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Agropur Software Engineer Interview Questions & Guide 2026

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

The 60-second brief

A Software Engineer at Agropur plays a vital role in bridging the gap between cutting-edge technology and the complex requirements of the dairy industry. You are not just writing code; you are building solutions that optimize production, streamline supply chain operations, and support the large-scale manufacturing processes that define the company’s success. Your work directly impacts how Agropur manages its resources, from raw material intake to the final distribution of products across its various plants.

The loop does not sample the job evenly, and arguing about that in the room costs you. Daily work is mostly incremental change inside code someone else wrote, while the loop samples narrow slices of it; prepare for the slices and save the realism argument for your questions at the end.

Focus areas

Reconcile a movement ledger against derived positions
Make every external side effect safe to retry
Fold late, out-of-order scan events by event time

Overview

This role requires a balance of technical proficiency and practical problem-solving. You will often find yourself collaborating with cross-functional teams, including plant operations and management, to turn business requirements into functional, scalable software. Whether you are automating packaging consolidation or improving system efficiencies, your contributions are critical to maintaining Agropur’s operational excellence in a competitive and fast-paced sector.

While the role is technical, Agropur places a significant emphasis on how you integrate into a team environment. Be prepared to discuss your ability to communicate complex ideas to non-technical stakeholders in plant or management roles.

The Agropur interview process

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

Initial Screening

The person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.

What to demonstrate

Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural

How to prepare

Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one

Behavioral Interview

This round is deciding whether a change you make without supervision can be allowed to reach production. It is scored on what you knew at the moment you decided, not on how it turned out, so a story that opens with the result and works backwards reads as luck retold as judgement. Say what the options were, what you did not know, what you did to shrink the unknown before committing, and what you accepted as the worst plausible case. The detail that separates answers is a bound: how many users, how much data, and for how long, if you had been wrong.

What to demonstrate

Whether the reasoning you give was available at the time you decided rather than after the result came in, since a story whose deciding evidence arrived later describes an outcome and not a judgement
Whether you can put units on the exposure (users, rows, minutes of degraded service) and whether the containment you chose actually bounded it: a canary bounds the request path it fronts, while a background job writing to a shared table reaches every user regardless of which version served their requests
Whether the reversal path existed before you shipped or was improvised during the incident, and whether it restores state or only stops further damage

How to prepare

For your three largest changes, write down the one thing you would have had to be wrong about for it to fail, and what your best estimate of it was on the day you shipped. If you never held an estimate, that is the gap the follow-up questions will find
Write the undo procedure for one of those changes as it existed at the time, then mark which steps restore data and which only stop new damage. Turning a flag off or reverting a deploy ends the new writes; rows already written come back only from a copy you kept, and a dropped column comes back empty unless something outside the schema holds the values
Rehearse one story from the decision point forward and stop before the outcome, then have someone ask what you would do next. If the story only works with the ending attached, it is an anecdote rather than a decision you can defend

Technical Assessment

The same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.

What to demonstrate

In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes

How to prepare

Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first

Feedback and Team Fit

Part of what is being decided is what your manager's week looks like once you are on the team: how they find out that something you shipped is broken, and whether they hear it from you or from a customer. The material that moves this round is therefore not the launch, it is the week after. Describe how you knew the change was working, which number you watched and for how long, and what you would have reverted to. A project that ends at the word shipped leaves the manager guessing about the part they care about most.

What to demonstrate

How a change of yours was verified in production, and whether that check existed before the deploy or was assembled afterwards once something looked wrong
Whether the rollback story is specific, including the case where a revert is not enough: once a migration has dropped a column or rewritten data, going back is its own change with its own risk
Whether you can describe an incident you caused, how it was found, and how much time passed between it starting and anyone noticing
Whether bad news in your account travels early and from you, or consistently arrives via somebody else

How to prepare

For your last significant change, write down what you watched after the deploy, for how long, and the number that would have made you revert. If nothing was watched, say that plainly instead of inventing a dashboard.
Work out the rollback answer for a change that touched stored data, and be able to say what made it reversible or what you would have had to do instead. That distinction is a level signal on its own.
Write the two-minute version of an incident you owned: what broke, how it surfaced, what you did in the first ten minutes, and the change that stopped it recurring. Get the timeline straight, because this is the story most likely to be interrupted with questions.

Where candidates lose points

Judging delivery windows and cut-offs in UTC

A delivery appointment, a receiving cut-off and a same-day promise are all statements about local wall-clock time at a specific stop, so evaluating them in UTC misclassifies on-time performance for any leg that crosses a zone, and misses cut-offs by an hour on the two days a year a region changes offset. The subtlety that catches experienced engineers is that converting a future local time to a UTC instant at write time bakes in the offset rules as they were known then: when a jurisdiction changes its rules, an appointment stored that way silently moves, which is why a future window is better stored as local wall time plus an IANA zone identifier and resolved at read time, while a past event is correctly stored as an instant. Duration arithmetic needs the same care, because a transit that spans a DST boundary is genuinely 23 or 25 hours long in local terms even though the elapsed instant count is unchanged. Storing only a date with no zone - common on partner feeds - is the version of this that cannot be fixed downstream, so the zone has to be captured at ingestion.

Treating the database as the truth about physical stock

Every system in this domain drifts from reality, through miscounts, unrecorded damage, mis-scans and theft, so a design with no adjustment path forces the drift somewhere worse. The usual shape is a CHECK constraint or application guard that clamps a position at zero: the negative that would have exposed a missing receipt is silently swallowed, and the first visible symptom is a picker standing at an empty location for an order the system promised. The correct posture is to let the invariant fail loudly - record the cycle-count adjustment as a first-class movement with its own reason code, surface the variance, and let planning see that this location is unreliable - because an adjustment that is indistinguishable from a normal pick destroys the only signal that would have found the root cause. A related version of the same mistake is deleting or updating a movement row to fix a mistake, which leaves the position correct and the history a lie.

Finishing a solution without stating its complexity

Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.

Listing technologies instead of trade-offs

Name the property the design needs first, such as ordered range scans, multi-entity transactions, cheap appends, or a predictable p99, then pick something that provides it and say what it gives up in exchange. Almost any component is defensible once you state the requirement it satisfies and the one it sacrifices.

Questions & practice

Parse partner quantity and date segments without floating point

Easy · parsing · unit-of-measure · time zones

A partner sends a delimited flat file whose segments look like QTY+17:4.500:CS and DTM+232:20260317. Quantities carry up to three decimal places in a named unit; each item has an integer pack factor to eaches that is versioned by effective date. Produce observation_event rows with an integer quantity in eaches, the unit recorded, and occurred_at resolved, rejecting any line whose conversion is not exact. Files reach 4 GB. Single pass, no floating point. State your bounds and what you do with the zoneless date.

Approach

Scan with indices, not a regex. Find the segment terminator, then split the payload on : by scanning forward and slicing — O(bytes) with no backtracking and no per-line allocation beyond the output row. A regex with nested quantifiers over 4 GB is where this job silently becomes CPU-bound.
Parse the decimal as an integer mantissa plus a scale. 4.500 becomes (mantissa = 4500, scale = 3): accumulate digits into an int64, count digits after the point, and never call the runtime's float parser. Reject a mantissa that would overflow int64 when multiplied by the pack factor, checking before the multiply rather than after.
Convert by integer arithmetic and demand exactness: numerator = mantissa * factor; emit numerator / 10^scale only when numerator % 10^scale == 0, otherwise reject the line with a reason code. Worked: 4.500 CS at 12 eaches per case gives 54000/1000 = 54, exact. 0.125 PL at 36 eaches per pallet gives 4500/1000 = 4.5, which is half a sealed unit and must fail at the boundary, not round three services away.
Select the pack factor by the file's effective date, not by the current item master. Factors are properties of a point in time — a supplier moving a case from 12 to 10 must not reinterpret last quarter's file — so the lookup is a range query on the factor's validity interval.
20260317 carries no zone and no time. Attach the IANA zone of the node the segment refers to, captured at ingestion, because it cannot be recovered downstream; then pick and document one convention for a date-only field (start of local day is the usual choice) and store the raw local string beside the resolved instant so the assumption stays visible. Complexity: O(total bytes) time, O(longest line) working space, one pass.

Follow-up

The same file is re-sent in full tomorrow with three lines changed. What makes the second ingest a no-op for the unchanged lines?
A line names a unit that is not in the ladder for that item. Reject the line, reject the file, or quarantine it — and who finds out?
The partner starts sending an offset like +0100 on some dates. Does that change your storage decision for a future appointment as well as a past event?

Find the busiest ten-minute window per reader after deduplication

Medium · sliding window · event time · deduplication

One day of observation_event from fixed readers: source_id (reader), subject_id (container), occurred_at, received_at, dedupe_key. 300 million rows, partner and device replays resend whole windows, and some events arrive six hours after they occurred. For each reader, return the 10-minute event-time window containing the most distinct containers, with the window's start. Target linear work per reader after sorting. State the window semantics you chose, and what the job does with an event that lands after its window was already reported.

Approach

Deduplicate before windowing, on dedupe_key, and keep the two concerns separate. A replayed batch is a no-op only if the duplicate is removed first; deduplicating inside the window logic counts a resend as traffic and reports a congestion peak that never happened. Dedupe is a hash set per reader-day or a sort-unique on the key — the same sort you need anyway.
Partition by source_id and sort each partition by occurred_at, never by received_at. Arrival order is an artefact of device connectivity and partner batch cadence; a reader that went offline for six hours would otherwise show its entire backlog as one impossible burst. Sorting dominates: O(p log p) per reader of size p, O(n log n) overall.
Two pointers over the sorted partition with a hash multiset container -> count and a separate distinct integer. Advance the right pointer, incrementing the count and bumping distinct only on a 0-to-1 transition; advance the left pointer while occurred_at[left] <= occurred_at[right] - 600s, decrementing and lowering distinct only on a 1-to-0 transition, and erasing the key from the map so the map's size and distinct never diverge. Each element enters and leaves once: O(p) amortised after the sort, O(window occupancy) space.
Justify the candidate set rather than scanning every possible start. With half-open windows [t, t+600), sliding a window right until its left edge coincides with an event can only add events and never removes one, so the maximum is attained at some event's occurred_at; checking the p windows anchored at events is sufficient and exhaustive. State the half-open convention explicitly — an event exactly at t+600 belongs to the next window, and flip-flopping on that produces off-by-one disagreements between this job and whatever reads its output.
Handle lateness by choosing a policy, not by hoping. Either run over a closed event-time batch with a stated lateness allowance — the day plus six hours, so the six-hour tail is inside the batch and the day-late tail is not — or emit under a watermark and restate the affected windows when a late event lands. Both are defensible; what is not defensible is silently dropping the late event, because that is how the busiest window quietly becomes the one with the best connectivity.

Worked solution (30 min)

Write the window semantics as one sentence — half-open [t, t+600), anchored at event timestamps — at the top of the function, and make the comparisons match it.
Implement dedupe, sort, then the two-pointer scan with the multiset and the explicit distinct counter.
Build a fixture for one reader: twelve events, one container seen three times inside a window, one duplicate dedupe_key, and one event at exactly t+600.
Hand-compute the expected maximum and window start, then run.
Re-run with the fixture shuffled and with the duplicate resent five more times, confirming the answer is unchanged.

Expected result: The container seen three times contributes 1 to the distinct count; the resent duplicates contribute 0; the event at exactly t+600 falls in the next window; the reported start is an event timestamp, not a rounded clock boundary.

Verify it yourself

A brute-force O(p^2) scan over all event-anchored windows agrees on a 1,000-event fixture.
The distinct counter equals the multiset's key count at every step — assert it in the loop while testing.
Sorting by received_at instead of occurred_at changes the answer on a fixture with a backlogged device, which is the test that proves event time is doing work.

Follow-up

Two events for one container arrive from two readers a second apart. Is that congestion, a misread, or one container passing two portals?
You need this continuously rather than as a daily batch. What changes, and where does the memory go?
A device's clock is three hours fast. What does that do to this result, and how would you detect it from clock_offset_ms?

Find the hottest write keys under a fixed memory budget

Medium · top-k · heap · streaming sketches

One day of writes to inventory_movement is 200 million rows; each carries (item_id, node_id). Distinct pairs reach 60 million. Return the 200 pairs with the most writes so they can be moved behind a single writer before the next peak. First give the exact answer assuming you may hold 60 million counters, then give an answer under a 200 MB budget, and state the error guarantee that second answer carries. Both must be one pass over the day's rows unless you argue for a second.

Approach

Exact version: one hash pass to build (item_id, node_id) -> int64, then a single scan of the map maintaining a min-heap of size K = 200, pushing and popping when the incoming count exceeds the heap root. O(n + d log K) with n = 2e8 and d = 6e7, versus O(d log d) if you sort all 60 million entries — a factor of roughly log d / log K, about 3.4x, for no benefit. The heap must be a min-heap: the root is the weakest survivor, so eviction is O(log K).
Cost the map honestly. 60 million entries at a 16-byte key, an int64 count and per-entry overhead is 2-3 GB in most runtimes, which is the whole reason for the second version rather than a theoretical aside.
Bounded version: Misra-Gries with m counters. Keep at most m keys with counts; on a key already present increment it; on a new key when fewer than m are held, insert at 1; otherwise decrement every held counter and drop those reaching zero. With m = 1,000,000 at roughly 40 bytes per counter, that is about 40 MB, comfortably inside 200 MB alongside the heap.
State the guarantee as a number, not as 'approximate'. Misra-Gries returns an estimate satisfying f - n/(m+1) <= f_hat <= f, so with n = 2e8 and m = 1e6 every count is an undercount by at most 200. If the 200th hottest pair has hundreds of thousands of writes, a 200-count error cannot reorder the top of the list — and if the counts near rank 200 are within 200 of each other, the ranking there was never meaningful for the decision being made. The sketch also merges: run one per shard, combine, prune back to m, and the bound holds with n the combined length.
If exactness is genuinely required, argue for two passes rather than more memory: the Misra-Gries pass yields a candidate superset of size m, and a second pass counts only those candidates exactly in O(m) memory. That is the standard way to get an exact top-K under a memory bound, and the cost is one extra read of the day's rows.

Follow-up

The decision downstream is which keys get a single writer. Does a rank error at position 190 change that decision, and what does the answer tell you about how much accuracy to buy?
Write volume on one key is bursty — a promotion concentrates it into twenty minutes. Does a daily top-K find it, and what window would?
How do you keep the sketch across a worker restart without replaying the day?

Fold late, out-of-order scan events into one status per leg

Medium · window functions · event time · monotonic fold

observation_event holds event_id, subject_type, subject_id, observed_status, occurred_at, received_at, dedupe_key (UNIQUE) and processing_status. A lookup table maps observed_status to status_rank SMALLINT along a progress lattice. Write one query returning the current status of every shipment leg that received an event in the last hour, such that a picked_up event arriving after a delivered event never changes the answer. Then explain exactly which part of your ordering makes a shuffled replay of the same batch produce identical output.

Approach

Separate the two jobs. Deduplication is the UNIQUE on dedupe_key at ingest, because partner feeds resend whole windows; ordering is the fold. Either alone leaves a defect: dedupe without a lattice still lets a stale event win, and a lattice without dedupe double-counts a replayed window in anything that aggregates.
Select the affected legs in one CTE by received_at >= now() - interval '1 hour', then fold over every event for those legs, not only the recent ones. Restricting the fold's input to the recent window is the failure: a lone late picked_up would be the only row considered and the leg would regress.
Fold as a maximum over the lattice: SELECT DISTINCT ON (subject_id) subject_id, observed_status FROM ... ORDER BY subject_id, status_rank DESC, occurred_at DESC, event_id DESC. The equivalent form is ROW_NUMBER() OVER (PARTITION BY subject_id ORDER BY status_rank DESC, occurred_at DESC, event_id DESC) = 1.
The determinism comes from the final event_id tiebreak. Rank and occurred_at alone leave two equal-rank events at the same instant to be resolved by heap order, which changes when rows are re-inserted, so the shuffled replay differs in exactly the cases nobody tests. With a total order over the tuple the fold is a max, which is commutative and associative and therefore order-insensitive.
Do not advance a leg on occurred_at alone. The timestamp is device-reported and can be wrong, which is why the lattice rank leads the ORDER BY and why clock_offset_ms is worth storing: a future-dated event with a low rank then cannot promote itself past a delivered.
Keep superseded rows and mark processing_status = 'superseded' rather than deleting them. The discarded event is usually the one that explains a delivery dispute, and deleting it makes the trace unanswerable.

Follow-up

An event arrives a week late for a leg that is already invoiced. Does the fold apply it, reject it as late, or apply it and emit a correction?
The stream is partitioned by source_id rather than by subject_id. What ordering guarantee do you actually have, and what breaks?
How would you materialise this fold so the customer-facing status read is not running a window function per request?

Model the movement ledger so a retried write is a no-op

Easy · append-only ledger · idempotency · audit history

inventory_movement is append-only: movement_id, item_id, node_id, lot_id, state, delta_qty (signed BIGINT), uom_code, reason_code, ref_type, ref_id, idempotency_key, occurred_at, recorded_at, reversed_by. inventory_position is keyed (item_id, node_id, lot_id, state) with qty, version, last_movement_id. Give the DDL constraints and the exact statements a pick confirmation runs so that a client which times out after the commit and retries changes nothing. Then say how you correct a pick recorded against the wrong lot two days ago, and why the position carries CHECK (qty >= 0) rather than clamping at zero.

Approach

Put the guarantee in the schema, not the handler: UNIQUE (idempotency_key) on inventory_movement. A SELECT-then-INSERT cannot work, because two concurrent retries both read nothing before either commits and both then insert.
Claim with INSERT ... ON CONFLICT (idempotency_key) DO NOTHING RETURNING movement_id, and branch on rowcount. Zero rows means the first attempt already committed, so the handler returns success without touching the position. This branch is the whole point: applying the position update on the conflict path is what makes the ledger and the shelf disagree by one pick forever.
Keep the insert and the position update in one transaction: UPDATE inventory_position SET qty = qty + :delta, version = version + 1, last_movement_id = :id WHERE item_id = :i AND node_id = :n AND lot_id = :l AND state = :s AND version = :v, then test the rowcount and retry the transaction on zero.
Use lot_id BIGINT NOT NULL DEFAULT 0 rather than NULL. In a default unique index NULLs are distinct from each other, so a NULL lot would let two position rows exist for the same untracked item and split the running sum in half.
Correct the mis-keyed pick with two compensating movements, never an UPDATE or a deleted_at flag: one reversal restoring the wrong lot, one fresh pick against the right lot, both with reason_code = 'reversal' or 'pick' and ref pointing at the original movement_id. The original row keeps its values; reversed_by is a back-pointer written in the same transaction so a reader does not need a reverse scan to see that the row was undone.
CHECK (qty >= 0) aborts the transaction and surfaces a real defect. GREATEST(qty + :delta, 0) writes a plausible number instead, so the missing receipt that caused the negative is never investigated and the first symptom is a picker at an empty location.

Worked solution (20 min)

Write the two CREATE TABLE statements including UNIQUE (idempotency_key), the composite primary key on the position, CHECK (qty >= 0) and lot_id NOT NULL DEFAULT 0.
Write the handler as one transaction: the ON CONFLICT DO NOTHING RETURNING insert, the rowcount branch, then the versioned conditional UPDATE with its own rowcount test.
Send the same pick twice with one idempotency key and assert one movement row and a position that moved by delta once, not twice.
Apply the lot correction as two new movements and show that summing delta_qty per lot now matches the shelf while both the wrong and the right lot retain a legible history.

Expected result: One movement row per idempotency key, a position equal to the signed sum of its movements, and a corrected pick represented as three rows (original, reversal, re-pick) with no row ever updated except reversed_by and the position aggregate.

Verify it yourself

SUM(delta_qty) per (item_id, node_id, lot_id, state) equals inventory_position.qty after the double-send and after the correction.
Drop the UNIQUE constraint and confirm the double-send test now fails; if it still passes, the retry is not actually racing.
Force a negative by replaying a pick against an empty position and confirm the transaction aborts rather than storing zero.
Query the history of the mis-keyed pick and confirm the original quantity and its reason code are still readable.

Follow-up

Two retries of the same pick arrive concurrently on different application instances. Walk through both transactions statement by statement and say which one commits a position change.
occurred_at from a handheld is five minutes ahead of recorded_at because the device clock drifted. Which of your queries break, and which column should each of them actually be using?
How long do you keep idempotency_key unique for, given the table grows by tens of millions of rows a month?

How would you approach getting seven plants to consolidate packaging?

Medium

Approach

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

Follow-up

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

How would you go about dealing with a superior who, because of your lack of seniority, will not listen to your opinions?

Medium

Approach

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

Follow-up

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

Specify the idempotency key contract for allocation requests

Medium · idempotency · concurrency · api contract

The Fulfilment Orchestrator calls POST /allocations with order_line_id, item_id, node_id, lot_id, qty, uom_code, tier and expires_at, and retries from a durable saga step after any timeout. Specify the Idempotency-Key contract: key ownership and derivation, key scope and lifetime, what is fingerprinted, the response when a retry arrives while the first request is still in flight, and what a replay returns given that allocation rows move through held, committed, consumed, released and expired. Relate it to the partial index UNIQUE (order_line_id, node_id, lot_id) WHERE status IN (held, committed).

Approach

Scope the key to (client_id, endpoint, key) and insert the key row in the same transaction as the allocation, with a unique constraint doing the work. A read-then-write check against a cache is not idempotency - it is a race with a smaller window.
Fingerprint the semantically significant fields (canonical serialisation of item, node, lot, qty, uom_code, tier) and store the hash. Same key with a different fingerprint is a caller defect, not a retry: answer 422 with its own code, following the convention in the IETF Idempotency-Key header draft, and change nothing.
Define the in-flight case rather than letting the database decide. A second request arriving while the first holds the key row either waits on the lock - bounded, and a 50 ms caller budget makes that unattractive - or fails fast with 409 plus Retry-After. Starting a second allocation is never an option.
Make the replay return the current representation of the referenced allocation with the first call's status code, not a frozen body. A soft claim that expired between the original call and the retry must not be replayed as live. This is a deliberate departure from byte-identical replay, and the contract must name the fields the caller re-checks: status and expires_at.
Size the lifetime against the business object, not a round number: the key row has to outlive the longest soft expiry plus the orchestrator's maximum retry horizon, or a late retry allocates a second claim against the same line. Retiring keys early is the failure that looks like double allocation months after ship.
Place the partial unique index correctly in the story: it is the backstop that stops two concurrent held claims for one (line, node, lot), not the mechanism. It cannot stop a re-allocation after the first was released, which is exactly why the key contract exists above it.

Worked solution (30 min)

Write the key table DDL: (client_id, endpoint, key) primary key, request_fingerprint, allocation_id, response_status, created_at, plus the retention rule.
Write the happy-path transaction in order: insert key row, insert allocation, decrement the position under its version CAS, commit.
Derive the key deterministically from persisted saga state (order_line_id plus the sourcing attempt number) and confirm a crashed process regenerates the same value instead of minting a fresh UUID.
Enumerate four retry timings - during, after commit, after soft expiry, after release - and write the exact response for each.
Check the fingerprint path by resending the key with qty changed and confirming nothing is written.

Expected result: A contract where the key is caller-derived from durable state, the key row is written in the allocation transaction, a mismatched fingerprint is a 422 that writes nothing, an in-flight duplicate is a 409 with Retry-After, and a replay returns the allocation's current status rather than a stale success.

Verify it yourself

N concurrent retries of one key produce exactly one allocation row: one call returns the create status, the other N-1 return 409 with Retry-After, and a retry sent after that call commits replays its status code and allocation_id instead of a 409.
A retry arriving after the soft claim expired does not present an expired allocation as held.
The partial unique index, dropped in a test, changes nothing about the duplicate-retry outcome - proving the key is the mechanism and the index is the backstop.

Follow-up

A line short-ships and is re-sourced to a second node. Does that call reuse the key, and what is the key derived from now?
The key store and the allocation rows are in different shards. What does the write path look like, and what have you given up?
What does a request carrying an expired key do, and how does the caller tell that apart from a key it never sent?

Duplicate bookings and second trailers after a partner slowdown

Hard · idempotency · partial failure · retries

A partner's booking API ran at 45 s p99 for ninety minutes. Afterwards 214 shipment_leg rows carry two booking references, 41 stops received two trailers, and finance holds duplicate invoices. booking_idempotency_key is generated and persisted on the leg before the call and is sent to the partner, whose documented dedupe window is 60 seconds. The HTTP client retries timeouts three times, and a sweeper re-tenders any leg still in status 'planned' after ten minutes. Explain how duplicates happened despite the key, and give the corrected flow.

Approach

Establish that the duplicates are timeouts rather than double-clicks. Join the gateway request log to shipment_leg for the 214 legs and check whether each duplicate pair carries the same idempotency key and how far apart the calls landed. Same key, calls minutes apart, first attempt ending in a client timeout is the ambiguous-outcome signature; different keys would mean the key is being regenerated and is a different bug.
Name why the key did not help. The key was correct and persisted, but deduplication was delegated to the partner, whose window is 60 seconds. A retry at 45 s may land inside it; the third retry does not, and the ten-minute sweeper re-tender certainly does not, so the partner sees a new request and books a second truck. A short partner window plus a long retry horizon is an at-least-once system wearing an at-most-once label.
Identify the missing state. The leg goes planned to tendered only on a 2xx, so a timeout leaves it exactly as if nothing was sent, which is false: the partner may have committed before the connection dropped. The state machine needs an explicit in-doubt status between requested and confirmed, and nothing may issue a new tender for a leg in that state.
Replace re-issuing with asking. On any ambiguous outcome, move the leg to in-doubt and let a sweeper resolve it by querying the partner for that key or for bookings against our shipment reference, and only tender again when the query definitively returns nothing. Where the partner offers no query, the manifest or the status feed is the authority and resolution is slower, so the in-doubt state must be allowed to persist rather than being timed out into a retry.
Move retry ownership out of the process. A retry decorator around an HTTP client disappears on deploy or crash, so the persisted state is the only thing that can drive recovery: a sweeper reading legs in requested or in-doubt, with bounded attempts, backoff and an alert on in-doubt depth. The call site should not retry an irreversible effect at all.
Deal with the 214 that already exist, since code cannot undo them. Reconcile against the partner's manifest, cancel the duplicate where the cancellation window still allows it, and for legs already executed record the second cost as a compensating entry and dispute the invoice. State plainly that this is a business process with a cost, which is exactly why the guard belongs before the call.

Follow-up

The partner offers no lookup by your key and no search by reference. What is your resolution path and how long may a leg stay in doubt?
Which other effects in this system have the same shape and the same missing state?
How do you test this without a degraded partner, and what does the fault injection have to simulate that a plain timeout does not?

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.

Disagreement needs an outcome you did not choose

Conflict answers where you were right and everyone came round are the weakest ones. Stronger: the evidence you went and collected, what would have changed your mind, and what you did in the weeks after the call went against you. Implementing a design you argued against, properly, is a specific and checkable behaviour.

What did you do when you had a conflict with a team member?

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

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

What was a time when you lost your temper or got super mad, and how did you handle it?

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.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

Tell me about your experience in teamwork.

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

Describe your leadership style.

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
Give the blast radius: what could have broken, and what you measured.
Name the disagreement and how you resolved it with evidence.

Follow-up

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