Crunchyroll Machine Learning Engineer Interview Questions & Guide 2026

The 60-second brief

As a Machine Learning Engineer at Crunchyroll, you play a pivotal role in enhancing user experiences through personalized content recommendations and advanced fraud detection algorithms. This position is vital for driving the strategic initiatives that enable Crunchyroll to deliver tailored content to its millions of users, ensuring they discover shows and movies that resonate with their preferences. By applying machine learning techniques, you'll help shape how users interact with the platform, ultimately impacting user engagement and satisfaction.

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
• Make every write idempotent under retry
• Trace a symptom to a mechanism under load

Overview

The complexity and scale of this role are significant. You will work with large datasets, employing state-of-the-art algorithms to tackle challenges in recommendation systems and fraud detection. Your contributions will not only improve user retention but also safeguard the integrity of the platform. Collaborating with cross-functional teams, you'll engage in projects that directly influence the growth and functionality of Crunchyroll's services, making this role both critical and exciting for those passionate about technology and entertainment.

The Crunchyroll interview process

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

Phone Screen

Half of this call is the part candidates treat as small talk: start date, notice period, work authorisation and its timing, location and time zone, on-call, and the number. Those are what kill offers late, after several engineers have each spent a day. Surfacing a hard constraint now costs you nothing and occasionally buys you something, since a loop compressed to fit a competing deadline can usually only be arranged if it is asked for early. The common failure is deflecting the compensation question twice, then discovering at offer stage that the band never reached your number.

What to demonstrate

• Whether your hard constraints are compatible with the role before a loop gets booked: earliest start, notice period, what authorisation you hold and when it needs action, days on site, willingness to carry a pager
• Whether you give a compensation range with something behind it, such as current total compensation or a competing timeline, rather than leaving the band untested
• Whether your stated timeline is real, since a competing deadline raised now is something scheduling can sometimes work around and the same deadline raised at offer stage usually is not

How to prepare

• Write each constraint down in one line before the call and state them as facts rather than negotiating them live under a question you were not expecting
• Set your range from two or three current data points for that level and location, and name the structure you are quoting in, so the number is comparable to the one they are holding
• If another process is running, say where it stands and by when, and ask directly whether this loop can be scheduled inside that window

Technical 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.

Final Interviews

A day like this is several different games in a row, and the expensive mistake is carrying the previous one into the next room. Coding rewards narrow precision and finishing inside a timer. Design rewards breadth, stated assumptions and naming what you are deliberately not building. Behavioural rewards specificity about people and decisions. Candidates who over-engineer a coding problem they were supposed to finish, or who start sketching class hierarchies before anyone has agreed what the system has to do, are usually still playing the last round. Between rooms, name out loud which game the next one is.

What to demonstrate

• Whether the coding round ends with something that runs and has been traced against a degenerate input, rather than an extensible design that was never finished
• Whether a design discussion opens by agreeing on traffic shape, read-to-write ratio and what is allowed to be stale, instead of proceeding from an architecture you arrived with
• Whether a behavioural answer names a person, a disagreement and what you did about it, rather than describing the system the story happened inside
• Whether the opening habits still appear late in the day: restating the problem, asking for constraints, saying the plan before typing

How to prepare

• Book three mocks of different types back to back on one afternoon and ask each interviewer afterwards which round you answered in the wrong mode
• Write a three-line opening script per round type — coding: restate, name the approach and its cost, then type; design: ask for scale, read-write mix and what must not break; behavioural: name the person, the stakes and the decision — and run it off a card so the switch is mechanical rather than remembered
• Practise coding with a timer you do not extend, stopping when it stops, so the trained reflex is to finish a correct solution rather than to keep improving one

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.

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.

A queue or buffer with no bound

Every producer-consumer boundary needs a capacity and a policy for reaching it: block the producer, shed load, or drop the oldest entry. Unbounded buffering converts a temporary slowdown into memory exhaustion and hides the backpressure signal that would have revealed the consumer was falling behind.

Hardcoding to the sample inputs

Solve the stated problem rather than the two examples; special-casing a literal to make a sample pass is obvious immediately and reads as either a misunderstanding or an attempt to fake progress. If you genuinely cannot generalise yet, say which part is a stub and what would replace it.

Questions & practice

Discuss the importance of feature engineering and how you would approach it.

Medium · machine learning fundamentals

Approach

1. Pick the metric from the cost of each error type, not from habit.
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?
• Where could label leakage enter this setup?

Discuss how you would ensure the robustness of your machine learning services.

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. Name the simplest model that could work and what would make you move past it.

Follow-up

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

How do you handle overfitting in a model?

Medium · machine learning fundamentals

Approach

1. Name the simplest model that could work and what would make you move past it.
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

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

Write a function to implement linear regression from scratch.

Medium · machine learning fundamentals

Approach

1. Pick the metric from the cost of each error type, not from habit.
2. Name the simplest model that could work and what would make you move past it.
3. Say how you would validate it, and where leakage could enter the split.

Follow-up

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

How would you optimize a sorting algorithm to handle large datasets?

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. Choose the data structure from the access pattern, not from familiarity.

Follow-up

• Which test case would catch an off-by-one here?
• How does this change if the input no longer fits in memory?

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

Easy · hashing · at-least-once · aggregation

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

Approach

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

Worked solution (20 min)

1. Write the pass: look up last_applied_version, skip if the event's version is not greater, otherwise upsert into the keep-map only when the incoming version exceeds the version already held, incrementing the discard counter on every skip.
2. Hand-trace one aggregate whose events arrive as v5, v3, v4, v5 with last_applied_version = 2, and confirm the output holds v5 once while the counter reads 3.
3. Compute the table footprint for 2,000,000 entries at 12 bytes of payload and a 0.7 load factor, then state the multiplier for a runtime that boxes keys and values.
4. Add the hash-partitioning fallback and say in one sentence why the per-partition results need no cross-partition merge logic.

Expected result: A single-pass O(n) reduction keyed on aggregate_id that keeps the maximum version rather than the last occurrence, O(d) space with the byte cost stated for 2,000,000 aggregates, a hash-partitioning fallback keyed on aggregate_id, and an explicit statement that batch-local deduplication does not replace a projection write keyed on (aggregate_id, version).

Verify it yourself

• Shuffling the batch order changes nothing in the output, which is the test that separates max-wins from last-wins.
• Feeding the same batch twice produces an identical keep-set and only doubles the discard counter.
• Every emitted event has a version strictly greater than the supplied last_applied_version for its aggregate.

Follow-up

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

Denormalise tenant onto revisions and backfill it live

Hard · denormalisation · online migration · backfill

resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.

Approach

1. Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
2. Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
3. Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
4. Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.
5. Step three, backfill: batch by primary key rather than by created_at so the cursor is dense and resumable — UPDATE resource_revision rr SET tenant_id = r.tenant_id FROM resource r WHERE r.resource_id = rr.resource_id AND rr.revision_id > $1 AND rr.revision_id <= $1 + 5000 AND rr.tenant_id IS NULL — committing per batch and persisting the cursor. Throttle on replica replay lag and on dead-tuple count, since each batch writes 5,000 new row versions. Run the backfill before the index exists so those updates can stay HOT.
6. Step four, index then enforce then contract: CREATE INDEX CONCURRENTLY (cannot run inside a transaction block, scans the table twice, waits on open transactions, and leaves an INVALID index to drop concurrently if it fails); ADD CONSTRAINT ... CHECK (tenant_id IS NOT NULL) NOT VALID, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE, after which SET NOT NULL uses the validated check instead of re-scanning on PostgreSQL 12 and later. Only then move the audit reads onto the column and, in a later deploy, delete the join path.

Follow-up

• The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
• How do you verify the backfill actually finished, given rows are still being inserted while it runs?
• A resource must now be movable between tenants. What does that do to the composite foreign key and to the revisions already written?

Keep soft-deleted accounts from blocking re-registration

Easy · soft delete · unique constraints · engine differences

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

Approach

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

Worked solution (20 min)

1. Create the PostgreSQL partial unique index, insert a live row, soft delete it, and insert the same address again.
2. Repeat the delete-and-reinsert cycle three times and confirm three deleted rows coexist with exactly one live row.
3. Write the MySQL form with the discriminator, then deliberately reverse the polarity so live rows carry NULL, and show two live duplicates commit.
4. Attempt a second live insert on both engines and map the resulting 23505 / ER_DUP_ENTRY to the 409 the handler should return.

Expected result: PostgreSQL rejects only the second live row and admits unlimited deleted ones. MySQL behaves identically with UNIQUE (tenant_id, email, deleted_marker) where live = 0; with the polarity reversed, two live rows commit and nothing afterwards can tell them apart.

Verify it yourself

• Group live rows by (tenant_id, lower(email)) after a randomised insert/delete run — no group exceeds one.
• Inserting 'User@x.test' after 'user@x.test' is rejected; if it is accepted, the case handling is in application code rather than in the index.
• EXPLAIN a login lookup and confirm the partial index is chosen — it is only usable by queries that carry the deleted_at IS NULL predicate, so a query without it falls back to a scan.

Follow-up

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

You are tasked with reducing the latency of a recommendation engine. What steps would you take?

Medium · ml system design and production

Approach

1. Say where features come from at serving time and how they match training.
2. Fix the product goal and the online metric before choosing any model.
3. Name what you would monitor after launch and what triggers a retrain.

Follow-up

• How would you roll the new model out safely?
• What happens when a feature is missing at serving time?

How would you design a recommendation system for a streaming service?

Medium · ml system design and production

Approach

1. Say where features come from at serving time and how they match training.
2. Fix the product goal and the online metric before choosing any model.
3. Name what you would monitor after launch and what triggers a retrain.

Follow-up

• How would you roll the new model out safely?
• What happens when a feature is missing at serving time?

Describe a microservices architecture you have worked on for machine learning.

Medium · system design

Approach

1. State the consistency you need, and where you are willing to be stale.
2. Name the read and write paths separately; they rarely have the same bottleneck.
3. Choose a partition key and say what query it makes expensive.

Follow-up

• What would you drop to keep the system up under load?
• What breaks first when traffic grows ten times?

Explain the difference between supervised and unsupervised learning.

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. Work from the requirement backwards to the design.

Follow-up

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

Design the bulk write endpoint a migration script retries blindly

Medium · batch api · partial failure · idempotency

A customer's migration script pushes 2 million resources through POST /v1/resources:batch, up to 500 items per call, and retries any call that errors or times out. Within a call some items fail validation, some collide with rows that already exist, and some succeed. Specify the request and response shape, whether a batch is atomic or per-item, how idempotency works for the call and for each item, the status code for a mixed outcome, the size and item-count limits with their error codes, and exactly what the script does after a timeout mid-batch.

Approach

1. Choose atomicity deliberately and price it. All-or-nothing means one transaction holding locks for the whole batch on a primary already absorbing about 1.2k writes/second, which bounds batch size by lock duration, and it turns one bad row into 499 rejections the script must resend. Per-item partial success is the right default for a migration, and the contract's job is then to make a partial outcome impossible to miss.
2. Require a client-supplied id on every item and echo it in every result. Deriving a per-item key from the array index breaks the first time the script resends a batch with the failures removed: the indices shift, previously-succeeded items acquire new keys, and they are created a second time.
3. Key the effects at two levels. The call's Idempotency-Key covers an exact resend of the same bytes; per-item keys of (tenant_id, client_item_id) make a partially-applied batch safe to resend whole. Resending an identical batch must reproduce the same per-item results, not 500 conflicts the script has to interpret.
4. Answer a mixed outcome with one status plus per-item detail: 200, or 207 borrowed from WebDAV if you prefer it - document whichever you pick - carrying an array of client_item_id, per-item status, and either resource_id or an error code from the same taxonomy the single-item endpoint uses. Reserve 4xx for the request as a whole: unparseable body, too many items, payload over the limit (413). Put a failed count at the top level so that even a script checking only the cheapest thing cannot conclude success while rows were dropped.
5. Cap the request before doing any of it: item count, total bytes, and concurrent batches per tenant, since one tenant's migration otherwise consumes write capacity everyone shares. Anything that cannot finish inside the request deadline belongs on job_run behind a 202, not in a synchronous call that will time out halfway.
6. Script behaviour after a timeout: the outcome is unknown and no results were received, so resend the identical batch with the same keys and read the results. Never resend 'only the items I have no result for' - a timeout yields no results at all, and that rule silently means resend everything anyway.

Worked solution (30 min)

1. Write the request schema with the per-item client id, and the response schema with per-item status and a top-level failed count.
2. Write the atomicity decision and the sentence of justification that names the lock cost or the resend cost.
3. Define both key levels and trace a resend of a half-applied batch through them, item by item.
4. List the whole-request rejections with their status codes and limits.
5. Write the script's timeout rule and the pacing it should apply between calls.

Expected result: A per-item batch with client-supplied item ids, per-item idempotency plus a call-level key, a mixed-outcome response that a status-only check cannot misread, explicit request caps, and a resend rule that is correct when no results were received.

Verify it yourself

• Resending an identical batch twice leaves exactly the same set of resource rows as sending it once.
• Resending a batch with the successful items removed does not create duplicates of anything.
• A caller that reads only the top-level fields can still tell that some items failed.

Follow-up

• At 500 items per call, what is the wall-clock time for 2 million rows, and what pacing do you publish so the migration does not become an incident?
• One item in every batch fails with the same code. How does the script discover that without a human reading logs?

Edge instances grow 400 MB per hour until the nightly restart

Hard · memory leak · heap analysis · metric cardinality · gc

Edge API instances start at 700 MB resident and grow about 400 MB/hour; a nightly rolling restart has hidden it for weeks. Growth continues unchanged when request rate halves overnight, p99 degrades in the last hours before an instance is recycled, and heap used immediately after a forced full GC rises monotonically. The service holds no product state. Name the discriminating measurement that separates the plausible causes, give the most likely cause, and give the fix and how you would verify it.

Approach

1. Separate resident memory from live heap first, because they fail differently. Resident size can grow from fragmentation, native buffers or thread stacks while the heap is flat; heap used after a full GC rising monotonically is the measurement that says objects are reachable and not being released. You already have it, so this is retention, not fragmentation, and that closes off half the candidate list.
2. Use the rate's independence from traffic as the discriminator. Growth that continues at half the request rate rules out per-request objects that are merely slow to collect and points at a structure that grows with distinct values observed rather than with call volume. Write the candidates that have that property: a metrics registry keyed on a high-cardinality label, an unevicted cache, an interner, a per-key lock map.
3. Take two heap snapshots an hour apart and diff by retained size, reading the dominator tree, not by allocation count or instance count. Expect one root holding a map with millions of entries, then follow the reference chain to the code that inserts and never removes. Allocation profilers point at churn, which is the wrong signal here.
4. The candidate that fits this service is an observability label carrying an identifier, such as a request path recorded before templating so that /v1/resources/48213 becomes its own metric series. That grows with distinct ids seen, is independent of rate, and explains the late p99 degradation, since GC cost rises with the size of the live set.
5. Fix by bounding cardinality at the source: template the path to /v1/resources/{id} before it becomes a label, move tenant id from a label to a log field or an exemplar, and cap the registry with a bounded map that evicts. Add a cardinality ceiling that fails loudly in a lower environment rather than growing quietly in production.
6. Verify with a soak rather than a restart. Hold one instance out of the nightly recycle for 48 hours with the fix and compare post-GC heap and series count against an unfixed control taking the same traffic.

Follow-up

• Post-GC heap is now flat but resident size still creeps. What are you looking at, and does it matter?
• How would you have detected this before an OOM, given the nightly restart masked the trend?
• That label is what makes one dashboard useful. How do you keep the dashboard and lose the leak?

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.

Tell me about a time you had to persuade a team to adopt a new technology.

Medium · behavioural and collaboration

Approach

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

Follow-up

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

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?

Turn a code review disagreement into a decision

Easy · code review · optimistic concurrency · communication

A colleague's change updates a row with UPDATE resource SET version = version + 1 WHERE resource_id = $1 AND version = $2 and treats an affected-row count of zero as a successful no-op. You read that as a silently lost update; they think returning 200 is friendlier to clients than returning a conflict. Describe how you have handled a review disagreement of this shape: what goes in the comment, when you leave the thread, and who decides. Then write the comment you would leave here, in under 80 words.

Approach

1. Sort the disagreement before writing anything. A silently discarded write is a correctness claim about data; the choice between 409 and 412 is taste. Only the first justifies blocking a merge, and saying which one you are doing is most of the value of the comment.
2. Make the claim reproducible in the comment itself with an interleaving rather than a principle: A reads version 7, B reads version 7, B commits version 8, A's predicate matches zero rows, A is told it succeeded and A's edit is gone.
3. Offer the alternative with its cost attached: return 409 carrying the current version and the revision that won, so the client can re-read and re-apply. Note that automatic retry is not the fix, because a retry re-reads the winner's state and reapplies an intent formed against data that no longer exists.
4. Apply an escalation rule you can state: two round trips on the thread, then a call, and the service's owner decides rather than the reviewer. A reviewer who cannot be overruled is a bottleneck with extra steps.
5. Close in writing wherever the decision lands, so the next reader finds the reasoning in the code or the ticket instead of in a collapsed review thread.

Follow-up

• Where would you put the test that fails if someone reintroduces the swallowed zero rowcount?
• The author says clients cannot handle a 409. How do you check whether that is true?
• How do you handle the same review comment when the author is more senior than you and in a hurry?