A Software Engineer at Mican Technologies designs, builds, and maintains the systems that support the company's business operations. Whether you focus on front-end user experiences, conversational AI platforms, or cloud infrastructure, your work directly affects the reliability, scalability, and performance of Mican Technologies' technology stack.
You will operate at the intersection of complex problem-solving and strategic execution, collaborating closely with architects, product stakeholders, and cross-functional teams. This role is not just about writing code; it is about engineering resilient solutions that protect data integrity, optimize business processes, and provide smooth experiences for the company's users. You can expect an environment that values technical rigor, continuous improvement, and the ability to navigate both high-level system architecture and detailed production troubleshooting.
Initial Assessment
reportedInput 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
Technical Discussions
reportedWhat this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.
What to demonstrate
- Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
- Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
- Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly
How to prepare
- Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
- Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
- Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live
Software Development Lifecycle
reportedAn unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.
What to demonstrate
- Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
- Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
- Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating
How to prepare
- Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
- Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
- Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes
Collaborative Evaluation
reportedThe same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.
What to demonstrate
- In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
- In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
- Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes
How to prepare
- Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
- Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
- For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first
Final Evaluation
reportedNobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.
What to demonstrate
- Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
- Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
- Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
- Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience
How to prepare
- Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
- For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
- Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed
PracHub editorial advice for the preparation topics above.
Checking a quota with a select and then writing
Under read-committed isolation, two concurrent transactions both observe a count below the limit and both insert, so the limit is exceeded by exactly the concurrency. Repeatable read does not rescue it either: it provides a stable snapshot, and this is write skew, which snapshot isolation permits by design. The options are serialisable isolation, which detects the conflict and aborts one transaction with a serialisation failure and therefore obliges the caller to retry; a single statement with the predicate inside the write; or a constraint that makes the surplus insert fail outright. The reason this pattern survives review is that it is correct in every test that runs one request at a time.
Holding money in a floating-point type, or rounding it more than once
Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.
Check-then-act on shared state
Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.
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.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Parse and verify a timestamped multi-signature webhook header
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. Accepttexactly once and treat a secondtas a reject rather than last-wins. Push everyv1onto a short list and ignore any other key, so av2can 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 testacc == 0at 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| > 300seconds. 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.
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?
Hold a tenant to a trailing sixty-second request limit
The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.
Approach
- Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while
front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request. - Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
- Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate,
prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact. - Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (
tokens,last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual. - Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
- Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.
Follow-up
- One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
- Quotas rather than rate limits: the check is
select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes. - How do you return an accurate
Retry-Afterfrom the exact algorithm without a second scan?
Order a job dependency graph and find its critical path
A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.
Approach
- Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
- Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
- Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order:
earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E). - Second pass in reverse topological order for
latest_finish, thenslack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain.slack[v] = 0is exactly the statement that some longest path runs through v; equivalently, the longest path through v has lengthT - slack[v]. - The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by
min(d, T - L_avoid(v)), whereL_avoid(v)is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, whilemin(10, 15 - 14)predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan ismax(T - d, L_avoid(v)). - Compute
L_avoid(v)the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, sinceL_avoid(v)only ever matters through that max: setduration[v] := 0, recompute the makespan asT0(v) = max(T - duration[v], L_avoid(v)), and the gain ismin(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.
Worked solution 30 min
- Build four fixtures. A: 12 jobs, two branches of 100 s and 95 s that share no job. B: fixture A plus one back edge. C: two disjoint paths tied at 100 s. D: the shared-prefix case, one job of 10 s feeding a 5 s job and a 4 s job, so the longest path is 15 s and the runner-up is 14 s.
- Run Kahn; on fixture B confirm it emits fewer than V nodes, then run the residual-subgraph DFS and print the actual cycle.
- Compute
earliest_finishforward andlatest_finishbackward, and list the zero-slack set for each fixture. - For each zero-slack job v, recompute the makespan with
duration[v] := 0to getT0(v), and record both the correct boundT - T0(v)and the wrong one,T - second_longest_path, side by side. - Apply the shortening for real (20 s off the critical branch of A, 10 s off the shared prefix of D) and diff the recomputed makespan against each prediction.
Follow-up
- Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
- Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
- Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?
Decide which facts an invoice line copies instead of joining
invoice_line_item already denormalises tenant_id, which is reachable through invoice_id, and stores amount_minor even though quantity times unit_price_micros would recompute it. A reviewer asks you to normalise both away, and separately asks whether the tenant's legal name and billing address should be copied onto the invoice header. Decide each case. For every field you keep denormalised, name the read pattern or the invariant that justifies it, the anomaly the copy can develop, and the mechanism that prevents that anomaly here.
Approach
- Split the question into two kinds of copy, because they fail differently. A copy of a currently mutable fact is a cache: it drifts and needs invalidation. A copy of a fact frozen at write time is not a cache at all, it is the record of what happened, and normalising it away destroys information the source no longer holds.
- Keep tenant_id on the line. It costs 8 bytes, it leads every index on the table so no read is ever accidentally cross-tenant, and it turns a wrong join into an empty result rather than another tenant's money. Prevent the drift structurally: a unique constraint on invoice (invoice_id, tenant_id) plus a composite foreign key from the line on (invoice_id, tenant_id) makes a mismatched pair impossible, so the database enforces agreement instead of a code review.
- Keep amount_minor. Rounding must happen exactly once, at a named site, with a stated mode (half-even here). If readers recompute from quantity and unit_price_micros, every reader owns a rounding decision, and half-up and half-even diverge systematically across thousands of lines rather than cancelling out. A check constraint can bound the stored value but deliberately cannot re-derive it.
- Copy the legal name and billing address onto the invoice header, written once and never updated. The statement must show what was true when it was sealed, and the tenant record will change afterwards. This is a snapshot for the same reason
source_rollup_watermarkis stored per line: without it, nobody can reconstruct what the customer was told. - Name the read pattern that pays for all of it. Rendering, dispute response and export are per-tenant, per-period reads over thousands of lines that would otherwise join back to slowly changing dimensions that no longer hold the historical value. The write side is a once-per-period batch, so the extra columns cost nothing that matters.
- Concede the case where the reviewer is right: a mutable operational attribute such as the tenant's current plan name has no business on a line. If a report wants it, join. If a statement needs the plan as of the period, that is another snapshot and it belongs on the header with the rest.
Follow-up
- Write the composite foreign key and the unique constraint it requires on the parent. What does it cost on every line insert, and what does it do to a bulk load?
- A tenant is renamed after being invoiced. Which rows change, and what does the customer see on last quarter's PDF?
- Where does currency live, and what breaks if a tenant's billing currency changes between two periods?
Find the join that inflates every invoice total
invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.
Approach
- Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row:
sum(l.amount_minor)is the true total multiplied by the attempt count, andcount(p.attempt_id)is attempts times lines. Three lines and two attempts report double the money and six attempts. - Reject the reflex repair.
count(distinct p.attempt_id)does fix the count, because attempt_id is unique.sum(distinct l.amount_minor)does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match. - Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
- Keep invoices with no attempts by making the attempt branch a LEFT JOIN with
coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about. - Push each filter to its own grain:
where l.voided_at is nullbelongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants. - Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.
Worked solution 25 min
- Create one invoice with three lines of 1000, 1000 and 500 minor units and two payment attempts, then run the original query.
- Confirm it reports 5000 and 6 rather than 2500 and 2.
- Apply
sum(distinct l.amount_minor)and confirm the total becomes 1500, which is worse rather than better. - Write the two-CTE version with a LEFT JOIN and coalesce, and confirm 2500 and 2.
- Add a second invoice with lines and no attempts and confirm it still appears.
Follow-up
- Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with
filter (where ...)be better? - Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
- The same report is needed per tenant per month. What index makes the line CTE cheap?
What strategies do you employ for monitoring and alerting to ensure hi…
What strategies do you employ for monitoring and alerting to ensure high availability in production environments?
Approach
- State the consistency you need, and where you are willing to be stale.
- 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.
Follow-up
- How does this behave when that dependency is down for an hour?
- What would you drop to keep the system up under load?
How would you design a load-balanced, fault-tolerant system that requi…
How would you design a load-balanced, fault-tolerant system that requires strict data consistency?
Approach
- State the consistency you need, and where you are willing to be stale.
- Choose a partition key and say what query it makes expensive.
- Fix the scope first: who calls this, how often, and what they do when it fails.
Follow-up
- What breaks first when traffic grows ten times?
- How does this behave when that dependency is down for an hour?
How do you utilize Terraform modules and state management to ensure co…
How do you utilize Terraform modules and state management to ensure consistent cloud infrastructure deployments?
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
- What assumption would you test first?
- How would you know your answer was wrong?
Can you explain your approach to managing application state and data f…
Can you explain your approach to managing application state and data flow in large-scale front-end projects?
Approach
- Clarify what is being asked and what a complete answer contains.
- State your assumptions explicitly before working the problem.
- Say what you would check first and why it is the highest-information step.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Design the batch ingest endpoint metering agents retry into
A customer-run agent posts usage events in batches of up to 1,000 to metering-ingest with a 30-second timeout and at-least-once retry of the whole batch. Each event carries event_id, idempotency_key, sku, quantity and occurred_at; the server adds ingested_at, and usage_event is partitioned daily on ingested_at with unique (ingested_day, tenant_id, idempotency_key). Design the endpoint: the request shape, what the response says when 900 events are new, 90 are duplicates and 10 are malformed, the status code, and the agent's algorithm on timeout. Then state the deduplication horizon and justify it against that unique constraint.
Approach
- Fix the per-item outcome taxonomy first, because the status code follows from it: accepted, duplicate, and rejected with a permanent code. A duplicate is a success; reporting it as an error makes the agent either re-send revenue it already delivered or drop it.
- Allow only permanent failures per item. A transient per-item failure inside a 200 invites the agent to discard that event, so anything transient escalates to a 5xx for the entire batch. A 200 is then a promise that every event not marked rejected is committed and durable.
- Return 200 with a results array aligned by index and carrying the event id, so the agent can retry precisely the subset that needs it and quarantine the ten malformed events instead of hard-looping a poison batch forever. Cap the batch at 1,000 items and a byte size, with 413 beyond it and 429 with Retry-After for backpressure.
- Deduplicate per event, never per batch: the agent may split, merge or reorder a retried batch, so a batch-level key matches nothing on the second attempt. The key is (tenant_id, idempotency_key), and the tenant comes from the resolved credential; a tenant id present in the body is compared against it, never trusted.
- Size the horizon as a correctness parameter. The unique index includes the partition key, so it deduplicates only within one day: a retry that crosses midnight, or a replay run a week later, passes straight through it. A separate dedup store keyed (tenant_id, idempotency_key) with a TTL exceeding the agent's maximum retry window plus the longest replay you intend to support is what actually enforces the invariant, which makes its retention a correctness setting rather than a cost knob.
- Order the commit against the response and the acknowledgement: commit then respond at the endpoint, and downstream commit the fold then acknowledge the message. Acknowledging first turns a crash into silently lost revenue with no error raised anywhere.
Worked solution 40 min
- Write the request body schema with the batch envelope and one event, and state which fields the server assigns rather than accepts.
- Write the 200 response for the 900/90/10 case, showing three result entries, one of each outcome, with the rejected one carrying a permanent code.
- Write the rule separating per-item rejection from whole-batch failure, and list which conditions fall on each side.
- Compute the dedup horizon from the agent's retry window plus the replay window you support, and say where the dedup state lives and how it ages out.
- Write the agent's pseudocode for timeout, 5xx, 429 and 200-with-rejections, four branches, and mark which branch may drop an event.
- Trace the crash between commit and response, and between fold and acknowledgement, and say what each produces.
Follow-up
- The agent times out at 30 seconds having received nothing. What exactly does it do next, and what in your design makes that safe?
- Ten events are rejected every hour for a week and nobody notices. What does the endpoint owe the customer beyond a per-item 4xx code?
- A replay pushes 40 million events through this endpoint in an hour. Which part of your design degrades first?
Invoice detail latency triples after an ORM relationship refactor
An invoice detail endpoint returned in 40 ms at p99 last week. After a refactor replaced a hand-written join with ORM relationship access it returns in 1.4 s, and the regression grows with the number of invoice_line_item rows on the invoice. Database CPU rose, but no statement in the slow-query log exceeds 3 ms. You have request traces with per-span SQL, the ORM statement log, and a staging copy of the data. Produce an ordered diagnostic checklist, the measurement that confirms the cause before any code change, and the fix.
Approach
- Count statements per request before reading any statement duration. A slow-query log hides this class by construction, because every individual query is fast and only their number is wrong; take one trace and count SQL spans.
- Establish proportionality rather than asserting it: sample invoices with 5, 20, 60 and 200 line items and plot statements per request against line count. A straight line of slope 1 through an intercept of one or two identifies a lazy relationship load, and no index or cache would move that line.
- Locate the emitting attribute access in the refactored code and check whether the same shape repeats one level deeper, for instance a tax or adjustment collection hanging off each line, which turns the cost quadratic.
- Fix with a bounded statement count: either one join that fetches invoice and lines together, or two statements where the second is WHERE invoice_id = $1 AND tenant_id = $2. Keep tenant_id in the predicate so the read stays tenant-scoped even though invoice_id already implies it.
- Choose between the two deliberately: the join duplicates the wide parent row across N children on the wire, the two-statement form avoids that for one extra round trip. Prefer the join for narrow parents and the split for wide ones.
- Pin it with a per-request statement-count assertion in a test that varies line count, because a latency assertion passes on a small fixture and would not have caught this.
Follow-up
- The endpoint now also needs per-line tax rows. Show the shape that keeps statement count constant instead of reintroducing the same defect one level down.
- How does this change if a transaction-pooling proxy sits between the service and the database, so each statement may land on a different backend session?
- The same page paginates invoices with LIMIT and OFFSET. Why is that a second, independent defect, and what replaces it?
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.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Coding, 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.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Design, 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.
Practice prompt ↗Practice prompt ↗03Fundamentals 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.
Practice prompt ↗Practice prompt ↗04The 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.
Practice prompt ↗Practice prompt ↗Worked solution ↗05First 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.
Practice prompt ↗Practice prompt ↗06Second 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.
Practice prompt ↗Practice prompt ↗07Reassemble 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.
Practice prompt ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Bring the two or three numbers the story rests on and know how they were collected. A p99 whose timer starts inside your handler excludes the time a request spent queued, so it can sit flat while users wait longer. Give the window, the percentile and what the measurement left out, or drop the number.
How do you maintain alignment and collaboration when working with mult…
How do you maintain alignment and collaboration when working with multiple dependency teams on a large integration project?
Approach
- Name the disagreement and how you resolved it with evidence.
- State the situation in two sentences and spend the rest on the reasoning.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
Tell us about a situation where you had to troubleshoot a critical pro…
Tell us about a situation where you had to troubleshoot a critical production outage under pressure.
Approach
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
Can you describe a time you identified and resolved a significant perf…
Can you describe a time you identified and resolved a significant performance bottleneck in a production application?
Approach
- Name the disagreement and how you resolved it with evidence.
- 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.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
- 01
How do you maintain alignment and collaboration when working with multiple dependency teams on a large integration project?
- 02
Tell us about a situation where you had to troubleshoot a critical production outage under pressure.
- 03
Can you describe a time you identified and resolved a significant performance bottleneck in a production application?
Is this an official Mican Technologies interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Mican Technologies. Rounds and questions reflect what candidates have reported, not a process Mican Technologies 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 typically take?
The timeline can vary, but the company aims for an efficient process. From your initial application to a final decision, candidates usually move through the stages over several weeks.
PracHub interview research ↗What differentiates a good candidate from a great one?
A great candidate demonstrates "ownership." This means they don't just write code; they consider the monitoring, security, and long-term maintenance of the systems they build.
PracHub interview research ↗Is the role fully remote?
Yes, these positions are remote-based, though the company operates on East Coast business hours to stay aligned with its core teams.
PracHub interview research ↗Should I prepare for whiteboard coding?
While the interviews focus on real-world scenarios, you should be prepared to discuss code structure, algorithm efficiency, and system design patterns in a collaborative, live setting.
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
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