A Software Engineer at Micro Control serves as a vital bridge between complex high-power burn-in test systems and the customers who rely on them for mission-critical semiconductor validation. You are not just writing code; you are building the software infrastructure that ensures hardware reliability, identifies bugs in high-stakes environments, and keeps industrial test systems running at peak performance.
This role is highly collaborative, requiring you to interact with cross-functional teams, including hardware engineers, manufacturing, and external customers. Whether you are developing test plans, integrating software onto hardware, or providing direct technical support, your work directly influences the uptime and success of Micro Control products. You will find this position intellectually stimulating if you enjoy solving complex technical puzzles, working at the intersection of software and hardware, and taking ownership of the full software lifecycle.
Initial Introduction
reportedWhen 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
Formal Interviews
reportedYou cannot drill a format you do not know, so put the preparation into material that travels. Three pieces of your own work, each rehearsed until you can take a follow-up you did not anticipate, will carry a conversation or a code walkthrough equally well. Specificity is what separates that from filler. A number needs its definition before it means anything: a p99 is over some window and measured at some hop, and a server-side figure excludes the queueing and network time a client would see. The number you cannot qualify is the one to leave out.
What to demonstrate
- Whether your examples carry detail only someone who did the work would hold, such as what the binding constraint actually was, which alternative you rejected and why it was worse, and what you measured on each side of the change
- Whether a number survives one follow-up, meaning you can say what it was measured over and whether it moved because of your change or merely alongside it
- Whether a failure is described with the specific change that followed it, rather than a lesson stated in general terms
- Whether your part in a team effort is stated accurately, including what other people did
How to prepare
- Write a page on each of three projects covering the constraint, the option you rejected, the measurement before and after, and what went wrong. Cut any line you cannot take a follow-up on, since you are writing the parts you will be pressed on rather than a summary.
- Recover the real figures while you still have access: request volume, data size, latency with its percentile and window, team size, timeline. Note where each came from, whether a dashboard, a design document or memory, and mark the estimates so you can say which they are out loud.
- Take your weakest project story to someone who works in a different area and have them ask why four times in succession. The point where you run out of answer is the part to go and re-read before the round.
Behavioral Fit Assessment
reportedYour 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
PracHub editorial advice for the preparation topics above.
Letting a slow dependency consume unbounded concurrency
The failure that takes a service down is usually not an error but a delay. A dependency answering in thirty seconds instead of fifty milliseconds holds each request's worker or connection six hundred times longer, and since required concurrency is arrival rate times latency, a fleet sized for sixty in-flight requests now needs thirty-six thousand to sustain the same rate - so it queues, and requests whose clients have already abandoned them still occupy resources. Retries make it precisely worse: a policy of three attempts triples the load on a dependency at the exact moment it is least able to serve, which is how one slow dependency becomes an outage of everything sharing that pool. Containment is four specific things - a timeout on every outbound call shorter than the caller's remaining budget, a bounded pool per dependency so one cannot starve the others, backoff with full jitter rather than a fixed delay so retries do not resynchronise, and a circuit that stops sending once the failure rate makes an attempt pointless.
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.
Sharing mutable state with no stated owner
Say which thread, request or task owns each mutable structure, and what protects it when the answer is more than one: a lock, a queue that hands ownership across, or an immutable copy per reader. A structure documented as safe for concurrent reads is usually not safe for a concurrent write alongside those reads.
Comparing floating-point values for equality, or holding money in them
Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Track a rolling failure rate per destination for circuit decisions
The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.
Approach
- Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
- Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
- State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
- Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
- Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
- Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.
Follow-up
- The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
- A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
- How would you make the window survive a process restart, and is it worth the cost?
Archive a resource graph without breaking live references or recursing
Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.
Approach
- Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
- Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
- Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
- Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
- Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
- Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.
Follow-up
- The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
- The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
- An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?
Find overlapping job attempts and peak concurrency from lease records
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.
Worked solution 30 min
- Write the interval derivation with the COALESCE and state in one line which of the two end sources is observed and which is assumed.
- Write the concurrency sweep: the endpoint tuples, the sort key including the end-before-start tie-break, and the per-job_type counter.
- Hand-trace four attempts of one job, two disjoint and two overlapping by three seconds, and confirm the overlap detector fires exactly once.
- Add the skew filter as a minimum overlap duration, state the value you chose, and justify it from how the timestamps are written.
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?
Replace offset paging on the resource feed with keyset
resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.
Approach
- Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
- Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
- Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
- Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
- State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
- Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.
Worked solution 25 min
- Seed one tenant with 500k active resources, 2% of them sharing an identical updated_at.
- Time LIMIT 50 OFFSET 0 against OFFSET 20000 and record rows-read from EXPLAIN (ANALYZE, BUFFERS) for each.
- Page the whole set with the keyset query while an insert-only writer adds 100 rows/second, collecting resource_ids, and repeat the run with OFFSET.
- Repeat both runs under a second writer profile that also deletes 20 rows/second from pages already returned and bumps updated_at on 20 more, and diff each collected id set against the rows that existed for the whole run.
- Remove resource_id from the cursor so the seek degrades to updated_at < $2, and re-run the tie-heavy section of the feed.
Follow-up
- The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
- Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
- What does the cursor do when the row it points at has since been deleted?
Stop tag and share joins from fanning out a page
resource_tag is (resource_id, tag_id) with PK (resource_id, tag_id); resource_share is (resource_id, shared_with_user_id, permission). The tagged-and-shared listing inner-joins resource to both, filters tenant_id, tag_id = ANY($2) and shared_with_user_id = $3, orders by updated_at DESC and takes 50. Pages come back with fewer than 50 distinct resources and the total in the header is far too high. Explain the row multiplication, rewrite both the page query and the count query so each is correct, and name the index each one needs. PostgreSQL 16.
Approach
- Do the arithmetic against the predicates that are actually there. An inner join emits one row per matching child row, and both joins are filtered: tag_id = ANY($2) admits only the requested tags, shared_with_user_id = $3 admits one user's share rows. So a resource holding three of the requested tags and shared with $3 once yields three rows, not one — the multiplier is its count of matching tags times its share rows for that single user, and that second factor is 1 unless the table admits duplicate (resource_id, shared_with_user_id) pairs. LIMIT 50 then limits rows rather than resources, and COUNT(*) counts pairs — the header is the product, not the population.
- Reject DISTINCT as the fix. It deduplicates after the product has been built, so the planner must materialise and sort the fanned-out set before the LIMIT can apply, and it leaves any SUM or AVG in the same select list wrong.
- Rewrite both filters as semi-joins, keeping resource as the only row source: AND EXISTS (SELECT 1 FROM resource_tag rt WHERE rt.resource_id = r.resource_id AND rt.tag_id = ANY($2)) and the same shape against resource_share. A semi-join stops at the first match per resource and preserves the driving index order, so ORDER BY updated_at DESC, resource_id DESC LIMIT 50 still stops after 50 rows.
- Count with the same predicates and no join at all: SELECT count(*) FROM resource r WHERE r.tenant_id = $1 AND r.status = 'active' AND EXISTS (...) AND EXISTS (...). Nothing multiplies a resource, so the number is the population.
- Attach the tags for display after the page has been cut — LEFT JOIN LATERAL (SELECT array_agg(rt.tag_id) FROM resource_tag rt WHERE rt.resource_id = p.resource_id) ON TRUE over the 50 returned rows. Aggregate over the page, never over the tenant.
- Index both directions and say which query each serves: PK (resource_id, tag_id) serves the lateral lookup, (tag_id, resource_id) serves the EXISTS probe by tag, and resource_share needs (shared_with_user_id, resource_id) for the same reason. An index covering one direction only leaves the other as a scan.
Follow-up
- The filter changes from 'any of these tags' to 'all of these tags'. Rewrite it and state what it costs relative to the ANY form.
- A resource can be shared with the same user twice under different permissions. Does your count change, and should it?
- Where does the correct total come from when the tenant holds 4M resources and the header must not cost 200 ms?
What experience do you have with hardware-software integration?
What experience do you have with hardware-software integration?
Approach
- Say what you would check first and why it is the highest-information step.
- Work from the requirement backwards to the design.
- State your assumptions explicitly before working the problem.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
How do you approach identifying and reproducing a complex software bug…
How do you approach identifying and reproducing a complex software bug?
Approach
- Work from the requirement backwards to the design.
- State your assumptions explicitly before working the problem.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Make resource creation safe for a client that retried a timeout
POST /v1/resources creates a resource row and, in the same transaction, a resource_revision and an outbox_event. A partner's HTTP client times out at 2 seconds and retries twice; under load the first request is still executing when the retry arrives, and duplicate resources are appearing. Using idempotency_key (tenant_id, idempotency_key, request_fingerprint, state, response_status, response_body, resource_id, locked_until, expires_at), specify the whole contract: first call, concurrent retry, retry after completion, a key reused with a different body, and what the client does with each response.
Approach
- Name why the plain REST shape fails here. POST is not idempotent, a timeout leaves the client unable to distinguish a lost request from a lost response, so it must retry - and the retry is a second create. Checking whether the key exists and then inserting it is check-then-act: under exactly the concurrency that caused the timeout, both attempts look, both see nothing, both proceed.
- Let the unique constraint arbitrate instead. INSERT the key row with state 'in_flight' and locked_until set to now plus the request deadline, and commit that reservation on its own before any work is attempted - an uncommitted row is invisible to the concurrent retry, which is the entire point of writing it first. One caller wins the insert and proceeds; the other takes the unique violation and reads the row rather than falling through to the work.
- Branch the loser on what it found. 'succeeded' replays response_status and response_body verbatim, with a header marking it a replay. 'in_flight' with locked_until in the future returns 409 and a Retry-After - do not park a request thread waiting on another request, that is how a concurrency bound is consumed by retries. 'in_flight' past locked_until is reclaimable by a conditional UPDATE predicated on the old locked_until, so exactly one reclaimer wins. A fingerprint mismatch is 422 and never the stored response: a different body under the same key is a client defect, and serving the old response hides it forever.
- Commit the effect and the key's completion together - resource insert, revision row, outbox row, and the UPDATE of the reserved key to 'succeeded' with the stored response, in one transaction. That makes a crash binary: either all of it is durable and the key reads 'succeeded', or the transaction rolls back and the only durable trace is the reservation row, still 'in_flight', with no resource behind it. The reclaim branch finishes that case - once locked_until passes, one retry takes the row by conditional UPDATE and runs the work for the first time. Splitting the two commits is what creates the window with no recovery: the effect durable, the key still 'in_flight', and a reclaimer that cannot distinguish that from work never started, so it creates a second resource.
- State the lifetime explicitly. Keys expire - 24 hours is a common choice - and are swept, so the client's total retry window must be shorter than the TTL, or a late retry re-executes with no record that it is a duplicate. The key is scoped by tenant so two tenants choosing the same string do not collide, and the client generates one key per logical attempt, not per network send.
- Write the client rules: on timeout, retry the same key with capped backoff; on 409, wait Retry-After and retry the same key; on 422, stop and fix the caller; on 2xx, treat the body as authoritative whether it was executed or replayed.
Worked solution 30 min
- Write the two-request interleaving that defeats read-then-insert, with the line each request is executing at each step.
- Write the insert-first sequence and the four branches taken on unique violation, as a table of state, locked_until and response.
- Put the completion UPDATE inside the same transaction as the resource, revision and outbox writes, then state what a crash before that commit leaves durable and which branch cleans it up.
- Define the key's scope, lifetime and generation rule, then bound the client's retry budget by the TTL.
- Fire two concurrent requests carrying one key and count rows in resource and resource_revision.
Follow-up
- Response bodies are pruned at 24 hours but resource_id is kept. What do you return to a replay that arrives after pruning?
- The endpoint now also sends a welcome email. Which part of this contract prevents a second send, and which part cannot?
- What exactly breaks if the key is scoped globally rather than per tenant?
Listing latency scales with page size, not with filters
The tenant listing endpoint reads resource filtered by tenant_id and status, ordered by updated_at DESC, and returns each row plus the owner's display name from app_user and the actor of that resource's latest resource_revision. p99 is 55 ms at 10 rows per page and 1.4 s at 200. Database telemetry shows 401 statements per request, each under 1 ms, and nothing in the slow-query log. Diagnose the cause and give the fix, stating the statement count per request and the p99 you expect afterwards.
Approach
- Read the counters before forming a theory. 401 statements for 200 rows is one driver query plus two per row, and sub-millisecond execution with an empty slow-query log rules out a bad plan. The time is round trips, which is why it is invisible in every per-query metric and scales with rows returned rather than with filter selectivity.
- Name the two per-row statements from their normalised text: a single-row app_user lookup by user_id, and a resource_revision lookup by resource_id ordered by version DESC LIMIT 1. Confirm by dropping those two response fields and watching the statement count fall to one. That locates the calls in the serialisation layer, not the repository.
- Check that the arithmetic accounts for the whole gap. Measure one round trip to the replica in isolation; 400 trips at roughly 3 ms of network plus 0.2 ms of execution is about 1.3 s on top of a 55 ms baseline, which matches. If the multiplication had fallen short, the N+1 would only be part of the story and you would keep looking.
- Batch both lookups. Collect owner_user_ids and resource_ids from the driver query, then issue WHERE tenant_id = $1 AND user_id = ANY($2) for the users, and PostgreSQL's SELECT DISTINCT ON (resource_id) ... WHERE resource_id = ANY($2) ORDER BY resource_id, version DESC for the latest revision, which the UNIQUE (resource_id, version) index serves directly. On an engine without DISTINCT ON, use a lateral join or a row_number window. Three statements per request at any page size.
- Keep the tenant predicate in the batched query. The per-row version was implicitly scoped because its ids came from tenant-scoped rows; a batched user_id = ANY(...) with no tenant_id is an unscoped read that behaves correctly only as long as the id list is trustworthy.
- Re-measure at 10, 50 and 200 rows and confirm the statement count is constant. Latency should now track bytes returned.
Follow-up
- The page size is capped at 200 today. What breaks first if it is raised to 2,000, and is it still this bug?
- How do you stop the next N+1 from reaching production, given that no individual query is slow and the endpoint's tests pass?
- The latest-revision actor is only used to render an avatar. Make the case for denormalising it onto resource, and name the write anomaly that introduces.
For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Rebuild the primitives by implementing them
- Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
- Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
- For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.
Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Arrays under an invariant: two pointers, sliding window, binary search
- Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
- Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
- Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.
Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.
Practice prompt ↗Practice prompt ↗03Sorting, heaps, and the greedy argument that has to be proved
- Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
- Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
- Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.
Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.
Practice prompt ↗Practice prompt ↗04Recursion, memoisation, and the step to a table
- Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
- Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
- Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.
Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Graphs, where most of the work is choosing the traversal
- Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
- Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
- Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.
Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.
Practice prompt ↗Practice prompt ↗06One day for everything that is not an algorithm
- Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
- Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
- Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.
Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.
Practice prompt ↗Practice prompt ↗07Solve out loud, under time
- Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
- Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
- Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.
Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.
Tell me about a time you had to explain a complex technical issue to a…
Tell me about a time you had to explain a complex technical issue to a non-technical stakeholder.
Approach
- Give the blast radius: what could have broken, and what you measured.
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
What is your communication style?
What is your communication style?
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
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
How do you handle conflict or differing opinions when working in a cro…
How do you handle conflict or differing opinions when working in a cross-functional team?
Approach
- Give the blast radius: what could have broken, and what you measured.
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
How do you ensure your code or documentation is maintainable for other…
How do you ensure your code or documentation is maintainable for other team members?
Approach
- Close with what you would do differently, concretely.
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
- 01
Tell me about a time you had to explain a complex technical issue to a non-technical stakeholder.
- 02
What is your communication style?
- 03
How do you handle conflict or differing opinions when working in a cross-functional team?
- 04
How do you ensure your code or documentation is maintainable for other team members?
Is this an official Micro Control interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Micro Control. Rounds and questions reflect what candidates have reported, not a process Micro Control has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long does the interview process take?
The timeline varies. From your initial screen to a final decision, you can typically expect the process to span a few weeks, depending on interview availability.
PracHub interview research ↗What differentiates successful candidates?
Successful candidates demonstrate a "hands-on" mindset. They don't just talk about code; they show they understand how software interacts with physical hardware and how to communicate effectively with customers.
PracHub interview research ↗Is there a lot of travel involved?
While the role is primarily based at Micro Control's facility, there may be occasional travel to customer sites. Flexibility in this regard is an asset.
PracHub interview research ↗What is the culture like at Micro Control?
Candidates report that Micro Control values collaboration, initiative, and a strong work ethic. You will be working in an environment that prizes technical accuracy and customer satisfaction above all else.
PracHub interview research ↗Sources & methodology 3 sources ↗
No official company page is cited. Rounds and questions come from candidate reports and PracHub editorial material; each source shows the date it was read.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
PracHub page · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-22