Interview guides · DMC

DMC Software Engineer Interview Questions & Guide 2026

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

The 60-second brief

A Software Engineer at DMC serves as a vital bridge between high-level software architecture and physical-world implementation. You will not be confined to pure web or application development; instead, you will work on complex, multidisciplinary projects that often integrate custom software with hardware, industrial automation, and control systems. Your work directly impacts how physical systems operate, ranging from factory automation and robotics to specialized data-acquisition platforms.

For a design discussion, a catalogue of architectures is worth less than the ability to turn a vague requirement into a data model and an API contract. A box diagram with no schema under it collapses at the first follow-up question.

Focus areas

Bound every outbound call with a timeout
Paginate large result sets with keyset cursors
Choose indexes from the query's access path

Overview

This role is critical to DMC because it requires a rare blend of deep technical curiosity and practical problem-solving. You will be expected to thrive in an environment where the "system" includes not just code, but sensors, actuators, and mechanical constraints. If you enjoy understanding the "why" behind a system's failure and have the drive to architect robust, scalable, and elegant solutions to messy real-world problems, you will find this position highly rewarding.

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.

Letting a slow dependency consume unbounded concurrency

The failure that takes a service down is usually not an error but a delay. A dependency answering in thirty seconds instead of fifty milliseconds holds each request's worker or connection six hundred times longer, and since required concurrency is arrival rate times latency, a fleet sized for sixty in-flight requests now needs thirty-six thousand to sustain the same rate - so it queues, and requests whose clients have already abandoned them still occupy resources. Retries make it precisely worse: a policy of three attempts triples the load on a dependency at the exact moment it is least able to serve, which is how one slow dependency becomes an outage of everything sharing that pool. Containment is four specific things - a timeout on every outbound call shorter than the caller's remaining budget, a bounded pool per dependency so one cannot starve the others, backoff with full jitter rather than a fixed delay so retries do not resynchronise, and a circuit that stops sending once the failure rate makes an attempt pointless.

Assuming the input fits in memory

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

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.

Questions & practice

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.

Follow-up

The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
How would you make the window survive a process restart, and is it worth the cost?

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.

Approach

Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.

Follow-up

The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?

Identify the heaviest tenants in a five-minute window under memory pressure

Medium · top-k · heavy hitters · streaming

The edge service handles about 3,000 requests per second across roughly 50,000 tenants, peaking near 9,000. Expose the 50 heaviest tenants by request count over the trailing five minutes so limits can be tightened before one tenant's backfill starves the fleet. You may not retain five minutes of raw records. Give the exact solution and its memory, then the bounded-memory approximation with its error stated as a formula, and say which you would ship and at what tenant cardinality that choice changes.

Approach

Do the exact version first, because it is affordable at this cardinality: a ring of 300 one-second counters per tenant, advanced lazily, is 1,200 bytes of counters per tenant and roughly 60 to 90 MB for 50,000 tenants with overhead. Carry a running total and subtract the bucket you overwrite so a window read is O(1) rather than 300 adds.
Extract the top 50 with a size-k min-heap over the tenant sums: O(d log k) for d tenants, against O(d log d) to sort them all. Maintaining the heap continuously instead of on query requires a tenant-to-heap-index map, because incrementing a count already inside the heap means sifting from a known position, and without that map you rebuild the heap on every request.
State the approximation precisely rather than gesturing at sketches. Misra-Gries with m counters retains every item whose true count exceeds N/(m+1), and each retained count underestimates the truth by at most N/(m+1). With m = 1,000 and N = 900,000 requests in the window the error is roughly 900 requests, which is fine for spotting a tenant sending 50,000 and useless for ranking two tenants 200 apart.
Say what breaks when the window slides: Misra-Gries and Space-Saving are insert-only and cannot be decremented as records age out. The workable construction is one summary per sub-window, say ten seconds, with 30 summaries merged at query time, and the merged error is the sum of the per-summary errors, so the bound degrades linearly in the number of sub-windows.
Choose and defend it: at 50,000 tenants the exact rings cost under 100 MB in a process that already holds more, so ship exact. Keep the sketch for the case that actually motivates it, a per-principal or per-IP key where cardinality runs to millions and is not bounded by anything you control.
Raise the fleet problem before it is asked: each of 20 to 40 instances sees only its share, and the top 50 of one shard is not the top 50 of the fleet. Either aggregate counts centrally or accept that a per-instance threshold multiplied by instance count is the limit you are really enforcing.

Worked solution (25 min)

Size the exact structure: 300 one-second counters per tenant across 50,000 tenants, plus the running-total trick that makes a window read O(1).
Write the top-k extraction with a size-50 min-heap and compare its complexity against sorting all 50,000 sums.
Substitute N = 900,000 and m = 1,000 into N/(m+1) and state in requests what the sketch can and cannot distinguish.
Write the sub-window merge for the sliding case and state the resulting bound for 30 merged summaries.

Expected result: An exact per-tenant ring of 300 one-second counters at roughly 60 to 90 MB for 50,000 tenants with O(1) window reads, top-50 extraction by a size-k min-heap in O(d log k), a Misra-Gries bound of N/(m+1) with the numbers substituted, the sub-window merge needed to slide it, and a decision to ship exact at this cardinality with the sketch reserved for unbounded keys.

Verify it yourself

The exact window sum for one tenant equals a replay of its raw records over the same five minutes.
The heap result matches a full sort of the tenant sums on a sample.
The sketch's count for a known heavy tenant is at most the true count and no less than the true count minus N/(m+1).

Follow-up

The heaviest tenant is heavy because of one export job rather than user traffic. Should the limiter treat those as the same tenant?
Two tenants sit tied at the boundary of the top 50. Does your answer flap, and does the flapping matter?
You switch to per-principal keys and cardinality goes to 10 million. Walk through what changes.

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

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.
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.
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.
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.
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.
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.

Worked solution (25 min)

Write the three statements plus the rowcount branch and confirm they sit in one BEGIN/COMMIT with no outbound call between them.
Run two clients that both read version 7 and apply their updates 5 ms apart; assert one 200 and one 409.
Assert resource.version = 8, exactly one resource_revision row at version 8, and one outbox_event row at aggregate_version 8.
Repeat at REPEATABLE READ and record the different failure shape (SQLSTATE 40001) the handler must also map to 409.
Delete the version predicate and re-run: both writes commit and the first edit disappears with no error raised anywhere.

Expected result: Exactly one of the two commits. version = 8 with one revision row and one outbox row at version 8, the loser gets 409 carrying version 8, and no state exists in which the resource moved but the revision or the outbox row is missing.

Verify it yourself

COUNT(*) of resource_revision for that resource equals resource.version, with no gap in the version sequence.
Kill the process between the UPDATE and the outbox INSERT — the transaction rolls back and no event describes a change that never happened.
Under a 50-way race, resource.version equals the number of 2xx responses rather than the number of requests.

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?

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

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.
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.
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.
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.
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.
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.

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?

Explain the buoyancy force on a brick and a balloon inside water.

Medium

Approach

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

Follow-up

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

Plot velocity and acceleration based on a position graph.

Medium

Approach

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

Follow-up

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

Describe the purpose and uses of an inductor and a capacitor.

Medium

Approach

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

Follow-up

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

Analyze an encoder waveform to determine the number of ticks.

Medium

Approach

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

Follow-up

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

Specify the outbound webhook contract a customer endpoint can verify

Medium · webhooks · hmac · at-least-once · ordering

Egress delivery sends resource.created, resource.updated and resource.archived to roughly 40k customer endpoints at about 1.5k deliveries/second, with a per-destination concurrency cap of 4 and a 10 second connect-plus-read timeout, at-least-once from the relay. The receiver is code you do not control, often behind a proxy that may re-encode JSON. Specify the payload envelope, how the receiver verifies authenticity and rejects replays, what ordering it may assume, how it deduplicates, which status codes it should return and what each means to us, and the retry, circuit and replay behaviour.

Approach

Send the fact, not a pointer: event_id, event_type, aggregate_type, aggregate_id, aggregate_version, occurred_at and the payload copied from outbox_event. A receiver told only that resource 42 changed has to call back, and the call back returns a newer state than the event describes - that is precisely how a consumer applies changes out of order while believing it is following them.
Sign the exact bytes on the wire: HMAC-SHA256 over the timestamp, a separator, and the raw body, in a header, compared in constant time before the body is parsed. Instruct receivers to verify the raw bytes, never a re-serialised copy, since a proxy that reorders keys or changes whitespace invalidates a signature computed over re-encoded JSON. Reject a timestamp outside a tolerance - five minutes is typical - and keep a short seen-set of event_ids, because a capture replayed inside the tolerance still verifies.
Support key rotation by sending several signature values in one header during the overlap window and publishing that any one that verifies is sufficient. Without that rule a key rotation is a synchronised outage across every destination at once.
State the guarantee as it actually is: at-least-once and unordered, including within a single aggregate_id. The outbox is ordered per aggregate at the source, but the relay holds up to 4 requests in flight per destination and backs off each retry independently, so a version 6 delivery that needs one retry lands after a version 7 that succeeded on its first attempt. Publishing per-aggregate ordering we do not enforce is worse than publishing none, because a receiver that trusts it overwrites new state with old. Receivers deduplicate on event_id and apply conditionally - UPDATE ... WHERE aggregate_version < :incoming, or the same comparison under a row lock on the aggregate - which discards stale redeliveries and stays correct when two deliveries for one aggregate are processed at the same time. A duplicate and an out-of-order delivery then have the same outcome: no-op.
Define the response contract in the receiver's terms. 2xx means durably accepted, and accepting means writing it down - a receiver that does its processing inline inside our 10 second timeout will be retried while it is still working. 4xx is permanent and we stop. 429, 5xx and timeouts are retried with capped exponential backoff and full jitter within the per-destination concurrency cap, and a circuit opens after a run of consecutive failures and half-opens on a schedule, so a permanently dead endpoint costs one probe per interval instead of a growing queue.
Give them a recovery path that does not involve us holding a queue for everyone: a visible parked state, and a replay endpoint keyed by event_id or time range, so a receiver that was down for two hours asks for the gap on its own schedule.

Worked solution (30 min)

Write the envelope fields and mark which are needed for deduplication and which for discarding a stale delivery.
Write the signature string construction and the receiver's verification steps in order, including the constant-time compare and the tolerance check.
Write the receiver's idempotent apply: dedupe on event_id, then the version-conditional write that discards stale and concurrently delivered redeliveries.
Write the status-to-action table from our side: which codes retry, which stop, and what opens the circuit.
Describe the parked state and the replay call, including what the customer must supply.

Expected result: A self-contained signed envelope, verification over raw bytes with a replay window and a seen-set, an at-least-once guarantee that states delivery is unordered even within one aggregate, a version-conditional apply, a status table that distinguishes permanent from transient, and a customer-driven replay path.

Verify it yourself

Delivering the same event twice leaves the receiver's state identical to delivering it once.
Delivering version 7 and then version 6 for one aggregate leaves version 7 applied.
A destination that always times out costs one probe per circuit interval rather than a retry queue.

Follow-up

A customer insists an event never arrived. What do you check, in order, and which table settles the argument?
A receiver cannot store a version and asks us to deliver their events in order. What would the relay have to give up to offer that, and what does one failing event then do to the rest of their stream?

One log partition stops advancing while the others drain

Easy · poison message · head-of-line blocking · consumer lag · projections

Search results for a subset of tenants are hours stale; the rest are current. The projection consumer reports lag of zero on 15 of 16 partitions and 400,000 on one. Its error rate is flat and its CPU is idle. outbox_event has no pending rows older than a second, so the relay has published everything it holds. Identify the mechanism, give the ordered checks, and state what you do in the first ten minutes versus what you change permanently.

Approach

Read the lag distribution first. A slow consumer lags everywhere; zero on fifteen partitions and 400,000 on one is not throughput. Idle CPU on the stuck partition means the consumer is not advancing its offset at all, which points at one message it cannot get past rather than at a rate problem.
Exonerate the producer before touching the consumer. No pending outbox rows older than a second means the relay published, so the event exists in the log. This separates never sent from sent and never applied, which are different code paths and usually different owners.
Read the message at the stuck offset and the handler's log lines for its event_id. A flat error rate with no progress has two explanations and you must distinguish them: the handler is throwing and the retry loop is swallowing it, or the handler is blocking on something and never returning. Idle CPU with no error lines favours the second.
Mitigate before diagnosing further. Move the offending event to a dead-letter store and commit the offset past it. Adding consumers does nothing here, because a partition is consumed by exactly one member of the group, and the blast radius is every aggregate hashed to that partition, not only the aggregate that produced the bad event.
Fix permanently by bounding handler attempts and dead-lettering on exhaustion, so no single message can stop a partition. Then replay the dead-lettered event once the handler is fixed: it carries aggregate_id and aggregate_version, so a consumer that discards versions it has already applied can absorb the replay, and resource_revision is the fallback if the event itself is unusable.

Follow-up

The dead-lettered event carried aggregate_version 7 and the projection had applied 6. What must the replay do differently if 8 and 9 landed in the meantime?
How do you show staleness to the user while the partition is behind, given the API already returns the projection's watermark?
What changes if the message is poison because a previous deploy wrote a payload shape the current code cannot parse?

Suggested practice checklist

Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.

Day 1 — Coding, one pass at shallow depth

Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
Fix nothing today. The value of the pass is the unfixed record.

Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.

Day 2 — Design, one pass at shallow depth

Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.

Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.

Day 3 — Fundamentals and the practical rounds

Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.

Deliverable: Eight scored short answers and one written reading of unfamiliar code.

Day 4 — The rounds that are about you, and the map

Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.

Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.

Day 5 — First chosen area, to the depth you skipped

Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.

Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.

Day 6 — Second chosen area, where the gap is coverage rather than speed

Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.

Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.

Day 7 — Reassemble the loop

Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
Reduce the week to one page holding only the rules you can state without reading them.

Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.

Know where your numbers came from

Bring the two or three numbers the story rests on and know how they were collected. A p99 whose timer starts inside your handler excludes the time a request spent queued, so it can sit flat while users wait longer. Give the window, the percentile and what the measurement left out, or drop the number.

Estimate work you have never done and defend the range

Hard · estimation · backfills · expand-contract

You are asked to estimate a change you have never attempted: add a column to a 100-million-row table, populate it, move reads across, and drop the old shape. Give a range with the assumptions that generate it, including batch size, the signal your backfill throttles on, and wall-clock hours, and name the three unknowns that would move the number most. Then describe a real estimate you gave under comparable ignorance: how you expressed its uncertainty, what you committed to, and how wrong you turned out to be.

Approach

Decompose into independently deployable steps before estimating anything: add the column nullable, write both shapes, backfill in batches, verify, move reads, stop writing the old shape, drop it. That is four deploys spread over days, and the calendar estimate is dominated by them rather than by the loop's runtime.
Do the arithmetic aloud for the part that has arithmetic in it: batch size times number of batches times per-batch duration, at a write rate the primary can absorb alongside roughly 1.2k writes per second of production traffic. The loop is throttled by replication lag and lock waits, not by how fast it can issue statements.
Price the schema step by its lock rather than its statement duration. In PostgreSQL an ALTER TABLE taking ACCESS EXCLUSIVE waits for every open transaction on that table while later queries queue behind it, so a millisecond change issued during a thirty-second analytics query stalls that table for thirty seconds. Adding a nullable column with a non-volatile default avoids a rewrite from version 11; a new index wants CREATE INDEX CONCURRENTLY, which cannot run inside a transaction block and leaves an invalid index behind if it fails.
Express the answer as a range whose endpoints each trace to a stated assumption, then name the cheapest experiment that collapses it, which is almost always running one real batch against the real table and multiplying.
Commit to a checkpoint rather than a completion date: the day you report a measured number from that first batch. That is a promise you can keep under uncertainty, and it is what the asker actually needs in order to plan.

Follow-up

How do you verify the backfill genuinely finished, given rows written by production traffic while it ran?
Where does the backfill resume from after a worker is killed mid-batch, and what makes that resume point trustworthy?
Your first batch comes back ten times slower than assumed. What do you tell the person waiting on the estimate, and when?

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

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.
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.
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.
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.
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?

Ship under a deadline and bound the debt you chose

Medium · pagination · technical debt · tradeoffs

You have four days to ship a tenant-facing listing endpoint. The version you would defend uses keyset pagination over (tenant_id, status, updated_at DESC, resource_id DESC); the version you can finish uses LIMIT/OFFSET with no matching index. Describe a deadline call you actually made of this shape: what you shipped, what you knowingly deferred, how you bounded the damage with a mechanism rather than an intention, and the specific numeric condition that would force the follow-up. Name who you told and where you wrote it down.

Approach

Name the deferred failure precisely instead of calling it slow. OFFSET n makes the database produce and discard n rows, so cost grows with page depth; without an index matching the sort, every matching row is read and sorted before the limit applies; and rows inserted between two page fetches shift across the boundary so items are skipped or repeated with nothing in the response to signal it.
Bound the blast radius with something mechanical rather than a promise: cap maximum page depth, cap page size, restrict the endpoint to one internal caller, or keep it behind a flag. State which failure each cap removes and which it leaves standing.
Attach a number to the trigger and wire it to an alarm: the first tenant crossing N resources, or the endpoint's p99 crossing its share of the 400 ms budget, so the debt announces itself instead of waiting to be remembered.
Write it where the next engineer looks, which is the code and the ticket, not a chat message: what was deferred, why, the cap, and the trigger.
Report what actually happened in your real example, including the case where the trigger never fired and the debt was correctly never repaid.

Follow-up

At what page depth does the offset version breach your latency budget, given your page size and row counts?
What breaks first when you switch to keyset pagination later, and what does a client holding an old page token see?
Who would have overruled you if you had asked for two more days, and did you ask?