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Davidson Technologies Software Engineer Interview Questions & Guide 2026

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

The 60-second brief

As a Software Engineer at Davidson Technologies, you help deliver advanced, intelligent technology solutions for national defense. Operating within the aerospace and missile defense industry, this role directly contributes to high-impact programs like the Sentinel program, focusing on software-intensive systems that meet emerging mission-critical needs. Your work ensures that complex, real-time systems function reliably under rigorous demands, bridging the gap between high-level architectural design and robust tactical deployment.

The loop does not sample the job evenly, and arguing about that in the room costs you. Daily work is mostly incremental change inside code someone else wrote, while the loop samples narrow slices of it; prepare for the slices and save the realism argument for your questions at the end.

Focus areas

Keep money in integer minor units
Scope every query and cache key by tenant
Make every write idempotent under client retries

Overview

The position requires you to navigate both technical complexity and collaborative cross-functional environments. You will work side-by-side with systems engineers, chief architects, and Scrum teams across multiple sites, contributing to software design, development, and unit testing. What makes this role particularly engaging is the scale of the engineering challenges: you are not just writing standard application code, but building software that interfaces with operational weapon systems, embedded environments, and real-time operating systems.

You can expect a fast-paced yet supportive culture that prioritizes technical excellence, ownership, and continuous innovation. Davidson Technologies is widely recognized for its employee well-being and exceptional workplace environment, allowing you to focus on solving hard engineering problems while enjoying a balanced work schedule, such as the standard 9/80 schedule providing every other Friday off. Success in this role requires a self-starting mentality, a readiness to take ownership of difficult challenges, and a commitment to building consensus within high-performing teams.

The Davidson Technologies interview process

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

Initial Screening

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 Assessment

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

Behavioral Interview

Your first answer is not really what is scored. It buys the follow-up questions, and those decide the round. An interviewer with fifteen minutes takes one thread and pushes on it four or five times, so a story you can only tell at a single level of detail collapses under the third why. That is an argument for fewer stories known deeply rather than one prepared per prompt. Four or five pieces of work you can still explain down to the code you changed and the argument you had about it will cover nearly anything asked in this round.

What to demonstrate

Whether a story holds as the questioning moves from what you did to why that instead of the alternative, and then to what you would change knowing what you know now
Whether you can re-cut a project to answer the question actually asked rather than delivering a rehearsed block that answers an adjacent one
Whether your level of detail is chosen rather than habitual: going down to the schema when the question is about the data model, staying out of it when the question is about the person who disagreed with you

How to prepare

Pick four projects and write the chain out four levels deep for each: what you did, why that, why not the alternative, and what would have to be true for the alternative to have won. Where you cannot reach the fourth level, you have a placeholder rather than a story
Have someone ask why three times in a row on a single thread with nothing else added, and mark the point where you start repeating a sentence you already said. That point is where the interviewer stops learning anything
Build a one-page index instead of an answer bank: the common prompts in this round (disagreement, a failure that was yours, thin requirements, a deadline you missed, work you inherited) mapped to which of your four projects you would use for each, so the choosing is done now rather than while an interviewer waits

Team Discussions

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

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.

Treating a timed-out write as a failed write

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

Tests that assert on the implementation rather than the behaviour

Assert on what a caller can observe, not on the number of internal calls or the shape of a private field. A test that breaks on every refactor but still passes when the answer is wrong costs more than it protects.

Retrying a write that is not safe to repeat

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

Questions & practice

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

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.
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.
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.
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.
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.
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.

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?

Fold a deduplicated usage stream into hourly rollups

Easy · aggregation · deduplication · watermarks · exact-arithmetic

You are given one day of usage_event rows, up to 250 million, each carrying event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at and ingested_at. Produce usage_rollup_hourly cells keyed (tenant_id, workspace_id, sku, hour_start) with quantity_sum, event_count and source_max_ingested_at. An event counts once per (tenant_id, idempotency_key). The rollup grain has no environment column, so state your filter. One pass. Give your time and space bounds, and say what the deduplication actually costs in memory.

Approach

Bucket on occurred_at, never ingested_at: hour_start = date_trunc('hour', occurred_at at time zone 'UTC'). The two columns answer different questions. occurred_at says which hour the customer is billed for; ingested_at says how current the fold is. Using the second for the first makes late data invisible instead of correctable.
The fold is trivial and the deduplication is the entire cost, so price it before designing anything clever. An exact set over (tenant_id, idempotency_key) at 250M entries, stored as a 16-byte 128-bit hash in an open-addressed table at 0.7 load factor, needs about 357M slots at 16 bytes each, roughly 5.7 GB. The fix is partitioning by hash(tenant_id) % P so each shard holds 1/P of the set and no tenant's keys straddle shards.
Rule out a Bloom filter as a replacement, in the right direction: a false positive reports 'already seen' for an event never seen, so you drop a real event and lose revenue with no error raised. It is usable only as a negative pre-filter in front of the exact set, where a miss is conclusive and a hit must fall through to the real lookup.
Accumulate in scaled integers, not binary floating point. numeric(20,6) admits values below 10^14, so one event scaled to micro-units can reach 10^20, past int64's 9.22 x 10^18; use a 128-bit or arbitrary-precision accumulator unless you first bound the per-event maximum. binary64 represents integers exactly only to 2^53, about 9.01 x 10^15, and cannot represent 0.1 at all, so two runs that sum in different orders disagree.
Carry source_max_ingested_at = max(ingested_at) over the events folded into each cell, and count event_count over accepted, post-dedup events. Without that watermark there is no way to prove later what a number did and did not include, which is the first question any reconciliation asks.
State the environment filter explicitly, because the rollup grain cannot record it. A fold that quietly includes staging bills non-production traffic; one that quietly excludes it loses a cost signal. Production-only is the billing answer, and either way it belongs in the job name and the output metadata. Complexity: O(n) time, O(distinct dedup keys) space, dominated by the dedup set rather than by the cells.

Follow-up

A producer retries at 23:59:59 and the retry lands at 00:00:01. The unique index on the daily-partitioned table must include the partition key. What gets double-counted, and what is the smallest change that fixes it?
The consumer acknowledges its batch before committing the fold. Which failure loses revenue now, and which arrangement duplicates instead?
What makes a re-run over the same day produce byte-identical rollups?

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

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.
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.
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.
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.
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.
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.

Worked solution (15 min)

Write the grammar on one line before coding: header := field (',' field)*, field := key '=' value, split on the first = only.
Implement the parser to return {t: int, v1: [hex, ...]}, rejecting a missing t, a duplicate t, any v1 that is not 64 hex characters, and a header over 1 KiB, all before any cryptography runs.
Implement the verifier: for each active secret compute HMAC-SHA256(secret, f'{t}.'.encode() + raw_body), compare it in constant time against each parsed v1, and OR the results with no early exit.
Test with a valid signature; the same body with t moved 400 seconds into the past; the same body with t 400 seconds into the future; a header carrying an unknown v2= alongside a valid v1; and a body re-serialised with different JSON key order.

Expected result: The valid case accepts. Both out-of-tolerance cases reject, including the future one. The unknown v2 field is ignored and the v1 still verifies. The re-serialised body fails, which is correct and is exactly why the raw bytes must be retained.

Verify it yourself

Timing the comparison over 10,000 mismatching signatures that differ at the first character versus the last shows no separation you can measure.
Removing t from the MAC input makes an old captured body with its original signature pass. Write that test, watch it pass, then put t back.
Every rejection except the signature mismatch itself returns before an HMAC is computed.

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?

Model credential revocation so history survives the delete

Easy · soft delete · partial indexes · check constraints · audit history

tenant_api_key stores key_id, tenant_id, workspace_id, name, key_prefix, secret_hash, scopes text[], status (active, revoked, expired, compromised), auth_version, created_at, expires_at, last_used_at, revoked_at, revoked_reason. Rotation inserts a new row and revocation never deletes, because an incident review asks which credential served a request last quarter. Write the constraints that enforce: a label is unique only among a tenant's live keys, revoked_at and status can never disagree, and scopes is never empty. Then write the authentication lookup predicate, and name one column in this table that must stay out of it.

Approach

Reach for a partial unique index rather than a plain UNIQUE: create unique index on tenant_api_key (tenant_id, name) where revoked_at is null. Any number of revoked rows may share a label, the live namespace stays unique per tenant, and the revoked majority is not in the index at all, so it stays small on a table that only grows.
Tie the nullable timestamp to the enum so the two cannot drift: check ((revoked_at is not null) = (status in ('revoked','compromised'))) and check ((revoked_at is null) = (revoked_reason is null)). A revocation that records no reason is the one an incident review cannot use.
Write the emptiness check as check (cardinality(scopes) > 0), not array_length(scopes, 1) > 0. array_length returns NULL for an empty array, a CHECK constraint passes when its expression is NULL, so the array_length version accepts exactly the value it was written to reject.
Make the lookup a single index probe with every liveness condition inside it: where secret_hash = $1 and revoked_at is null and (expires_at is null or expires_at > now()) and auth_version = $2, backed by a unique index on secret_hash. Nothing is filtered in application code, so there is no path that forgets a clause.
Keep last_used_at out of that predicate. It is written asynchronously and is allowed to lag by a minute, so it is a usage signal; feeding it into an authorisation decision makes the decision depend on a write that may be late, batched away or lost.
Flag the modelling smell while you are here: expired is derivable from expires_at < now(), so storing it as a status obliges a job to keep it true and guarantees the column is wrong between the expiry instant and that job's next run. Derive it in the predicate; keep the stored status for states that are decisions rather than clock readings.

Worked solution (20 min)

Create the table with all three constraints on a scratch database and insert two revoked rows sharing (tenant_id, name); the partial index should accept both.
Insert a second live row with that same name and confirm the violation names the partial index.
Run update tenant_api_key set revoked_at = now() leaving status = 'active' and confirm the CHECK rejects it; then try insert ... scopes = '{}' against both the cardinality and the array_length forms and note that only one rejects it.
Run explain (analyze, buffers) on the lookup predicate for a live key and confirm an index scan on secret_hash with rows removed by filter equal to zero.

Expected result: A partial unique index on (tenant_id, name) where revoked_at is null; check ((revoked_at is not null) = (status in ('revoked','compromised'))); check ((revoked_at is null) = (revoked_reason is null)); check (cardinality(scopes) > 0); a single-probe lookup on secret_hash plus revoked_at, expires_at and auth_version; last_used_at excluded, with a note that 'expired' should be derived rather than stored.

Verify it yourself

Two revoked keys with the same name insert cleanly; a second live one does not.
The array_length form of the emptiness check accepts an empty array and the cardinality form rejects it.
No statement anywhere in the design deletes or updates a revoked row.
Revoking a key touches exactly one row and takes effect on the next lookup with no application-side filtering.

Follow-up

Rotation issues a replacement while the old key stays live for a 30-day overlap. What does the uniqueness rule become, and what does the UI show to tell two same-named keys apart?
A password reset bumps the principal's auth_version. No row in this table changed. How does the next request fail, and what query counts how many keys that bump just killed?
A key turns up in a public repository. Which columns let you find it, and what do you write to the row?

Explain why the metering dashboard scans every daily partition

Medium · indexing · partition pruning · sargability · tenant skew

usage_event is range-partitioned daily on ingested_at and holds tenant_id, workspace_id, environment, sku, quantity numeric(20,6), occurred_at and ingested_at. The only relevant index is on (occurred_at). A dashboard runs select sku, sum(quantity) from usage_event where tenant_id = $1 and date_trunc('hour', occurred_at) >= $2 and environment = 'production' group by sku, and EXPLAIN shows a sequential scan of every partition. Give each distinct reason, rewrite the predicate so an index can serve it, propose the index, and state the write cost its column order adds.

Approach

Separate the three causes rather than blaming one. First, date_trunc('hour', occurred_at) wraps the column, so the predicate is not sargable against a btree on the bare column. Second, pruning keys off ingested_at while the query constrains occurred_at, so no partition can be excluded. Third, even made sargable, (occurred_at) is not tenant-leading, so for one tenant among thousands the scan reads the whole time range and discards nearly all of it.
Rewrite the bound carefully, because the obvious rewrite is only conditionally equivalent. date_trunc('hour', x) >= $2 equals x >= $2 only when $2 is already hour-aligned; for an arbitrary $2 it means x >= date_trunc('hour', $2) + interval '1 hour'. Normalise the parameter in the caller and leave the column bare.
Restore pruning with a second, redundant predicate on the partition key: ingested_at >= $2 - interval '<late-data horizon>'. State both sides of it. It prunes to a handful of partitions, and it silently omits any event whose ingest lagged past that horizon, which is precisely what a producer replay produces. Either document the horizon as a stated bound, or partition on occurred_at and move the problem into the dedup window instead.
Propose (tenant_id, occurred_at) include (sku, quantity) per partition. A partial index where environment = 'production' mostly saves size rather than selectivity, since production dominates the three environments; take it if non-production is a meaningful share and skip it otherwise.
Price the write path honestly. At roughly 250M rows/day each extra index is another insert plus WAL per row, and a tenant-leading key scatters inserts across one hot leaf per active tenant instead of appending to a single rightmost leaf, so page dirtying and random I/O both rise. An INCLUDE payload widens every leaf entry and enlarges the index accordingly.
Add the index-only-scan caveat before someone reports it as a regression: on a freshly appended table the visibility map is not yet set for recent pages, so the INCLUDE columns still cost heap fetches until autovacuum has been through, and the newest hour is exactly the data the dashboard reads.

Follow-up

CREATE INDEX CONCURRENTLY is not supported on a partitioned parent. Give the sequence that gets this index onto 400 existing partitions without blocking ingest.
One tenant holds 200 times the median row count and the dashboard still times out for them with the index in place. What changes?
Should this read hit usage_rollup_hourly instead? State what that costs in freshness and what the watermark lets you promise.

How do you utilize modeling languages like SysML or UML to represent system behavior and architecture?

Medium · system design and architecture

Approach

Choose a partition key and say what query it makes expensive.
Name the failure you are designing for, then the recovery path.
State the consistency you need, and where you are willing to be stale.

Follow-up

What breaks first when traffic grows ten times?
What would you drop to keep the system up under load?

How do you interface with chief architects to ensure technical requirements are fully met?

Medium · system design and architecture

Approach

Fix the scope first: who calls this, how often, and what they do when it fails.
Name the read and write paths separately; they rarely have the same bottleneck.
Name the failure you are designing for, then the recovery path.

Follow-up

What breaks first when traffic grows ten times?
What would you drop to keep the system up under load?

Walk through how you develop technical documentation, such as activity and sequence diagrams, for a software-intensive system.

Medium · system design and architecture

Approach

Name the failure you are designing for, then the recovery path.
Choose a partition key and say what query it makes expensive.
State the consistency you need, and where you are willing to be stale.

Follow-up

What would you drop to keep the system up under load?
What breaks first when traffic grows ten times?

How do you approach designing, developing, and troubleshooting complex C++ applications?

Medium

Approach

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

Follow-up

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

How do you self-organize and participate effectively in an Agile development lifecycle?

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

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

Metering ingest that survives a six-hour producer replay

Hard · idempotency · dedup horizon · partitioning · watermarks

metering-ingest consumes usage events at-least-once - 250M/day, 10-40k/second at peak - and folds them into usage_rollup_hourly keyed (tenant_id, workspace_id, sku, hour_start). usage_event is partitioned daily on ingested_at with unique (ingested_day, tenant_id, idempotency_key). A producer outage ends in a six-hour replay that re-sends events already ingested, some of whose originals crossed midnight. Design the consumer: partitioning, where the acknowledgement sits relative to the commit, the deduplication horizon and its storage cost, and how the rollup watermark advances. Nothing may be double-counted and nothing may be silently dropped.

Approach

Choose the acknowledgement position deliberately and name what each choice costs. Acknowledging after the fold commits makes the consumer at-least-once: a crash between the two replays the batch and produces duplicates, which are ordinary and absorbable. Acknowledging first makes it at-most-once: a crash between the two drops revenue with no error raised anywhere and no way to detect it later. Take at-least-once and design everything downstream to absorb duplicates.
Put the dedup and the fold in one transaction so there is no window between them. Insert the batch into usage_event with ON CONFLICT DO NOTHING, take the rows actually inserted, and fold only those into usage_rollup_hourly with an upsert on (tenant_id, workspace_id, sku, hour_start) bucketed by occurred_at, not ingested_at. A batch of about 2,000 rows is one round trip and one index probe per event.
Attack the partition-key flaw head on: the unique index includes ingested_day because a unique index on a partitioned table must contain the partition key, so the same (tenant_id, idempotency_key) re-sent after midnight is a different index entry and passes. Deduplicate instead against a store keyed (tenant_id, idempotency_key) with no date component, whose horizon exceeds the producer's maximum retry window plus the longest replay you intend to support. At 14 days that is 250M x 14 = 3.5 billion keys, which is a dedicated key-value store, not a larger index on the same table. The alternative - partitioning usage_event on (tenant_id, occurred_day) so the natural key is stable - fixes dedup but loses pruning on ingest time and makes retention by dropping partitions awkward.
Partition the consumer by hash of tenant_id so one tenant's replay stalls only its own partitions, and give replay traffic a separate lower-priority lane so live ingest keeps its latency. The cost is explicit: that tenant's watermark lags while the replay drains, and everything gated on the watermark waits for it.
Define the watermark as a property of committed work, not of wall-clock time: per partition it is the largest occurred_at such that every event with a smaller occurred_at has committed, and the sealing decision uses the minimum across partitions. Record source_max_ingested_at on every rollup row so any number can prove what it did and did not include, and keep restatement legal only while status = 'open' - after sealed_at the value is frozen and a late event becomes an invoice adjustment instead.

Worked solution (40 min)

Write the consumer loop in pseudocode with the acknowledgement after the commit, then annotate each line with what is lost or duplicated if the process dies exactly there.
Size the dedup store: events/day x horizon_days keys, bytes per key including the tenant prefix, and the resulting memory or disk. Compare that cost against simply extending retention on the partitioned table and say why the latter does not fix the problem.
Take one event ingested at 23:59:58 and replayed at 00:00:04 and work out its fate under (a) the partitioned unique index alone and (b) the separate dedup store.
Write the per-partition watermark formula, then what the seal uses, then what a single stalled partition does to sealing.

Expected result: An at-least-once consumer acknowledging after the commit; deduplication on (tenant_id, idempotency_key) in a store whose horizon of at least 14 days exceeds max retry window plus supported replay length, sized at roughly 3.5 billion keys; tenant-hashed partitions with replay on a separate lane; a per-partition watermark with sealing gated on the minimum across partitions; source_max_ingested_at written on every rollup row.

Verify it yourself

The midnight-crossing retry must be rejected by whatever you chose. If the table's unique index is your only dedup, it is accepted and the invoice is wrong by exactly that event.
You should be able to name a distinct failure for acknowledge-before-commit and for acknowledge-after-commit. If both sound the same, the position was not actually chosen.
quantity_sum for any (tenant, workspace, sku, hour) must be reproducible by re-aggregating usage_event filtered to ingested_at <= source_max_ingested_at. If it is not reproducible, the rollup is an opinion rather than a projection.

Follow-up

The dedup store is lost entirely. What can you still guarantee, and how do you rebuild it from what remains?
A replay delivers events for an hour that is already sealed. Trace exactly what happens to them, row by row.
One partition is stuck on a poison message, so the minimum-across-partitions watermark never advances and no tenant can be sealed. What is your escape hatch and what does it cost in correctness?

Gateway p99 spikes on a five-minute cadence

Medium · cache stampede · ttl · invalidation

edge-gateway caches each credential-to-authorisation-context decision for five minutes. p99 sits at 6 ms except for a spike to 900 ms roughly every five minutes, worst in the region with the most pods, and control-plane CPU and read latency rise in step with it. The error rate stays near zero. Customers are told a revoked credential stops authorising within 60 seconds. Give the ordered checklist that identifies the mechanism, and a fix that removes the spike without weakening the 60-second bound.

Approach

Test periodicity before anything else: take the spike timestamps modulo the TTL in seconds. A tight cluster at a fixed offset means expiry phase, while traffic-driven spikes scatter.
Overlay pod start times. Entries filled at first request inherit the phase of the pod that filled them, so a cohort of pods deployed together expires together and the amplitude should track cohort size rather than tenant count.
Separate a herd from a capacity shortfall by measuring control-plane requests per second during a spike against baseline. A stampede shows a step of roughly (pods x hot keys) for one interval with hit rate collapsing to near zero, not a gradual climb that would indicate the dependency is simply undersized.
Apply three independent controls: randomise each key's TTL by a factor drawn uniformly from something like 0.8 to 1.0 so cohorts de-phase; coalesce concurrent misses per key per pod so exactly one refresh is in flight; and serve the stale value while that refresh runs so a miss costs the stale read rather than the dependency's queue.
Bound staleness against the published contract rather than against comfort: serve-stale is admissible only up to the 60-second revocation bound, so the TTL floor and the stale window together must stay inside it, and the published invalidation must delete the entry rather than schedule a refresh.
Decide in advance what a miss does when the control plane is unreachable, because that is now the only uncached path: failing closed converts a dependency outage into a total outage, while extending stale service past the bound breaks the revocation promise. Pick one and configure it explicitly.

Follow-up

Publish-subscribe invalidation is lossy under a partition. Given that, what actually enforces the 60-second bound, and what number would you put in the contract if asked to defend it?
One tenant's key is hot enough that a single pod's coalesced refresh still matters. What changes?
Would a shared cache tier in front of the control plane help or shift the problem, and what new failure does it add?

Suggested practice checklist

Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.

Day 1 — Coding, one pass at shallow depth

Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
Fix nothing today. The value of the pass is the unfixed record.

Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.

Day 2 — Design, one pass at shallow depth

Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.

Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.

Day 3 — Fundamentals and the practical rounds

Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.

Deliverable: Eight scored short answers and one written reading of unfamiliar code.

Day 4 — The rounds that are about you, and the map

Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.

Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.

Day 5 — First chosen area, to the depth you skipped

Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.

Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.

Day 6 — Second chosen area, where the gap is coverage rather than speed

Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.

Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.

Day 7 — Reassemble the loop

Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
Reduce the week to one page holding only the rules you can state without reading them.

Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.

'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.

How do you build consensus when technical disagreements arise during sprint planning or design phases?

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Pick a story where you made the decision, not one where you watched it.
Name the disagreement and how you resolved it with evidence.

Follow-up

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

How do you handle conflicting design inputs from systems engineers and software developers?

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
Name the disagreement and how you resolved it with evidence.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

What would you do differently if you ran that again?
What did you decide not to do, and why?

What are your greatest professional weaknesses, and how do you actively work to overcome them?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

What would you do differently if you ran that again?
What did you decide not to do, and why?