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Evince Development Software Engineer Interview Questions & Guide 2026

Timed coding practice for Evince Development software engineer candidates: full attempts under a clock, then scored against a rubric.

The 60-second brief

The Software Engineer role at Evince Development is a pivotal position focused on delivering robust, scalable, and high-quality software solutions for a diverse range of clients. As a member of the engineering team, you will translate complex business requirements into functional code, maintain existing systems, and contribute to the overall technical strategy of Evince Development's projects.

If the loop includes an asynchronous take-home, treat it as a code review of you rather than as a puzzle: structure, what you chose to test, and what you wrote down about the constraint you were working under. Hold the stated time box and say what you would have done with more of it.

Focus areas

  • Make every write idempotent under client retries
  • Scope every query and cache key by tenant
  • Bound blast radius with per-tenant concurrency limits

Overview

This role is critical to the company’s success because it serves as the bridge between conceptual design and tangible product delivery. You will work within an environment that prizes technical precision and practical problem-solving. Whether you are developing web applications, managing database structures, or optimizing backend performance, your work directly influences the efficiency and reliability of the products Evince Development provides to its partners.

Where to focus your preparation

Treat the sections below as preparation areas and confirm the format with your recruiter.

Preparation focus

No round sequence has been reported for this company, so work the categories below and confirm the format with your recruiter.

What to demonstrate

  • Breadth across SQL, experimentation and product reasoning
  • Ability to state assumptions before choosing a method

How to prepare

  • Drill the practice exercises below and time yourself
  • Prepare three quantified stories about decisions you drove

Where candidates lose points

Serialising a tenant's writes through select ... for update on a single counter row

It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.

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.

Quoting amortised or average cost as if it were a worst-case guarantee

Appending to a dynamic array is amortised O(1), but the append that triggers a resize copies every element, and hash lookup is constant only while the hash spreads the actual keys. Say which guarantee you are offering when the caller cares about the latency of one call rather than the total over many.

Comparing floating-point values for equality, or holding money in them

Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.

Questions & practice

Schedule ordered webhook retries with a heap of subscription queues

Medium · heap · scheduling · backoff · head-of-line-blocking

Design the in-memory scheduler for webhook delivery. Up to 20 million rows sit in status pending or failed_retryable across 200,000 subscriptions, each row carrying next_attempt_at and attempt_count, and each endpoint having a circuit breaker. Deliveries for one subscription must be attempted in order, so at most one attempt per subscription may be in flight. Support due(now), complete(delivery, outcome) and insert(delivery) in O(log S), where S is the subscription count rather than the delivery count. Give the backoff formula you schedule retries with.

Approach

  1. Key the global heap by subscription, not by delivery. Each subscription owns a FIFO of its due deliveries in event order; the heap holds one entry per eligible subscription, keyed by its head's next_attempt_at. That is 200,000 heap entries instead of 20 million, and it makes the one-in-flight rule structural rather than a check somebody can forget.
  2. due(now): peek the minimum. If its key is in the future, sleep until then instead of spinning. Otherwise pop it, move the subscription into an in-flight set, and do not re-push it. A subscription absent from the heap cannot be dispatched twice, which is precisely how ordering is preserved.
  3. complete: on success, drop the head and re-push the subscription keyed by its new head, or leave it out when the queue empties. On a retryable failure, increment attempt_count and set next_attempt_at = now + uniform(0, min(cap, base * 2^attempt)), sampled uniformly across the whole interval. That is full jitter; deterministic backoff re-synchronises the herd you just created.
  4. Circuit breaker: park the subscription in a second heap keyed by its half-open time, so an endpoint dead for six hours costs one heap entry and zero attempts rather than consuming worker slots. Admit exactly one probe at half-open and close the breaker only on its success.
  5. Say the price of the ordering guarantee out loud. One in-flight attempt per subscription means an endpoint answering in 10 seconds drains at 0.1 deliveries/second however many workers you run, and its backlog grows until it recovers. If the customer does not need order, allow k in flight and document delivery as unordered; that is the trade, and it is a product decision.
  6. All three operations are O(log S) with O(S) resident heap memory and the queues themselves backed by the store. The database-backed equivalent is a partial index on (subscription_id, next_attempt_at) where status in ('pending','failed_retryable') claimed with FOR UPDATE SKIP LOCKED, and the write-back must be fenced on lease_token so a worker that stalled and resumed cannot overwrite a newer attempt.

Worked solution (30 min)

  1. Define the four structures explicitly: queues: subscription_id -> deque[delivery], ready: min-heap of (next_attempt_at, subscription_id), inflight: set[subscription_id], breaker: min-heap of (half_open_at, subscription_id).
  2. Write down the invariant you will assert after every operation: a subscription appears in at most one of ready, inflight and breaker, never in two.
  3. Implement due, complete and insert, then simulate 200,000 subscriptions with Zipf-distributed queue depths totalling 20 million deliveries.
  4. Add one endpoint that always times out after 10 seconds and one that always answers in 20 ms, then measure the fast endpoint's throughput with and without the per-endpoint breaker.
  5. Instrument heap size across the run.

Expected result: Heap size stays at or below 200,000 regardless of the 20 million deliveries. The fast endpoint's throughput is unaffected by the dead one once the breaker trips. The dead endpoint's deliveries accumulate in their own deque and cost exactly one heap entry.

Verify it yourself

  • The asserted invariant never fires under concurrent workers.
  • Replaying the same delivery set twice produces identical per-subscription ordering, even though the interleaving across subscriptions differs.
  • Replacing full jitter with fixed backoff makes retry attempts cluster visibly in a histogram of attempt timestamps.

Follow-up

  • One subscription has 4 million queued deliveries. What stops it from starving the other 199,999, and what does your heap look like under that load?
  • A customer requests redelivery of last Tuesday's events. Where do those rows enter your structure, and what keeps them from reordering live traffic?
  • The process restarts. How much state do you rebuild, and what stops every subscription from being attempted in the same second?

Hold a tenant to a trailing sixty-second request limit

Medium · sliding-window · two-pointer · rate-limiting · tenant-skew

The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.

Approach

  1. Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request.
  2. Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
  3. Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate, prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact.
  4. Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (tokens, last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual.
  5. Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
  6. Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.

Follow-up

  • One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
  • Quotas rather than rate limits: the check is select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes.
  • How do you return an accurate Retry-After from the exact algorithm without a second scan?

Parse and verify a timestamped multi-signature webhook header

Easy · parsing · hmac · constant-time-compare · replay-protection

An inbound webhook carries a signature header of at most 1 KiB shaped t=<unix seconds>,v1=<64 hex chars>, with up to five v1 values during secret rotation and possibly unknown scheme keys. You hold the raw request body bytes and the currently active signing secrets. Write the parser and the verifier: accept when any active secret reproduces a signature and the timestamp is within a five-minute tolerance in either direction, reject otherwise. Single left-to-right pass over the header, no regular expression. State what is inside the MAC and why.

Approach

  1. Parse in one scan: split on ,, then on the first = only, since a value may itself contain = under a future scheme. Accept t exactly once and treat a second t as a reject rather than last-wins. Push every v1 onto a short list and ignore any other key, so a v2 can be introduced later without breaking this verifier.
  2. Say what is signed: HMAC-SHA256 over the exact byte string <t>.<raw body bytes>, yielding 32 bytes or 64 hex characters. The timestamp sits inside the MAC because otherwise an attacker replays yesterday's body with its still-valid signature and only has to edit the header timestamp.
  3. Hash the bytes as received. Verifying against a re-serialised JSON body is the usual defect: key order, whitespace and number formatting all change the bytes while the parsed objects compare equal, so signatures fail for honest senders and the popular 'fix' is to stop checking.
  4. Compare in constant time over fixed-length digests. Decode the hex to 32 bytes, accumulate acc |= a[i] ^ b[i] across the whole length, and test acc == 0 at the end. Evaluate every candidate without an early exit; at five candidates that is five HMACs over the body, linear in body size and negligible beside the network.
  5. Apply the tolerance as a two-sided bound, rejecting when |now - t| > 300 seconds. A sender whose clock runs ahead of yours is an ordinary case, and an unbounded future timestamp is a free replay window.
  6. Complexity: O(L) over the header producing k candidates, plus k HMACs at O(|body|) each. Space is O(k) beyond the body itself. Do the cheap rejections, including the tolerance check, before any cryptography runs.

Follow-up

  • The body is 40 MB. What changes about where you verify, and what can you do before the whole body has arrived?
  • A customer reports that signatures fail for exactly the requests whose body contains a non-ASCII character. What is your first hypothesis?
  • How do you rotate the signing secret with no failed deliveries, and how long do both secrets stay live?

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

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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)

  1. 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.
  2. Write the dedup CTE alone and confirm its row count drops by exactly the injected duplicate count.
  3. Aggregate and compare quantity_sum against a ground truth computed outside SQL from the generator's own records.
  4. Add the two window and filtered aggregates, and confirm the final hour's running total equals the ungrouped day total for each (workspace, sku).
  5. Run the whole query twice over static data and diff the two outputs.
  6. 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

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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?

Can you explain the importance of indexing in DBMS?

Medium

Approach

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

Follow-up

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

How do you manage data flow between controllers in your applications?

Medium

Approach

  1. Clarify what is being asked and what a complete answer contains.
  2. State your assumptions explicitly before working the problem.
  3. 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?

Describe your process for debugging a complex issue in a production environment.

Medium

Approach

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

Follow-up

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

Explain the core pillars of OOPs (Object-Oriented Programming).

Medium

Approach

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

Follow-up

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

Specify webhook signature verification a customer can implement

Easy · webhooks · hmac · replay protection

The webhook-delivery service signs each payload before POSTing it to a customer endpoint. Write the signature specification a customer implements in their own language: the header format, exactly which bytes are signed, the algorithm, how replay is bounded, and how a signing secret rotates without a delivery gap. Then write the verification steps the customer performs, in order, including what they compare and what they return on failure. Constraint: most customers reach for their web framework's parsed JSON body by default. Deliverable: the spec section plus reference pseudocode.

Approach

  1. Sign the concatenation of the timestamp and the raw body, t + "." + body, and emit a header of the form t=<unix seconds>,v1=<hex>. The timestamp has to be inside the MAC, or an attacker re-stamps a captured body and the tolerance window buys nothing.
  2. Require the raw request bytes. A framework that parses JSON and re-serialises it changes key order, whitespace and number formatting, so the spec must tell the customer to capture the body before the parser runs and give the middleware note for each common framework.
  3. Use HMAC-SHA256, not sha256(secret || body): SHA-256 is a Merkle-Damgard construction, so the naive form admits length extension. Require a constant-time comparison as well, since a short-circuiting byte compare leaks the expected prefix under repeated probing.
  4. Bound replay in two layers: reject when |now - t| exceeds a stated tolerance such as 300 seconds, then deduplicate on the event identifier header. The tolerance is what makes the customer's dedup store finite rather than unbounded.
  5. Rotate by allowing two live secrets and emitting both signatures in one header (v1=<old>,v1=<new>); the customer accepts if any candidate matches, so neither side needs an instantaneous cutover. A failed verification returns 400 and the body is not processed.

Worked solution (15 min)

  1. Write the header grammar and one real example line with a plausible timestamp and hex digest.
  2. Write the signed string construction explicitly as a byte concatenation, and add the sentence telling the customer where in their framework to obtain the raw body.
  3. Write the five verification steps in order: extract t and candidates, check the tolerance, recompute the HMAC over t + '.' + raw body, compare in constant time against each candidate, then deduplicate on the event identifier.
  4. Add the rotation paragraph: two active secrets, both signatures sent, overlap window stated in the dashboard.
  5. State the failure response and the fact that the payload is not processed, plus what the sender does with that 400.

Expected result: A spec where the signed input is the timestamp joined to the exact raw bytes, the algorithm is HMAC-SHA256 compared in constant time, replay is bounded by a stated tolerance plus event-id deduplication, and rotation is an overlap of two accepted signatures rather than a cutover.

Verify it yourself

  • The timestamp is inside the signed string, not merely sent alongside it.
  • The spec names raw bytes and says where the customer gets them before parsing.
  • Rotation requires no coordinated instant on either side.
  • The reason for HMAC over sha256(secret || body) is stated as length extension, not as a preference.

Follow-up

  • A customer's verification passes locally and fails in production behind a proxy that re-encodes the response body. Where do you look first?
  • Why sign with a per-endpoint secret rather than the tenant's API key?

Regional error rate explodes after a dependency merely slows

Hard · retry storm · circuit breaker · partial outage

A control-plane read replica in one region degrades from 4 ms to 120 ms. Within ninety seconds that region's gateway error rate rises from 0.01% to 40% and its p99 becomes bimodal, one mode near the old p99 and one at the client timeout. The other two regions are unaffected. The gateway retries control-plane reads three times with exponential backoff and no jitter. Give an ordered checklist that separates trigger from amplifier, the offered-load arithmetic, and the controls that break the loop.

Approach

  1. Split the incident into three questions before touching a control: what started it, what amplified it, and what would make recovery slow. Here they are the replica slowdown, the retry policy interacting with queueing, and a synchronised unjittered herd at recovery. They are different mechanisms and each needs its own control.
  2. Read the distribution rather than the mean. A bimodal p99 with one mode pinned at the client timeout is two populations, not one degraded path; split latency by cache hit and miss and confirm the fast mode is hits and the timeout mode is misses that reached the replica.
  3. Do the load arithmetic. Three retries turn one client request into up to four upstream requests, so offered load reaches roughly 4x on a dependency that is already slower, and it arrives at the worst moment. With utilisation approaching one, queueing delay grows superlinearly, which is why a 30x latency increase upstream does not produce a 30x increase downstream, it produces timeouts.
  4. Break the loop with controls that bound offered load rather than with more attempts: a concurrency limit on the control-plane client so at most N calls are in flight and the remainder fail fast, a circuit breaker scoped per dependency and region, and a retry budget capping retries at a small fraction of base traffic so amplification has a ceiling that does not depend on how many clients are retrying.
  5. Add full jitter to whatever retries survive, sleeping uniformly in [0, min(cap, base x 2^attempt)], so attempts de-correlate instead of arriving in waves aligned to the moment of failure.
  6. Decide the unreachable-dependency behaviour in advance, because it is the actual product decision underneath: serving from an expired credential cache keeps the product available while extending a revoked key's life past the stated bound, and failing closed converts a dependency degradation into a total outage. State the mode and the staleness number rather than letting the timeout choose.

Follow-up

  • The replica recovers. Describe what happens in the first ten seconds with your controls in place versus without them.
  • Which single metric would have paged before the error rate moved, and why is upstream latency by itself not it?
  • Requests that fail fast under the concurrency limit still need an answer. What does the gateway return, and what does it do to the usage event it would otherwise have emitted?

Suggested practice checklist

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

Day 1 — Fix the scope and take a cold baseline

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

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

Day 2 — One pattern, written three times from blank

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

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

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

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

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

Day 4 — Only the fundamentals you will have to defend

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

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

Day 5 — Your own work, timed

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

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

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

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

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

Day 7 — Taper

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

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

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

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

Reverse a webhook ordering decision after measuring its cost

Medium · reversing decisions · head-of-line blocking · at-least-once · api contracts

You argued for strict per-subscription ordering in webhook-delivery, which means one in-flight attempt per subscription. It shipped. Three months later a single unresponsive endpoint holds one subscription's queue at a six-hour backlog, and two customers report events arriving out of order anyway once their own retries are counted. Describe a decision you reversed: what you originally optimised for, the measurement that changed your mind, what the reversal cost in engineering time and customer change, and how you told the people who had already built on the original guarantee.

Approach

  1. State the original decision as a trade you made knowingly. Ordering across a network requires a single in-flight attempt per subscription, and its price is head-of-line blocking whenever one endpoint is slow. 'We priced it wrong' is a much stronger opening than 'we did not realise', and it is usually the true one.
  2. Bring the measurement that flipped it, not the anecdote: backlog age at the ninety-ninth percentile per subscription, the share of subscriptions where one slow endpoint gated an otherwise healthy queue, and the delivery throughput lost to serialisation. A reversal justified by complaints is indistinguishable from a reversal justified by fatigue.
  3. Name what you learned about the guarantee itself, which is the engineering content of this story. At-least-once delivery means a retried event already arrives after newer ones and the consumer already must be idempotent, so a guarantee the customer has to defend against anyway was never worth what it cost to provide.
  4. Describe the migration, because reversing a published contract is the hard half and the part candidates skip. Parallel attempts behind a per-subscription flag, a monotonically increasing sequence number added to the envelope so order-sensitive consumers can sort or discard, documentation that states at-least-once and unordered in those words, and a deprecation measured in quarters because the client is a pinned SDK inside a build pipeline you cannot see or redeploy.
  5. Give the cost in the two currencies that matter: engineer-weeks, and how many customers had to change code. Then say who you told before it shipped rather than in a changelog afterwards, and which large customer you left on the old behaviour and for how long.
  6. Close with the signal you now weight differently, stated as something you would do earlier next time: measuring the blocking cost on the slowest decile of endpoints before committing to the guarantee, rather than after a customer noticed.

Follow-up

  • A customer insists they need ordering. What do you offer them that is not global serialisation?
  • How did you choose the deprecation window given that you cannot see or redeploy the clients?
  • What would have to be true for you to reverse back?

Resolve a review disagreement over a quota check

Easy · code review · isolation levels · write skew · disagreement

A colleague's pull request enforces a per-tenant quota by selecting the current count and then inserting when it is under the limit. You flag it as a race. They reply that the transaction already runs at repeatable read, so the snapshot makes it safe, and the tests pass. Walk through taking that disagreement to a resolution: what you write in the review, what you demonstrate rather than assert, which fix you propose and why, and what you do if they still disagree after all of it.

Approach

  1. Answer the claim precisely instead of restating your objection, because they have made a specific technical argument. In PostgreSQL, repeatable read is snapshot isolation; this is write skew, which snapshot isolation permits by design. Both transactions read a count that is stable within their own snapshot, insert disjoint rows that the other cannot see, and both commit, so the limit is exceeded by exactly the concurrency.
  2. Demonstrate rather than cite. Two psql sessions, both BEGIN ISOLATION LEVEL REPEATABLE READ, both select the count, both insert, both commit: it succeeds. Repeat at SERIALIZABLE and the second commit fails with serialization_failure, SQLSTATE 40001. That takes two minutes, ends the argument without anyone conceding a position, and leaves an artefact for the next reviewer.
  3. Offer the options with their costs rather than a verdict. Serialisable plus a retry loop on 40001 is correct but obliges every caller to retry and degrades under contention. An increment-and-compare on a counter row — update tenant_quota set used = used + 1 where tenant_id = $1 and used < limit returning used — is safe even at read committed, because a blocked updater re-evaluates the WHERE clause against the row version it finally locks, and zero rows returned means full. A unique or exclusion constraint that makes the surplus write fail is the third.
  4. Name the plausible non-fix explicitly, since it is what usually gets merged instead: folding the count into the insert as insert ... select ... where (select count(*) ...) < limit is still racy under read committed, because the subquery cannot see the other transaction's uncommitted rows. It looks atomic and is not.
  5. Say what you do if they still disagree: escalate the decision rather than the disagreement. Attach the reproduction, hand it to the service owner or a third reviewer, and state that you will not block the merge if the owner accepts the risk knowingly — and that you want that acceptance written down.
  6. Close with the general lesson worth leaving in the review thread: a passing suite is weak evidence for a concurrency claim because it runs one request at a time. Ask for a test that runs two.

Follow-up

  • Write the counter-row version. Does your answer change if the quota counts child rows rather than a column?
  • Under serialisable, who performs the retry, and what does the API client see if the retry also fails?
  • This is the third disagreement with the same reviewer this month. What changes in how you review?

Unblock an engineer on a job run that finished twice

Easy · mentoring · fencing tokens · lease expiry · debugging method

An engineer two weeks into the team brings you a job_run row showing status succeeded with an exit_code written by a worker declared dead ten minutes earlier; the retry attempt also shows succeeded. They have spent a day adding logging and are no closer. You have twenty minutes and you do not want to take the keyboard. Describe how you unblock someone: the question you ask first, what you let them find themselves, the concept you name and when, and how you check the next day that they own the fix rather than having watched you produce it.

Approach

  1. Ask what they expect rather than what they see: which statement set status to succeeded, and what did it check before writing? That question points directly at the update's WHERE clause, which is where the answer lives, and it costs them nothing to answer, so it does not read as a test.
  2. Let them build the timeline themselves from the row: queued_at, started_at, leased_until, finished_at and worker_id, on both the original run and the retry. Two different worker_ids with a lease expiry between them tells the whole story, and they will see it before you say it.
  3. Name the concept once the evidence has earned it. A lease bounds time; it does not prevent a write. The store has to reject a stale writer, which means the update carries a fencing token the row compares — update job_run set status = 'succeeded' where run_id = $1 and lease_token = $2 and status = 'running' — and a long garbage-collection pause or a brief partition is enough to produce what they are looking at.
  4. Point at the second, less obvious half and let them decide it: 'lost' exists in the status enum precisely so a run whose worker vanished is not recorded as failed, because failed asserts an outcome nobody observed and the system then bills and retries on that assertion. Ask them what these two rows should have said.
  5. Leave them with the next step rather than the patch — a test that kills the first worker after the sandbox exits and before the row is written — and say when you are available again, so the offer is real rather than polite.
  6. Check ownership the next day by what they produced, not by asking if it went well: a test that reproduces the window proves they understood it; a test that only asserts the new WHERE clause proves they copied it. Ask them to explain it to a third person and listen for whether the explanation is theirs.

Follow-up

  • They propose a longer lease instead of a token. What do you say, and what breaks when legitimate runs last thirty minutes?
  • How can you tell whether your explanation landed or they simply deferred to you?
  • The same engineer hits a variant of this next month. What did you fail to teach the first time?