Interview guides · Advania

Advania Software Engineer Interview Questions & Guide 2026

Software engineer interview prep for Advania: original coding, system design and debugging practice, each with a worked solution.

The 60-second brief

As a Software Engineer at Advania, you are at the forefront of driving digital transformation for a diverse range of enterprise clients. This role is not just about writing code; it is about acting as a technical consultant who bridges the gap between complex business requirements and robust, scalable cloud solutions. You will often work within the Microsoft ecosystem, helping organizations modernize their infrastructure, improve operational efficiency, and leverage cloud-native technologies.

Prepare in one language you know well enough to debug in rather than the one you think reads best. Under a clock an unfamiliar language costs you standard-library lookups and iteration mechanics, and that time comes out of your thinking budget, not your typing budget.

Focus areas

Make every write idempotent under client retries
Build at-least-once pipelines with explicit deduplication horizons
Evolve APIs without breaking pinned SDK clients

Overview

The impact of this role is significant. Whether you are working as a Microsoft Cloud Consultant, an Azure Developer, or a Support Services Engineer, your contributions directly influence the stability and innovation of client businesses. You will navigate high-stakes environments where your ability to troubleshoot, design, and implement systems directly affects how these companies serve their own end users. It is a position for those who thrive on technical challenge and enjoy the collaborative nature of professional services.

The Advania interview process

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

Screening Phase

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

Technical Interviews

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 Round 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

One shared connection pool for every tenant and every query class

A single tenant with a large table and a missing index can occupy every connection with slow queries, and every other tenant then waits in connection acquisition -- a queue invisible in database metrics, because the database itself looks healthy while the application starves. Containment is bulkheads: separate pools or per-tenant concurrency caps for interactive requests, background jobs and exports, a statement timeout low enough that a pathological query dies before it accumulates, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks indefinitely. One caveat worth knowing in advance: if a transaction-pooling proxy sits in front of the database, session-scoped behaviour changes, so session-level advisory locks and settings applied outside a transaction do not survive the way they do on a direct connection.

Holding money in a floating-point type, or rounding it more than once

Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.

Starting work without saying what you are about to spend time on

State the plan before executing it: the approach, roughly how long it will take, and what you intend to leave hand-waved. That gives the interviewer a chance to redirect you in ten seconds rather than watching you spend fifteen minutes on the wrong sub-problem.

Issuing one query per row of a result set

Fetch related rows in a single batched query keyed by the ids you already hold, or join them into the original query. A per-row round trip multiplies network latency by the row count, and it looks perfectly fine against the ten rows in your development database.

Questions & practice

What would you do in order to solve a problem which you are unfamiliar with?

Medium · data structures 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.
State the target complexity and say which constraint rules the naive version out.

Follow-up

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

Seal an hour under late data with bounded memory

Hard · watermarks · late-data · quantile-sketch · conditional-write

Metering ingest reads 256 partitions at 10,000 to 40,000 events/second. Events carry occurred_at and ingested_at, and during a producer replay the gap between them is hours. Seal each UTC hour once no more than 50 parts per million of that hour's eventual quantity can still arrive, using memory that does not grow with the size of the replay. Define the watermark, the lateness parameter and how you measure it, the structure holding open hours, and the write that performs the seal. State what an idle partition does to your watermark.

Approach

Two clocks, two jobs. Bucket by occurred_at, because that is the hour the customer is billed for, and advance the watermark on ingested_at, because that is what the fold has consumed and what source_max_ingested_at records. Conflating them is what makes late data invisible.
The global watermark is the min over partitions of each partition's committed ingested_at, not the max: the fold is trustworthy only as far as the slowest partition. The consequence is that one idle partition pins the watermark forever and nothing seals, so an idle partition must promote its watermark to wall clock after a stated idle timeout, and that timeout becomes a correctness parameter, because a partition that is slow rather than idle gets sealed past.
Choose the lateness L from the measured distribution of ingested_at - occurred_at, weighted by quantity rather than by event count. The target is 50 ppm of the hour's quantity, and a replay is rare in events while carrying disproportionate mass, so an event-weighted quantile picks an L that is comfortably wrong at exactly the moment it matters.
Measure that quantile in bounded memory. A Greenwald-Khanna summary gives epsilon-approximate quantiles in O((1/epsilon) log(epsilon n)) space; a t-digest costs more per merge but has relative error that tightens at the tails, which is the half of the distribution you are reading at p99.99. Keep a separate summary per tenant class, because one tenant's batch importer is not the population.
Hold open hours in a min-heap keyed by hour_start. When the watermark advances, pop every hour with hour_end + L < W and seal it: O(log H_open) per advance and O(1) amortised per event to touch its bucket. Memory is open hours multiplied by distinct (tenant, workspace, sku) keys, so cap the number of simultaneously open hours and spill the oldest into usage_rollup_hourly as status='open' with a revision bump. While an hour is open the row is upsertable, so the store is your overflow.
The seal itself is a conditional write: update ... set status='sealed', sealed_at=now() where status='open' returning .... Two sealers race on every restart, and the loser must see zero rows and stop rather than write a second value. After the seal, an event for that hour is not an upsert but an adjustment, and source_max_ingested_at is what proves it arrived afterwards.

Follow-up

A replay starts during the sealing window for a period you are about to close. What do you do, and what is the customer-visible consequence of each option?
Your measured quantity-weighted p99.99 lateness is six hours and the invoice must be issued at 02:00 UTC on the first. How do you reconcile those two numbers?
How would you detect that L has drifted before it costs you an hour's quantity?

Order a job dependency graph and find its critical path

Medium · topological-sort · dag-longest-path · cycle-detection · critical-path

A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.

Approach

Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order: earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E).
Second pass in reverse topological order for latest_finish, then slack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain. slack[v] = 0 is exactly the statement that some longest path runs through v; equivalently, the longest path through v has length T - slack[v].
The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by min(d, T - L_avoid(v)), where L_avoid(v) is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, while min(10, 15 - 14) predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan is max(T - d, L_avoid(v)).
Compute L_avoid(v) the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, since L_avoid(v) only ever matters through that max: set duration[v] := 0, recompute the makespan as T0(v) = max(T - duration[v], L_avoid(v)), and the gain is min(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.

Worked solution (30 min)

Build four fixtures. A: 12 jobs, two branches of 100 s and 95 s that share no job. B: fixture A plus one back edge. C: two disjoint paths tied at 100 s. D: the shared-prefix case, one job of 10 s feeding a 5 s job and a 4 s job, so the longest path is 15 s and the runner-up is 14 s.
Run Kahn; on fixture B confirm it emits fewer than V nodes, then run the residual-subgraph DFS and print the actual cycle.
Compute earliest_finish forward and latest_finish backward, and list the zero-slack set for each fixture.
For each zero-slack job v, recompute the makespan with duration[v] := 0 to get T0(v), and record both the correct bound T - T0(v) and the wrong one, T - second_longest_path, side by side.
Apply the shortening for real (20 s off the critical branch of A, 10 s off the shared prefix of D) and diff the recomputed makespan against each prediction.

Expected result: Fixture A: makespan 100 s, and shortening by 20 s leaves 95 s, a gain of 5 s. Both formulas agree here, because the 95 s branch avoids the shortened job. Fixture D: makespan 15 s, and shortening by 10 s leaves 5 s, a gain of the full 10 s, which T - T0(v) = 15 - 5 = 10 predicts and T - second_longest = 1 does not. Fixture C: the zero-slack set covers both tied paths, and shortening a job on one of them alone gains nothing, since the other path still runs 100 s.

Verify it yourself

The cycle printed by the DFS is a real cycle: every consecutive pair is an edge present in the input.
Summing durations along any zero-slack path equals the makespan exactly.
For every candidate, the predicted gain equals the makespan recomputed from scratch with the shortened duration. Assert the equality; do not eyeball it.
Fixture D is the regression test for the bound, and a fixture whose branches are disjoint is not: it passes under either formula and proves nothing.

Follow-up

Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?

Paginate a tenant's delivery export without skipping rows

Easy · keyset pagination · indexing · multi-tenancy

A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.

Approach

Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
Replace the position with a value predicate over a stable, unique, indexed ordering: where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window.
Index (tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit.
Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.

Worked solution (20 min)

Load about 50k deliveries for one tenant, then walk them with the OFFSET query while a writer inserts 10 rows/second, collecting every returned delivery_id.
Compare the distinct ids collected against the set of ids that existed when the walk started, and record the shortfall.
Repeat the walk with the keyset query and confirm every pre-existing id is returned exactly once.
Run explain (analyze, buffers) on page 1 and page 500 of each form and compare shared buffer hits.

Expected result: A keyset query using the row comparison (created_at, delivery_id) < ($2, $3) with order by created_at desc, delivery_id desc limit 100, backed by (tenant_id, created_at, delivery_id) scanned backwards; plus the statement that OFFSET drops rows under concurrent insert and costs time proportional to the offset, and that updated_at is disqualified because it mutates on retry.

Verify it yourself

Buffer counts for page 500 are the same order of magnitude as page 1.
Every id present before the walk is returned exactly once when only inserts run concurrently.
The query plan shows no Sort node above the index scan.
Two rows sharing a created_at value to the microsecond are both returned across the page boundary.

Follow-up

The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added created_at <= $snapshot bound, and a materialised export table.
A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?

Explain why the metering dashboard scans every daily partition

Medium · indexing · partition pruning · sargability · tenant skew

usage_event is range-partitioned daily on ingested_at and holds tenant_id, workspace_id, environment, sku, quantity numeric(20,6), occurred_at and ingested_at. The only relevant index is on (occurred_at). A dashboard runs select sku, sum(quantity) from usage_event where tenant_id = $1 and date_trunc('hour', occurred_at) >= $2 and environment = 'production' group by sku, and EXPLAIN shows a sequential scan of every partition. Give each distinct reason, rewrite the predicate so an index can serve it, propose the index, and state the write cost its column order adds.

Approach

Separate the three causes rather than blaming one. First, date_trunc('hour', occurred_at) wraps the column, so the predicate is not sargable against a btree on the bare column. Second, pruning keys off ingested_at while the query constrains occurred_at, so no partition can be excluded. Third, even made sargable, (occurred_at) is not tenant-leading, so for one tenant among thousands the scan reads the whole time range and discards nearly all of it.
Rewrite the bound carefully, because the obvious rewrite is only conditionally equivalent. date_trunc('hour', x) >= $2 equals x >= $2 only when $2 is already hour-aligned; for an arbitrary $2 it means x >= date_trunc('hour', $2) + interval '1 hour'. Normalise the parameter in the caller and leave the column bare.
Restore pruning with a second, redundant predicate on the partition key: ingested_at >= $2 - interval '<late-data horizon>'. State both sides of it. It prunes to a handful of partitions, and it silently omits any event whose ingest lagged past that horizon, which is precisely what a producer replay produces. Either document the horizon as a stated bound, or partition on occurred_at and move the problem into the dedup window instead.
Propose (tenant_id, occurred_at) include (sku, quantity) per partition. A partial index where environment = 'production' mostly saves size rather than selectivity, since production dominates the three environments; take it if non-production is a meaningful share and skip it otherwise.
Price the write path honestly. At roughly 250M rows/day each extra index is another insert plus WAL per row, and a tenant-leading key scatters inserts across one hot leaf per active tenant instead of appending to a single rightmost leaf, so page dirtying and random I/O both rise. An INCLUDE payload widens every leaf entry and enlarges the index accordingly.
Add the index-only-scan caveat before someone reports it as a regression: on a freshly appended table the visibility map is not yet set for recent pages, so the INCLUDE columns still cost heap fetches until autovacuum has been through, and the newest hour is exactly the data the dashboard reads.

Follow-up

CREATE INDEX CONCURRENTLY is not supported on a partitioned parent. Give the sequence that gets this index onto 400 existing partitions without blocking ingest.
One tenant holds 200 times the median row count and the dashboard still times out for them with the index in place. What changes?
Should this read hit usage_rollup_hourly instead? State what that costs in freshness and what the watermark lets you promise.

How do you see yourself in 10 years?

Medium

Approach

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

Follow-up

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

How do you contribute to company culture?

Medium

Approach

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

Follow-up

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

Build a resumable usage export the customer can reconcile against

Medium · pagination · cursors · late data

Customers reconcile invoices against usage_event (event_id uuid, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at, source_service, request_id), partitioned daily on ingested_at. The current export is ?page=N&per_page=1000 ordered by occurred_at, and a customer syncing hourly reports rows that never appear in their export but do appear on their invoice. Design the replacement: the ordering, the cursor's contents, the index it requires, and the rule deciding where a page stops. State what the client does on a timeout and on a cursor older than retention.

Approach

Separate the two defects in the current shape. Offset makes the database produce and discard N * per_page rows, so page cost grows linearly and a deep page degrades from milliseconds to seconds. Concurrent inserts also shift the window between requests, so a walker skips rows with no error raised anywhere, which for a customer sync is silent data loss.
Order by ingestion, not by occurrence. During a replay events arrive hours out of occurred_at order, so a consumer holding an occurred_at high-water mark can never see a late event that falls below it; (ingested_at, event_id) is the only ordering under which 'everything after my cursor' is a complete statement.
Page with a row comparison: where tenant_id = $1 and (ingested_at, event_id) > ($2, $3) order by ingested_at, event_id limit $4, backed by an index on (tenant_id, ingested_at, event_id). That seeks directly to the resume point, so every page costs the same regardless of depth.
Trail the head of the table. With ingested_at defaulting to now(), which is transaction start time, a long insert transaction receives an earlier timestamp and becomes visible after a reader has already passed it. Cap each page at ingested_at <= now() - delta, with delta larger than the longest write transaction as bounded by statement_timeout and idle_in_transaction_session_timeout, or the export skips exactly the rows written under load.
Make the cursor opaque and self-describing: base64 of the timestamp, the event id and a fingerprint of the filters, rejected when the filters differ from the current request. Return 410 with a cursor_expired code once the cursor's partition has been dropped, so the client restarts from a known time instead of resuming into a hole.
Keep a page a pure GET with no server-side consumption, so a timeout is resolved by retrying the identical cursor.

Worked solution (30 min)

Construct the failing case on paper: an event with occurred_at at 09:00 ingested at 14:00, and a consumer that read up to 10:00 at 11:00.
Write the keyset query with the row comparison and the exact index it needs, then say which column of the index each predicate uses.
Add the trailing-head predicate and pick delta from a named timeout setting rather than a round number.
Define the cursor's encoded contents and the two error cases: filter mismatch and expired partition, with their status codes.
Write the client's algorithm in four lines: request, persist cursor after processing the page, retry the same cursor on timeout, restart from a time on 410.

Expected result: Keyset pagination over (ingested_at, event_id) with an index on (tenant_id, ingested_at, event_id), a trailing head predicate sized above the longest write transaction, an opaque filter-bound cursor, and 410 for a cursor past retention.

Verify it yourself

A late event with an old occurred_at is still delivered by the export.
Page cost is independent of how many pages have already been read.
The cursor persists across a client restart and survives retrying the same page.
The tie-break column is unique, so no row can be skipped or repeated at a page boundary.

Follow-up

The customer asks for a total count alongside the first page. What do you offer instead, and why is an exact count both expensive here and wrong by the time it is read?
How would you let a customer re-read a window they have already consumed without giving up the forward-only cursor?

How do you approach debugging complex code or system failures?

Medium · debugging and production issues

Approach

Say what evidence would prove you wrong, then go and look for it.
Separate the trigger from the cause; the deploy is rarely the bug.
Check the instrumentation before believing the symptom.

Follow-up

What would you add now so this is faster to diagnose next time?
What would you look at first, and what would it rule out?

Person moved desk. No Internet via LAN, even though the phone is powered on over Ethernet. How do you troubleshoot?

Medium · debugging and production issues

Approach

Check the instrumentation before believing the symptom.
Separate the trigger from the cause; the deploy is rarely the bug.
Pick a bisection that eliminates candidates whichever way it turns out.

Follow-up

What would you add now so this is faster to diagnose next time?
What would you look at first, and what would it rule out?

Suggested practice checklist

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

Day 1 — Coding, one pass at shallow depth

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

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

Day 2 — Design, one pass at shallow depth

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

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

Day 3 — Fundamentals and the practical rounds

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

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

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

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

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

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

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

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

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

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

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

Day 7 — Reassemble the loop

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

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

'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 your experience with JavaScript or other core development languages used in our stack.

Medium · behavioural and engineering judgement

Approach

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.
Name the disagreement and how you resolved it with evidence.

Follow-up

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

Describe a time you worked collaboratively to resolve a team issue.

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Name the disagreement and how you resolved it with evidence.
Pick a story where you made the decision, not one where you watched it.

Follow-up

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

Explain your experience with Azure cloud environments.

Medium · behavioural and engineering judgement

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?
What did you decide not to do, and why?