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Accrete Machine Learning Engineer Interview Questions & Guide 2026

Accrete Machine Learning Engineer interview guide: reported rounds, practice questions with worked solutions and a 7-day prep plan.

The 60-second brief

As a Machine Learning Engineer at Accrete, you are at the intersection of complex data processing and high-stakes decision-making. This role is fundamental to the company’s mission of building sophisticated AI solutions that transform how organizations interpret vast amounts of information. You will not just be building models; you will be architecting systems that solve real-world, high-dimensional problems.

The shape of the workload matters more for your prep than the industry label does. Read-heavy serving, write-heavy ingestion and scheduled batch processing have different binding constraints and fail in different places, so find out which one the team lives in before picking design topics.

Focus areas

  • Make every write idempotent under retry
  • Detect concurrent edits instead of losing writes
  • Choose indexes from the query's access path

Overview

In this position, you will contribute directly to the development of robust machine learning pipelines, specifically focusing on Natural Language Processing (NLP) and predictive modeling. Because Accrete operates in domains requiring high precision, your work has a direct impact on the efficacy of the products and the strategic value provided to clients. You will collaborate with scientists and engineers to translate complex requirements into scalable, reliable production code.

Expect a fast-paced environment that demands both academic rigor and practical software engineering discipline. Success here requires a balance of deep theoretical understanding and the ability to navigate the complexities of data pipelines. You will be challenged to maintain high standards for model performance while ensuring your solutions are maintainable and efficient.

The Accrete interview process

Accrete candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Deep-Dive Conversation

An unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.

What to demonstrate

  • Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
  • Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
  • Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating

How to prepare

  • Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
  • Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
  • Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes

Practical Assessment

Most of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.

What to demonstrate

  • Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
  • Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
  • Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
  • Whether a failing case is isolated and explained before any edit is made to the code

How to prepare

  • From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
  • Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
  • Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.

Where candidates lose points

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.

Running a schema change as though the lock lasts as long as the statement

In PostgreSQL an ALTER TABLE that needs an ACCESS EXCLUSIVE lock must first wait for every open transaction touching that table, and while it waits, later queries needing a conflicting lock queue behind it rather than overtaking it. A DDL statement that would execute in milliseconds, issued while a thirty-second analytics query is open, therefore stalls all traffic on that table for thirty seconds: the outage length is set by the longest open transaction, not by the change. The defences are specific and worth knowing by name - set lock_timeout low and retry rather than queue, add columns without a volatile default so no table rewrite occurs (from version 11 a non-volatile default is a metadata-only change), build indexes with CREATE INDEX CONCURRENTLY while accepting that it cannot run inside a transaction block and leaves an invalid index behind if it fails, and add constraints as NOT VALID followed by a separate VALIDATE CONSTRAINT, which takes a weaker lock.

Treating a network call as though it were a local function call

A remote call can be slow, fail, or return after you stopped waiting, so name the timeout, the retry policy, and what the caller sees while the dependency is down. A call with no timeout turns one slow dependency into an exhausted thread or connection pool in every service upstream of it.

Retrying a write that is not safe to repeat

A timeout tells you nothing about whether the server applied the write, so a blind retry of a create or a charge can duplicate it. Either make the operation idempotent, with a caller-supplied key the server deduplicates on or a conditional update, or do not retry it; and use exponential backoff with jitter so the retries of many clients do not synchronise into a second outage.

Questions & practice

Describe the bias-variance tradeoff.

Medium · machine learning fundamentals

Approach

  1. Say how you would validate it, and where leakage could enter the split.
  2. State the learning problem: the label, the unit of prediction and how the model is used.
  3. Name the simplest model that could work and what would make you move past it.

Follow-up

  • What changes if the classes are heavily imbalanced?
  • Where could label leakage enter this setup?

Describe the main idea behind boosting along with well-known algorithms such as AdaBoost and XGBoost.

Medium · machine learning fundamentals

Approach

  1. Say how you would validate it, and where leakage could enter the split.
  2. Pick the metric from the cost of each error type, not from habit.
  3. State the learning problem: the label, the unit of prediction and how the model is used.

Follow-up

  • What changes if the classes are heavily imbalanced?
  • Where could label leakage enter this setup?

Describe several options for embedding text data into a feature along with pros and cons of each.

Medium · machine learning fundamentals

Approach

  1. Name the simplest model that could work and what would make you move past it.
  2. Say how you would validate it, and where leakage could enter the split.
  3. State the learning problem: the label, the unit of prediction and how the model is used.

Follow-up

  • What changes if the classes are heavily imbalanced?
  • How would you know the model is overfitting?

What is an example of a problem where you would favor recall over precision and vice versa?

Medium · machine learning fundamentals

Approach

  1. Say how you would validate it, and where leakage could enter the split.
  2. State the learning problem: the label, the unit of prediction and how the model is used.
  3. Pick the metric from the cost of each error type, not from habit.

Follow-up

  • What changes if the classes are heavily imbalanced?
  • Where could label leakage enter this setup?

Describe your favorite sorting algorithm.

Medium · coding and algorithms

Approach

  1. State the target complexity and say which constraint rules the naive version out.
  2. Walk one small example through your approach before writing the whole thing.
  3. Name the brute-force solution and its complexity before improving on it.

Follow-up

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

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

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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.

Worked solution (30 min)

  1. Write the edge-loading query with the tenant predicate on both endpoints and state what it does with a cross-tenant edge.
  2. Implement iterative Tarjan with an explicit stack and confirm on a three-node cycle that it emits one component of size three.
  3. Build the transpose restricted to the visited set and mark every node with an in-edge from outside it as refused, carrying the referrer id.
  4. Run Kahn over the condensation and verify the emitted order against the referrer-before-referenced rule.
  5. Size the CSR arrays for 2,000,000 nodes and 8,000,000 edges and compare against a boxed adjacency map.

Expected result: An iterative O(V+E) traversal over a tenant-scoped CSR subgraph, SCC condensation so cycles archive atomically as one component, a transpose-based refusal list naming the external referrer for each blocked resource, and a Kahn topological order over the condensation, with recursion replaced by an explicit stack because of graph depth rather than style.

Verify it yourself

  • A two-node cycle with no external referrers archives as one unit, and adding a single external referrer moves both nodes to the refusal list.
  • No resource appears in the archive order before something that references it, except within its own component.
  • A counter on edge visits matches the loaded edge count exactly, confirming each edge is traversed once.

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?

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?

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?

Evolve the resource contract without breaking integrations you cannot upgrade

Hard · versioning · backward compatibility · deprecation

GET /v1/resources/{id} returns status from a four-value enum ('draft','active','archived','deleted'), a numeric version, and the body inline. Consumers are a browser app you deploy and roughly 300 server-side integrations, some untouched for two years, that switch exhaustively on status and parse ids as JSON numbers. You must add a 'pending_review' status, move bodies over 256 KB to a body_ref pointer, and expose per-field change history from resource_revision. Specify the compatibility policy, the wire changes, how both generations are served, and the evidence that lets you remove the old shape.

Approach

  1. Write the policy first and date it: fields are added, never retyped or repurposed; consumers ignore unknown fields; an unknown enum value maps to a documented fallback; nothing is removed until telemetry shows no caller reads it. Then say the uncomfortable part out loud - v1 shipped without the unknown-value rule, so 300 running integrations have no fallback, and no server change can install one into code that is already deployed.
  2. That single fact forces per-request negotiation rather than a server-side default. Keep one internal model and select a serialiser from an explicit version in the request, and default a caller that sends nothing to the oldest supported version. Defaulting to the newest is the change that breaks every integration that never asked for anything, on the day you ship.
  3. Downgrade 'pending_review' for old callers to the nearest state they already handle, 'draft', and state the loss explicitly: those integrations cannot see review state and will treat the resource as editable. If that is unacceptable for one integration, the remedy is moving it to the new version, not a cleverer projection - there is no mapping that invents a state the client has no code for.
  4. Make the body change additive. Old callers keep body inline; the new shape adds body_ref and a size field, and resources over the limit are served to old callers by resolving the pointer server-side or by refusing with a documented code, chosen once and published. Never repurpose body to carry the pointer: a client that renders it shows a storage key to a user, and that failure is silent, where a missing field would have been loud. While you are here, serialise BIGINT ids as strings in the new shape - a browser parsing JSON numbers gets IEEE-754 doubles, exact only to 2^53 - and treat that as its own breaking change requiring the same negotiation, not a quiet fix.
  5. Add change history as a separate sub-resource, GET /v1/resources/{id}/revisions, keyset-paginated over (resource_id, version) rather than as an array inside the resource. A field added to a hot response is paid for by every caller including those that never read it, and an unbounded array inside a cached object destroys the size assumptions the cache was configured with.
  6. Retire on evidence rather than on a date alone: count requests per negotiated version per credential, publish a Sunset header (RFC 8594) with the removal date and a link to the migration, contact the credentials still on the old version, then answer 410 Gone once it is removed. Keep each version's serialiser under snapshot tests so a refactor cannot change v1's bytes by accident.

Worked solution (40 min)

  1. Write and date the compatibility policy as five rules, then mark which of them v1 callers cannot honour.
  2. Choose the negotiation mechanism and the default for an unversioned request, and justify the default in one sentence.
  3. Write the downgrade table: new state to old state, new body shape to old body shape, and what is lost in each direction.
  4. Specify the revisions sub-resource: its URL, its cursor, and why it is not a field on the resource.
  5. Write the sunset plan: the per-version per-credential metric, the header, the lead time, and the terminal status code.

Expected result: A dated additive policy, version negotiation over one internal model with the oldest version as the default, a downgrade table with its losses named, history as a paginated sub-resource, and a removal gated on per-credential version telemetry rather than a calendar entry.

Verify it yourself

  • A two-year-old integration receives a resource that is really in 'pending_review' and neither crashes nor renders a blank state.
  • Deleting the legacy body field from the internal model breaks exactly one serialiser and no business logic.
  • Every v1 response is byte-stable under snapshot tests after an unrelated refactor.

Follow-up

  • An old integration submits a status transition while the resource is really in 'pending_review'. What does the write path accept, and what does it reject?
  • Two years on you want to delete the v1 serialiser. What evidence makes that safe, and who must be contacted before it happens?
  • How would you test against a two-year-old integration rather than against today's source?

Design the async export contract a client can resume safely

Hard · async operations · idempotency · leases

A tenant asks for a CSV of every resource. The work runs for minutes on the worker fleet through a job_run row carrying a lease, an attempt count and a dedupe_key, far past the edge's 400 ms budget. Callers are a browser that polls and a script that walks away and checks later. Specify what the submit call returns, the operation resource and its states, how a duplicate submit is handled, how a client learns about completion, what cancellation means given that a lease can expire mid-run, and how the result is fetched and when it expires.

Approach

  1. Split the API in two. Submit returns 202 with an operation id and a location to poll, and never blocks on the work. The operation is a real resource with its own lifecycle - queued, running, succeeded, failed, cancelled - plus attempt, a monotonic progress figure, and a terminal error drawn from the same code taxonomy the synchronous endpoints use, so a client needs one error vocabulary rather than two.
  2. Deduplicate at submit using job_run.dedupe_key, unique over (job_type, dedupe_key) while status is 'queued' or 'running': a repeat submit of the same logical export returns 200 with the existing operation instead of 202 with a new one, and the partial index deliberately permits a legitimate re-run once the first has finished. Pair it with the request's idempotency key so an HTTP-level retry of the submit is exact rather than merely similar.
  3. Tell the poller how to poll: Retry-After on the polling response, a minimum interval enforced at the edge, and a documented maximum lifetime after which an operation is reaped. Polling is the contract of record; the webhook is the fast path, and both must lead to the same terminal state, so a client that receives the completion event and then polls anyway sees no contradiction.
  4. Be exact about cancellation. A cancel request records intent; it cannot stop work already executing. The handler reads the flag at checkpoints, and because a lease expires on a clock that cannot distinguish a dead worker from a slow one, a second copy may start after the cancel was recorded - so the handler re-reads the flag immediately after claiming the lease. 'cancelled' becomes terminal only when no lease is outstanding; reporting it earlier shows a client a stopped job while a worker is still writing output.
  5. Make the handler safe to run twice, because the lease guarantees that it will be. Write output to a deterministic object key derived from the operation id so a second copy overwrites its own work instead of appending a second file, and record completion with a conditional update that only the copy holding the current lease can win.
  6. Treat result fetch as a separate authorised read: a short-lived signed URL, the tenant checked when it is issued rather than only when the file was produced, and a documented retention after which the operation remains terminal but the bytes are gone - a state the client must be able to tell apart from a failure.

Follow-up

  • An operation has said 'running' for 40 minutes and the worker is gone. What does the client see, and which columns in job_run decide that?
  • Two tenants each submit 50 exports at once. What in this contract stops one of them delaying the other?
  • The customer wants the export emailed instead. What changes, and what becomes harder to make exactly-once?

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 fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.

Day 1 — Measure before reasoning

  • Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
  • Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
  • Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.

Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.

Day 2 — References, copies, and the bugs they produce

  • Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
  • Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
  • Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.

Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.

Day 3 — Types, once, in a language that checks them

  • Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
  • Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
  • Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.

Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.

Day 4 — Concurrency, starting with what actually runs at the same time

  • Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
  • Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
  • Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.

Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.

Day 5 — Debugging as a procedure rather than an instinct

  • Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
  • Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
  • Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.

Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.

Day 6 — Tests that catch the bug you are about to write

  • Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
  • Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
  • Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.

Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.

Day 7 — Debug something broken, out loud

  • Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
  • Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
  • Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.

Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.

Bounding a vague ticket before writing code

When the requirements were thin, the interesting part is how you fenced the problem off: the assumption you wrote down, who you got to confirm it, the narrow version you shipped first so the rest stayed cheap to change. Guessing and being right is luck. Guessing in writing, where someone could correct you, is method.

Why do you want to change your career?

Medium · behavioural and collaboration

Approach

  1. Name the disagreement and how you resolved it with evidence.
  2. Close with what you would do differently, concretely.
  3. State the situation in two sentences and spend the rest on the reasoning.

Follow-up

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

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?

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

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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?