Interview guides · Quid

Quid Machine Learning Engineer Interview Questions & Guide 2026

Quid 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 Quid, you will play a pivotal role in leveraging data to drive innovation and enhance product functionality. Your primary responsibility will be to build and optimize machine learning models that enable Quid to derive insights from vast datasets, thereby improving decision-making processes for users. The products you will work on directly influence how clients interact with data, making your contributions crucial to the overall success and growth of the company.

Ask whether any round happens inside an existing repository instead of a blank file. Reading unfamiliar code, isolating a fault and making the smallest correct change is a different skill from writing a function from scratch, and it needs its own practice.

Focus areas

Bound every outbound call with a timeout
Detect concurrent edits instead of losing writes
Paginate large result sets with keyset cursors

Overview

This role is distinctively engaging because it operates at the intersection of cutting-edge technology and real-world application. You will be tasked with tackling complex problems that require not only technical expertise but also a strategic understanding of business needs. Your work will impact various teams, including product management and engineering, as you collaborate to develop solutions that enhance user experience and drive business objectives.

By joining Quid, you will be part of a dynamic environment that values experimentation and rapid iteration. Expect to engage with significant challenges in data science, model deployment, and algorithm optimization, all while contributing to products that are transforming how organizations understand and interact with their data.

The Quid interview process

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

Phone Screen

Before anything technical happens, someone has to decide which rung of the ladder your loop is calibrated to, and that decision sets the bar for every round after it. It comes from how you describe scope, not from your title, because titles do not convert cleanly between companies. The weak version of the answer is team size and years. The strong version names the largest change you shipped where nobody reviewed the design, what would have broken if you had been wrong, and what you were paged for. Get the level said out loud on this call, because the range and the loop both follow from it.

What to demonstrate

Whether the scope in your own account maps onto a level the team actually has an opening at, so a mismatch ends the process cheaply rather than after four interviewers have spent a day
Whether your title needs re-mapping: the same word describes very different amounts of independent decision-making at a twenty-person company and a ten-thousand-person one
Whether your compensation expectation can be filled at that level in the structure the role pays in, which is why the number gets asked for before any engineer is scheduled

How to prepare

Write down two changes from the last two years: the largest one you designed with nobody reviewing the design, and the largest one where someone more senior did. Lead with the first when scope comes up, and be ready to say which parts of the second were yours
Ask which level the loop is calibrated to and what changes at the level above it, then plan your weeks from that answer rather than from the posting
Settle a total-compensation range beforehand with the split named, base against bonus against equity and its vesting period, so a question about numbers gets a number instead of the word market

Take-Home Test

Nobody is in the room when this is read, so the submission stands in for what an ordinary pull request from you would cost to review. That makes navigability the first thing being judged. A reader should reach the code that does the actual work within a minute of cloning, without tracing three layers of indirection to get there. What most separates a strong submission from a weak one is separation itself: the logic that implements the prompt kept apart from file reading, argument parsing and printing, so each piece can be read, and tested, on its own.

What to demonstrate

Whether the entry point makes the shape of the program obvious: what is read, what is computed, what is written, in that order
Whether the core logic can be called without touching the filesystem, a socket or stdin, since that is what makes it testable at all
Whether names hold up on a second reading; a function called parse that also writes, or a total that means something different from the README, costs the reader more than a longer name would have
Whether the commit history reads as a sequence of steps someone could follow, rather than one commit containing everything

How to prepare

Clone your own finished submission into an empty directory and read it in the order a stranger would, README then entry point then downward, noting every place you rely on something the code never told you, and fix those places
Take an exercise you have already solved and restructure it so the module holding the rules imports no file, CLI or network library, leaving input parsing and output as a thin shell around it
Build the history while you work, committing at points where the tests pass, so the log shows the order you actually solved the problem in

Technical Interviews

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

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.

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.

Designing for a scale nobody asked for

Ask for request rate, data size, read-to-write ratio and expected growth, then size the simplest option first; one relational instance on current hardware covers a large share of real workloads. Reaching for shards, queues and a cache tier before any number has been quoted reads as pattern-matching rather than judgement.

Trusting input because it came from your own front end

Anything crossing a trust boundary is hostile: parameterise queries instead of building SQL by concatenation, validate against an allow-list rather than a deny-list, and bound the size of anything you allocate from a request. Raising this unprompted in an API or design question is a cheap and unusually strong signal.

Questions & practice

What is overfitting, and how can it be prevented?

Medium · machine learning fundamentals

Approach

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

Follow-up

How would you know the model is overfitting?
Where could label leakage enter this setup?

Discuss a project where you implemented a machine learning model from start to finish.

Medium · machine learning fundamentals

Approach

Pick the metric from the cost of each error type, not from habit.
Name the simplest model that could work and what would make you move past it.
State the learning problem: the label, the unit of prediction and how the model is used.

Follow-up

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

How would you improve the accuracy of a model that is underperforming?

Medium · machine learning fundamentals

Approach

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

Follow-up

How would you know the model is overfitting?
Where could label leakage enter this setup?

Describe your methodology for evaluating the performance of a machine learning model.

Medium · machine learning fundamentals

Approach

Pick the metric from the cost of each error type, not from habit.
State the learning problem: the label, the unit of prediction and how the model is used.
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?

Explain the time complexity of your algorithm.

Medium · coding and algorithms

Approach

Name the brute-force solution and its complexity before improving on it.
Choose the data structure from the access pattern, not from familiarity.
Walk one small example through your approach before writing the whole thing.

Follow-up

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

Solve a problem related to data manipulation using a programming language of your choice.

Medium · coding and algorithms

Approach

Name the brute-force solution and its complexity before improving on it.
Choose the data structure from the access pattern, not from familiarity.
Walk one small example through your approach before writing the whole thing.

Follow-up

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

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

Hard · reconciliation · range hashing · throttling

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

Approach

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

Worked solution (35 min)

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

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

Verify it yourself

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

Follow-up

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

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?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

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

Approach

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

Follow-up

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

How would you approach building a recommendation system for a new product?

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?

Cache the tenant listing feed with a bounded staleness window

Easy · caching · cache invalidation · keyset pagination

GET /v1/resources returns one tenant's resources ordered by updated_at DESC, 20 per page, at 14k requests/second peak against a 120 ms p99. The table carries the index (tenant_id, status, updated_at DESC, resource_id DESC) and writes land on the primary at 1.2k/second. Design the read path: the pagination contract, the cache key and value, what a write invalidates, and the staleness a user can observe. State the request rate that actually reaches the database, and the one repopulation race that deleting on write does not close.

Approach

Settle the pagination contract first, because it decides what is cacheable. OFFSET makes the database produce and discard the skipped rows, so page 500 costs five hundred pages of work, and rows inserted between two fetches shift across the boundary and are skipped or repeated with nothing in the response to reveal it. The cursor is the row value of the last row returned: WHERE tenant_id = $1 AND status = $2 AND (updated_at, resource_id) < ($3, $4) ORDER BY updated_at DESC, resource_id DESC LIMIT 21. That is a row-value comparison, not updated_at < $3 AND resource_id < $4, which is a different and wrong predicate.
Confirm the index actually serves it: equality on the two leading columns, then a range on the pair that follows in exactly the index's sort order, so the plan is an index scan that touches 21 entries with no sort node. Requesting 21 to return 20 is how has_more is answered without a count. resource_id is not decoration - updated_at is not unique, and without the tie-break two rows sharing a timestamp at a page boundary are the skip that keyset pagination was adopted to remove.
Key the cache on every value the predicate reads: tenant_id, status, cursor and page size. A key that omits tenant_id is a cross-tenant disclosure, and no test running against a single tenant's data will show it.
Be honest that a write does not invalidate one key. An update moves its row to the head of the ordering, so it invalidates the first page and every cursor page whose range spans the row's old and new position, which is not enumerable. Cache the first page per (tenant_id, status) - that is where the traffic is - with a short TTL, invalidate it on write, and serve deep cursor pages uncached from a replica, since each is already a 21-row index scan and they are rare.
Name the residual race and the real bound. A reader that loaded rows before the write can populate the cache after the invalidation deleted the key, so the delete is not a staleness bound; the TTL is. Choose the TTL as the staleness a listing can tolerate, and jitter expiries so a busy tenant's keys do not all expire together and stampede the replica. Then do the arithmetic: at an 85% hit rate, 14k requests/second is about 2.1k database reads/second across two replicas, and that is the number capacity planning uses.

Worked solution (20 min)

Write the keyset predicate and match it column by column against the index, marking which columns are equality, which is the range, and which satisfies the ORDER BY.
Compare rows examined for page 1 and page 500 under OFFSET and under keyset, and state both numbers.
Write the cache key template and the first-page invalidation the write path performs.
Write the interleaving in which a stale value is written into the cache after the invalidation, and identify what bounds it.

Expected result: Keyset pagination on the row value (updated_at, resource_id) served as a 21-entry index scan with no sort at any page depth. The first page per (tenant_id, status) is cached under a short jittered TTL and deleted on write; deeper cursor pages go to a replica uncached. Observable staleness is the TTL, not the invalidation, because a concurrent reader can repopulate the key afterwards. At an 85% hit rate roughly 2.1k reads/second reach the two replicas.

Verify it yourself

The query plan shows an index scan with no Sort node, and rows examined equal to the limit, at page 1 and at page 500.
Two rows sharing an updated_at value across a page boundary are returned exactly once across the two pages.
Every cache key contains tenant_id.
Removing the TTL leaves staleness unbounded even though every write still invalidates.

Follow-up

The tenant writes and immediately lists. What does it see, and what is the smallest change that makes its own write visible without sending all 14k requests/second to the primary?
A tenant has 4 million resources and a client walks every page nightly. What does that do to the cache hit rate, and should that traffic share this path at all?
Sort order becomes configurable - by title, by created_at. What happens to the index set and to the cache key space?

Exports duplicate a row range about once a week

Hard · leases · clock skew · idempotency · fencing

Roughly once a week an export writes a file containing a duplicated range of rows. The affected job_run rows show attempt = 1, status = succeeded, one started_at, and a lease_owner naming a different host from the one whose logs show the job starting. Leases last 30 seconds and are heartbeated every 10 from inside the handler; lease_expires_at is computed on the worker and compared against the database's now(). Find the mechanism, and give a fix that holds even if you cannot fix the clocks.

Approach

Start from the fact that eliminates the obvious answer. attempt = 1 means no retry was recorded, so this is not a re-run after failure; two workers ran the same row concurrently and the takeover path never touched the counter. lease_owner naming a host other than the one that started the job is the same statement from the other side.
Enumerate the mechanisms that cause a premature takeover, then find the signal that separates them. Either the lease genuinely expired because the heartbeat did not fire, which is what happens when the heartbeat runs on the handler's own thread and the handler makes a long blocking call, or it only appeared expired because two clocks disagree, since lease_expires_at is written from the worker's clock and evaluated against the database's. The discriminator is the distribution: incidents clustered on the longest exports indict the heartbeat, incidents clustered on one host indict skew. Measure both, and measure each host's offset against the database directly.
Read the reclaim query precisely. In PostgreSQL now() is transaction start time, not statement time, so a reclaimer holding a long transaction compares against an older timestamp than expected; clock_timestamp() is the statement-time function. This is worth ruling in or out before you redesign anything, because it changes which rows look expired.
Remove the second clock rather than trying to synchronise it. Issue and extend the lease in the database, with lease_expires_at = now() + interval '30 seconds' in both the claim and the heartbeat, so exactly one clock is ever compared and worker skew stops mattering to this predicate.
Accept that a lease can still expire under a slow worker, because a lease cannot distinguish slow from dead, and fence the work. Carry a monotonically increasing lease generation and make every write the handler performs conditional on still holding it, as UPDATE ... WHERE job_run_id = $1 AND lease_owner = $2 AND lease_generation = $3, so a displaced worker's writes affect zero rows and it aborts instead of duplicating.
Make the handler's writes idempotent independently of all that: give each exported chunk a natural key of (job_run_id, batch_start) with a unique constraint so a second copy conflicts rather than appends, move the heartbeat off the handler's thread, and increment attempt on takeover so the event is visible in a metric.

Follow-up

The displaced worker has already streamed half the file to object storage. What makes that side effect safe to repeat?
You now count takeovers. What alert fires on that counter, and at what threshold?
What breaks if you simply raise the lease to five minutes?

Suggested practice checklist

Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.

Day 1 — Diagnostic, scored before you study anything

Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.

Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.

Day 2 — Largest gap: find the boundary rather than the subject

Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
Re-attempt one of them from blank four hours later with nothing open.

Deliverable: A sub-skill map with the two blocking sub-skills circled.

Day 3 — Drill the blocking sub-skill by repeating the shape

Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
Have someone else read your one-sentence rule and find the precondition you left out.

Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.

Day 4 — Second gap, plus maintenance on the strongest area

Run the same sub-skill decomposition on the second-largest gap in half the time.
Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.

Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.

Day 5 — The gap that is not a skill

Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.

Deliverable: Two recordings with a counted reduction in time-to-first-question.

Day 6 — Retest under day-one conditions

Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
Write down which single block you would still lose the offer on.

Deliverable: A second scored rubric placed beside the first, with one named remaining risk.

Day 7 — Full loop under interview conditions

Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.

Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.

Disagreement needs an outcome you did not choose

Conflict answers where you were right and everyone came round are the weakest ones. Stronger: the evidence you went and collected, what would have changed your mind, and what you did in the weeks after the call went against you. Implementing a design you argued against, properly, is a specific and checkable behaviour.

Discuss a situation where you had to collaborate with a team to achieve a goal.

Medium · behavioural and collaboration

Approach

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

Follow-up

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

How do you handle constructive criticism?

Medium · behavioural and collaboration

Approach

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

Follow-up

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

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

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.
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.
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.
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.
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?