Corbins Software Engineer Interview Questions & Guide 2026

The 60-second brief

At Corbins, a Software Engineer—often operating within the context of Project Engineering or Field Engineering—plays a vital role in bridging the gap between digital design and physical construction. You are not just writing code in a vacuum; you are developing and leveraging technical solutions that drive efficiency across massive infrastructure and electrical construction projects. Your work directly impacts how complex data is translated into actionable field instructions, ensuring that large-scale projects are delivered with precision and safety.

If the seat owns a service boundary, scope your preparation toward failure behaviour rather than topology. Retrying over an at-least-once channel produces duplicates by construction, so a retry policy is only as safe as the idempotency key underneath it.

Focus areas

• Make every write idempotent under retry
• Bound every outbound call with a timeout
• Paginate large result sets with keyset cursors

Overview

This role is unique because it requires a blend of high-level technical aptitude and a practical understanding of engineering environments. You will be expected to utilize industry-standard software to model, analyze, and troubleshoot real-world systems. It is an intellectually stimulating position for those who want to see their technical contributions manifest as tangible, high-impact results in the field.

While the role is titled Software Engineer, your day-to-day work will frequently intersect with tools like AutoCAD and Revit. Expect your technical evaluation to focus heavily on your ability to model and interpret engineering parameters rather than pure algorithm development.

Where to focus your preparation

Treat the sections below as preparation areas and confirm the format with your recruiter.

Preparation focus

No round sequence has been reported for this company, so work the categories below and confirm the format with your recruiter.

What to demonstrate

• Breadth across SQL, experimentation and product reasoning
• Ability to state assumptions before choosing a method

How to prepare

• Drill the practice exercises below and time yourself
• Prepare three quantified stories about decisions you drove

Where candidates lose points

Shipping a migration and the code that depends on it as a single change

During any rolling deploy, and for as long as a rollback remains possible, old and new code execute against the same schema at the same time. A migration that drops or renames a column breaks every instance that has not restarted yet, and code that requires a column the migration has not applied breaks every instance that restarted early. The discipline is expand then contract: add the new column nullable, write both shapes, backfill in batches, move reads across once the backfill is verified, and only then stop writing the old shape and drop it - four deploys, usually spread over days. It feels disproportionate until the first rollback, at which point it is the only reason the previous version still runs.

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.

Sorting when the problem never required a total order

Match the algorithm to the guarantee actually needed: the top k comes from a size-k heap in O(n log k) time and O(k) space, distinctness needs a set rather than an ordering, and a small bounded integer key range admits a linear counting pass. A full O(n log n) sort is the right default only when you genuinely need everything in order.

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.

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.

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.

Worked solution (35 min)

1. Write the heap comparator on (occurred_at, event_id) and the per-partition head refill.
2. Write the watermark computation and the emit-loop condition, then list which buffered events are held at a chosen instant.
3. Compute the buffer at 4,000 events per second, 30 seconds and 1 KB per event, and state what fraction of a worker's heap that represents.
4. Add the idle-partition marker and trace the watermark with one silent partition, both with and without the marker.
5. Write the late-event path and name the key that makes applying it safe.

Expected result: A 64-way min-heap merge at O(n log P) with a deterministic (occurred_at, event_id) comparator, a watermark of the per-partition minimum less 30 seconds gating emission, a stated buffer of 120,000 events and roughly 120 MB, idle markers so a silent partition cannot freeze the watermark, and a late-event policy justified by the projection's idempotency on (aggregate_id, aggregate_version).

Verify it yourself

• Feeding the partitions in an adversarial interleaving produces the same output sequence as feeding them in timestamp order.
• Silencing one partition stalls output without the idle marker and does not stall it with the marker.
• Applying a late event twice leaves the projection byte-identical, which is what permits the out-of-order path to exist.

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?

Canonicalise a request body into a stable idempotency fingerprint

Medium · parsing · canonicalisation · hashing

idempotency_key.request_fingerprint is a SHA-256 over the method, path and canonicalised body, and a retry whose fingerprint differs must be rejected with 422 rather than served the stored response. Write the canonicaliser. Bodies are JSON up to 256 KB nested at most 32 levels; clients vary key order, whitespace and unicode escaping, and some send 64-bit ids as JSON numbers. Produce a deterministic byte string such that semantically identical bodies match and any semantic difference does not. State your complexity and name two normalisations you refuse to perform.

Approach

1. Parse once into a tree, then re-serialise under fixed rules: object keys sorted, array order preserved, one escaping convention, no insignificant whitespace. Parsing is O(n) and sorting keys is O(k log k) per object, so O(n log n) overall with O(depth) stack, and the 32-level cap is enforced during parsing because hostile nesting is how a canonicaliser becomes a stack overflow.
2. Sort keys by their UTF-8 bytes and say why the obvious implementation is wrong in some runtimes: a default string comparison that orders by UTF-16 code units places surrogate pairs, meaning code points from U+10000 up, below U+E000 to U+FFFF, which is not UTF-8 byte order, so two services written in different languages disagree on the same document.
3. Do not re-encode numbers through a double. IEEE-754 binary64 represents integers exactly only up to 2^53, so normalising a 19-digit id through a float changes it, and 1 against 1.0 cannot be reconciled without deciding whether they are the same value. Preserve the literal token, and require ids as strings at the API boundary if you want them comparable.
4. Reject duplicate keys rather than picking one. JSON permits them and parsers disagree, most keeping the last, so any choice you make ties the fingerprint to a parser detail that the code handling the request does not necessarily share.
5. Frame the hash preimage so concatenation cannot collide: delimit or length-prefix the method, path and body, otherwise one request's fields can be rearranged into another request with the same byte stream and the same fingerprint.
6. Name the refusals and their consequence: no case folding, no dropping of null-valued keys, no Unicode normalisation. Each makes two different requests fingerprint alike, and the resulting failure is the worst one this table has, since the second request is answered with the first one's stored response and its effect never happens.

Follow-up

• A client sends the same logical request with an extra field your API ignores. Same key, different fingerprint, so you return 422. Is that the right answer?
• Where does the fingerprint get computed relative to request decompression and the body-size limit?
• The endpoint takes 1,000 requests per second with 256 KB bodies. What does hashing cost, and does it belong at the edge or in the core service?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

resource carries version INT NOT NULL DEFAULT 1. resource_revision holds revision_id, resource_id, version, actor_user_id, change_kind, patch JSONB, request_id, created_at with UNIQUE (resource_id, version). outbox_event holds aggregate_type, aggregate_id, aggregate_version, event_type, payload, status. A PUT carries the version the client read. Write the exact statements for the single transaction that applies the edit, records the revision and enqueues 'resource.updated', and give the handler's branch on zero affected rows. Then say what PostgreSQL 16 does under READ COMMITTED when two of these updates hit one row at once.

Approach

1. One transaction, three writes, no network call inside it: UPDATE resource SET title = $3, version = version + 1, updated_at = now() WHERE resource_id = $1 AND tenant_id = $4 AND version = $2; then INSERT the resource_revision row at version $2 + 1; then INSERT the outbox_event row at the same aggregate_version. The event goes to a table rather than a broker because no transaction spans both.
2. Branch on the affected-row count before doing anything else. Zero has three causes — stale version, wrong tenant, row gone — so re-read once and map to 409 carrying the current version, or 404 for an id outside the caller's tenant, which also stops the endpoint confirming that another tenant's id exists.
3. State the engine behaviour instead of assuming it. Under READ COMMITTED the second UPDATE blocks on the row lock, and when the first commits PostgreSQL re-evaluates the WHERE clause against the newly committed row, so the version predicate now fails and the statement reports zero rows. Under REPEATABLE READ the identical collision raises SQLSTATE 40001 instead, so the handler must fold both shapes into one conflict response.
4. Keep UNIQUE (resource_id, version) even though the predicate already serialises writers. It is what makes a lost update unwritable if any other path ever reaches the revision table, and it converts a logic bug into 23505 rather than into a silently missing history row.
5. Refuse to auto-retry the whole PUT. A retry re-reads the winner's state and reapplies an intent formed against data that no longer exists — the silent overwrite the version token was added to detect. Return the conflict; merge field-wise only if the patches are provably disjoint.
6. Note that now() is the transaction timestamp in PostgreSQL, so resource.updated_at, the revision's created_at and the outbox row share one instant, which is what later makes reconciliation between the three tables unambiguous.

Follow-up

• A client sends the version it read ten minutes ago and the resource has moved three versions. What is in your 409 so it can resolve the conflict without a full re-fetch?
• Two editors, two disjoint fields, no overlap. Does your answer still refuse the second write, and should it?
• Every write now touches a second hot table. How do you keep the outbox insert and its partial index from becoming the write bottleneck at 1.2k writes/second?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.

Approach

1. Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
2. For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
3. Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
4. Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.
5. Answer the two negatives explicitly. owner_user_id alone is not a left prefix of the new index, so it degrades to a full scan of the index at best. Ordered by updated_at, the query still seeks on the (tenant, owner) pair but must sort, because only created_at is ordered within that pair.
6. Measure both sides with EXPLAIN (ANALYZE, BUFFERS) and compare estimated against actual rows at the lowest node — a 2M-versus-200 misestimate there is usually what chose the plan, and adding an index will not fix a statistics problem.

Worked solution (25 min)

1. Load 2M resource rows across 5k owners in one tenant, run the query under EXPLAIN (ANALYZE, BUFFERS), and record the node reading the most rows plus the Sort Method line.
2. Create (tenant_id, owner_user_id, created_at DESC) concurrently and re-run, confirming the Sort node is gone and actual rows fall to about 20.
3. Run the two negative cases and capture the plan for each.
4. Re-run the original tenant listing query to confirm the new index has not displaced the index that query depends on.

Expected result: With (tenant_id, owner_user_id, created_at DESC) the plan is an Index Scan with no Sort, roughly 20 rows and a handful of buffers, sub-millisecond. owner_user_id alone still scans; ORDER BY updated_at seeks on the pair and then sorts that owner's rows.

Verify it yourself

• Actual rows at the leaf node is on the order of the LIMIT, not the tenant's row count.
• No Sort node appears above the index scan.
• Buffers read stays constant as one owner's row count grows; if it grows with it, the ORDER BY is not being satisfied by the index.
• pg_stat_user_indexes shows idx_scan climbing on the new index and not on the listing index for this endpoint.

Follow-up

• 90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
• A dashboard runs this for 40 owners in one page load. What changes about the design?
• How do you roll this index out on a table taking 1.2k writes/second, and what does it cost on every insert from then on?

Do you have experience with AutoCAD?

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

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

Can you draw a relationship diagram based on the parameters provided?

Medium

Approach

1. Work from the requirement backwards to the design.
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?

Can you explain a time you used technical software to solve a specific engineering problem?

Medium

Approach

1. Say what you would check first and why it is the highest-information step.
2. State your assumptions explicitly before working the problem.
3. Work from the requirement backwards to the design.

Follow-up

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

How proficient are you in Revit?

Medium

Approach

1. Clarify what is being asked and what a complete answer contains.
2. Work from the requirement backwards to the design.
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?

Relay committed events to the log without gaps or reordering

Medium · transactional outbox · at-least-once · ordering · leases

outbox_event rows are written in the same transaction as the state change and carry aggregate_type, aggregate_id, aggregate_version, payload and status, with a partial index on (created_at, event_id) WHERE status = 'pending'. The relay publishes about 4k events/second to a partitioned append-only log keyed by aggregate_id, with one leader per partition range holding a lease. Consumers must never miss an event; they may see one twice. Design the claim-publish-mark loop, and state exactly what consumers observe when a leader's lease expires while it is mid-batch.

Approach

1. Claim with SELECT ... WHERE status='pending' ORDER BY created_at, event_id LIMIT $batch FOR UPDATE SKIP LOCKED inside a transaction. SKIP LOCKED lets several relay workers share a range without serialising on each other's rows, and the partial index keeps the claim proportional to the backlog rather than to a table that is overwhelmingly published rows. At 4k/second a batch of 500 is eight claims per second, each an index scan of 500 entries.
2. Publish before marking, never the reverse, and say why it is a choice. Marking first loses the event outright if the process dies in the gap, and the loss is silent - nothing remains to retry, and it surfaces later as a projection missing a row. Publishing first can repeat the event, and repetition is what every consumer is already built to survive. That single ordering is the whole at-least-once guarantee.
3. Preserve the only ordering on offer. Partition by aggregate_id and never publish two events for one aggregate concurrently: claim in (created_at, event_id) order and publish sequentially within an aggregate, or hash aggregate_id to a worker slot. Order across aggregates is not available at any price here, which is why the event carries aggregate_version and the full fact rather than a delta - a consumer can then discard what it has already applied without coordinating with anyone.
4. State the failover behaviour precisely, because it is the consistency-versus-availability decision in this design. A lease expires because the holder is slow, and no mechanism distinguishes that from dead, so for the length of the lease window two leaders can publish the same claimed batch. The system accepts duplicates to avoid stalling publication for every aggregate in the range whenever one worker pauses. Consumers deduplicate on (aggregate_id, aggregate_version) and drop anything at or below what they have applied.
5. Bound the failure paths and pick the right alarm. A row that fails to publish increments attempts, records last_error, and moves to 'dead' after a limit so one poison payload cannot block the backlog behind it. Alert on the age of the oldest pending row, not on the relay's error rate: the failure worth catching is a relay reporting itself healthy while nothing is being published.

Worked solution (25 min)

1. Write the claim statement and check it against the partial index: which columns it seeks on, how many entries it touches, and what two concurrent workers do to each other.
2. Write both orderings of publish and mark, and for each state what exists after a crash at every point in the loop.
3. Write the consumer's dedupe rule on (aggregate_id, aggregate_version) and test it against a replayed batch of 500.
4. Compute the backlog after a 40-minute outage and the batch rate needed to drain it while 4k/second continues to arrive.

Expected result: A claim using FOR UPDATE SKIP LOCKED over the partial index in (created_at, event_id) order, publish to the aggregate_id partition, then mark published - in that order, accepting duplicates on lease expiry. Per-aggregate order is preserved by never publishing an aggregate concurrently; cross-aggregate order is explicitly not guaranteed. Failures increment attempts and terminate at 'dead'. The alarm is the age of the oldest pending row.

Verify it yourself

• Killing the relay mid-batch produces duplicates on restart and no gap.
• The claim query's plan uses the partial index and reads at most batch-size entries regardless of table size.
• Two relay workers on the same range never block on each other's rows.
• Events for one aggregate_id arrive in aggregate_version order, and no consumer relies on order across aggregates.

Follow-up

• The relay is down 40 minutes and 9.6 million rows are pending. What does catch-up do to the primary, and what changes in the claim loop to survive it?
• A consumer insists it never received an event. Which single query settles whether the relay lost it, and what does each answer look like?
• Delivery is at-least-once. What would exactly-once require end to end, and why is that a property of the consumer rather than of the relay?

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

Day 1 — Rebuild the primitives by implementing them

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Day 7 — Solve out loud, under time

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

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

Review and mentoring: what you let through

Counting review comments or mentees proves nothing. The useful version is a specific change you approved with a reservation you stated, or one you blocked and the delay that cost. Say which standard you were holding and why it was worth the friction. A mentoring story needs the thing the other person can now do without you.

Tell callers you do not own that their integration breaks

Medium · deprecation · compatibility · stakeholders

A field in a write endpoint's response must change shape. You own the endpoint; you do not own the four internal callers or the outbound webhook consumers who read it. Describe a deprecation you were responsible for: what you shipped first, how you established who was actually reading the field, the window you gave and what set its length, what you did about the consumer who never moved, and how you decided removal was safe. Name the signal you used, not the announcement you sent.

Approach

1. Establish the reader set empirically rather than from a wiki of owners: per-field usage counters keyed by principal, or access logs attributed to a consumer. State the blind spot of whichever you pick, since a consumer that reads the field only on a monthly job will not appear in a week of logs.
2. Ship additive first. Populate the new field alongside the old one so no reader is forced to move, which is also what keeps a rolling deploy safe, because old and new instances answer the same requests at the same time and a rollback must still find the old shape present.
3. Set the window from the slowest legitimate consumer's release cadence, not from your calendar, and decide separately what to do for a consumer with no release process at all, such as an external webhook endpoint you can only email.
4. Convert silence into evidence before you rely on it: a short, low-traffic removal window that makes a still-dependent consumer fail visibly and loudly while you are watching, rather than at three in the morning after you have moved on.
5. State the removal criterion as a measurement with a duration attached, such as observed reads at zero across a full billing cycle, and keep the change reversible for one release after removal.

Follow-up

• How would you detect a consumer that reads the field only during a monthly export?
• One caller refuses to move and has a commercial relationship behind it. What changes in your plan and what does not?
• After removal, what makes the change irreversible, and how long before you cross that line?

Unblock an engineer without taking the keyboard

Easy · mentoring · leases · at-least-once

A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.

Approach

1. Ask before diagnosing, and ask for things answerable from data they already have: the attempt count on the job rows that produced duplicates, the handler's observed duration against its lease expiry, and whether the duplicate rows share a natural key that a unique constraint could have caught.
2. Teach the shape rather than the answer. A lease cannot distinguish a dead worker from a slow one, so a handler that outruns its lease is running twice by design, and deploys deliver the other half by killing handlers mid-run on every rollout. Both of their candidate theories produce identical duplicate rows, which is why the evidence has to come from timings rather than from argument.
3. Hand over a checklist they execute: a natural key on every write the handler performs so the second copy collides rather than appends, the record of intent written before any external effect, a lease heartbeat while running, and the metric that shows it working.
4. Keep ownership with them deliberately. Pair on the first write, then step back; if you finish it yourself you have closed one ticket and left the same person stuck on the next redelivery.
5. Close on the systemic gap that let two days pass, which is usually a missing dashboard for attempt counts or an undocumented at-least-once contract, and fix that rather than only the bug.

Follow-up

• How would you distinguish a genuine double-delivery from a lease expiry using only the data already stored?
• Their handler calls an external endpoint before recording that it did. What do you tell them to change first?
• What do you do the third time the same person brings you the same class of bug?

Reverse your own decision and price the reversal

Medium · reversibility · measurement · migrations

Describe a technical decision you made and later reversed. Pick one that cost something: a service you split and merged back, a cache you added and removed, an index you created that pushed the planner onto a worse plan, a projection you rebuilt from scratch. State what you believed when you decided, the measurement that changed your mind, how long the wrong version ran in production, and what the reversal cost in migrations, dual writes, and a deprecation window for callers you did not own.

Approach

1. State the original rationale without irony, in the version you would still defend given what was known then. If it is not defensible, the story is about carelessness rather than judgement, and a different example serves you better.
2. Give the measurement that moved with a before and after: the p99 that did not improve, the cache hit rate that sat at 40%, the plan that flipped to a sequential scan once the table passed a size you can name.
3. Cost the reversal in steps, not adjectives: expand-and-contract deploys, the dual-write window, the callers who had to be notified, the rows already written in the wrong shape that had to be backfilled or abandoned.
4. Distinguish reversal from rewrite by naming what you kept. Most good reversals preserve the schema or the interface and undo one decision inside it, which is also why they were affordable.
5. Finish on the process change: the smallest experiment that would have produced the same measurement in a day, and why you did not run it the first time.

Follow-up

• What in that decision was irreversible, and did you know it was irreversible when you made it?
• How did you tell the people who had already built on top of the original decision?
• What do you now measure before committing to a change of this size?