Akita Software Engineer Interview Questions & Guide 2026

The 60-second brief

The Software Engineer role at Akita—often categorized within their support and engineering hierarchy as a 1st Line or 2nd Line Engineer—is a critical position focused on maintaining the operational integrity of client systems. You are the primary technical interface for users, responsible for diagnosing complex issues, implementing solutions, and ensuring that technical infrastructure remains stable and efficient.

Treat capacity estimation as a conversion skill rather than a table to memorise: turn a user count and an action rate into requests per second and bytes per day, then name the component that number breaks first. The figure only matters if it changes the design.

Focus areas

• Trace a symptom to a mechanism under load
• Make every write idempotent under retry
• Choose indexes from the query's access path

Overview

This role is not just about fixing bugs; it is about providing high-quality technical stewardship. You will contribute to the ongoing success of Akita by bridging the gap between sophisticated technical architecture and the practical, day-to-day needs of clients. Whether you are performing initial triage as a 1st Line Engineer or diving into deeper technical resolution as a 2nd Line Engineer, your work directly impacts client satisfaction and the reputation of the company.

The Akita interview process

Akita candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Screening Call

Before anything technical happens, someone has to decide which rung of the ladder your loop is calibrated to, and that decision sets the bar for every round after it. It comes from how you describe scope, not from your title, because titles do not convert cleanly between companies. The weak version of the answer is team size and years. The strong version names the largest change you shipped where nobody reviewed the design, what would have broken if you had been wrong, and what you were paged for. Get the level said out loud on this call, because the range and the loop both follow from it.

What to demonstrate

• Whether the scope in your own account maps onto a level the team actually has an opening at, so a mismatch ends the process cheaply rather than after four interviewers have spent a day
• Whether your title needs re-mapping: the same word describes very different amounts of independent decision-making at a twenty-person company and a ten-thousand-person one
• Whether your compensation expectation can be filled at that level in the structure the role pays in, which is why the number gets asked for before any engineer is scheduled

How to prepare

• Write down two changes from the last two years: the largest one you designed with nobody reviewing the design, and the largest one where someone more senior did. Lead with the first when scope comes up, and be ready to say which parts of the second were yours
• Ask which level the loop is calibrated to and what changes at the level above it, then plan your weeks from that answer rather than from the posting
• Settle a total-compensation range beforehand with the split named, base against bonus against equity and its vesting period, so a question about numbers gets a number instead of the word market

Technical Assessments

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

Where candidates lose points

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

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

Letting a slow dependency consume unbounded concurrency

The failure that takes a service down is usually not an error but a delay. A dependency answering in thirty seconds instead of fifty milliseconds holds each request's worker or connection six hundred times longer, and since required concurrency is arrival rate times latency, a fleet sized for sixty in-flight requests now needs thirty-six thousand to sustain the same rate - so it queues, and requests whose clients have already abandoned them still occupy resources. Retries make it precisely worse: a policy of three attempts triples the load on a dependency at the exact moment it is least able to serve, which is how one slow dependency becomes an outage of everything sharing that pool. Containment is four specific things - a timeout on every outbound call shorter than the caller's remaining budget, a bounded pool per dependency so one cannot starve the others, backoff with full jitter rather than a fixed delay so retries do not resynchronise, and a circuit that stops sending once the failure rate makes an attempt pointless.

Not asking what the system looks like if it dies halfway through

For any multi-step write, say what state remains if the process stops between step two and step three, and what brings it back: a single transaction, a saga with compensating actions, an outbox, or a reconciliation job. Partial failure is routine at any real call volume, so 'that shouldn't happen' is an answer with nothing behind it.

Abandoning working code to chase the optimal solution

Get the straightforward version correct, state its complexity, and only then optimise, keeping the working version until the faster one passes the same cases. A correct quadratic solution with a stated path to linear beats a half-written optimal one that never ran.

Questions & practice

Canonicalise a request body into a stable idempotency fingerprint

Medium · parsing · canonicalisation · hashing

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

Approach

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

Follow-up

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

Find overlapping job attempts and peak concurrency from lease records

Medium · sweep line · intervals · leases

A day of job_run history yields about 50,000,000 attempt records: (job_run_id, job_type, attempt, started_at, finished_at which is NULL when the worker died, lease_expires_at). Leases expire on a clock, so a job that outran its lease ran twice. Produce (a) every job_run_id whose attempts overlapped in wall-clock time and (b) the peak number of simultaneously running attempts per job_type with the minute it occurred. Target O(n log n). State how you treat a NULL finished_at and what clock skew does to your answer.

Approach

1. Define the interval before sorting anything: an attempt occupies [started_at, COALESCE(finished_at, lease_expires_at)). finished_at is observed and lease_expires_at is only a promise, so every attempt without a finish contributes an estimate and the whole result is a lower bound on overlap rather than an exact count.
2. For peak concurrency, sweep: emit 2n endpoints, sort by (timestamp, kind) with ends ordered before starts at equal timestamps, then walk the sequence maintaining a counter per job_type and record each type's maximum with its timestamp. O(n log n) dominated by the sort, O(n) space, or O(1) extra if the sort is external and the walk streams.
3. For overlap detection, do not compare attempts pairwise. A single global sort by (job_run_id, started_at) gives both the grouping and the order; within a group, keep the maximum end seen so far and report an overlap exactly when the next start is less than that running maximum, which is one linear pass after the sort.
4. Half-open intervals matter and are easy to get wrong: with closed intervals an attempt ending at the same millisecond another begins reads as concurrency two, and across 50,000,000 records that artefact swamps the real signal.
5. State the clock caveat: started_at and finished_at are written by different workers, so under skew of a few hundred milliseconds an apparent overlap shorter than that bound is not evidence. Filter reported overlaps by a minimum duration, or prefer timestamps written by whichever component heartbeats the lease.
6. Scale the sort rather than assuming it fits: the sweep emits two endpoints per attempt, so 50,000,000 records become 100,000,000 endpoints, and at roughly 24 bytes each, an 8-byte timestamp plus a 4-byte job_type plus a kind flag padded to alignment, that is about 2.4 GB of sort keys before any scratch space. Either push the ordering into the database behind an index on (job_type, started_at) or run an external merge sort in chunks; the overlap pass sorts n records rather than 2n, so it is the cheaper of the two.

Worked solution (30 min)

1. Write the interval derivation with the COALESCE and state in one line which of the two end sources is observed and which is assumed.
2. Write the concurrency sweep: the endpoint tuples, the sort key including the end-before-start tie-break, and the per-job_type counter.
3. Hand-trace four attempts of one job, two disjoint and two overlapping by three seconds, and confirm the overlap detector fires exactly once.
4. Add the skew filter as a minimum overlap duration, state the value you chose, and justify it from how the timestamps are written.

Expected result: One global sort by (job_run_id, started_at) driving a running-maximum-end check for overlaps, plus a sweep over 2n endpoints with ends ordered before starts for per-job_type peak concurrency, both O(n log n), with half-open intervals, NULL finished_at falling back to lease_expires_at, and overlaps shorter than the clock-skew bound excluded.

Verify it yourself

• Two attempts touching at exactly the same instant report concurrency 1, not 2.
• A job with a single attempt never appears in the overlap output regardless of its lease.
• The reported peak timestamp for a job_type lies inside at least one of that type's intervals.

Follow-up

• A handler is not idempotent and you have found 400 overlapping jobs. Which of them actually caused damage, and what would you query to find out?
• Peak concurrency for one job_type is 4 against a configured cap of 4. Is the cap working, or is the data hiding attempts that never started?
• How would you compute both answers incrementally as records arrive rather than in a daily batch?

Diff a projection against the primary without per-row point reads

Hard · reconciliation · range hashing · throttling

The listing projection has drifted and some rows show a stale version. The primary holds 40,000,000 resource rows across 12,000 tenants while serving 1,200 writes and 14,000 reads per second. The obvious repair, reading each resource row and comparing its version against the projection, is correct and would eventually finish. Explain precisely why it is unacceptable here, then give a diff that finds the differing rows, state its complexity, and make it safe to run against a live primary. Replication lag is usually under 100 ms and is not bounded.

Approach

1. Quantify the naive cost rather than calling it slow: 40,000,000 point reads at even 0.5 ms each is over five hours serialised, and the only lever is concurrency, which is exactly what you cannot spend. The primary's pool is sized for the write path, and 40,000,000 random reads evict the buffer cache that sustains the 85 percent cache hit rate, so the audit degrades the system it is auditing.
2. Replace random access with one ordered pass per side. Both sides can be read in (tenant_id, resource_id) order, which is a sequential scan on each and a merge join in O(n) time and O(1) memory. For a dense diff that is the whole answer, and it reads the primary once instead of 40,000,000 times.
3. For the expected sparse case, compare range hashes instead of rows: partition the key space, compute per range an order-independent aggregate over hash(resource_id, version), compare aggregates, and descend only into ranges that differ. With d differing rows and branching factor B, at most d ranges mismatch per level, so the drill-down examines O(d log_B(n/d)) ranges and reads full rows only in mismatching leaves.
4. Aggregate with a sum modulo 2^64 or a multiset hash, never XOR. XOR is order-independent but self-cancelling, so two rows wrong in the same way, or a row duplicated on one side, leave the range aggregate matching and the range is declared clean.
5. Pin the comparison to a point in time or it reports lag as drift: consider only rows whose updated_at is older than now minus a lag margin, and re-check each candidate mismatch individually before repairing. At 1,200 writes per second a diff without this reports thousands of false positives, and an unattended repairer would then overwrite live rows with stale values.
6. Make the run resumable and throttled: batch by range key, persist the last completed range, and watch a signal such as replica lag or primary CPU, pausing rather than pressing on. A reconciliation that cannot be stopped and resumed gets killed halfway and restarted from zero, which is how a repair becomes an incident.

Follow-up

• The diff reports 900 stale rows. How do you decide between patching those rows and rebuilding the projection from resource_revision?
• Same job, but the projection lives in a search index that cannot be scanned in key order. What changes?
• How would you run this continuously at low cost instead of only as incident response?

Replace offset paging on the resource feed with keyset

Easy · keyset pagination · composite index · sort order

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

Approach

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

Follow-up

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

Keep soft-deleted accounts from blocking re-registration

Easy · soft delete · unique constraints · engine differences

app_user holds user_id, tenant_id, email CITEXT, password_hash (NULL for SSO principals), email_verified_at, auth_version, status ('invited','active','suspended','deactivated'), created_at, updated_at, deleted_at. Two live accounts for one address inside a tenant must be impossible, but an address freed by a soft delete must be reusable, and the same tenant may delete and re-register it repeatedly. Write the uniqueness DDL for PostgreSQL 16, then the equivalent for MySQL 8 where partial indexes do not exist, and say what each permits once three deleted rows already hold that address.

Approach

1. Start from what is actually unique: not (tenant_id, email), but (tenant_id, email) among live rows. PostgreSQL says that directly — CREATE UNIQUE INDEX app_user_live_email ON app_user (tenant_id, email) WHERE deleted_at IS NULL. A full constraint over the same two columns burns the address permanently the first time someone deletes an account.
2. Keep case-insensitivity in the type or the index, never in the application: CITEXT as given, or UNIQUE (tenant_id, lower(email)) as an expression index where the extension is unavailable. A case-sensitive unique column is exactly how two accounts for one human appear.
3. For MySQL 8 the predicate has to move inside the key: add a discriminator column that is a constant 0 while the row is live and is set to user_id on delete, with UNIQUE (tenant_id, email, deleted_marker). Live rows share the constant and still collide; deleted rows differ from each other and stop colliding.
4. State the NULL variant and its dependency: leaving the marker NULL for deleted rows also works, because a unique index treats NULLs as distinct — true in MySQL, and true in PostgreSQL only under the default NULLS DISTINCT, which PostgreSQL 15 lets you reverse. Check the polarity against the three existing deleted rows: constant-on-live is what preserves the collision you want, and reversing it silently admits duplicate live accounts.
5. Say what a soft delete must do besides setting deleted_at: increment auth_version so existing tokens stop validating, leave resource.owner_user_id and resource_revision.actor_user_id intact, and accept that the address is retained — erasure is a different requirement answered by scrubbing the column, not by a DELETE that would break those references.

Worked solution (20 min)

1. Create the PostgreSQL partial unique index, insert a live row, soft delete it, and insert the same address again.
2. Repeat the delete-and-reinsert cycle three times and confirm three deleted rows coexist with exactly one live row.
3. Write the MySQL form with the discriminator, then deliberately reverse the polarity so live rows carry NULL, and show two live duplicates commit.
4. Attempt a second live insert on both engines and map the resulting 23505 / ER_DUP_ENTRY to the 409 the handler should return.

Expected result: PostgreSQL rejects only the second live row and admits unlimited deleted ones. MySQL behaves identically with UNIQUE (tenant_id, email, deleted_marker) where live = 0; with the polarity reversed, two live rows commit and nothing afterwards can tell them apart.

Verify it yourself

• Group live rows by (tenant_id, lower(email)) after a randomised insert/delete run — no group exceeds one.
• Inserting 'User@x.test' after 'user@x.test' is rejected; if it is accepted, the case handling is in application code rather than in the index.
• EXPLAIN a login lookup and confirm the partial index is chosen — it is only usable by queries that carry the deleted_at IS NULL predicate, so a query without it falls back to a scan.

Follow-up

• A deleted account re-registers with the same address the next day. Do the old resource rows follow the new user_id, and how does the API keep the two principals apart?
• How do you honour an erasure request while resource_revision.actor_user_id still references this table?
• What changes if a user may hold membership in two tenants?

How do you balance the need for speed with the need for a permanent, high-quality fix?

Medium

Approach

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

Follow-up

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

Provide an example of a time you had to explain a technical solution to someone without an engineering background.

Medium

Approach

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

Follow-up

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

Cache the tenant listing feed with a bounded staleness window

Easy · caching · cache invalidation · keyset pagination

GET /v1/resources returns one tenant's resources ordered by updated_at DESC, 20 per page, at 14k requests/second peak against a 120 ms p99. The table carries the index (tenant_id, status, updated_at DESC, resource_id DESC) and writes land on the primary at 1.2k/second. Design the read path: the pagination contract, the cache key and value, what a write invalidates, and the staleness a user can observe. State the request rate that actually reaches the database, and the one repopulation race that deleting on write does not close.

Approach

1. Settle the pagination contract first, because it decides what is cacheable. OFFSET makes the database produce and discard the skipped rows, so page 500 costs five hundred pages of work, and rows inserted between two fetches shift across the boundary and are skipped or repeated with nothing in the response to reveal it. The cursor is the row value of the last row returned: WHERE tenant_id = $1 AND status = $2 AND (updated_at, resource_id) < ($3, $4) ORDER BY updated_at DESC, resource_id DESC LIMIT 21. That is a row-value comparison, not updated_at < $3 AND resource_id < $4, which is a different and wrong predicate.
2. Confirm the index actually serves it: equality on the two leading columns, then a range on the pair that follows in exactly the index's sort order, so the plan is an index scan that touches 21 entries with no sort node. Requesting 21 to return 20 is how has_more is answered without a count. resource_id is not decoration - updated_at is not unique, and without the tie-break two rows sharing a timestamp at a page boundary are the skip that keyset pagination was adopted to remove.
3. Key the cache on every value the predicate reads: tenant_id, status, cursor and page size. A key that omits tenant_id is a cross-tenant disclosure, and no test running against a single tenant's data will show it.
4. Be honest that a write does not invalidate one key. An update moves its row to the head of the ordering, so it invalidates the first page and every cursor page whose range spans the row's old and new position, which is not enumerable. Cache the first page per (tenant_id, status) - that is where the traffic is - with a short TTL, invalidate it on write, and serve deep cursor pages uncached from a replica, since each is already a 21-row index scan and they are rare.
5. Name the residual race and the real bound. A reader that loaded rows before the write can populate the cache after the invalidation deleted the key, so the delete is not a staleness bound; the TTL is. Choose the TTL as the staleness a listing can tolerate, and jitter expiries so a busy tenant's keys do not all expire together and stampede the replica. Then do the arithmetic: at an 85% hit rate, 14k requests/second is about 2.1k database reads/second across two replicas, and that is the number capacity planning uses.

Worked solution (20 min)

1. Write the keyset predicate and match it column by column against the index, marking which columns are equality, which is the range, and which satisfies the ORDER BY.
2. Compare rows examined for page 1 and page 500 under OFFSET and under keyset, and state both numbers.
3. Write the cache key template and the first-page invalidation the write path performs.
4. Write the interleaving in which a stale value is written into the cache after the invalidation, and identify what bounds it.

Expected result: Keyset pagination on the row value (updated_at, resource_id) served as a 21-entry index scan with no sort at any page depth. The first page per (tenant_id, status) is cached under a short jittered TTL and deleted on write; deeper cursor pages go to a replica uncached. Observable staleness is the TTL, not the invalidation, because a concurrent reader can repopulate the key afterwards. At an 85% hit rate roughly 2.1k reads/second reach the two replicas.

Verify it yourself

• The query plan shows an index scan with no Sort node, and rows examined equal to the limit, at page 1 and at page 500.
• Two rows sharing an updated_at value across a page boundary are returned exactly once across the two pages.
• Every cache key contains tenant_id.
• Removing the TTL leaves staleness unbounded even though every write still invalidates.

Follow-up

• The tenant writes and immediately lists. What does it see, and what is the smallest change that makes its own write visible without sending all 14k requests/second to the primary?
• A tenant has 4 million resources and a client walks every page nightly. What does that do to the cache hit rate, and should that traffic share this path at all?
• Sort order becomes configurable - by title, by created_at. What happens to the index set and to the cache key space?

Describe your process for handling a critical system outage reported by a client.

Medium · debugging and production issues

Approach

1. Say what evidence would prove you wrong, then go and look for it.
2. Check the instrumentation before believing the symptom.
3. Establish what changed and when, before forming any theory.

Follow-up

• What would you add now so this is faster to diagnose next time?
• How would you tell a cause from a coincidence here?

Explain the difference between [specific protocol or service] and how you would troubleshoot a failure in that area.

Medium · debugging and production issues

Approach

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

Follow-up

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

What steps do you take when you encounter a technical issue you have never seen before?

Medium · debugging and production issues

Approach

1. Separate the trigger from the cause; the deploy is rarely the bug.
2. Pick a bisection that eliminates candidates whichever way it turns out.
3. Say what evidence would prove you wrong, then go and look for it.

Follow-up

• How would you tell a cause from a coincidence here?
• What would you add now so this is faster to diagnose next time?

Suggested practice checklist

For a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.

Day 1 — Numbers before diagrams

• Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
• Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
• Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.

Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.

Day 2 — One system, from requirements to schema

• Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
• Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
• Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.

Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.

Day 3 — The consistency you are actually buying

• Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
• Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
• Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.

Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.

Day 4 — Failure is the design

• For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
• Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
• Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
• Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.

Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.

Day 5 — Scaling the hot path

• Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
• Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
• Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
• Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.

Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.

Day 6 — Keep the coding hand in, at the bar that applies to you

• Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
• Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
• Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.

Deliverable: One problem hardened to review standard, with its test list and one written review comment.

Day 7 — Defend it while being interrupted

• Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
• Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
• Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.

Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.

What the behavioural round is actually measuring

Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.

How do you handle a frustrated client who is experiencing downtime?

Medium · behavioural and engineering judgement

Approach

1. Give the blast radius: what could have broken, and what you measured.
2. Name the disagreement and how you resolved it with evidence.
3. State the situation in two sentences and spend the rest on the reasoning.

Follow-up

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

Reverse your own decision and price the reversal

Medium · reversibility · measurement · migrations

Describe a technical decision you made and later reversed. Pick one that cost something: a service you split and merged back, a cache you added and removed, an index you created that pushed the planner onto a worse plan, a projection you rebuilt from scratch. State what you believed when you decided, the measurement that changed your mind, how long the wrong version ran in production, and what the reversal cost in migrations, dual writes, and a deprecation window for callers you did not own.

Approach

1. State the original rationale without irony, in the version you would still defend given what was known then. If it is not defensible, the story is about carelessness rather than judgement, and a different example serves you better.
2. Give the measurement that moved with a before and after: the p99 that did not improve, the cache hit rate that sat at 40%, the plan that flipped to a sequential scan once the table passed a size you can name.
3. Cost the reversal in steps, not adjectives: expand-and-contract deploys, the dual-write window, the callers who had to be notified, the rows already written in the wrong shape that had to be backfilled or abandoned.
4. Distinguish reversal from rewrite by naming what you kept. Most good reversals preserve the schema or the interface and undo one decision inside it, which is also why they were affordable.
5. Finish on the process change: the smallest experiment that would have produced the same measurement in a day, and why you did not run it the first time.

Follow-up

• What in that decision was irreversible, and did you know it was irreversible when you made it?
• How did you tell the people who had already built on top of the original decision?
• What do you now measure before committing to a change of this size?

Tell callers you do not own that their integration breaks

Medium · deprecation · compatibility · stakeholders

A field in a write endpoint's response must change shape. You own the endpoint; you do not own the four internal callers or the outbound webhook consumers who read it. Describe a deprecation you were responsible for: what you shipped first, how you established who was actually reading the field, the window you gave and what set its length, what you did about the consumer who never moved, and how you decided removal was safe. Name the signal you used, not the announcement you sent.

Approach

1. Establish the reader set empirically rather than from a wiki of owners: per-field usage counters keyed by principal, or access logs attributed to a consumer. State the blind spot of whichever you pick, since a consumer that reads the field only on a monthly job will not appear in a week of logs.
2. Ship additive first. Populate the new field alongside the old one so no reader is forced to move, which is also what keeps a rolling deploy safe, because old and new instances answer the same requests at the same time and a rollback must still find the old shape present.
3. Set the window from the slowest legitimate consumer's release cadence, not from your calendar, and decide separately what to do for a consumer with no release process at all, such as an external webhook endpoint you can only email.
4. Convert silence into evidence before you rely on it: a short, low-traffic removal window that makes a still-dependent consumer fail visibly and loudly while you are watching, rather than at three in the morning after you have moved on.
5. State the removal criterion as a measurement with a duration attached, such as observed reads at zero across a full billing cycle, and keep the change reversible for one release after removal.

Follow-up

• How would you detect a consumer that reads the field only during a monthly export?
• One caller refuses to move and has a commercial relationship behind it. What changes in your plan and what does not?
• After removal, what makes the change irreversible, and how long before you cross that line?