. Name the disagreement and how you resolved it with evidence.

Follow-up

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

Argue against failing open when the control plane is unreachable

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

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

Approach

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

Follow-up

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