As a Software Engineer at Navvyasa Consulting, you are at the heart of delivering robust technological solutions for a diverse client base. Whether you are operating as a Dot Net Full Stack Developer, a Dot Net Developer, or a Service Engineer, your work directly impacts the efficiency and scalability of the systems Navvyasa Consulting's clients rely on to manage their business operations. You are not just writing code; you are architecting solutions that solve complex technical hurdles in real-world scenarios.
This role requires a high degree of technical agility and a mindset geared toward problem-solving. You will be expected to balance the demands of modern software development—such as building responsive front-end interfaces and maintaining high-performance back-end services—with the necessity of providing reliable, long-term technical support. It is a position that offers significant ownership, allowing you to influence the technical direction of projects while working in a collaborative, fast-paced consulting environment.
Technical Screen
reportedMost of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.
What to demonstrate
- Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
- Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
- Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
- Whether a failing case is isolated and explained before any edit is made to the code
How to prepare
- From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
- Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
- Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.
Architectural Discussion
reportedBecause the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.
What to demonstrate
- Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
- Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
- Whether you can be redirected in the middle of an answer without restarting it from the beginning
- Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it
How to prepare
- Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
- Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
- Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing
Behavioral Discussion
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.
Forking the product per client instead of building a configuration or extension point
A fork is the cheapest possible answer for one client and the most expensive possible answer for the estate. Maintenance cost grows as O(clients x changes): a single security patch now needs one port per fork, each with its own conflicts, its own test run and its own release window, and the forks drift further apart with every port. The honest rule is that a required fork is a product gap, so it should either become a supported extension point with a stable interface, or be an explicit, time-boxed and separately priced artefact that is never expected to receive upstream fixes.
Long-lived shared credentials for client systems
One static key used by the connector runtime, a debugging script and three engineers cannot be attributed to a person in an audit, cannot be expired at engagement close, and cannot be rotated without breaking an unknown number of consumers at once. The expensive part of the eventual incident is not the leak; it is that rotation requires finding every holder under time pressure, and nothing recorded who they were. Issue per-principal, per-target, short-lived grants from a broker so rotation is a no-op, attribution is automatic, and closure is a TTL rather than a search.
Reading the constraints as preamble rather than as part of the problem
The bounds are usually there to eliminate the obvious approach: n up to 10^5 makes an O(n^2) scan roughly 10^10 operations, far outside any per-test time budget, and an input larger than memory rules out loading it at all. When a bound is not given, ask for it, then say out loud which approach it kills.
Optimising an axis nobody named
Ask which resource is actually scarce here: wall-clock latency, throughput, memory footprint, cost per request, or engineering time. Shaving a constant factor off an in-memory step is wasted effort when the same function makes a blocking remote call inside the loop.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Measure credential exposure past engagement close without double counting
For one engagement you have credential_grant rows: (grant_id, principal_id, target_system_id, issued_at, not_after, revoked_at which may be NULL). The cutoff instant T is the engagement's ends_on plus access_grace_days. A grant is live over the half-open interval [issued_at, min(not_after, revoked_at)). There are up to 200,000 grants across target systems. For each target_system_id, return the total wall-clock time after T during which at least one grant was live, plus the disjoint segments that make it up. Grants that overlap must be counted once.
Approach
- Clip each grant to [max(issued_at, T), min(not_after, coalesce(revoked_at, +inf))) and discard any interval whose start is not strictly before its end. O(n), and it removes every grant that already expired inside the engagement window.
- Bucket the survivors by target_system_id and sort each bucket by start. O(n log n) overall, and sorting dominates the whole algorithm.
- Sweep each bucket once, holding one open segment [s,e): if the next start is greater than e, emit [s,e) and open a new segment, otherwise set e = max(e, next_end). O(n) after the sort, O(1) working state, O(number of emitted segments) output.
- Total exposure is the sum of emitted segment lengths, which is the measure of the union. Summing per-grant durations instead answers a different question and is unbounded above by the elapsed wall clock.
- Produce a second, conservative figure that ignores revoked_at and uses not_after alone. revoked_at records that you asked for revocation; whether the client's identity provider honoured it is not something this table knows, and a self-contained token validated offline stays valid to its own expiry regardless.
- If a bucket does not fit in memory, sort externally and sweep the stream, or push end timestamps into a min-heap keyed by end and pop those below the current start. Same O(n log n), memory proportional to the maximum number of concurrently live grants.
Follow-up
- Three grants overlap and one of them belongs to an offboarded contractor. What must the sweep emit so exposure can be attributed per principal?
- Which of your two numbers is the real bound on access if the client system validates tokens offline, and what does that imply about issuing TTLs in the first place?
- Run this across 20,000 engagements as a nightly job with a fixed memory budget. What changes in the shape of the computation?
Resolve effective-dated engagement versions for two million access checks
The engagement table holds 500,000 effective-dated rows: (engagement_id, version, effective_from, effective_to which is NULL for the open version). A database exclusion constraint guarantees that per engagement_id the tstzrange(effective_from, effective_to) values never overlap, and the ranges are half-open. You must answer 2,000,000 access checks of the form (engagement_id, at_time) in one process with no database, returning the version open at that instant or none. Also report, per engagement, any coverage gap between adjacent versions. Give the complexity of the scan-per-check approach and say why it is not an option.
Approach
- State the naive cost first: scanning the table per check is 2e6 x 5e5 = 1e12 range comparisons. Even at 1e9 simple comparisons per second that is roughly fifteen minutes of pure compare work, against a budget measured in seconds. It is correct, just unaffordable, and the fix is an index built once rather than a faster comparison.
- Build a hash map engagement_id -> array of versions sorted by effective_from, once. O(V log V) to sort, O(V) memory. The build is amortised over 2e6 lookups, which is what makes it worth doing at all.
- Answer each check by binary-searching for the last effective_from <= at_time, then confirming effective_to IS NULL OR at_time < effective_to. If the confirmation fails, the instant lies in a gap and the answer is none, not the neighbouring version. O(log v) per check, O(V log V + Q log v) total.
- Name the precondition: predecessor search is only valid because the ranges per engagement do not overlap, which the exclusion constraint enforces with btree_gist for the equality operator. Without that guarantee you need an interval tree returning all stabbed intervals, at O(log n + k). Use the same half-open convention as the database, where at_time equal to effective_to belongs to the next version.
- Find gaps with one O(V) pass over each sorted array comparing prev.effective_to against next.effective_from. The exclusion constraint forbids overlaps, not holes, so a hole is entirely possible and is where an access check fails closed for a reason nobody can see in the contract.
- If the checks can be buffered, sort them by (engagement_id, at_time) and merge against the version arrays: O(Q log Q + V log V), usually a large constant-factor win over 2e6 random binary searches because the merge walks memory in order.
Follow-up
- The open version has effective_to NULL. How do you keep both the sort and the comparison total without a sentinel that overflows your timestamp type?
- A gap exists for one engagement and a check lands in it. What should the access path return, and what should it emit so someone fixes the amendment?
- Amendments are being written while you answer checks. What read model keeps an answer self-consistent, and what does a check mean if the snapshot is a second stale?
Collapse connector retry attempts into logical jobs and failure counts
You are given one UTC day of connector_run rows: (connector_id, idempotency_key, attempt_no, status, failure_class, records_read, records_applied), unique on (connector_id, idempotency_key, attempt_no), up to 5,000,000 rows in arbitrary order. Retries of one logical job share the idempotency_key; the job's outcome is the status of its highest attempt_no. Return, per connector_id: the number of logical jobs, a count of terminally failed jobs by failure_class, and the dedupe hit total, meaning sum(records_read - records_applied) over terminal-succeeded attempts only. One pass over the input; state your memory bound.
Approach
- Key a hash map on (connector_id, idempotency_key) and keep only the highest attempt_no seen so far plus that row's status, failure_class and record counts. One pass, O(n) time, O(d) space where d is the number of distinct logical jobs, not O(n).
- Treat a repeated (key, attempt_no) as corrupt input and raise, since the table's uniqueness constraint says it cannot happen; silently overwriting hides a double-insert in the runtime.
- Fold the d surviving entries into per-connector counters in a second pass over the map, not over the rows. The failure_class histogram is a small fixed-width map per connector because failure_class is an enum.
- Compute dedupe hits only from terminal-succeeded attempts. A failed attempt's records_read is work attempted, not work deduplicated, and adding it counts the same source records once per retry.
- If d does not fit in memory, partition the input by hash(connector_id, idempotency_key) into p files and aggregate each partition independently. Same O(n) total work, p sequential passes, memory traded for I/O; the partition function must use the full key or a job's attempts split across files.
Worked solution 20 min
- Take six rows: (c=7,K=A,1,failed,client_429,0,0), (c=7,K=A,2,succeeded,null,500,480), (c=7,K=B,1,ambiguous,client_timeout,300,300), (c=7,K=B,2,succeeded,null,300,0), (c=9,K=C,1,failed,schema_mismatch,0,0), (c=9,K=C,2,failed,schema_mismatch,0,0).
- Build the map: A -> attempt 2 succeeded, B -> attempt 2 succeeded, C -> attempt 2 failed/schema_mismatch. Three entries from six rows.
- Fold per connector: connector 7 has 2 logical jobs, 0 terminal failures; connector 9 has 1 logical job, 1 terminal failure classed schema_mismatch.
- Dedupe hits from terminal-succeeded attempts only: (500-480) + (300-0) = 320 for connector 7, 0 for connector 9.
- Compute the wrong answer deliberately by counting attempt rows: 6 jobs and 3 failures, which is what a naive GROUP BY on run rows returns.
Follow-up
- A job's highest attempt is 'cancelled' but an earlier attempt succeeded. What is the job's outcome, and what does that say about who writes the cancel?
- Produce this incrementally as runs land instead of as a daily batch: what state do you keep per key, and what happens when an attempt arrives out of order?
- records_applied is written by your runtime, not by the client. What would you reconcile it against before anyone trusts the dedupe-hit number?
Return the latest attempt per connector idempotency key
connector_run stores one row per attempt: run_id, connector_id, idempotency_key, attempt_no SMALLINT, status (queued, running, succeeded, failed, ambiguous, cancelled, abandoned), failure_class, watermark_from, watermark_to, started_at, finished_at, unique on (connector_id, idempotency_key, attempt_no). For one connector_id, write the query returning the highest-numbered attempt per idempotency_key, keeping only those keys whose latest attempt is not succeeded, ordered by watermark_from. Give the index that serves it, and say what the plan does without that index.
Approach
- Use DISTINCT ON (idempotency_key) with ORDER BY idempotency_key, attempt_no DESC. PostgreSQL requires the ORDER BY to lead with exactly the DISTINCT ON expressions, and that requirement is what makes the row picked deterministic rather than whichever the scan happened to reach first.
- Apply the not-succeeded filter outside, in a wrapping SELECT. Putting it in the inner WHERE changes the question from 'keys whose latest attempt failed' to 'keys that have any non-succeeded attempt', which reports a key that failed twice and then succeeded, and is the single most common wrong answer here.
- Know the portable alternative: row_number() OVER (PARTITION BY idempotency_key ORDER BY attempt_no DESC) in a subquery, filtered = 1 outside. A window function cannot be filtered in WHERE because WHERE is evaluated before the window, and PostgreSQL has no QUALIFY clause, so the subquery is mandatory rather than stylistic.
- Index (connector_id, idempotency_key, attempt_no DESC). With connector_id fixed by equality, the index supplies the exact order DISTINCT ON needs and the Sort node disappears. It does not reduce rows read: through PostgreSQL 17 there is no loose index scan, so the executor still walks every attempt in range rather than skipping to each next distinct key.
- Give the complexity honestly: without the index the plan is a scan plus an O(n log n) sort over the connector's attempts; with it, O(n) ordered reads and constant working memory, where n is attempts in range, not distinct keys. If the distinct keys are a tiny fraction of attempts, a recursive skip-scan is the next step.
Worked solution 15 min
- Create the table with a handful of rows across three idempotency keys, including one key with attempts 1 failed, 2 ambiguous, 3 succeeded.
- Write the inner DISTINCT ON query alone and confirm it returns one row per key.
- Wrap it and add WHERE status <> 'succeeded' plus ORDER BY watermark_from in the outer query.
- Create the index and compare EXPLAIN output before and after.
Follow-up
- attempt_no is not monotonic in time because a retry was queued out of order. Does your query still pick the right row, and which column should it actually order by?
- How would you return the latest attempt per key across all connectors in one environment without the query degrading to a full sort?
- An ambiguous latest attempt means the target may or may not have been written. What does this query feed, and should ambiguous be treated as failed here?
Explain why the failed-run backlog query ignores three indexes
connector_run holds roughly six million rows over three years. Its indexes are (connector_id, started_at DESC); (started_at) WHERE status = 'failed'; and ((finished_at AT TIME ZONE 'UTC')::date). The operator backlog query is SELECT run_id, connector_id, failure_class, started_at FROM connector_run WHERE environment_id = $1 AND status = ANY($2) AND started_at >= now() - interval '7 days' ORDER BY started_at DESC LIMIT 50, with $2 bound to {failed, ambiguous}. It sequentially scans. Explain index by index why each is unusable, then give the one index you would add.
Approach
- Index one leads on connector_id, which this query never constrains. A B-tree can only be probed from a prefix of its key columns, so with no equality on connector_id it degenerates into a full index scan plus heap fetches, which the planner will rightly cost above a sequential scan. PostgreSQL 18 added B-tree skip scan for an unconstrained prefix, so state your server version before assuming this is fatal; it helps only when that prefix has few distinct values, and hundreds of connectors is already too many to rely on it.
- Index two is partial on status = 'failed', and the rows the query wants are partly not in it at all. Predicate implication runs one way: the query's predicate must imply the index's. status IN ('failed','ambiguous') is weaker than status = 'failed' and implies nothing, so no amount of inlining the parameter helps. It would serve only the failed branch of a UNION ALL.
- Index three is an expression index on a function of finished_at, while the query filters started_at as a plain range. Even on the right column, a range predicate on the bare column cannot use an index on a function of it. Note also why the AT TIME ZONE 'UTC' wrapper is present: finished_at::date on a timestamptz is STABLE rather than IMMUTABLE because it depends on the session TimeZone, so PostgreSQL refuses to index it without a fixed zone.
- Add (environment_id, started_at DESC) WHERE status IN ('failed','ambiguous'): equality on the scoping column first so the scan starts in the right place, the ordering column second so ORDER BY started_at DESC LIMIT 50 stops after fifty index entries instead of sorting the week, and partial on the status set so the index covers the small failing minority rather than six million rows.
- Confirm the partial predicate is provable for the query's parameterised form. Under a custom plan the array parameter is substituted as a constant and the prover handles the ScalarArrayOpExpr; under a generic plan it stays a Param and the proof fails, so check plan_cache_mode or the sixth execution of a prepared statement before declaring victory.
- Say what you deliberately did not do. Adding INCLUDE (run_id, connector_id, failure_class) to chase an index-only scan pays only while the visibility map is current, which ties the query's latency to whether autovacuum is keeping up with the write rate — a dependency worth accepting consciously, not by accident.
Follow-up
- The same query is also run without the environment filter, for an estate-wide backlog. Does your index still help, and what would you add instead?
- How does row-level security on this table interact with your index, given that policy quals are applied before non-leakproof user quals?
- Six months from now the partial index covers 30 percent of the table. What signal tells you to change it?
Explain the difference between an interface and an abstract class in C…
Explain the difference between an interface and an abstract class in C#.
Approach
- State your assumptions explicitly before working the problem.
- Say what you would check first and why it is the highest-information step.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Explain your approach to debugging a slow-loading web application.
Explain your approach to debugging a slow-loading web application.
Approach
- State your assumptions explicitly before working the problem.
- Clarify what is being asked and what a complete answer contains.
- 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 a run API where the outcome is genuinely unknown
connector_run holds run_id, connector_id, environment_id, idempotency_key, attempt_no, status, failure_class, watermark_from, watermark_to, records_read, records_applied, source_schema_version, started_at and finished_at; idempotency_key is derived from the connector and the watermark range, and (connector_id, idempotency_key, attempt_no) is unique. Workers claim a range, write to a client-owned target that offers no idempotency support of its own, then report back. A worker can die at any point, and a write can time out leaving the target's state unknown. Design the claim, report and settle endpoints and the status transitions a scheduler and an operator drive.
Approach
- Derive the run's identity from the work rather than from the caller. POST /connectors/{id}/runs carrying a watermark range computes idempotency_key deterministically, so a scheduler retrying a claim receives the same logical run instead of a second one covering the same rows. Mint a new attempt_no only on an explicit retry of an attempt that has already settled as failed, which keeps the attempt history readable and stops a crashed worker from quietly opening a parallel attempt over the same range.
- Give the claim a lease with an expiry rather than a lock. The worker heartbeats; an expired lease makes the run reclaimable. Without an expiry, a dead worker wedges the binding until someone notices. With a lease and no fencing, two workers can both believe they hold the range, so carry a monotonically increasing lease token and reject any settle whose token is stale — that is what stops a zombie's late report overwriting the live attempt's.
- Make
ambiguousa first-class status with its own transitions, separate fromfailed. A timeout, or a reset after the request bytes went out, leaves the target's state unknown: settling as failed and retrying duplicates, settling as succeeded and advancing the watermark loses data silently. The settle endpoint therefore accepts succeeded, failed with a class, or ambiguous with evidence, and only succeeded advances the watermark. - Keep the ledger write and the watermark advance in one local transaction, and have settle carry the dedupe keys actually applied. The target write and your commit cannot be made atomic without a distributed transaction no client will grant, so ambiguity is resolved by a reconciliation read: re-read the target by dedupe key over the claimed range and settle from what you find. Store the key's input fields and a derivation version alongside the hash, because the client will eventually rename or retype one of those fields and every previously computed key still has to be reproducible.
- Define each caller's behaviour precisely. The scheduler re-claims expired leases and retries only definite failures whose class is retryable; it never retries an ambiguous run. The operator has an explicit resolve endpoint used after reconciliation, recording who decided and on what evidence. An ambiguous run that auto-retries is the design bug — it converts an unknown into a duplicate inside a system you do not own and cannot correct.
- Close the loop with the counters. records_applied differs from records_read by the dedupe hit count, so a retry over a partially applied range should report a high hit count and little net change. That difference is the observable proving the dedupe key works: if a retry reports applied equal to read, the key is not matching and you are duplicating on the client's side right now.
Worked solution 40 min
- Write the state machine — queued, running, succeeded, failed, ambiguous, cancelled, abandoned — with every legal transition and who triggers it.
- Write the claim endpoint's request and response, showing how idempotency_key is derived and what a repeated claim returns.
- Inject a crash between each adjacent pair of claim, fetch credential, validate, write, ledger, advance, and say what the next claim of that range does in each case.
- Write the settle payload for the ambiguous case, listing the evidence a later reconciliation will need.
- Trace a zombie worker settling ninety seconds after its lease expired, and show the token check rejecting it.
- Write the operator's resolve endpoint and the audit fields it records.
Follow-up
- The client's endpoint offers no read-back by your dedupe key. How do you settle an ambiguous run then?
- A worker's lease expires while its write is still in flight and a second worker claims the range. What exactly prevents the double apply?
- How long do you keep the dedupe ledger, and what happens when a record arrives after its ledger row was pruned?
Connector workers are OOM-killed every day and a half
Connector runtime workers are OOM-killed after 18 to 40 hours and a restart clears it. RSS grows monotonically at about 1.9 MB per completed run, flat regardless of how many records that run moved, and workers serving more distinct bindings do not grow faster than workers serving one. Live heap measured immediately after a forced full collection rises with the same slope as RSS. Traffic is unchanged week over week. Give the ordered diagnosis, the specific object you expect at the root, and the fix.
Approach
- Prove retention before hunting for it. RSS and the post-collection live set rising with the same slope means the objects are reachable, which is a leak in the ordinary sense. RSS rising while the live set stays flat would be allocator fragmentation or native buffers, and that is a different investigation with different tools.
- Find the unit the growth is keyed on by division rather than by guessing. 1.9 MB per completed run, flat across record counts and flat across distinct bindings, points at something retained exactly once per run. Per-record growth would implicate an accumulating buffer; per-binding growth would implicate a legitimate cache that is merely unbounded.
- Diff two heap snapshots an hour apart by retained size and walk the dominator tree to a single root. Expect a module-level dictionary whose entry count equals the number of runs completed since process start, which is the observation that confirms step two.
- Read the key and the value separately, because they are two different bugs. The key contains something run-scoped, an idempotency key or a correlation id, which is why the hit rate is zero and the map only ever grows. The value is megabytes rather than kilobytes because the compiled schema validator closes over the first parsed page, so every entry pins a client payload as well.
- Fix both halves. Re-key on (connector_id, source_schema_version), whose cardinality is in the hundreds and stable. Cap the map with a fixed-size LRU so an unforeseen key can never make it unbounded again. Stop the closure capturing the page, so an entry costs a compiled schema rather than a schema plus a payload.
- Make the class of bug visible next time: export entry count and hit rate for every process-level cache. A zero hit rate is the unmistakable signature of a run-scoped key and appears within minutes, whereas this one took 18 hours to present itself as an OOM kill.
Follow-up
- Those retained pages are client payloads sitting in a long-lived shared process. What does that mean beyond memory?
- A worker killed mid-run leaves connector_run rows in 'running'. Who reconciles them, and how do you avoid applying that range twice?
- Same symptom, but the post-collection live set is flat and only RSS climbs. What are you looking at instead, and which tool do you reach for?
Roughly ninety minutes on weeknights with one longer weekend block. The plan cuts scope rather than compressing everything, on the assumption that one thing finished per night beats four half-started.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Fix the scope and take a cold baseline
- Read the role description and write the three things the loop will almost certainly test, then write an explicit not-doing list and keep it visible all week.
- Take one twenty-five-minute coding problem and one fifteen-minute design prompt cold, and write the single sentence naming what blocked each, because those two sentences decide where the remaining evenings go.
- Set the week's rule: one thing finished every night, including the night you only have forty minutes.
Deliverable: A one-page scope with a not-doing list and two cold attempts, each carrying one sentence on what blocked it.
Practice prompt ↗Practice prompt ↗Worked solution ↗02One pattern, written three times from blank
- Choose the single pattern most likely to appear in your loop and write it three times from an empty file rather than editing the previous attempt.
- On the third pass, write the invariant as a comment before the loop body and the complexity before the first line of code.
- Stop at ninety minutes even if the third version is imperfect, and write the one thing you would fix given another hour.
Deliverable: Three independent implementations of the same pattern plus a note on what changed between them.
Practice prompt ↗Practice prompt ↗03One design, only to the depth you can defend
- Take one system shape and go only as far as requirements, interface and data model, refusing to draw a box you could not survive a follow-up about.
- Attach one number to each non-functional requirement, deriving it rather than asserting it, and write the assumption the number rests on.
- Write the one tradeoff you are choosing against and the observation that would make you reverse it.
Deliverable: One design at interface-and-schema depth with derived numbers and one written reversible tradeoff.
Practice prompt ↗Practice prompt ↗04Only the fundamentals you will have to defend
- Write, in under two hundred words each, the answers to the two questions that follow almost any implementation: why this structure and not the obvious alternative, and what happens to this code at a hundred times the input.
- Write what an index actually costs: faster lookups on the indexed columns against a write that now maintains a second structure, plus the cases where the planner declines to use it anyway, low selectivity, or a predicate wrapping the column in a function.
- Delete any answer you cannot deliver aloud in under a minute, since an answer that needs reading is not an answer you have.
Deliverable: Three written answers, each under two hundred words and each timed aloud.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Your own work, timed
- Write a ninety-second and a four-minute version of your main project and time both aloud rather than reading them.
- Prepare the two follow-ups that always come: what you would do differently, and how you knew it worked.
- Put one number in the first sentence and be ready to say exactly where it came from and what it excludes.
Deliverable: Two timed narratives with one defensible number in the opening line.
Practice prompt ↗Practice prompt ↗06The one full rehearsal, in the weekend block
- Run a sixty-minute mock covering a coding round and a design round in one sitting with no break, because sustained attention is the thing evenings have not trained.
- Immediately afterwards, and before hearing any feedback, write the three moments you lost the thread.
- Spend the rest of the block only on those three moments, and on nothing you merely feel shaky about.
Deliverable: Mock notes naming three failure moments with a specific fix written under each.
Practice prompt ↗Practice prompt ↗07Taper
- Write the twenty-minute warm-up you will actually do on the morning: one problem you can already solve from a blank file, one design you can narrate, and nothing you have never seen.
- Re-read only your own notes from this week and open no new material.
- Write the logistics down: the editor or shared document you will be working in, whether execution and lookups are permitted, and the sentence you will use when you do not know something.
Deliverable: A one-page card holding the design structure, the project numbers, and the logistics.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
A migration is a cost you chose to pay, not an achievement. The story is what the old system made expensive, what you measured before committing, what kept serving traffic during the cutover, and what you would have done if the numbers had come back flat. Without those, a rewrite reads as taste.
Describe a challenging technical problem you solved under a tight dead…
Describe a challenging technical problem you solved under a tight deadline.
Approach
- Pick a story where you made the decision, not one where you watched it.
- Name the disagreement and how you resolved it with evidence.
- 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?
How do you handle dependency injection in a.NET Core application?
How do you handle dependency injection in a.NET Core 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.
- Close with what you would do differently, concretely.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
Describe a time you had to integrate a third-party API into a web appl…
Describe a time you had to integrate a third-party API into a web application.
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
Tell me about a time you had to learn a new technology quickly to comp…
Tell me about a time you had to learn a new technology quickly to complete a project.
Approach
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
- 01
Describe a challenging technical problem you solved under a tight deadline.
- 02
How do you handle dependency injection in a.NET Core application?
- 03
Describe a time you had to integrate a third-party API into a web application.
- 04
Tell me about a time you had to learn a new technology quickly to complete a project.
Is this an official Navvyasa Consulting interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Navvyasa Consulting. Rounds and questions reflect what candidates have reported, not a process Navvyasa Consulting has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long should I spend preparing for the technical rounds?
Plan for at least two weeks of focused preparation, specifically reviewing .NET internals and system design patterns. The goal is to reach a level of fluency where you can discuss your previous projects in technical detail without hesitation.
PracHub interview research ↗What is the company culture like?
Navvyasa Consulting prides itself on a culture of continuous learning and collaboration. Interviewers value engineers who are not afraid to ask questions and who prioritize delivering high-quality, reliable solutions for the company's clients.
PracHub interview research ↗Is the interview process mostly remote?
Yes, the interview process is designed to be remote-friendly, reflecting the company's commitment to flexible work arrangements for its engineering team.
PracHub interview research ↗What differentiates a successful candidate?
Successful candidates don't just know the "how"; they understand the "why." They can explain why a specific architectural choice was made and how it impacts the long-term maintainability of the software.
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
Explore related engineering environments and compare the preparation focus.
Inizio Partners Software Engineer Interview Questions & Guide 2026
Code-reading and review practice for Inizio Partners candidates: tracing unfamiliar control flow, finding the invariant a function quietly depends on, and separating a correctness defect from a matter of taste.
MetaDesign Solutions Software Engineer Interview Questions & Guide 2026
Code-reading and review practice for MetaDesign Solutions candidates: tracing unfamiliar control flow, finding the invariant a function quietly depends on, and separating a correctness defect from a matter of taste.
Kran Consulting Private Software Engineer Interview Questions & Guide 2026
Data structures and algorithms practice for Kran Consulting Private candidates, where stating a bound means stating its precondition: constant-time hash lookup is an average case over keys that spread.
CAP Digisoft Solutions Software Engineer Interview Questions & Guide 2026
Systems practice for CAP Digisoft Solutions software engineer candidates, weighted toward consulting and client services work: the load that actually arrives, the failure modes that matter there, and the invariants a partial outage must not