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Compare The Market Software Engineer Interview Questions & Guide 2026

Compare The Market software engineer prep: algorithm drills where every complexity claim has to name the assumption under it.

The 60-second brief

A Software Engineer at Compare The Market plays a pivotal role in maintaining the company’s position as a leader in the price comparison industry. You are not just writing code; you are building the digital infrastructure that millions of users rely on to make critical financial and lifestyle decisions. Your work directly influences complex commercial journeys and high-traffic platforms, ensuring that the user experience remains seamless, fast, and reliable.

The loop does not sample the job evenly, and arguing about that in the room costs you. Daily work is mostly incremental change inside code someone else wrote, while the loop samples narrow slices of it; prepare for the slices and save the realism argument for your questions at the end.

Focus areas

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

Overview

Whether you are working on Comparison Capabilities, Commercial Journeys, or emerging AI initiatives, you will be expected to balance technical excellence with a deep understanding of user needs. The environment is fast-paced and collaborative, requiring engineers who can thrive in cross-functional teams, navigate technical ambiguity, and contribute to a culture that values ambition, authenticity, creativity, and unity. This role is ideal for those who want to solve high-scale engineering challenges while directly impacting the company’s core business outcomes.

The Compare The Market interview process

Compare The Market candidates report 5 rounds · ≈ 4-6 weeks. The stages below are what candidates describe, not a published process.

Recruiter 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

Online 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.

Technical Deep Dives

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

Competency-Based Interview

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

Assessment Center/Final Presentation

Nobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.

What to demonstrate

Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience

How to prepare

Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed

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.

Paginating with LIMIT/OFFSET over a set that changes while the client is reading it

OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.

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.

Naming no test cases at all

State what you would test before being asked: empty input, a single element, all elements equal, the maximum permitted size, and the input that exercises the branch you just wrote. It costs thirty seconds and is much of what separates someone who has shipped code from someone who has only solved puzzles.

Questions & practice

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

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

Approach

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

Worked solution (30 min)

Write the edge-loading query with the tenant predicate on both endpoints and state what it does with a cross-tenant edge.
Implement iterative Tarjan with an explicit stack and confirm on a three-node cycle that it emits one component of size three.
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.
Run Kahn over the condensation and verify the emitted order against the referrer-before-referenced rule.
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?

Canonicalise a request body into a stable idempotency fingerprint

Medium · parsing · canonicalisation · hashing

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

Approach

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

Follow-up

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

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.

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?

Stop tag and share joins from fanning out a page

Medium · join fan-out · semi-joins · many-to-many

resource_tag is (resource_id, tag_id) with PK (resource_id, tag_id); resource_share is (resource_id, shared_with_user_id, permission). The tagged-and-shared listing inner-joins resource to both, filters tenant_id, tag_id = ANY($2) and shared_with_user_id = $3, orders by updated_at DESC and takes 50. Pages come back with fewer than 50 distinct resources and the total in the header is far too high. Explain the row multiplication, rewrite both the page query and the count query so each is correct, and name the index each one needs. PostgreSQL 16.

Approach

Do the arithmetic against the predicates that are actually there. An inner join emits one row per matching child row, and both joins are filtered: tag_id = ANY($2) admits only the requested tags, shared_with_user_id = $3 admits one user's share rows. So a resource holding three of the requested tags and shared with $3 once yields three rows, not one — the multiplier is its count of matching tags times its share rows for that single user, and that second factor is 1 unless the table admits duplicate (resource_id, shared_with_user_id) pairs. LIMIT 50 then limits rows rather than resources, and COUNT(*) counts pairs — the header is the product, not the population.
Reject DISTINCT as the fix. It deduplicates after the product has been built, so the planner must materialise and sort the fanned-out set before the LIMIT can apply, and it leaves any SUM or AVG in the same select list wrong.
Rewrite both filters as semi-joins, keeping resource as the only row source: AND EXISTS (SELECT 1 FROM resource_tag rt WHERE rt.resource_id = r.resource_id AND rt.tag_id = ANY($2)) and the same shape against resource_share. A semi-join stops at the first match per resource and preserves the driving index order, so ORDER BY updated_at DESC, resource_id DESC LIMIT 50 still stops after 50 rows.
Count with the same predicates and no join at all: SELECT count(*) FROM resource r WHERE r.tenant_id = $1 AND r.status = 'active' AND EXISTS (...) AND EXISTS (...). Nothing multiplies a resource, so the number is the population.
Attach the tags for display after the page has been cut — LEFT JOIN LATERAL (SELECT array_agg(rt.tag_id) FROM resource_tag rt WHERE rt.resource_id = p.resource_id) ON TRUE over the 50 returned rows. Aggregate over the page, never over the tenant.
Index both directions and say which query each serves: PK (resource_id, tag_id) serves the lateral lookup, (tag_id, resource_id) serves the EXISTS probe by tag, and resource_share needs (shared_with_user_id, resource_id) for the same reason. An index covering one direction only leaves the other as a scan.

Worked solution (30 min)

Build a tenant where each resource carries 0-5 tags from a 20-tag vocabulary and is shared with 0-4 distinct users, then bind $2 to three tags and $3 to a user holding shares on about half the resources. Run the joined query and compare its row count to the distinct resource count on page one.
Run COUNT(*) on the joined shape and on the EXISTS shape and compare both to a ground truth computed from distinct ids; then give $3 a second permission row on 10% of resources and record which of the two counts moves.
EXPLAIN both page queries and compare rows-read plus the presence of a Sort or HashAggregate node above the join.
Add (tag_id, resource_id), re-run the EXISTS probe, and record the plan change on the inner side.

Expected result: The EXISTS page returns exactly 50 distinct resources and reads about 50 driving rows with no aggregate above the join. The EXISTS count matches the ground truth. The joined COUNT(*) is larger by roughly the average number of matching tags per qualifying resource — the tags it holds from $2, not its total tag count — times its share rows for $3, which is 1 until the duplicate permission rows exist and rises with them afterwards. The EXISTS count is unchanged by those duplicates.

Verify it yourself

The page's resource_id list has no duplicates and is 50 long whenever at least 50 rows match.
count(*) from the rewrite equals count(DISTINCT r.resource_id) over the joined shape.
EXPLAIN shows a semi-join node on the inner side rather than a full join followed by a unique or aggregate step.

Follow-up

The filter changes from 'any of these tags' to 'all of these tags'. Rewrite it and state what it costs relative to the ANY form.
A resource can be shared with the same user twice under different permissions. Does your count change, and should it?
Where does the correct total come from when the tenant holds 4M resources and the header must not cost 200 ms?

Replace offset paging on the resource feed with keyset

Easy · keyset pagination · composite index · sort order

resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.

Approach

Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.

Follow-up

The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
What does the cursor do when the row it points at has since been deleted?

How do you approach building an AI-heavy system?

Medium

Approach

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

Follow-up

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

What is one of your strengths that will help you succeed in this role?

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.
State your assumptions explicitly before working the problem.

Follow-up

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

Convert the listing endpoint from offset pages to stable cursors

Medium · pagination · cursors · index access path

GET /v1/resources returns a tenant's resources newest-updated first, today with page and per_page, backed by index (tenant_id, status, updated_at DESC, resource_id DESC). Callers are a browser feed and a nightly sync job that walks every page. Users report items appearing twice or vanishing between pages, and page 400 is slow. Design the cursor contract: what the cursor contains and how it is encoded, the exact WHERE and ORDER BY, what happens when a row's updated_at changes mid-walk, how a client detects the end, and what the sync job does when a cursor is rejected.

Approach

Separate the two defects, because they have different fixes. Cost: OFFSET n makes the engine produce and discard n rows, so price grows with page depth rather than page size and page 400 pays for 400 pages of work. Correctness: while the set shifts, rows cross the offset boundary and are skipped or repeated, and nothing in the response lets the client detect it.
Write the seek: WHERE tenant_id = $1 AND status = $2 AND (updated_at, resource_id) < ($k, $id) ORDER BY updated_at DESC, resource_id DESC LIMIT n. The tie-break is not decoration - updated_at is not unique, and two rows sharing a timestamp across a page boundary reintroduce exactly the skip this was adopted to remove. PostgreSQL seeks the composite index on the row-value comparison directly; on an engine that does not optimise a row constructor, expand it into the equivalent OR form or the plan quietly degrades to a scan.
Encode the cursor as an opaque token carrying the sort key, the id, and a fingerprint of the filter and sort order, signed or at minimum validated. A cursor replayed against a different sort or filter must be a 400 with its own code, not a silently wrong page - the sync job cannot notice the difference otherwise.
State the guarantee precisely rather than generously. Keyset is stable against inserts and deletes elsewhere in the set, because the position is a value and not a count. It is not a snapshot: updated_at is mutable, so a row that is edited during the walk re-sorts and may be seen twice or not at all. If the sync job needs exactly-once coverage, order on an immutable key such as (created_at, resource_id), or walk resource_revision by revision_id and treat updated_at as data.
Define termination and limits in the response, not in the client's inference: fetch n+1 rows, return n, and emit next_cursor only when the extra row existed. Absence of next_cursor is the sole end signal, because a page can legitimately come back short when rows are filtered after retrieval. Cap n and document the cap rather than honouring per_page=10000.
Migrate without a flag day: keep page and per_page working, add the cursor, count usage per credential, and remove the offset path only once the sync job's traffic on it is zero.

Worked solution (25 min)

Write the old and new queries side by side and state the rows examined for page 400 under each.
Define the cursor payload field by field, including the filter and sort fingerprint, and choose the encoding.
Write the end-of-results rule and the cap, then the 400 response for a cursor that does not match the current query.
Construct the mid-walk edit case: a row updated between page two and page three, and say exactly what the client sees.
Write the deprecation plan for page and per_page, including the signal that says removal is safe.

Expected result: A keyset query that seeks the existing composite index with no sort step, an opaque cursor bound to its filter and sort order, a single documented end condition, and an explicit statement that mutable sort keys make the walk non-snapshot.

Verify it yourself

The plan for a page fetch shows an index seek and no sort, and reads about n rows regardless of depth.
Two rows sharing an updated_at value that straddle a boundary are both returned exactly once.
A cursor issued under one filter and replayed under another is rejected rather than served.

Follow-up

The client wants a total count and the ability to jump to page 400. What can you honestly offer instead, and what does each option cost?
How do you paginate backwards, and what does that require of the index?

Read latency spikes on a sixty-second sawtooth

Medium · cache stampede · single-flight · ttl · tail latency

The cached listing read path serves about 14k reads/second at an 85% hit rate. p99 sits at 35 ms for 57 seconds, jumps to 900 ms for 3, and repeats. During each spike the primary shows several hundred identical listing queries starting within the same millisecond, all carrying one large tenant's id. Cache entries use a 60-second TTL. Give the mechanism, the ordered checks, the fix, and the correctness hazard your fix must not introduce.

Approach

Match the period to a configured number before theorising about load. A spike every 60 seconds against a 60-second TTL is an entry expiring, and you confirm it by correlating spike timestamps with the entry's write time rather than with the traffic curve. If the period had matched a cron or a GC interval instead, this is a different investigation.
Establish the concurrency of the miss. Several hundred identical queries in one millisecond means the miss path has no coalescing: every request that arrives between expiry and repopulation recomputes. The herd size is that key's arrival rate times its recompute time, so at 1.2k reads/second for the hot key and a 250 ms recompute you expect about 300 concurrent misses, which matches what is observed.
Add single-flight on the miss path so one caller per key recomputes under a short-lived lock while the rest wait for its result. Prefer stale-while-revalidate where the read tolerates it: return the expired value immediately and refresh asynchronously, which removes the latency spike rather than serialising it into a queue of waiters.
De-synchronise the keys. Write TTLs with jitter, for example 60 seconds plus or minus 10%, so a deploy or a mass invalidation does not align every key on the same second and turn a per-key herd into a fleet-wide one.
Name the hazard the fix must not introduce. Serving a stale listing is acceptable only because the API reports the projection watermark, and a reader that loaded the old value before a write can repopulate the entry after the invalidation, so the bounded TTL is what actually caps staleness rather than the delete. Keep read-after-write pinned to the primary for the writing session regardless.
Verify on miss concurrency, not hit rate. The hit rate barely moves, because the herd is one miss multiplied; the number that must change is distinct origin queries per key per minute.

Follow-up

The same sawtooth appears on a key that is invalidated on write rather than expired. Is that the same bug?
How does your answer change if the recompute takes 4 seconds instead of 250 ms?
What exactly does a client see during a stale-while-revalidate window, and how does the watermark let them tell?

Suggested practice checklist

Roughly ninety minutes on weeknights with one longer weekend block. The plan cuts scope rather than compressing everything, on the assumption that one thing finished per night beats four half-started.

Day 1 — Fix the scope and take a cold baseline

Read the role description and write the three things the loop will almost certainly test, then write an explicit not-doing list and keep it visible all week.
Take one twenty-five-minute coding problem and one fifteen-minute design prompt cold, and write the single sentence naming what blocked each, because those two sentences decide where the remaining evenings go.
Set the week's rule: one thing finished every night, including the night you only have forty minutes.

Deliverable: A one-page scope with a not-doing list and two cold attempts, each carrying one sentence on what blocked it.

Day 2 — One pattern, written three times from blank

Choose the single pattern most likely to appear in your loop and write it three times from an empty file rather than editing the previous attempt.
On the third pass, write the invariant as a comment before the loop body and the complexity before the first line of code.
Stop at ninety minutes even if the third version is imperfect, and write the one thing you would fix given another hour.

Deliverable: Three independent implementations of the same pattern plus a note on what changed between them.

Day 3 — One design, only to the depth you can defend

Take one system shape and go only as far as requirements, interface and data model, refusing to draw a box you could not survive a follow-up about.
Attach one number to each non-functional requirement, deriving it rather than asserting it, and write the assumption the number rests on.
Write the one tradeoff you are choosing against and the observation that would make you reverse it.

Deliverable: One design at interface-and-schema depth with derived numbers and one written reversible tradeoff.

Day 4 — Only the fundamentals you will have to defend

Write, in under two hundred words each, the answers to the two questions that follow almost any implementation: why this structure and not the obvious alternative, and what happens to this code at a hundred times the input.
Write what an index actually costs: faster lookups on the indexed columns against a write that now maintains a second structure, plus the cases where the planner declines to use it anyway, low selectivity, or a predicate wrapping the column in a function.
Delete any answer you cannot deliver aloud in under a minute, since an answer that needs reading is not an answer you have.

Deliverable: Three written answers, each under two hundred words and each timed aloud.

Day 5 — Your own work, timed

Write a ninety-second and a four-minute version of your main project and time both aloud rather than reading them.
Prepare the two follow-ups that always come: what you would do differently, and how you knew it worked.
Put one number in the first sentence and be ready to say exactly where it came from and what it excludes.

Deliverable: Two timed narratives with one defensible number in the opening line.

Day 6 — The one full rehearsal, in the weekend block

Run a sixty-minute mock covering a coding round and a design round in one sitting with no break, because sustained attention is the thing evenings have not trained.
Immediately afterwards, and before hearing any feedback, write the three moments you lost the thread.
Spend the rest of the block only on those three moments, and on nothing you merely feel shaky about.

Deliverable: Mock notes naming three failure moments with a specific fix written under each.

Day 7 — Taper

Write the twenty-minute warm-up you will actually do on the morning: one problem you can already solve from a blank file, one design you can narrate, and nothing you have never seen.
Re-read only your own notes from this week and open no new material.
Write the logistics down: the editor or shared document you will be working in, whether execution and lookups are permitted, and the sentence you will use when you do not know something.

Deliverable: A one-page card holding the design structure, the project numbers, and the logistics.

'We rewrote it in a better framework' is not a story

A migration is a cost you chose to pay, not an achievement. The story is what the old system made expensive, what you measured before committing, what kept serving traffic during the cutover, and what you would have done if the numbers had come back flat. Without those, a rewrite reads as taste.

Describe a time when you showed effective risk management.

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
Close with what you would do differently, concretely.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

How would your friends describe you?

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Pick a story where you made the decision, not one where you watched it.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

Tell me about a time you had to deal with a difficult situation or a conflict with a teammate.

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Close with what you would do differently, concretely.
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 me about a project you are proud of and why.

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
Pick a story where you made the decision, not one where you watched it.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

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