Interview guides · Abacode

Abacode Software Engineer Interview Questions & Guide 2026

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

The 60-second brief

A Software Engineer at Abacode plays a pivotal role in the technical infrastructure that supports the company’s mission. This position is not merely about writing code; it is about building scalable, efficient, and robust solutions that allow the organization to maintain its edge in a competitive and rapidly evolving landscape. You will be tasked with solving complex problems that directly impact the reliability and functionality of internal and client-facing systems.

The shape of the workload matters more for your prep than the industry label does. Read-heavy serving, write-heavy ingestion and scheduled batch processing have different binding constraints and fail in different places, so find out which one the team lives in before picking design topics.

Focus areas

Bound every outbound call with a timeout
Trace a symptom to a mechanism under load
Choose indexes from the query's access path

Overview

The work you do at Abacode requires a deep understanding of core programming fundamentals, specifically within the Python ecosystem. Whether you are working on network deployment logic or broader software architecture, your contributions will be central to how the team manages data, automates processes, and delivers high-quality results. Expect a role that demands both technical precision and a strong grasp of object-oriented principles to ensure that systems remain maintainable as they scale.

While the core technical focus is on Python and OOP, be prepared to explain how these concepts apply to real-world network or deployment challenges.

The Abacode interview process

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

Initial Screening

The person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.

What to demonstrate

Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural

How to prepare

Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one

Technical Discussions

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.

Running a schema change as though the lock lasts as long as the statement

In PostgreSQL an ALTER TABLE that needs an ACCESS EXCLUSIVE lock must first wait for every open transaction touching that table, and while it waits, later queries needing a conflicting lock queue behind it rather than overtaking it. A DDL statement that would execute in milliseconds, issued while a thirty-second analytics query is open, therefore stalls all traffic on that table for thirty seconds: the outage length is set by the longest open transaction, not by the change. The defences are specific and worth knowing by name - set lock_timeout low and retry rather than queue, add columns without a volatile default so no table rewrite occurs (from version 11 a non-volatile default is a metadata-only change), build indexes with CREATE INDEX CONCURRENTLY while accepting that it cannot run inside a transaction block and leaves an invalid index behind if it fails, and add constraints as NOT VALID followed by a separate VALIDATE CONSTRAINT, which takes a weaker lock.

Never running a concrete value through the code

Trace one small input and one edge input by hand, index by index, out loud. Re-reading your own code catches design mistakes; walking a real value through it catches the off-by-one, the uninitialised accumulator and the loop that never advances.

Finishing a solution without stating its complexity

Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.

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

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.
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.
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.
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.
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.
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.

Worked solution (25 min)

Write the serialiser: recursive emit with a depth counter, objects sorted by UTF-8 key bytes, arrays in order, strings escaped by one fixed rule, numbers emitted as their original token.
Run it over three bodies: the same object with keys reordered, the same object with \u0041 written as A, and one with a nested array reversed. The first two must produce identical bytes and the third must not.
Take the id 9007199254740993, round-trip it through a double, show it returns as 9007199254740992, then state the rule that prevents this.
Define the hash preimage explicitly with its delimiters, and construct a pair of (path, body) inputs that would collide without them.

Expected result: A canonicaliser that is O(n log n) in body size with an enforced depth cap, sorts keys by UTF-8 byte order, preserves array order, keeps number literals verbatim, rejects duplicate keys, and feeds a delimited preimage to SHA-256, together with a stated list of normalisations deliberately not performed and the failure each would cause.

Verify it yourself

Key reordering and whitespace changes produce identical fingerprints while reordering an array does not.
A body containing an integer above 2^53 fingerprints identically after a parse and re-serialise round trip.
Two requests differing only in a field the endpoint ignores produce different fingerprints, and you can say why that is the safe direction to err.

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?

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

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.
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.
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.
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.
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.
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?

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

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.
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.
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.
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.
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.
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.

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?

Denormalise tenant onto revisions and backfill it live

Hard · denormalisation · online migration · backfill

resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.

Approach

Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.
Step three, backfill: batch by primary key rather than by created_at so the cursor is dense and resumable — UPDATE resource_revision rr SET tenant_id = r.tenant_id FROM resource r WHERE r.resource_id = rr.resource_id AND rr.revision_id > $1 AND rr.revision_id <= $1 + 5000 AND rr.tenant_id IS NULL — committing per batch and persisting the cursor. Throttle on replica replay lag and on dead-tuple count, since each batch writes 5,000 new row versions. Run the backfill before the index exists so those updates can stay HOT.
Step four, index then enforce then contract: CREATE INDEX CONCURRENTLY (cannot run inside a transaction block, scans the table twice, waits on open transactions, and leaves an INVALID index to drop concurrently if it fails); ADD CONSTRAINT ... CHECK (tenant_id IS NOT NULL) NOT VALID, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE, after which SET NOT NULL uses the validated check instead of re-scanning on PostgreSQL 12 and later. Only then move the audit reads onto the column and, in a later deploy, delete the join path.

Follow-up

The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
How do you verify the backfill actually finished, given rows are still being inserted while it runs?
A resource must now be movable between tenants. What does that do to the composite foreign key and to the revisions already written?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.

Approach

Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.
Answer the two negatives explicitly. owner_user_id alone is not a left prefix of the new index, so it degrades to a full scan of the index at best. Ordered by updated_at, the query still seeks on the (tenant, owner) pair but must sort, because only created_at is ordered within that pair.
Measure both sides with EXPLAIN (ANALYZE, BUFFERS) and compare estimated against actual rows at the lowest node — a 2M-versus-200 misestimate there is usually what chose the plan, and adding an index will not fix a statistics problem.

Worked solution (25 min)

Load 2M resource rows across 5k owners in one tenant, run the query under EXPLAIN (ANALYZE, BUFFERS), and record the node reading the most rows plus the Sort Method line.
Create (tenant_id, owner_user_id, created_at DESC) concurrently and re-run, confirming the Sort node is gone and actual rows fall to about 20.
Run the two negative cases and capture the plan for each.
Re-run the original tenant listing query to confirm the new index has not displaced the index that query depends on.

Expected result: With (tenant_id, owner_user_id, created_at DESC) the plan is an Index Scan with no Sort, roughly 20 rows and a handful of buffers, sub-millisecond. owner_user_id alone still scans; ORDER BY updated_at seeks on the pair and then sorts that owner's rows.

Verify it yourself

Actual rows at the leaf node is on the order of the LIMIT, not the tenant's row count.
No Sort node appears above the index scan.
Buffers read stays constant as one owner's row count grows; if it grows with it, the ORDER BY is not being satisfied by the index.
pg_stat_user_indexes shows idx_scan climbing on the new index and not on the listing index for this endpoint.

Follow-up

90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
A dashboard runs this for 40 owners in one page load. What changes about the design?
How do you roll this index out on a table taking 1.2k writes/second, and what does it cost on every insert from then on?

What are the core pillars of object-oriented programming?

Medium

Approach

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

Follow-up

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

How does Python handle memory management in the context of objects?

Medium

Approach

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

Follow-up

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

Can you explain the difference between overloading and overriding?

Medium

Approach

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

Follow-up

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

How do you apply inheritance in your code?

Medium

Approach

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

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

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.
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.
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.
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.
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)

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.
Compare rows examined for page 1 and page 500 under OFFSET and under keyset, and state both numbers.
Write the cache key template and the first-page invalidation the write path performs.
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?

One customer endpoint stalls deliveries to every other destination

Medium · bulkhead · timeouts · saturation · retry amplification

The egress service delivers about 1.5k webhooks/second across 40,000 destinations, with a per-destination concurrency cap of 4 and a 10-second connect-plus-read timeout. Throughput falls to 300/second, queue depth climbs, and p99 delivery latency for unaffected destinations goes from 200 ms to minutes, while the error rate barely moves. One tenant holds 900 destination rows whose URLs share a hostname that now answers in 9.5 seconds. Explain the mechanism with the arithmetic, then give the containment in the order you would apply it.

Approach

Look at saturation before errors. A flat error rate with collapsing throughput says nothing is failing, things are waiting, so the first signal to pull is in-flight request count or pool wait time rather than the error counter. This is the distinction that decides the whole investigation.
Group in-flight work by resolved host, not by destination id. The cap is keyed per destination row, so 900 rows sharing one hostname buy 3,600 concurrent slots against a single host, each held for 9.5 seconds. The bulkhead was never a bulkhead for that host, and grouping by the wrong dimension is why the dashboard looked healthy.
Do the arithmetic in both directions. Required concurrency is arrival rate times latency, so 1.5k/second at 200 ms needs about 300 in flight, which is entirely consumed by 3,600 slow slots; conversely whatever concurrency is left sustains rate equals concurrency divided by 9.5 seconds, which is the 300/second you are seeing. Matching both numbers is what promotes this from a plausible story to the mechanism.
Explain why the circuit breaker never helped. It opens on consecutive failures, and a 9.5-second response inside a 10-second timeout is a success. Slow is not failing, so an error-rate breaker cannot see this; you need a slow-call ratio, a deadline propagated from the caller's remaining budget, or a concurrency limiter.
Contain in order: park the offending host so the shared pool drains, add a per-resolved-host concurrency cap alongside the per-destination one, give slow hosts their own queue so they cannot occupy the general pool, derive the timeout from the delivery deadline rather than a round number, and check the retry policy is not tripling load on a host that is already slow. Use backoff with full jitter so retries do not resynchronise on recovery.
State the invariant you are restoring: one tenant's endpoints degrade only that tenant's deliveries. That is a property to load-test for, not to assume from a config value.

Follow-up

The host recovers to 80 ms. How long does the queue take to drain, and what does the drain do to the recovered host?
Where should the 10-second timeout number actually come from?
If that tenant had one destination row instead of 900, would the cap of 4 have saved you? What would you measure to be sure?

Suggested practice checklist

Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.

Day 1 — Diagnostic, scored before you study anything

Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.

Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.

Day 2 — Largest gap: find the boundary rather than the subject

Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
Re-attempt one of them from blank four hours later with nothing open.

Deliverable: A sub-skill map with the two blocking sub-skills circled.

Day 3 — Drill the blocking sub-skill by repeating the shape

Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
Have someone else read your one-sentence rule and find the precondition you left out.

Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.

Day 4 — Second gap, plus maintenance on the strongest area

Run the same sub-skill decomposition on the second-largest gap in half the time.
Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.

Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.

Day 5 — The gap that is not a skill

Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.

Deliverable: Two recordings with a counted reduction in time-to-first-question.

Day 6 — Retest under day-one conditions

Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
Write down which single block you would still lose the offer on.

Deliverable: A second scored rubric placed beside the first, with one named remaining risk.

Day 7 — Full loop under interview conditions

Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.

Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.

Know where your numbers came from

Bring the two or three numbers the story rests on and know how they were collected. A p99 whose timer starts inside your handler excludes the time a request spent queued, so it can sit flat while users wait longer. Give the window, the percentile and what the measurement left out, or drop the number.

Argue against a design, lose, and commit anyway

Medium · disagreement · service boundaries · decision records

Describe a design you argued against and lost. State the failure you predicted as a named mechanism, not a feeling about complexity: two services that would need one transaction, a projection with no rebuild path, a write path with no idempotency key. Say what evidence you brought, what the decision maker weighed instead, and what you did after the decision was made: what you instrumented, what you wrote down, and whether the prediction came true. Five minutes.

Approach

State the prediction in falsifiable form up front: the mechanism, the condition that triggers it, and the observable outcome. A prediction that cannot be checked also cannot be credited to you later.
Show the evidence you had at the time and label each piece honestly as measured, analogous, or intuition. Keeping the intuition is fine; disguising it as data is the thing that erodes your standing in the next argument.
Represent the opposing case at full strength, including the constraint you did not control: a fixed date, a team boundary, or the fact that the decision was cheap to reverse and yours was not.
Make disagree-and-commit concrete. Name the artefact you left behind so the prediction could be settled without you: the alert and its threshold, the counter on the dashboard, the decision note that recorded the trade-off and the condition that would revisit it.
Report the outcome without editing it. If the design held and your predicted mechanism never fired, say so and say what you had mis-weighted, which is more persuasive than a vindication story.

Follow-up

What threshold on that alert would have proved you right, and did anyone ever look at it?
If the same proposal arrived tomorrow with the same deadline, would you argue it the same way?
How did you behave toward the design once it shipped and started failing in a different way than you predicted?

Narrate an outage you owned from page to postmortem

Hard · incident response · blast radius · postmortems

Pick an incident you personally drove, ideally one where writes were affected rather than reads. In six to eight minutes: state the symptom as it first appeared on a dashboard, the blast radius you established before you knew the cause, the mitigation you applied and when, the mechanism you eventually proved, and the follow-up that would prevent a repeat. Bring numbers: error rate, tenants affected, minutes to mitigate, minutes to resolve. If you cannot name what you measured, choose a different incident.

Approach

Open on the signal rather than the cause: which metric at which percentile moved, on which service, at what time, so the listener follows the same evidence you had rather than a conclusion you already reached.
Separate mitigation from diagnosis out loud. State what you did to stop the bleeding (flag off, shed traffic, drain a lease, roll back a deploy) and say plainly that you did it before the mechanism was known, because those are two jobs with different deadlines.
Establish blast radius in countable terms: how many tenants, how many writes, and crucially whether the effect was loss or only delay. An append-only revision table or a pending outbox row means the change survived and the projection was merely behind, which is a repair rather than a data-loss incident.
Prove the mechanism instead of asserting it. Name the trace span that grew, the plan that flipped to a sequential scan, the lease that expired, plus one alternative you ruled out and the signal that stayed flat while you ruled it out.
Close on the durable fix and its cost, distinguishing what landed that week from what needed an expand-and-contract migration across several deploys, and say which of the two you actually finished.

Follow-up

What would you do differently in the first five minutes, given the same dashboard and no more information?
Which follow-up action did you deliberately not take, and why was dropping it the right call?
How did you convince yourself the mitigation was safe to apply while the cause was still unknown?

Turn a code review disagreement into a decision

Easy · code review · optimistic concurrency · communication

A colleague's change updates a row with UPDATE resource SET version = version + 1 WHERE resource_id = $1 AND version = $2 and treats an affected-row count of zero as a successful no-op. You read that as a silently lost update; they think returning 200 is friendlier to clients than returning a conflict. Describe how you have handled a review disagreement of this shape: what goes in the comment, when you leave the thread, and who decides. Then write the comment you would leave here, in under 80 words.

Approach

Sort the disagreement before writing anything. A silently discarded write is a correctness claim about data; the choice between 409 and 412 is taste. Only the first justifies blocking a merge, and saying which one you are doing is most of the value of the comment.
Make the claim reproducible in the comment itself with an interleaving rather than a principle: A reads version 7, B reads version 7, B commits version 8, A's predicate matches zero rows, A is told it succeeded and A's edit is gone.
Offer the alternative with its cost attached: return 409 carrying the current version and the revision that won, so the client can re-read and re-apply. Note that automatic retry is not the fix, because a retry re-reads the winner's state and reapplies an intent formed against data that no longer exists.
Apply an escalation rule you can state: two round trips on the thread, then a call, and the service's owner decides rather than the reviewer. A reviewer who cannot be overruled is a bottleneck with extra steps.
Close in writing wherever the decision lands, so the next reader finds the reasoning in the code or the ticket instead of in a collapsed review thread.

Follow-up

Where would you put the test that fails if someone reintroduces the swallowed zero rowcount?
The author says clients cannot handle a 409. How do you check whether that is true?
How do you handle the same review comment when the author is more senior than you and in a hurry?