A Software Engineer at Kpit plays a pivotal role in driving innovation within the automotive software and mobility engineering space. You will be responsible for designing, developing, and maintaining high-performance software solutions that power next-generation vehicle systems. Your work directly impacts the safety, efficiency, and intelligence of modern transportation, bridging the gap between complex hardware systems and user-facing software applications. This position demands a blend of deep technical expertise and a passion for engineering excellence. Whether you are working on frontend interfaces that manage vehicle diagnostics or backend systems that process critical sensor data, you will be tackling challenges at the intersection of scale and reliability. You will collaborate with multidisciplinary teams, including systems engineers and domain experts, to deliver robust software that meets the rigorous standards of the automotive industry. ##### Tip Given the technical nature of automotive software, ensure your preparation emphasizes not just coding proficiency, but also the ability to write clean, maintainable, and highly reliable code.
Preparation focus
editorialNo round sequence has been reported for this company, so confirm the format with your recruiter and work the reported questions below.
What to demonstrate
- Breadth across the topics this company reports testing
- Whether you confirm the format before preparing for it
How to prepare
- Ask the recruiter for the sequence, the duration of each stage and whether you will be writing code
- Work the reported questions below and time yourself
PracHub editorial advice for the preparation topics above.
Own your resume
Every line on your resume is fair game. Be prepared to explain every technology and project mentioned in detail.
Think aloud
During technical questions, explain your thought process. It helps the interviewer understand your problem-solving logic.
Prepare for follow-ups
If you mention a technology or concept, expect the interviewer to ask for a deeper explanation.
Stay calm
If you hit a challenging question, take a moment to collect your thoughts before responding.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
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.
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?
Identify the heaviest tenants in a five-minute window under memory pressure
The edge service handles about 3,000 requests per second across roughly 50,000 tenants, peaking near 9,000. Expose the 50 heaviest tenants by request count over the trailing five minutes so limits can be tightened before one tenant's backfill starves the fleet. You may not retain five minutes of raw records. Give the exact solution and its memory, then the bounded-memory approximation with its error stated as a formula, and say which you would ship and at what tenant cardinality that choice changes.
Approach
- Do the exact version first, because it is affordable at this cardinality: a ring of 300 one-second counters per tenant, advanced lazily, is 1,200 bytes of counters per tenant and roughly 60 to 90 MB for 50,000 tenants with overhead. Carry a running total and subtract the bucket you overwrite so a window read is O(1) rather than 300 adds.
- Extract the top 50 with a size-k min-heap over the tenant sums: O(d log k) for d tenants, against O(d log d) to sort them all. Maintaining the heap continuously instead of on query requires a tenant-to-heap-index map, because incrementing a count already inside the heap means sifting from a known position, and without that map you rebuild the heap on every request.
- State the approximation precisely rather than gesturing at sketches. Misra-Gries with m counters retains every item whose true count exceeds N/(m+1), and each retained count underestimates the truth by at most N/(m+1). With m = 1,000 and N = 900,000 requests in the window the error is roughly 900 requests, which is fine for spotting a tenant sending 50,000 and useless for ranking two tenants 200 apart.
- Say what breaks when the window slides: Misra-Gries and Space-Saving are insert-only and cannot be decremented as records age out. The workable construction is one summary per sub-window, say ten seconds, with 30 summaries merged at query time, and the merged error is the sum of the per-summary errors, so the bound degrades linearly in the number of sub-windows.
Follow-up
- The heaviest tenant is heavy because of one export job rather than user traffic. Should the limiter treat those as the same tenant?
- Two tenants sit tied at the boundary of the top 50. Does your answer flap, and does the flapping matter?
Collapse a redelivered event batch into per-aggregate high-water marks
You drain a batch of up to 5,000,000 events, each (aggregate_id BIGINT, aggregate_version INT, event_type, payload). The log guarantees order within one aggregate only; the batch merges 64 partitions, and a relay failover has redelivered a range, so an older version for an aggregate can appear after a newer one. Given a map of last_applied_version per aggregate, produce the events worth applying, at most one per (aggregate_id, version), plus the count discarded. Target O(n) time. State the memory for 2,000,000 distinct aggregates and what you do when it does not fit.
Approach
- One pass, one hash map from aggregate_id to the highest version kept, and a discard counter. An event whose version is at or below last_applied_version for its aggregate is dropped without further work, which is the whole reason the event carries its version rather than a delta. O(n) expected time, O(d) space in distinct aggregates.
- Keep the maximum, never the last occurrence. The redelivered range means the final appearance of an aggregate in the batch can be an older version than one seen earlier in the same batch, so last-wins applies stale state over newer state and the projection regresses with no error anywhere.
- Cost the memory instead of calling it large: an 8-byte key plus a 4-byte version is 12 bytes of payload, and an open-addressed table held at a 0.7 load factor costs roughly 17 bytes per entry before per-slot metadata, so 2,000,000 aggregates is tens of megabytes in a native layout and several times that in a runtime that boxes both key and value.
- If the distinct set exceeds memory, partition on hash(aggregate_id) mod P and reduce each partition independently. Every event for one aggregate hashes to the same partition, so the per-partition result is exact and the merge is concatenation rather than a second reduction.
Follow-up
- The payload is a patch rather than a snapshot, so applying only the highest version loses the intermediate changes. What changes in your reduction?
- How do you detect that version 7 arrived while version 6 was never delivered, and what should the consumer do about the gap?
Keep soft-deleted accounts from blocking re-registration
app_user holds user_id, tenant_id, email CITEXT, password_hash (NULL for SSO principals), email_verified_at, auth_version, status ('invited','active','suspended','deactivated'), created_at, updated_at, deleted_at. Two live accounts for one address inside a tenant must be impossible, but an address freed by a soft delete must be reusable, and the same tenant may delete and re-register it repeatedly. Write the uniqueness DDL for PostgreSQL 16, then the equivalent for MySQL 8 where partial indexes do not exist, and say what each permits once three deleted rows already hold that address.
Approach
- Start from what is actually unique: not (tenant_id, email), but (tenant_id, email) among live rows. PostgreSQL says that directly — CREATE UNIQUE INDEX app_user_live_email ON app_user (tenant_id, email) WHERE deleted_at IS NULL. A full constraint over the same two columns burns the address permanently the first time someone deletes an account.
- Keep case-insensitivity in the type or the index, never in the application: CITEXT as given, or UNIQUE (tenant_id, lower(email)) as an expression index where the extension is unavailable. A case-sensitive unique column is exactly how two accounts for one human appear.
- For MySQL 8 the predicate has to move inside the key: add a discriminator column that is a constant 0 while the row is live and is set to user_id on delete, with UNIQUE (tenant_id, email, deleted_marker). Live rows share the constant and still collide; deleted rows differ from each other and stop colliding.
- State the NULL variant and its dependency: leaving the marker NULL for deleted rows also works, because a unique index treats NULLs as distinct — true in MySQL, and true in PostgreSQL only under the default NULLS DISTINCT, which PostgreSQL 15 lets you reverse. Check the polarity against the three existing deleted rows: constant-on-live is what preserves the collision you want, and reversing it silently admits duplicate live accounts.
Follow-up
- A deleted account re-registers with the same address the next day. Do the old resource rows follow the new user_id, and how does the API keep the two principals apart?
- How do you honour an erasure request while resource_revision.actor_user_id still references this table?
Explain why the owner filter ignores the listing index
The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.
Approach
- Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
- For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
- Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
- Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.
Follow-up
- 90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
- A dashboard runs this for 40 owners in one page load. What changes about the design?
Explain the lifecycle and state management of a Promise in JavaScript.
Explain the lifecycle and state management of a Promise in JavaScript.
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.
- Work from the requirement backwards to the design.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Describe the difference between microtasks and macrotasks in the event loop.
Describe the difference between microtasks and macrotasks in the event loop.
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.
- Work from the requirement backwards to the design.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
How would you optimize the performance of a complex, data-heavy dashboard?
How would you optimize the performance of a complex, data-heavy dashboard?
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.
- Work from the requirement backwards to the design.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Exports duplicate a row range about once a week
Roughly once a week an export writes a file containing a duplicated range of rows. The affected job_run rows show attempt = 1, status = succeeded, one started_at, and a lease_owner naming a different host from the one whose logs show the job starting. Leases last 30 seconds and are heartbeated every 10 from inside the handler; lease_expires_at is computed on the worker and compared against the database's now(). Find the mechanism, and give a fix that holds even if you cannot fix the clocks.
Approach
- Start from the fact that eliminates the obvious answer. attempt = 1 means no retry was recorded, so this is not a re-run after failure; two workers ran the same row concurrently and the takeover path never touched the counter. lease_owner naming a host other than the one that started the job is the same statement from the other side.
- Enumerate the mechanisms that cause a premature takeover, then find the signal that separates them. Either the lease genuinely expired because the heartbeat did not fire, which is what happens when the heartbeat runs on the handler's own thread and the handler makes a long blocking call, or it only appeared expired because two clocks disagree, since lease_expires_at is written from the worker's clock and evaluated against the database's. The discriminator is the distribution: incidents clustered on the longest exports indict the heartbeat, incidents clustered on one host indict skew. Measure both, and measure each host's offset against the database directly.
- Read the reclaim query precisely. In PostgreSQL now() is transaction start time, not statement time, so a reclaimer holding a long transaction compares against an older timestamp than expected; clock_timestamp() is the statement-time function. This is worth ruling in or out before you redesign anything, because it changes which rows look expired.
- Remove the second clock rather than trying to synchronise it. Issue and extend the lease in the database, with lease_expires_at = now() + interval '30 seconds' in both the claim and the heartbeat, so exactly one clock is ever compared and worker skew stops mattering to this predicate.
Follow-up
- The displaced worker has already streamed half the file to object storage. What makes that side effect safe to repeat?
- You now count takeovers. What alert fires on that counter, and at what threshold?
Built from the topics and questions Kpit candidates report; no round sequence has been reported.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Establish the Kpit format
- No round sequence has been reported, so ask your recruiter for the sequence, the duration of each stage and whether you will write code.
Deliverable: A written reply from your recruiter confirming the format.
02Work Java
- Spend the session on Java, which Kpit candidates report being tested on.
- Write one worked example in Java and time yourself on it.
Deliverable: One timed worked example in Java.
03Work Microservices
- Spend the session on Microservices, which Kpit candidates report being tested on.
- Write one worked example in Microservices and time yourself on it.
Deliverable: One timed worked example in Microservices.
04Work SQL
- Spend the session on SQL, which Kpit candidates report being tested on.
- Write one worked example in SQL and time yourself on it.
Deliverable: One timed worked example in SQL.
05Answer out loud: Core JavaScript and Frontend Engineering
- Answer aloud, timed: Explain the lifecycle and state management of a Promise in JavaScript.
- Answer aloud, timed: How do you handle error propagation in asynchronous chains?
Deliverable: Spoken answers to 2 reported Core JavaScript and Frontend Engineering question(s), under time.
06Rehearse your own examples
- Prepare three examples from your own work where you made the decision, each with the outcome you can quantify.
Deliverable: Three examples written out, each with a number attached.
07Dry run for Kpit
- Run one full mock under time, then write down the two questions you most want to ask your interviewers.
Deliverable: A completed timed mock and two questions to ask.
Expand any day for tasks and deliverables. Your progress is saved on this device.
Behavioural rounds judge the decision you made and what it cost.
How do you handle error propagation in asynchronous chains?
How do you handle error propagation in asynchronous chains?
Approach
- 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.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
Discuss your experience with state management libraries in a large-scale project.
Discuss your experience with state management libraries in a large-scale project.
Approach
- 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.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
Unblock an engineer without taking the keyboard
A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.
Approach
- Ask before diagnosing, and ask for things answerable from data they already have: the attempt count on the job rows that produced duplicates, the handler's observed duration against its lease expiry, and whether the duplicate rows share a natural key that a unique constraint could have caught.
- Teach the shape rather than the answer. A lease cannot distinguish a dead worker from a slow one, so a handler that outruns its lease is running twice by design, and deploys deliver the other half by killing handlers mid-run on every rollout. Both of their candidate theories produce identical duplicate rows, which is why the evidence has to come from timings rather than from argument.
- Hand over a checklist they execute: a natural key on every write the handler performs so the second copy collides rather than appends, the record of intent written before any external effect, a lease heartbeat while running, and the metric that shows it working.
- Keep ownership with them deliberately. Pair on the first write, then step back; if you finish it yourself you have closed one ticket and left the same person stuck on the next redelivery.
Follow-up
- How would you distinguish a genuine double-delivery from a lease expiry using only the data already stored?
- Their handler calls an external endpoint before recording that it did. What do you tell them to change first?
- 01
How do you handle error propagation in asynchronous chains?
- 02
Discuss your experience with state management libraries in a large-scale project.
- 03
A teammate has spent two days on a job handler that occasionally writes duplicate rows. They are certain the queue is delivering twice by mistake. You suspect a lease expiring under a slow handler, so the job is running concurrently with itself. Describe how you have unblocked someone in this position: what you asked before offering a hypothesis, what you showed them rather than told them, and what you left them owning. Then say what you would do if their theory turned out to be the right one.
How much time should I spend preparing?
Dedicate at least 2–3 weeks to reviewing your core technical concepts and preparing detailed stories about your past projects.
Kpit Software Engineer candidate reports ↗What differentiates a successful candidate?
Successful candidates don't just know the syntax; they understand the "why" behind their technical choices and can explain how those choices impacted the project's success.
Kpit Software Engineer candidate reports ↗Is the interview process very theoretical?
No, Kpit focuses heavily on practical application, so be prepared to discuss how you have applied your knowledge to real-world scenarios.
Kpit Software Engineer candidate reports ↗Can I expect a remote interview?
Yes, most technical rounds are conducted virtually, so ensure your environment is set up for clear communication and coding exercises.
Kpit Software Engineer candidate reports ↗What topics does Kpit test in interviews?
Kpit interviews most often cover Behavioral Interviewing, Stakeholder Communication, Problem Solving, Java, and Selenium. The exact emphasis depends on the specific role you apply for.
Kpit Software Engineer candidate reports ↗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.
- 01Kpit Software Engineer candidate reports ↗
Company-reported rounds, questions and FAQ.
Candidate reports · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
PracHub practice material, not company-reported.
PracHub page · Accessed 2026-09-22 - 03PracHub preparation framework ↗
PracHub preparation guidance.
PracHub page · Accessed 2026-09-22