Interview guides · Statflo

Statflo Software Engineer Interview Questions & Guide 2026

Take-home style practice for Statflo software engineer candidates: a small service, a fixed budget, a written tradeoff.

The 60-second brief

As a Software Engineer at Statflo, you build the scalable systems that power the company's customer-facing engagement platform. You will be instrumental in developing features that help businesses communicate more effectively with their customers, requiring a balance of clean, maintainable code and a deep understanding of backend architecture.

The title spans product, platform and infrastructure work, and which of those the seat actually is decides whether design or algorithms carries more weight in your preparation. The posting rarely settles it; what the team is on call for usually does.

Focus areas

Scope every query and cache key by tenant
Bound blast radius with per-tenant concurrency limits
Build at-least-once pipelines with explicit deduplication horizons

Overview

This role is critical to the Statflo mission of bridging the gap between businesses and their users. You will work within a collaborative, fast-paced environment where your technical decisions affect the reliability and performance of Statflo's services. Successful engineers here are not just coders; they are problem solvers who thrive on tackling complex integration challenges and delivering high-quality, user-focused software solutions.

The Statflo interview process

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

Initial 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

Take-Home Assignment

The README is read before the code, and a follow-up conversation is usually built from it, so treat every sentence you put there as a question you have agreed to answer. It needs the command that runs the thing, the assumptions you made where the prompt was ambiguous, and the limits of what you built stated with the preconditions that make them true. Overclaiming is the expensive mistake here. Writing that something is thread-safe, or constant-time, or handles files larger than memory invites a reader to check that exact line, and a claim the code cannot support costs more than silence would have.

What to demonstrate

Whether the run instructions work from a clean clone, naming the exact commands, the language version you tested on, and any environment variable the program expects
Whether ambiguities in the prompt are resolved in writing, with the interpretation you picked and the reason, rather than settled silently in the code
Whether documented limits match the implementation, so a stated input bound is one the code enforces or at least does not contradict
Whether the trade-offs you list come with the condition that would make you choose the other way, instead of reading as a list of alternatives you happened to consider

How to prepare

Write the README before the final hour, then read the code against it claim by claim and correct or delete every statement the implementation does not back
For each ambiguity in the prompt, write one sentence fixing your interpretation and keep it; those sentences become the assumptions section and your answer when someone asks why you did it that way
Give the repository to someone who has not seen the prompt and ask them to run it using only what is written down, treating every question they have to ask you as a gap in the document

Technical Assessments

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.

Cultural Evaluation

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

Where candidates lose points

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.

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.

Designing for a scale nobody asked for

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

Reading the constraints as preamble rather than as part of the problem

The bounds are usually there to eliminate the obvious approach: n up to 10^5 makes an O(n^2) scan roughly 10^10 operations, far outside any per-test time budget, and an input larger than memory rules out loading it at all. When a bound is not given, ask for it, then say out loud which approach it kills.

Questions & practice

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.

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?

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?

Find peak concurrent sandbox usage from run intervals

Medium · sweep-line · intervals · concurrency-caps · null-semantics

Given up to 5 million job_run rows for one tenant over one day, with run_id, started_at, finished_at, status and wall_clock_limit_seconds, report the maximum number of sandboxes running at once, the earliest instant that maximum is reached, and the first run_id that would breach a per-tenant cap of C. started_at is null while a run is queued; finished_at is null both for runs still executing and for runs in status lost. Treat a run as occupying [started_at, finished_at). Give the complexity and state how you handle each null.

Approach

Turn each run into two sweep events, (started_at, +1) and (end, -1), then sort the 2n events by timestamp with -1 ordered before +1 at equal timestamps. That tie-break is what makes the interval half-open, so a run finishing at 10:00:00 and one starting at 10:00:00 never overlap.
Decide each null out loud before sweeping, because each choice moves the answer. A null started_at means queued and contributes nothing. A null finished_at with status running or leased is clipped to the window end. Status lost has no observed end at all, so clip it at started_at + wall_clock_limit_seconds on the grounds that the supervisor owns the timeout, and record that you did. The table's check (finished_at is null or started_at is not null) guarantees you never see an end without a start.
Sweep once, maintaining a running counter, the maximum, and the timestamp at which the maximum was first attained (update peak_at only on a strict increase, or you will report the last such instant instead of the earliest). Capture the first run_id whose +1 takes the counter to C+1 during the same sweep rather than in a second pass.
Complexity: O(n log n) dominated by the sort, O(n) space. If rows already arrive ordered by started_at, a min-heap of end times gives O(n log k) time and O(k) space with k the peak concurrency, which is the better shape when the rows come from an index scan on (tenant_id, started_at).
If second resolution is acceptable, counting-sort the endpoints into an 86,400-slot delta array and prefix-sum it: O(n + T) time and O(T) space, which beats the comparison sort at 5 million rows. It answers only at second granularity, so state which resolution the cap is defined in.

Follow-up

Now report peak concurrency per tenant for 10,000 tenants from one globally sorted stream. What changes about memory and about the sort?
The cap has to be enforced at dispatch rather than reported afterwards. What does the admission check look like, and where does it race?
How would you answer 'peak concurrency within any 5-minute window' without re-sorting?

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.

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?

Migrate a live partitioned event table without blocking ingest

Hard · online migration · ddl locks · expand and contract · partitioning

usage_event is range-partitioned daily on ingested_at, holds roughly 250M rows per day across 400 live partitions, and is written at 10-40k rows/second. Two changes are required: quantity must move from double precision to numeric(20,6), and a new environment column must become NOT NULL with a default of 'production'. Ingest cannot stop. Give the ordered plan, naming for each step the lock it takes, what that lock blocks, and roughly how long it is held. Identify the one step that cannot be rolled back cleanly once traffic depends on it.

Approach

Classify the two changes before planning anything. Adding a column with a non-volatile default has been metadata-only since PostgreSQL 11, so it is cheap. Changing double precision to numeric is not binary-coercible, so alter column ... type rewrites every partition under ACCESS EXCLUSIVE and rebuilds its indexes; on this volume that is hours of blocked ingest and is simply not an option, which is why the plan is expand-and-contract rather than one statement.
Expand: add quantity_numeric numeric(20,6) and environment with its default on the parent. Both are catalogue-only but both take a brief ACCESS EXCLUSIVE that cascades to partitions, so run each with lock_timeout set to a second or two and retry on failure. A queued ACCESS EXCLUSIVE request blocks every reader behind it, which is how a metadata-only change turns into an outage.
Dual-write: deploy producer code that populates both columns on every insert, and leave it running before anything reads the new column. This is the step that cannot be reverted cleanly. Once readers depend on quantity_numeric, reverting the writer leaves rows with a null there, and the gap is only discoverable by re-reading the old column, which the readers have stopped doing.
Backfill older partitions in batches keyed on the primary key, oldest first, committing every few thousand rows with a pause between batches, and skipping the partition still receiving writes until it rotates. Each batch is an ordinary UPDATE taking row locks only. The cost is bloat and WAL rather than blocking, so watch dead tuples and let autovacuum keep pace instead of wrapping 400 partitions in one transaction.
Make NOT NULL cheap with the three-step form: add constraint ... check (environment is not null) not valid (brief ACCESS EXCLUSIVE, no scan), then validate constraint (SHARE UPDATE EXCLUSIVE, scans while reads and writes continue), then set not null, which from PostgreSQL 12 uses the validated check and skips its own full scan. Do this per partition, then on the parent.
Switch and contract: move reads to the new column behind a flag, verify over a full period that both columns agree on freshly written rows, drop the old column (metadata-only), and only then remove the dual-write. Any index on the new column goes on with CREATE INDEX CONCURRENTLY per partition, since CIC is not supported on a partitioned parent: create the parent index with ONLY, build each child concurrently, then ALTER INDEX ... ATTACH PARTITION until the parent index becomes valid.

Worked solution (45 min)

On a scratch cluster, build 10 partitions of 2M rows each and run a writer at a few thousand inserts/second.
Run the naive type change and measure how long writes stall and how far ingest lag grows before killing it.
Run the expand step with lock_timeout = '2s' while the writer runs, and observe a clean lock timeout and retry instead of a pile-up of blocked readers.
Backfill in 5k-row batches and chart dead tuples and WAL generated per batch.
Run the not-valid, validate, set-not-null sequence and confirm from pg_stat_activity and timings that nothing held an exclusive lock through a full scan.
Add an index with CIC per partition plus ATTACH PARTITION and confirm the parent index reports valid only after the last attach.

Expected result: An ordered expand, dual-write, backfill, switch, contract plan with the lock named per step: metadata-only ADD COLUMN under a brief ACCESS EXCLUSIVE taken with lock_timeout; batched UPDATE backfill under row locks; CHECK NOT VALID, then VALIDATE under SHARE UPDATE EXCLUSIVE, then SET NOT NULL; CREATE INDEX CONCURRENTLY per partition with ATTACH PARTITION; and the dual-write-to-read switch identified as the step that cannot be reverted cleanly.

Verify it yourself

No step in the plan holds ACCESS EXCLUSIVE for longer than the configured lock_timeout.
The writer's error rate stays at zero through the rehearsal apart from the deliberate lock timeout.
After VALIDATE, SET NOT NULL completes in milliseconds on a table where the naive form scanned every row.
Freshly written rows agree on both columns for a full verification period before the old column is dropped.

Follow-up

A CREATE INDEX CONCURRENTLY fails halfway through the partition list. What state is the table in, how do you detect it, and what do you run?
The producer computes quantity itself. What happens to a request already in flight when the dual-write deploy lands, and does it matter?
Give two queries that prove the backfill is complete: one cheap enough to run every minute, one authoritative.

Explain core Object-Oriented Programming (OOP) principles and how you apply them in the design process.

Medium

Approach

Work from the requirement backwards to the design.
State your assumptions explicitly before working the problem.
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?

What are the best practices for handling paging and performance in a data-heavy application?

Medium

Approach

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

Follow-up

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

Ship three breaking-looking changes without breaking pinned SDKs

Hard · versioning · backward compatibility · deprecation

GET /v1/runs/{id} returns status from the set queued, leased, running, succeeded, failed, timed_out, cancelled. You must add the terminal state lost, add an optional billable_seconds field, and start rejecting name values over 64 characters that were previously accepted. Clients are generated SDKs pinned inside customer build pipelines you cannot redeploy or reliably contact, some with strict deserialisers. Classify each change as compatible or breaking and justify it mechanically, then give the rollout: the versioning scheme, the transform old clients receive, the telemetry, and the sunset process.

Approach

Classify against a deserialiser rather than against intuition. A new enum value breaks a generated client that validates the enum or switches exhaustively; a new optional field breaks one generated with additionalProperties false; tightening validation breaks every caller already sending an 80-character name even though no field moved. All three are breaking for some real client, and that is the answer, not a technicality.
Choose the versioning axis on maintenance cost. A path version forks the handler, and each subsequent change multiplies the fork. A date-pinned version header, defaulted per credential at first use and pinned thereafter, keeps one internal model and a chain of ordered response transforms, so the handler stays single and the transforms compose.
Write the downgrade honestly. Older versions have no lost, so the transform maps it to the least-wrong existing value and the changelog says which; mapping it to failed asserts an outcome nobody observed, so pair it with a reason field the old version already exposes rather than inventing a field, and keep the mapping in one documented table instead of a conditional inside the serialiser.
Sequence the validation tightening separately from the payload changes: first measure how many tenants send names over 64 characters, then warn for a stated window with a Deprecation header and a non-fatal warning code, then reject only for versions dated after the change, never retroactively for a pinned client.
Deprecate from evidence, not from announcement: per-version, per-tenant, per-credential request counts from the gateway, a Sunset header carrying the date, direct contact for the handful of accounts that make up most of the tail, and short scheduled brownouts before the cutoff so the clients nobody can reach fail on your calendar rather than during their release.

Worked solution (40 min)

Write the three-row classification table with the client-side mechanism that breaks in each case, not just a verdict.
Pick the versioning scheme and write the header, the default-and-pin rule for a new credential, and where the transform chain sits relative to the handler.
Write the lost mapping for pre-change versions as an explicit table entry plus the changelog sentence a customer reads.
Write the validation timeline with four dated steps, and mark which one is visible to clients that never change their version.
Write the deprecation plan: the telemetry query by version and tenant, the headers, the brownout schedule, and the fallback for the largest holdout.

Expected result: All three changes classified breaking with the mechanism named, a date-pinned version header with one internal model and ordered downgrade transforms, a documented lost mapping, validation tightened only for versions after the change, and a sunset driven by per-version traffic with brownouts.

Verify it yourself

No change is called additive without naming the client behaviour that makes it safe.
A client that never changes its version header sees no behaviour change from any of the three.
The plan names a measurement that tells you when sunset is safe.
The lost mapping is stated once, in a table, not implied by serialiser branches.

Follow-up

Which step of your rollout cannot be rolled back cleanly once a client has depended on it?
A tenant pinned to a sunset version breaks their build the morning after the cutoff. What in your plan already lets you resolve it within an hour?
Your transform chain is now six versions deep. What stops it becoming the fork you were avoiding?

A rare job-run overwrite that logging makes disappear

Hard · fencing tokens · leases · concurrency

About one job run in fifty thousand records billable_seconds matching no observed sandbox lifetime, and a few rows show worker_id changing after finished_at was already set. It does not reproduce: debug logging around the terminal write made it vanish for two weeks before it returned. Runs last from 200 ms to 30 minutes, the lease is 60 seconds and is renewed while a run executes. Give an ordered checklist, the mechanism, and a fix that makes the illegal write impossible rather than merely rarer.

Approach

Mine the evidence instead of chasing a repro: select rows where updated_at is later than finished_at, or where a terminal status was written twice, and join them to attempt history to recover both writer identities. The defect has already happened tens of times and the rows are the recording.
State the signature before measuring it. If the cause is a lease that expired while the original worker was stalled, affected runs should cluster where the gap between the last renewal and the terminal write exceeds the lease, and should correlate with worker pause metrics rather than with workload shape.
Read the disappearance honestly. Logging inside the window changed the timing and lowered the probability; it is evidence about how narrow the window is, not a fix. Reproduce by widening the window on purpose, shortening the lease and injecting a pause between sandbox exit and the terminal write, rather than by adding more instrumentation.
Name the mechanism precisely: the lease expires during a stall such as a long garbage-collection pause or a brief partition, the run is re-dispatched, and the original worker then wakes and writes its terminal state over the new attempt's row. A lease alone cannot stop this, because the check and the write are separated by the stall.
Fix by fencing the write itself: UPDATE job_run SET status = $2, finished_at = $3, billable_seconds = $4 WHERE run_id = $1 AND status = 'running' AND lease_token = $5, with zero rows affected interpreted as having been fenced rather than as success. The token lives on the row so the store arbitrates, not the worker's memory.
Keep the state machine honest: a retry inserts a new row pointing at parent_run_id rather than resetting the old one, and a run whose worker vanished terminates as lost with billable_seconds null, because recording failure asserts an outcome nobody observed and then bills and retries on that assertion.

Follow-up

The supervisor also emits a usage event on completion. What does the fenced worker do about the event it already emitted, and how does metering absorb it?
Lease renewal is itself a network call. What happens when a renewal times out, and how does the worker decide whether it still holds the lease?
Why is lengthening the lease past the longest legitimate run the wrong lever, and what breaks if you do it anyway?

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.

Outage answers are graded on instruments

Nobody is scoring your stamina at three in the morning. What carries weight is which signal told you something was wrong, what you measured before touching anything, what you rolled back versus what you fixed forward, and why you picked one. 'We restarted it and it went away' is a story about not knowing.

Can you walk through your experience with the Spring Boot framework?

Medium · behavioural and engineering judgement

Approach

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.
Close with what you would do differently, concretely.

Follow-up

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

Describe a time you experienced a conflict with a peer; what was the situation and how did you resolve it?

Medium · behavioural and engineering judgement

Approach

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.
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 did you decide not to do, and why?

How do you see yourself fitting into the Statflo culture?

Medium · behavioural and engineering judgement

Approach

Close with what you would do differently, concretely.
Give the blast radius: what could have broken, and what you measured.
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 is one specific skill or technology you are aiming to learn in the next 6 months?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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