Interview guides · Applied Systems

Applied Systems Software Engineer Interview Guide

Applied Systems software engineer prep on operations: decide what you would need to see to know a release is going wrong.

The 60-second brief

As a Software Engineer at Applied Systems, you join a leading provider of cloud-based software for the global insurance industry. The company builds critical infrastructure and software suites—including platforms like EZLynx, Indio, and core Applied insurance agency management systems—that power modern insurance agencies, brokerages, and insurers. Engineers at Applied Systems design and deliver high-availability SaaS applications, real-time rating engines, data collection flows, and enterprise API integrations that process millions of insurance transactions every day.

Find out what you will be typing into. A shared plain-text editor with no autocomplete, compiler or test runner changes what you have to hold in your head, and practising inside your own configured environment hides exactly that gap.

Focus areas

Keep money in integer minor units
Make every write idempotent under client retries
Build at-least-once pipelines with explicit deduplication horizons

Overview

The role demands a balance of robust software craftsmanship, strong problem-solving skills, and a dedication to system reliability. Depending on your team alignment, you may build high-throughput microservices using C#.NET or Python (Django), develop modern web interfaces in Angular or React, or engineer scalable infrastructure using,, and cloud platforms like or. Because insurance operations require strict data accuracy and uninterrupted uptime, every engineer actively contributes to architectural design, automated testing, continuous integration, and production performance monitoring.

Working at Applied Systems offers you the chance to modernize a massive, complex industry. You will collaborate closely with cross-functional product managers, cloud operations specialists, and software architects to simplify legacy workflows into scalable, cloud-native solutions. The hiring team evaluates candidates not only on technical syntax, but also on your ability to write maintainable code, test thoroughly, and explain past project decisions in depth.

The Applied Systems interview process

Applied Systems candidates report 4 rounds · ≈ 3-5 weeks. The stages below are what candidates describe, not a published process.

Phone Screening

The title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.

What to demonstrate

Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year

How to prepare

Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter

Cognitive Assessment

The same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.

What to demonstrate

In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes

How to prepare

Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first

Technical Interviews

Input bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.

What to demonstrate

Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply

How to prepare

For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last

Executive Interview

When a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.

What to demonstrate

Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened

How to prepare

Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure

Where candidates lose points

Treating a timed-out write as a failed write

A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.

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.

Sorting when the problem never required a total order

Match the algorithm to the guarantee actually needed: the top k comes from a size-k heap in O(n log k) time and O(k) space, distinctness needs a set rather than an ordering, and a small bounded integer key range admits a linear counting pass. A full O(n log n) sort is the right default only when you genuinely need everything in order.

Retrying a write that is not safe to repeat

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

Questions & practice

Write a function to solve the classic FizzBuzz problem, discussing space and time complexity.

Medium · data structures and algorithms

Approach

Name the brute-force solution and its complexity before improving on it.
Restate the input: its shape, its size, and what is guaranteed about it.
State the target complexity and say which constraint rules the naive version out.

Follow-up

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

How would you implement a basic counter application in React or Angular while handling state updates properly?

Medium · data structures and algorithms

Approach

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

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?

How do you structure a unit test suite using frameworks like NUnit, xUnit, or Jest to mock dependencies?

Medium · data structures and algorithms

Approach

Walk one small example through your approach before writing the whole thing.
Restate the input: its shape, its size, and what is guaranteed about it.
Choose the data structure from the access pattern, not from familiarity.

Follow-up

What is the worst case, and how likely is it on real data?
Which test case would catch an off-by-one here?

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.

Worked solution (40 min)

Replay a day of events with a synthetic lateness distribution: 99.9% under two minutes, plus a 0.05% tail at four to six hours that carries 3% of total quantity.
Compute the p99.99 lateness two ways, event-weighted and quantity-weighted, and put the two numbers side by side.
Implement the min-heap of open hours with the watermark as the min over 256 partitions, then stall one partition for 20 minutes and observe what seals.
Set the idle-partition timeout to 60 seconds, repeat the stall, and measure how much quantity arrives after the seal.
Attempt the seal from two workers at once and confirm the conditional update lets exactly one through.

Expected result: The quantity-weighted p99.99 is hours larger than the event-weighted one. Choosing L from the event-weighted number lets roughly the tail's 3% of quantity land after the seal, 600 times the 50 ppm target. With the min watermark and no idle timeout, the stalled partition blocks all sealing; with a 60-second timeout, the stall is sealed past and its events arrive late.

Verify it yourself

Post-seal quantity as a fraction of the hour's total is at or below 50 ppm at the chosen L, measured rather than assumed.
The second sealer's update reports zero rows affected and writes nothing.
Heap size equals the number of open hours and does not grow with the replay's event count.

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?

Rebuild an hourly rollup with deduplication and late-arrival accounting

Hard · window functions · deduplication · event time vs ingest time · watermarks

From usage_event (event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at), produce the values usage_rollup_hourly should hold for one tenant over one day: per (workspace_id, sku, hour_start) the deduplicated quantity_sum, event_count and source_max_ingested_at, bucketed by occurred_at. Duplicates share (tenant_id, idempotency_key). Also report, per hour, the running total across the day and the share of quantity that arrived more than two hours after the hour began. Write the query, and state which duplicates a daily unique index cannot catch.

Approach

Deduplicate in its own CTE before any aggregation, because a SUM cannot be un-summed: row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id) = 1. Include the tiebreaker. Without it the surviving row is non-deterministic when two duplicates share an ingested_at, and a rollup described as deterministically recomputable then disagrees with itself between runs.
Bucket on occurred_at and nothing else, and pin the timezone explicitly. date_trunc('hour', timestamptz) truncates in the session's TimeZone setting, so the same query run by a session set to a non-UTC zone buckets differently; use the three-argument date_trunc('hour', occurred_at, 'UTC') on PostgreSQL 16 or later, or date_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC' before that. Filter environment = 'production' explicitly, since metering covers three environments and billing covers one.
Aggregate to the grain with sum(quantity), count(*) and max(ingested_at). The last is not decoration: it is the watermark the row consumed up to, and without it there is no way to prove afterwards what a number did and did not include.
Compute the late share inside the dedup-and-aggregate step as a conditional aggregate, sum(quantity) filter (where ingested_at > hour_start + interval '2 hours'), then divide by the hour's total. Compute the running total as a window over the already aggregated rows: sum(quantity_sum) over (partition by workspace_id, sku order by hour_start rows between unbounded preceding and current row). Running either over raw rows puts the duplicates back.
Answer the index question exactly. The unique constraint is on (ingested_day, tenant_id, idempotency_key), because a unique index on a partitioned table must contain the partition key. It therefore deduplicates only within one ingest day and admits a duplicate whose retry crosses midnight or whose replay runs a week later. That is why this CTE dedups across the whole window being recomputed, and why the dedup horizon is a correctness parameter rather than a retention cost.
Keep the numeric type all the way through. quantity is numeric so the sums are exact; a cast to double precision anywhere in this pipeline reintroduces drift that surfaces only as a few unreconcilable cents per tenant per month, long after the query is out of anyone's mind.

Worked solution (40 min)

Generate a day of events for one tenant across three workspaces and two SKUs, inject 2% duplicates sharing an idempotency_key with a later ingested_at, and inject 1% whose ingested_at is six hours past occurred_at.
Write the dedup CTE alone and confirm its row count drops by exactly the injected duplicate count.
Aggregate and compare quantity_sum against a ground truth computed outside SQL from the generator's own records.
Add the two window and filtered aggregates, and confirm the final hour's running total equals the ungrouped day total for each (workspace, sku).
Run the whole query twice over static data and diff the two outputs.
Re-run with the session TimeZone set to a non-UTC zone and confirm hour_start does not move.

Expected result: One query: a dedup CTE using row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id); aggregation by (workspace_id, sku, hour_start) producing quantity_sum, event_count and max(ingested_at) plus a filtered late-quantity aggregate; a running total as a window over the aggregated rows; and an explicit statement that the partition-scoped daily unique index cannot catch a duplicate that crosses the ingest-day boundary or arrives in a later replay.

Verify it yourself

Row count after dedup equals raw count minus injected duplicates, exactly.
The running total at the final hour equals the ungrouped day total for that (workspace, sku).
Two consecutive runs over static data return byte-identical rows.
Every hour_start is hour-aligned in UTC and unchanged by the session's TimeZone setting.

Follow-up

A dispute forces the same recompute over 40 days for one tenant. What changes about the dedup CTE's memory use and the chosen plan, and what would you do about it?
Two runs a minute apart return different quantity_sum values for an hour that is already closed. Give two mechanisms that produce that, and the single query that distinguishes them.
Express the same rollup incrementally so it does not re-scan the day each time the watermark advances. What does the incremental version stop being able to answer?

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.

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?

What strategies do you use for secure token authentication and role-based access control in API development?

Medium · api and integration design

Approach

Separate accepted, pending, failed and confirmed; they are different facts.
Say who the caller is and what they do when the call fails halfway.
Define the identity of a request so a retry cannot double-apply it.

Follow-up

How does a client discover it is on an old version of this contract?
What happens if the caller retries after a timeout?

What are the key differences between Entity Framework and lightweight ORMs like Dapper regarding performance and control?

Medium · api and integration design

Approach

Design the error taxonomy before the success shape; callers branch on it.
Define the identity of a request so a retry cannot double-apply it.
State how the contract changes without breaking existing clients.

Follow-up

What happens if the caller retries after a timeout?
What does a partial failure look like to the caller?

How do you design a RESTful API endpoint to ensure proper HTTP status code handling and payload structure?

Medium · api and integration design

Approach

Say who the caller is and what they do when the call fails halfway.
Define the identity of a request so a retry cannot double-apply it.
State how the contract changes without breaking existing clients.

Follow-up

What does a partial failure look like to the caller?
How does a client discover it is on an old version of this contract?

How do abstract classes differ from interfaces, and when would you choose one over the other?

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

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

An idempotent create endpoint that returns a one-time secret

Easy · idempotency · race conditions · api design

POST /v1/api-keys inserts a tenant_api_key row (tenant_id, workspace_id, name, key_prefix, secret_hash, scopes, status, auth_version, created_at, expires_at) and returns the plaintext secret exactly once, since only its SHA-256 is stored. Write volume is tens per second and clients retry on timeout. Design the idempotency mechanism: what the key is scoped by, where the record lives, its retention, what happens when the same key arrives with a different body, what happens when a retry arrives while the first request is still in flight, and what a replay returns for the secret.

Approach

Scope the key by tenant, not globally: uniqueness is on (tenant_id, idempotency_key), or one tenant's key collides with another's and the second caller receives a stored response for a request it never sent. Store a fingerprint of the request alongside it - method, path and a hash of the canonicalised body - so a mismatch is detectable.
Let the unique constraint decide the race instead of application logic. In the same transaction as the credential insert, run INSERT INTO idempotency_record (tenant_id, key, request_fingerprint, status) VALUES (...) ON CONFLICT DO NOTHING RETURNING id; no returned row means this is a replay, and the existing record is then read. A select-then-insert here loses to itself under concurrency in exactly the way this endpoint is meant to prevent.
Write the three replay outcomes as a decision table rather than as prose: same fingerprint and completed returns the stored response; same fingerprint and still in flight returns 409 with Retry-After, without blocking and without executing; different fingerprint returns 422, because replaying a stored response for a mutated body would tell the caller a credential was created for parameters it never sent.
Handle the secret as the part that makes this endpoint different from an ordinary idempotent create. The plaintext cannot be regenerated from secret_hash, so either the stored response holds it - making the idempotency record a secret at rest whose retention is now the secret's exposure window - or a replay returns the key metadata without the secret and the documentation says a lost response is resolved by listing keys and revoking the orphan. The second is the safer default precisely because credentials are listable and revocable.
Set retention from the client's retry budget, not from a round number: the record must outlive the SDK's maximum total retry duration, so twenty-four hours is defensible if that budget is minutes. After expiry the key is reusable and a very late retry creates a second credential, which is acceptable here only because the object is listable and revocable, and would not be for an unlistable side effect. Expire with a scheduled delete on an index over created_at.

Worked solution (20 min)

Write the DDL for the idempotency record, including the unique constraint that makes the concurrent case impossible rather than unlikely.
Write the four outcomes - fresh, replay-completed, replay-in-flight, fingerprint-mismatch - as a decision table with the HTTP status for each.
Decide what a replay returns for the plaintext secret and write the exact sentence the API reference has to carry about it.
Pick a retention and justify it from the client library's own maximum retry duration rather than from a round number.

Expected result: Unique (tenant_id, idempotency_key); the record inserted in the same transaction as the credential with ON CONFLICT DO NOTHING deciding the winner; 200 with the stored response, 409 for in-flight, 422 for a fingerprint mismatch; a documented rule that the plaintext secret is not replayed and a lost response is resolved by listing and revoking; retention exceeding the SDK retry budget, for example twenty-four hours.

Verify it yourself

Two concurrent requests with the same key must leave exactly one row in tenant_api_key. If the logic is select-then-insert, they will not.
A retry with a mutated body must not return 200. If it does, the fingerprint is not being compared or is not covering the fields that matter.
The retention you chose must be strictly longer than the client library's maximum total retry duration. Look that number up rather than assuming it.

Follow-up

Two requests with the same key arrive concurrently. Show the exact statements and say which one loses, and how it finds out.
The client receives a timeout, retries, and gets 409 in-flight. What should the client library do next, and for how long?
How does the design change if the created object is not listable - a one-off payout, say - so an orphan cannot be found afterwards?

Metering partition crash-loops and the sealing watermark freezes

Medium · poison message · at-least-once · dead-letter queue

One metering-ingest partition has stopped advancing. Lag grows linearly, the consumer restarts about every 40 seconds, and the same offset appears in every startup log while other partitions stay healthy. Events are committed in batches of a few thousand and the acknowledgement follows the commit. Sealing is six hours away and source_max_ingested_at for that partition's tenants is frozen. Give an ordered checklist, a containment action available within minutes, and the durable fix, saying what each does to exactly-once accounting.

Approach

Distinguish a poison record from a capacity problem in one measurement: compare the offset and the exception across restarts. An identical pair every time is deterministic failure on one record, whereas a throughput problem still advances the offset between crashes.
Read the record from a separate consumer group so the bytes can be inspected without perturbing the stuck consumer, then classify the defect: schema violation, a quantity failing the non-negative check, a null workspace, an unmappable SKU enum, or a payload past a size limit. That classification decides whether this is a producer bug or a missing consumer guard.
Account for batch granularity before acting. With commits of a few thousand, one bad record fails thousands of good ones, so the blast radius is the batch. Halve the batch around the offset to isolate the record, or move to per-record error isolation so the radius becomes the record.
Contain by diverting that record to a dead-letter store with its raw bytes and offset, then resume. This is safe here precisely because the acknowledgement follows the commit: the good records from the failed batch are re-consumed and absorbed by the uniqueness check on (tenant_id, idempotency_key) rather than counted twice.
Make the fix durable with per-record error isolation, a bounded poison counter, and an alert on dead-letter rate rather than on lag alone, since lag only reveals this after the sealing margin has already been eaten.
Check the horizon before replaying anything. The unique index lives on a daily-partitioned table and therefore includes the partition key, so it deduplicates within a day only; a replay landing on a later ingest day needs the separate dedup store or it double-counts into a tenant's bill.

Follow-up

Move the acknowledgement before the commit and describe exactly what is lost and what is duplicated in each of the two crash windows.
Sealing is in six hours and the partition will not drain in time. What do you seal on, and what does the invoice have to record so the difference is explainable later?
A producer replays two weeks of events next month. Which part of your fix stops holding, and what is the dedup horizon you would actually configure?

Suggested practice checklist

For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.

Day 1 — Rebuild the primitives by implementing them

Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.

Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.

Day 2 — Arrays under an invariant: two pointers, sliding window, binary search

Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.

Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.

Day 3 — Sorting, heaps, and the greedy argument that has to be proved

Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.

Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.

Day 4 — Recursion, memoisation, and the step to a table

Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.

Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.

Day 5 — Graphs, where most of the work is choosing the traversal

Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.

Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.

Day 6 — One day for everything that is not an algorithm

Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.

Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.

Day 7 — Solve out loud, under time

Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.

Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.

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.

Tell me about a time you had to debug a difficult production issue under strict time constraints.

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

Follow-up

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

What are your career expectations, and how do you stay current with evolving tools and software frameworks?

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

Estimate a tenant-leading index migration you have never run

Hard · estimation · online migration · index builds · uncertainty

Someone needs a date. usage_event carries an index on (occurred_at) and needs (tenant_id, occurred_at); the largest tenant holds roughly a hundred times the median tenant's rows, the table is partitioned daily with years of retention, and you have never run a migration on a table this large. Give an estimate you would defend: how you decompose the work, the two or three numbers you would go and measure first, the range and confidence you state, and what you commit to when the person asking needs a single date today.

Approach

Refuse the bare number and then give one anyway, in the form that is actually useful: a range plus the measurement that collapses it. 'Four to eleven days; one afternoon building this index on a restored copy of the largest partition takes that to within a day' is an answer, while 'it depends' is not.
Decompose by failure mode rather than into equal chunks, because that is where estimates go wrong. On a partitioned parent you create the index ON ONLY the parent, build each partition's index with CREATE INDEX CONCURRENTLY, then ALTER INDEX ... ATTACH PARTITION, at which point the parent index becomes valid. CONCURRENTLY does not block writes but scans each partition twice, waits out older transactions, cannot run inside a transaction block, and on failure leaves an invalid index you must drop concurrently and retry.
Name the two unknowns that dominate and price them: build time on one restored partition of realistic size, and whether the planner actually chooses the new index for the skewed tenant, since selectivity for a tenant holding most of the rows is a different question from selectivity for the median tenant. Both are half-day measurements against a replica, and both are cheaper than being wrong by a week.
State the assumptions the range is conditional on, because that is what makes a slip a re-estimate instead of a credibility event: no partition above a stated row count, one concurrent build at a time so it does not compete with ingest for I/O, and an ingest backlog that can absorb the added write amplification while both indexes exist.
Budget the step nobody budgets: verification and the old index's removal. Dropping the old index is fast, but deciding it is safe to drop means confirming no plan still uses it, and that confirmation waits on real traffic across a full weekly cycle rather than on your patience.
Answer the single-date request honestly. Commit to a date for the first checkpoint — the measured build number from the replica — and to re-estimating on that date, and say plainly what you are not committing to yet. A date with a scheduled re-estimate is worth more to the asker than a confident wrong one, and you should say why in those words.

Follow-up

The concurrent build fails half way through the largest partition. What is the state of the database and what do you do next?
Your estimate slips by sixty percent. Which assumption broke, and at what point would you have known?
The person asking needs the date for a customer commitment. Does your answer change?