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Air Products Software Engineer Interview Questions & Guide 2026

Code reading practice for Air Products software engineer prep: work out what unfamiliar code does before judging it.

The 60-second brief

A Software Engineer at Air Products functions at the critical intersection of industrial technology and digital innovation. You are not just writing code; you are building the digital infrastructure that powers global industrial gas production, supply chain logistics, and complex process controls. Your work directly impacts the efficiency, safety, and reliability of large-scale chemical and manufacturing facilities that operate 24/7.

The title spans product, platform and infrastructure work, and which of those the seat actually is decides whether design or algorithms carries more weight in your preparation. The posting rarely settles it; what the team is on call for usually does.

Focus areas

  • Paginate large result sets with keyset cursors
  • Choose indexes from the query's access path
  • Make every write idempotent under retry

Overview

This role is highly collaborative, requiring you to bridge the gap between software development and real-world engineering environments. You will work alongside process engineers, control systems experts, and operations teams to solve high-stakes problems. Whether you are optimizing process models, developing control system interfaces, or automating data pipelines, your contributions ensure that Air Products remains a leader in the industrial sector. Expect a role that demands technical rigor, a deep respect for industrial safety, and a passion for solving complex, physical-world challenges.

The Air Products interview process

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

HR Screen

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

What to demonstrate

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

How to prepare

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

Technical Rounds

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

Behavioral Questions

Many of these questions are about something that went wrong, and the grading sits mostly in the hours after you knew. Who found out first, whether that was you or an alert or a user, how long it took you to say it out loud, and whether the people who needed the news got it while they could still act on it. Engineers under-tell this part because it feels like confessing. The pattern it is looking for is the opposite: the quiet fix, an incident absorbed without telling anyone, after which nothing changed and the same failure is still available.

What to demonstrate

  • How the problem was found, and whether that route was one you had built or one that happened to you, since a user reporting it first means your instrumentation did not cover that failure
  • Whether time-to-detect and time-to-tell are separate numbers in your account and whether you know both, because a fast fix that nobody heard about until the retro is a different answer from a slow one that was announced immediately
  • Whether the resolution left something durable behind, a check that fires or a default that changed, rather than depending on people remembering to be careful
  • Whether you can say what the failure cost without either inflating it or waving it away

How to prepare

  • Reconstruct one incident you were part of as a timeline with clock times: first bad request, first signal, first person who knew, first message outside the team, mitigation, permanent fix. The gaps between those entries are what gets asked about
  • Look up the configuration of the signal that caught it, including its evaluation window and threshold. An alert defined on a five-minute aggregate cannot fire until the condition holds across that window, which puts a floor under time-to-detect that has nothing to do with how severe the failure was. Be able to say what that floor was and whether anyone had chosen it deliberately
  • Prepare one story where you escalated early and the severity turned out to be smaller than you thought, including what it cost the people you pulled in. Without it, every answer you give about raising alarms is unfalsifiable

Technical Deep-Dives

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

Paginating with LIMIT/OFFSET over a set that changes while the client is reading it

OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.

Choosing an index from the columns a query mentions rather than from how it filters and orders

A composite B-tree index on (a, b, c) can be seeked only as a left prefix: equality on a, then equality on b, then a range or an ordering on c. A query that filters on b alone cannot seek into it at all and at best gets a full scan of the index; a query that filters a and ranges on b gets no benefit from c, because the index is only sorted by c within a fixed (a, b) pair. The practical consequence is that one index per column is close to useless for multi-predicate queries while a single correctly ordered composite index turns a scan into a lookup. The ordering half is what gets missed: if the index cannot satisfy the ORDER BY, the database must read every matching row and sort before the limit can apply, so a LIMIT 20 over a million matching rows still reads a million rows.

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

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

Check-then-act on shared state

Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.

Questions & practice

Diff a projection against the primary without per-row point reads

Hard · reconciliation · range hashing · throttling

The listing projection has drifted and some rows show a stale version. The primary holds 40,000,000 resource rows across 12,000 tenants while serving 1,200 writes and 14,000 reads per second. The obvious repair, reading each resource row and comparing its version against the projection, is correct and would eventually finish. Explain precisely why it is unacceptable here, then give a diff that finds the differing rows, state its complexity, and make it safe to run against a live primary. Replication lag is usually under 100 ms and is not bounded.

Approach

  1. Quantify the naive cost rather than calling it slow: 40,000,000 point reads at even 0.5 ms each is over five hours serialised, and the only lever is concurrency, which is exactly what you cannot spend. The primary's pool is sized for the write path, and 40,000,000 random reads evict the buffer cache that sustains the 85 percent cache hit rate, so the audit degrades the system it is auditing.
  2. Replace random access with one ordered pass per side. Both sides can be read in (tenant_id, resource_id) order, which is a sequential scan on each and a merge join in O(n) time and O(1) memory. For a dense diff that is the whole answer, and it reads the primary once instead of 40,000,000 times.
  3. For the expected sparse case, compare range hashes instead of rows: partition the key space, compute per range an order-independent aggregate over hash(resource_id, version), compare aggregates, and descend only into ranges that differ. With d differing rows and branching factor B, at most d ranges mismatch per level, so the drill-down examines O(d log_B(n/d)) ranges and reads full rows only in mismatching leaves.
  4. Aggregate with a sum modulo 2^64 or a multiset hash, never XOR. XOR is order-independent but self-cancelling, so two rows wrong in the same way, or a row duplicated on one side, leave the range aggregate matching and the range is declared clean.
  5. Pin the comparison to a point in time or it reports lag as drift: consider only rows whose updated_at is older than now minus a lag margin, and re-check each candidate mismatch individually before repairing. At 1,200 writes per second a diff without this reports thousands of false positives, and an unattended repairer would then overwrite live rows with stale values.
  6. Make the run resumable and throttled: batch by range key, persist the last completed range, and watch a signal such as replica lag or primary CPU, pausing rather than pressing on. A reconciliation that cannot be stopped and resumed gets killed halfway and restarted from zero, which is how a repair becomes an incident.

Worked solution (35 min)

  1. Compute the naive cost explicitly at 40,000,000 reads and 0.5 ms each, then at 100 concurrent, and state what those connections do to a pool already carrying 1,200 writes per second.
  2. Write the merge-join version over (tenant_id, resource_id) and state its memory.
  3. Define the range aggregate: the range key, the per-row hash input, and the combining function, with one sentence excluding XOR.
  4. Work an example with 40,000,000 rows, branching factor 256 and 5 differing rows, and count the ranges examined.
  5. Add the watermark filter and the resume point, and name the throttle signal the loop watches.

Expected result: A rejection of per-row point reads backed by the time cost and the cache-eviction argument, a single ordered merge join as the dense-case answer at O(n) time and O(1) memory, a range-hash drill-down examining O(d log_B(n/d)) ranges using a sum or multiset hash rather than XOR, a watermark excluding recently written rows, and a resumable throttled run loop.

Verify it yourself

  • Injecting one known stale row makes exactly one leaf range mismatch and the drill-down reaches it.
  • Duplicating a row on one side is detected, which is precisely the case XOR would miss.
  • Running the diff twice against an idle system reports zero mismatches both times.

Follow-up

  • The diff reports 900 stale rows. How do you decide between patching those rows and rebuilding the projection from resource_revision?
  • Same job, but the projection lives in a search index that cannot be scanned in key order. What changes?
  • How would you run this continuously at low cost instead of only as incident response?

Merge partitioned event streams into one ordered feed with bounded lateness

Hard · k-way merge · watermarks · out-of-order streams

The read-model service consumes 64 log partitions carrying about 4,000 events per second in total. Each partition is ordered within itself, but partitions drift by up to 30 seconds, and the activity feed must present a tenant's events in occurred_at order. Produce the merge. State its complexity, the buffer it requires in events and in bytes, what happens when one partition is idle, and what you do with an event that arrives after you have already emitted its position. Payloads average 1 KB.

Approach

  1. Merge with a min-heap over the 64 partition heads keyed on (occurred_at, event_id): O(log P) per event and O(n log P) overall. The tie-break on event_id is what makes the output deterministic when two partitions carry the same millisecond, which matters because the feed is paginated and a non-deterministic order reorders pages under the reader.
  2. Emitting the heap head is only correct once every partition has produced everything up to that timestamp, so the emit condition is a watermark: the minimum across partitions of the highest occurred_at seen, less the allowed lateness. Events are held until the watermark passes them, which is what turns individually ordered streams into a jointly ordered one.
  3. Size the buffer from the lateness rather than guessing: 4,000 events per second times 30 seconds is 120,000 buffered events, and at 1 KB each about 120 MB of heap. That number is the real price of the ordering guarantee and belongs in front of whoever asked for it.
  4. Handle the idle partition explicitly, because it fails the feed rather than corrupting it: a partition with no traffic never advances its own maximum, so the watermark freezes and output stops entirely. Either every partition emits a periodic idle marker carrying the broker's current time, or the watermark falls back to wall clock for a partition silent beyond a threshold.
  5. Choose the late-event policy from what the projection is keyed on. The projection upserts on (aggregate_id, aggregate_version) and discards a version it has already applied, so a late event is safe to apply out of order and correctness never depended on the merge at all. Apply it, recompute the affected feed page, and count lateness so the 30-second budget can be re-derived from data rather than folklore.
  6. Say what the merge does not buy: ordering is guaranteed within one aggregate by the log's partitioning, and no watermark makes the cross-aggregate order authoritative. Two events from different aggregates in the same millisecond have no true order, so the feed's order is a presentation choice that must be stable rather than correct.

Follow-up

  • The lateness budget is raised to five minutes. What is the new buffer, and what besides memory changes?
  • The consumer restarts. Where does it resume from, and what does the feed look like for the first 30 seconds?
  • One partition is ten minutes behind because its producer is slow. Do you stall the feed or emit without it?

Collapse a redelivered event batch into per-aggregate high-water marks

Easy · hashing · at-least-once · aggregation

You drain a batch of up to 5,000,000 events, each (aggregate_id BIGINT, aggregate_version INT, event_type, payload). The log guarantees order within one aggregate only; the batch merges 64 partitions, and a relay failover has redelivered a range, so an older version for an aggregate can appear after a newer one. Given a map of last_applied_version per aggregate, produce the events worth applying, at most one per (aggregate_id, version), plus the count discarded. Target O(n) time. State the memory for 2,000,000 distinct aggregates and what you do when it does not fit.

Approach

  1. One pass, one hash map from aggregate_id to the highest version kept, and a discard counter. An event whose version is at or below last_applied_version for its aggregate is dropped without further work, which is the whole reason the event carries its version rather than a delta. O(n) expected time, O(d) space in distinct aggregates.
  2. Keep the maximum, never the last occurrence. The redelivered range means the final appearance of an aggregate in the batch can be an older version than one seen earlier in the same batch, so last-wins applies stale state over newer state and the projection regresses with no error anywhere.
  3. Cost the memory instead of calling it large: an 8-byte key plus a 4-byte version is 12 bytes of payload, and an open-addressed table held at a 0.7 load factor costs roughly 17 bytes per entry before per-slot metadata, so 2,000,000 aggregates is tens of megabytes in a native layout and several times that in a runtime that boxes both key and value.
  4. If the distinct set exceeds memory, partition on hash(aggregate_id) mod P and reduce each partition independently. Every event for one aggregate hashes to the same partition, so the per-partition result is exact and the merge is concatenation rather than a second reduction.
  5. Reject sorting the batch by (aggregate_id, version) as the default. It is O(n log n) and buys nothing, because max is associative and commutative and needs no ordering; sorting earns its cost only when the downstream consumer must receive the events in order rather than a per-aggregate winner.
  6. Separate the two mechanisms out loud: in-batch deduplication does not make the consumer idempotent, because the same event redelivered tomorrow arrives in a different batch entirely. The projection write itself still has to be keyed on (aggregate_id, version).

Follow-up

  • The payload is a patch rather than a snapshot, so applying only the highest version loses the intermediate changes. What changes in your reduction?
  • How do you detect that version 7 arrived while version 6 was never delivered, and what should the consumer do about the gap?
  • Two events for one aggregate carry the same version with different payloads. Which one is wrong, and how would you find out?

Keep soft-deleted accounts from blocking re-registration

Easy · soft delete · unique constraints · engine differences

app_user holds user_id, tenant_id, email CITEXT, password_hash (NULL for SSO principals), email_verified_at, auth_version, status ('invited','active','suspended','deactivated'), created_at, updated_at, deleted_at. Two live accounts for one address inside a tenant must be impossible, but an address freed by a soft delete must be reusable, and the same tenant may delete and re-register it repeatedly. Write the uniqueness DDL for PostgreSQL 16, then the equivalent for MySQL 8 where partial indexes do not exist, and say what each permits once three deleted rows already hold that address.

Approach

  1. Start from what is actually unique: not (tenant_id, email), but (tenant_id, email) among live rows. PostgreSQL says that directly — CREATE UNIQUE INDEX app_user_live_email ON app_user (tenant_id, email) WHERE deleted_at IS NULL. A full constraint over the same two columns burns the address permanently the first time someone deletes an account.
  2. Keep case-insensitivity in the type or the index, never in the application: CITEXT as given, or UNIQUE (tenant_id, lower(email)) as an expression index where the extension is unavailable. A case-sensitive unique column is exactly how two accounts for one human appear.
  3. For MySQL 8 the predicate has to move inside the key: add a discriminator column that is a constant 0 while the row is live and is set to user_id on delete, with UNIQUE (tenant_id, email, deleted_marker). Live rows share the constant and still collide; deleted rows differ from each other and stop colliding.
  4. State the NULL variant and its dependency: leaving the marker NULL for deleted rows also works, because a unique index treats NULLs as distinct — true in MySQL, and true in PostgreSQL only under the default NULLS DISTINCT, which PostgreSQL 15 lets you reverse. Check the polarity against the three existing deleted rows: constant-on-live is what preserves the collision you want, and reversing it silently admits duplicate live accounts.
  5. Say what a soft delete must do besides setting deleted_at: increment auth_version so existing tokens stop validating, leave resource.owner_user_id and resource_revision.actor_user_id intact, and accept that the address is retained — erasure is a different requirement answered by scrubbing the column, not by a DELETE that would break those references.

Follow-up

  • A deleted account re-registers with the same address the next day. Do the old resource rows follow the new user_id, and how does the API keep the two principals apart?
  • How do you honour an erasure request while resource_revision.actor_user_id still references this table?
  • What changes if a user may hold membership in two tenants?

Replace offset paging on the resource feed with keyset

Easy · keyset pagination · composite index · sort order

resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.

Approach

  1. Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
  2. Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
  3. Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
  4. Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
  5. State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
  6. Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.

Worked solution (25 min)

  1. Seed one tenant with 500k active resources, 2% of them sharing an identical updated_at.
  2. Time LIMIT 50 OFFSET 0 against OFFSET 20000 and record rows-read from EXPLAIN (ANALYZE, BUFFERS) for each.
  3. Page the whole set with the keyset query while an insert-only writer adds 100 rows/second, collecting resource_ids, and repeat the run with OFFSET.
  4. Repeat both runs under a second writer profile that also deletes 20 rows/second from pages already returned and bumps updated_at on 20 more, and diff each collected id set against the rows that existed for the whole run.
  5. Remove resource_id from the cursor so the seek degrades to updated_at < $2, and re-run the tie-heavy section of the feed.

Expected result: Keyset reads about 50 index entries per page and its latency is flat from page 1 to page 10,000; under both writer profiles the collected ids contain no duplicate and omit no row that existed for the whole scroll. OFFSET at page 400 reads 20,050 rows to return 50. Under the insert-only writer it repeats rows and skips none — an insert can only push a row down past the boundary — and only the second profile, whose deletes and updated_at bumps move rows up past the cursor, also makes it skip. Without the tie-break, the seek is updated_at < $2, which drops every still-unread row sharing the boundary timestamp.

Verify it yourself

  • EXPLAIN shows an Index Scan with no Sort node — a Sort means the ORDER BY is not satisfied by the index.
  • Buffers read per page stays constant as page depth grows.
  • Ids collected across the scroll are unique, and complete for rows untouched during the run.

Follow-up

  • The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
  • Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
  • What does the cursor do when the row it points at has since been deleted?

Can you explain the logic behind a specific technical design or project you previously completed?

Medium

Approach

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

Follow-up

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

How do you approach troubleshooting a complex technical issue in a live production environment?

Medium

Approach

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

Follow-up

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

Publish rate-limit and deadline semantics the edge actually enforces

Easy · rate limiting · backpressure · retries

The edge API serves about 3k requests/second steady and 9k at peak against a 400 ms p99 budget, with an explicit bounded concurrency limit per instance. Limits exist per principal and per tenant. Callers are a partner integration running nightly bulk loads and a browser app. Specify the counting algorithm and window, which limit a request is charged against, the headers a well-behaved client reads, the status and body when a limit is hit, how that differs from the response when an instance is shedding load, and what each caller does with each.

Approach

  1. Choose the counter and name its failure mode. Fixed windows admit nearly twice the limit across a boundary - a full burst at the end of one window and another at the start of the next. A token bucket states sustained rate and burst separately, which is exactly what a nightly bulk load needs. A sliding-window counter is more faithful and costs more state per key. State the choice and the burst it permits.
  2. Charge each request against both keys and reject on the stricter. The tenant limit protects the shared primary, which absorbs roughly 1.2k writes/second in total; the per-principal limit stops one credential inside a tenant from consuming that tenant's whole allowance. The tenant is the fairness unit for the same reason it is the leading column of every index.
  3. Advertise limit, remaining and reset for the binding key on every response, not only on rejections, so a client can pace before it is refused. Pick one naming scheme - the RateLimit-* draft fields or an X-prefixed set - document the units, and never change them afterwards.
  4. Separate two rejections that look identical to a naive client. 429 means this caller exceeded its own share and Retry-After is a real schedule it should obey. 503 means the instance is at its concurrency bound and shedding, which is a statement about the server; a fleet-wide 503 retried on a fixed delay resynchronises every client into one stampede, so full jitter is mandatory there and the delay is the client's guess, not ours.
  5. Make shedding cheap and early - before the token is verified against the database, before any downstream call - because a rejection that costs as much as the work relieves nothing. Drop requests whose client deadline has already elapsed rather than serving them; the caller has stopped listening and the work is pure cost.
  6. Write the caller behaviours down: the bulk loader paces against remaining and treats a 429 as a defect in its own pacing; the browser surfaces the wait and must never retry a 429 inside a render loop, which turns one limited user into a self-inflicted flood.

Worked solution (20 min)

  1. Write the bucket parameters for both keys: sustained rate, burst size, and the refill interval, with the arithmetic that ties them to the 3k/9k figures.
  2. Draft the three response headers and one example 429 body carrying a code, the limit that bound, and Retry-After.
  3. Write the 429-versus-503 decision as a two-line rule an on-call engineer can apply to a log line.
  4. State where in the request pipeline the rejection happens and which work it skips.

Expected result: A token bucket per principal and per tenant with named burst, headers on every response, 429 with Retry-After for share exhaustion, 503 with jittered client backoff for concurrency shedding, and rejection placed before authentication's database read.

Verify it yourself

  • A client that respects remaining never sees a 429 under steady traffic at its configured rate.
  • The cost of a shed request is bounded and does not include a database round trip.
  • Peak throughput times the limiter's per-request state cost is affordable at 9k requests/second.

Follow-up

  • One tenant stays under its limit and still degrades everyone else during a backfill. What changes - the limiter, the worker concurrency caps, or both?
  • How are counters kept correct across 20 to 40 stateless instances, and what does your answer cost per request?

Every query on one table stalls for forty seconds mid-deploy

Medium · lock contention · ddl · migrations · postgres

During a release on PostgreSQL, every query touching resource times out for about 40 seconds and then recovers with no intervention. The release ran one migration, ALTER TABLE resource ADD COLUMN archived_reason TEXT, and the migration log shows it completing in 6 ms. Unrelated tables showed no change in error rate. Explain how a 6 ms statement caused a 40-second stall, give the ordered checks you would run on a live system to confirm it, and give the migration procedure that prevents a repeat.

Approach

  1. Separate the statement's duration from the lock's duration. ADD COLUMN with no default is a catalogue-only change and genuinely runs in milliseconds, but it requires ACCESS EXCLUSIVE, and it cannot acquire that until every transaction already touching the table has finished.
  2. Account for the queueing, which is the part that surprises people. A lock request that is waiting blocks later requests for conflicting modes behind it rather than letting them overtake, so one long-open transaction holds the DDL and the DDL holds all the traffic. The stall length is set by the longest open transaction, not by the size of the change.
  3. Confirm on a live system in this order: pg_stat_activity for that table ordered by xact_start, looking for the oldest transaction and specifically for state = idle in transaction; then pg_locks where granted = false to find the waiter; then join them on pid to name blocker and blocked. pg_blocking_pids() does that join for you and is the fastest single call.
  4. Prevent rather than merely time it better. Set lock_timeout to a second or two on the migration session so the DDL abandons the queue after a bounded wait and is retried, instead of holding it for as long as the oldest transaction lives. Be exact about what that buys: queries arriving during the wait still queue behind the pending ACCESS EXCLUSIVE request, so each attempt costs them up to one lock_timeout of added latency. The outage goes from 40 seconds to about one second per attempt, not to zero. Also run migrations away from deploy-time peaks, and put a statement timeout and an idle-in-transaction timeout on the analytics role that opens the long transactions.
  5. Know the lock each change takes, since the mitigation differs by change. A column with a non-volatile default is a metadata-only change from PostgreSQL 11 and still needs the brief ACCESS EXCLUSIVE; an index needs CREATE INDEX CONCURRENTLY, which cannot run inside a transaction block and leaves an INVALID index to drop if it fails; a check or foreign key is added NOT VALID and then VALIDATE CONSTRAINT as a separate statement under a weaker lock.

Follow-up

  • The same release also wants NOT NULL on that column. What is the sequence that gets there without a long lock?
  • Your lock_timeout retry fails ten times in a row because the analytics transaction is always open. What do you change?
  • How does this differ on MySQL with InnoDB online DDL, and what is the equivalent of the waiting-lock queue there?

Suggested practice checklist

For a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.

Day 1 — Numbers before diagrams

  • Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
  • Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
  • Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.

Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.

Day 2 — One system, from requirements to schema

  • Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
  • Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
  • Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.

Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.

Day 3 — The consistency you are actually buying

  • Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
  • Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
  • Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.

Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.

Day 4 — Failure is the design

  • For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
  • Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
  • Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
  • Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.

Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.

Day 5 — Scaling the hot path

  • Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
  • Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
  • Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
  • Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.

Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.

Day 6 — Keep the coding hand in, at the bar that applies to you

  • Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
  • Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
  • Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.

Deliverable: One problem hardened to review standard, with its test list and one written review comment.

Day 7 — Defend it while being interrupted

  • Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
  • Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
  • Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.

Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.

The failure that was yours

Keep one story where the bad call was yours rather than a dependency's or a manager's. Name the check that would have caught it, whether you added that check afterwards, and whether it has fired since. Answers that route blame outward end the conversation early; answers that end in a guardrail someone still relies on tend to open it up.

Why are you interested in a career at Air Products?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

Describe a mistake you made in a recent project; how did you rectify it and what did you learn?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

What does safety mean to you in the context of engineering, and how do you promote it?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

What is your experience with P&IDs (Piping and Instrumentation Diagrams) or alarm management?

Medium · behavioural and engineering judgement

Approach

  1. Name the disagreement and how you resolved it with evidence.
  2. State the situation in two sentences and spend the rest on the reasoning.
  3. Pick a story where you made the decision, not one where you watched it.

Follow-up

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