As a Software Engineer at Technext Technosoft, you help the company build scalable digital solutions. Whether you are working on Senior Website Development, Desktop Engineering, or Web-based applications, your role is to translate complex technical requirements into high-performance, user-centric software. You will help shape the architecture and reliability of the products that Technext Technosoft's clients rely on daily.
This position demands a balance of technical rigor and creative problem-solving. At Technext Technosoft, you will collaborate with cross-functional teams to tackle challenges that span the full development lifecycle. You will have the opportunity to influence the company's technical roadmap, improve its internal tools, and ensure that its software stands up to the demands of a growing user base. It is a critical role that requires both deep domain expertise and the ability to adapt to a fast-paced environment.
Your work will directly impact project delivery, so focus on demonstrating both your technical proficiency and your ability to maintain code quality under pressure.
Initial Screening
reportedThe title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.
What to demonstrate
- Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
- Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
- Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year
How to prepare
- Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
- Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
- Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter
Technical Evaluations
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
Behavioral Interviews
reportedThis round is deciding whether a change you make without supervision can be allowed to reach production. It is scored on what you knew at the moment you decided, not on how it turned out, so a story that opens with the result and works backwards reads as luck retold as judgement. Say what the options were, what you did not know, what you did to shrink the unknown before committing, and what you accepted as the worst plausible case. The detail that separates answers is a bound: how many users, how much data, and for how long, if you had been wrong.
What to demonstrate
- Whether the reasoning you give was available at the time you decided rather than after the result came in, since a story whose deciding evidence arrived later describes an outcome and not a judgement
- Whether you can put units on the exposure (users, rows, minutes of degraded service) and whether the containment you chose actually bounded it: a canary bounds the request path it fronts, while a background job writing to a shared table reaches every user regardless of which version served their requests
- Whether the reversal path existed before you shipped or was improvised during the incident, and whether it restores state or only stops further damage
How to prepare
- For your three largest changes, write down the one thing you would have had to be wrong about for it to fail, and what your best estimate of it was on the day you shipped. If you never held an estimate, that is the gap the follow-up questions will find
- Write the undo procedure for one of those changes as it existed at the time, then mark which steps restore data and which only stop new damage. Turning a flag off or reverting a deploy ends the new writes; rows already written come back only from a copy you kept, and a dropped column comes back empty unless something outside the schema holds the values
- Rehearse one story from the decision point forward and stop before the outcome, then have someone ask what you would do next. If the story only works with the ending attached, it is an anecdote rather than a decision you can defend
PracHub editorial advice for the preparation topics above.
Letting a slow dependency consume unbounded concurrency
The failure that takes a service down is usually not an error but a delay. A dependency answering in thirty seconds instead of fifty milliseconds holds each request's worker or connection six hundred times longer, and since required concurrency is arrival rate times latency, a fleet sized for sixty in-flight requests now needs thirty-six thousand to sustain the same rate - so it queues, and requests whose clients have already abandoned them still occupy resources. Retries make it precisely worse: a policy of three attempts triples the load on a dependency at the exact moment it is least able to serve, which is how one slow dependency becomes an outage of everything sharing that pool. Containment is four specific things - a timeout on every outbound call shorter than the caller's remaining budget, a bounded pool per dependency so one cannot starve the others, backoff with full jitter rather than a fixed delay so retries do not resynchronise, and a circuit that stops sending once the failure rate makes an attempt pointless.
Paginating with LIMIT/OFFSET over a set that changes while the client is reading it
OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.
Naming no test cases at all
State what you would test before being asked: empty input, a single element, all elements equal, the maximum permitted size, and the input that exercises the branch you just wrote. It costs thirty seconds and is much of what separates someone who has shipped code from someone who has only solved puzzles.
Trusting input because it came from your own front end
Anything crossing a trust boundary is hostile: parameterise queries instead of building SQL by concatenation, validate against an allow-list rather than a deny-list, and bound the size of anything you allocate from a request. Raising this unprompted in an API or design question is a cheap and unusually strong signal.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Merge partitioned event streams into one ordered feed with bounded lateness
The read-model service consumes 64 log partitions carrying about 4,000 events per second in total. Each partition is ordered within itself, but partitions drift by up to 30 seconds, and the activity feed must present a tenant's events in occurred_at order. Produce the merge. State its complexity, the buffer it requires in events and in bytes, what happens when one partition is idle, and what you do with an event that arrives after you have already emitted its position. Payloads average 1 KB.
Approach
- Merge with a min-heap over the 64 partition heads keyed on (occurred_at, event_id): O(log P) per event and O(n log P) overall. The tie-break on event_id is what makes the output deterministic when two partitions carry the same millisecond, which matters because the feed is paginated and a non-deterministic order reorders pages under the reader.
- Emitting the heap head is only correct once every partition has produced everything up to that timestamp, so the emit condition is a watermark: the minimum across partitions of the highest occurred_at seen, less the allowed lateness. Events are held until the watermark passes them, which is what turns individually ordered streams into a jointly ordered one.
- Size the buffer from the lateness rather than guessing: 4,000 events per second times 30 seconds is 120,000 buffered events, and at 1 KB each about 120 MB of heap. That number is the real price of the ordering guarantee and belongs in front of whoever asked for it.
- Handle the idle partition explicitly, because it fails the feed rather than corrupting it: a partition with no traffic never advances its own maximum, so the watermark freezes and output stops entirely. Either every partition emits a periodic idle marker carrying the broker's current time, or the watermark falls back to wall clock for a partition silent beyond a threshold.
- Choose the late-event policy from what the projection is keyed on. The projection upserts on (aggregate_id, aggregate_version) and discards a version it has already applied, so a late event is safe to apply out of order and correctness never depended on the merge at all. Apply it, recompute the affected feed page, and count lateness so the 30-second budget can be re-derived from data rather than folklore.
- Say what the merge does not buy: ordering is guaranteed within one aggregate by the log's partitioning, and no watermark makes the cross-aggregate order authoritative. Two events from different aggregates in the same millisecond have no true order, so the feed's order is a presentation choice that must be stable rather than correct.
Worked solution 35 min
- Write the heap comparator on (occurred_at, event_id) and the per-partition head refill.
- Write the watermark computation and the emit-loop condition, then list which buffered events are held at a chosen instant.
- Compute the buffer at 4,000 events per second, 30 seconds and 1 KB per event, and state what fraction of a worker's heap that represents.
- Add the idle-partition marker and trace the watermark with one silent partition, both with and without the marker.
- Write the late-event path and name the key that makes applying it safe.
Follow-up
- The lateness budget is raised to five minutes. What is the new buffer, and what besides memory changes?
- The consumer restarts. Where does it resume from, and what does the feed look like for the first 30 seconds?
- One partition is ten minutes behind because its producer is slow. Do you stall the feed or emit without 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.
- Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
- Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.
Follow-up
- The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
- The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
- An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?
Track a rolling failure rate per destination for circuit decisions
The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.
Approach
- Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
- Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
- State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
- Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
- Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
- Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.
Follow-up
- The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
- A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
- How would you make the window survive a process restart, and is it worth the cost?
Replace offset paging on the resource feed with keyset
resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.
Approach
- Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
- Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
- Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
- Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
- State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
- Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.
Worked solution 25 min
- Seed one tenant with 500k active resources, 2% of them sharing an identical updated_at.
- Time LIMIT 50 OFFSET 0 against OFFSET 20000 and record rows-read from EXPLAIN (ANALYZE, BUFFERS) for each.
- Page the whole set with the keyset query while an insert-only writer adds 100 rows/second, collecting resource_ids, and repeat the run with OFFSET.
- Repeat both runs under a second writer profile that also deletes 20 rows/second from pages already returned and bumps updated_at on 20 more, and diff each collected id set against the rows that existed for the whole run.
- Remove resource_id from the cursor so the seek degrades to updated_at < $2, and re-run the tie-heavy section of the feed.
Follow-up
- The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
- Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
- What does the cursor do when the row it points at has since been deleted?
Write the update path that detects a concurrent edit
resource carries version INT NOT NULL DEFAULT 1. resource_revision holds revision_id, resource_id, version, actor_user_id, change_kind, patch JSONB, request_id, created_at with UNIQUE (resource_id, version). outbox_event holds aggregate_type, aggregate_id, aggregate_version, event_type, payload, status. A PUT carries the version the client read. Write the exact statements for the single transaction that applies the edit, records the revision and enqueues 'resource.updated', and give the handler's branch on zero affected rows. Then say what PostgreSQL 16 does under READ COMMITTED when two of these updates hit one row at once.
Approach
- One transaction, three writes, no network call inside it: UPDATE resource SET title = $3, version = version + 1, updated_at = now() WHERE resource_id = $1 AND tenant_id = $4 AND version = $2; then INSERT the resource_revision row at version $2 + 1; then INSERT the outbox_event row at the same aggregate_version. The event goes to a table rather than a broker because no transaction spans both.
- Branch on the affected-row count before doing anything else. Zero has three causes — stale version, wrong tenant, row gone — so re-read once and map to 409 carrying the current version, or 404 for an id outside the caller's tenant, which also stops the endpoint confirming that another tenant's id exists.
- State the engine behaviour instead of assuming it. Under READ COMMITTED the second UPDATE blocks on the row lock, and when the first commits PostgreSQL re-evaluates the WHERE clause against the newly committed row, so the version predicate now fails and the statement reports zero rows. Under REPEATABLE READ the identical collision raises SQLSTATE 40001 instead, so the handler must fold both shapes into one conflict response.
- Keep UNIQUE (resource_id, version) even though the predicate already serialises writers. It is what makes a lost update unwritable if any other path ever reaches the revision table, and it converts a logic bug into 23505 rather than into a silently missing history row.
- Refuse to auto-retry the whole PUT. A retry re-reads the winner's state and reapplies an intent formed against data that no longer exists — the silent overwrite the version token was added to detect. Return the conflict; merge field-wise only if the patches are provably disjoint.
- Note that now() is the transaction timestamp in PostgreSQL, so resource.updated_at, the revision's created_at and the outbox row share one instant, which is what later makes reconciliation between the three tables unambiguous.
Follow-up
- A client sends the version it read ten minutes ago and the resource has moved three versions. What is in your 409 so it can resolve the conflict without a full re-fetch?
- Two editors, two disjoint fields, no overlap. Does your answer still refuse the second write, and should it?
- Every write now touches a second hot table. How do you keep the outbox insert and its partial index from becoming the write bottleneck at 1.2k writes/second?
Walk us through the architectural choices you made in your most recent…
Walk us through the architectural choices you made in your most recent project.
Approach
- Choose a partition key and say what query it makes expensive.
- Name the failure you are designing for, then the recovery path.
- State the consistency you need, and where you are willing to be stale.
Follow-up
- What breaks first when traffic grows ten times?
- What would you drop to keep the system up under load?
How do you approach breaking down a large, monolithic feature into sma…
How do you approach breaking down a large, monolithic feature into smaller, manageable tasks?
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.
- Choose a partition key and say what query it makes expensive.
Follow-up
- What would you drop to keep the system up under load?
- What breaks first when traffic grows ten times?
How would you design a system to handle a sudden spike in user traffic…
How would you design a system to handle a sudden spike in user traffic?
Approach
- Name the failure you are designing for, then the recovery path.
- Fix the scope first: who calls this, how often, and what they do when it fails.
- Choose a partition key and say what query it makes expensive.
Follow-up
- What breaks first when traffic grows ten times?
- How does this behave when that dependency is down for an hour?
Describe your process for debugging a complex issue in a production en…
Describe your process for debugging a complex issue in a production environment.
Approach
- Clarify what is being asked and what a complete answer contains.
- Work from the requirement backwards to the design.
- Say what you would check first and why it is the highest-information step.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
What are the advantages and disadvantages of using a specific framewor…
What are the advantages and disadvantages of using a specific framework versus building from scratch?
Approach
- Say what you would check first and why it is the highest-information step.
- State your assumptions explicitly before working the problem.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Keep one unresponsive destination from stalling all webhook delivery
Egress delivery sends about 1.5k webhooks/second to 40k destinations, with a per-destination concurrency cap of 4 and a 10-second connect-plus-read timeout. One destination begins accepting connections and never responding; within the hour 150 destinations behave the same way. Design the delivery path so unrelated destinations are unaffected: the pool structure, the timeouts, the retry policy, the per-destination circuit, and what is recorded so a retry is not a second effect at the receiver. State how many in-flight slots the degraded destinations hold and why that number decides the design.
Approach
- Start with the number, and with the law that produces it. In-flight work is arrival rate times time in service, so 1.5k/second against a healthy 200 ms response needs about 300 concurrent slots. Per destination the same product applies, ceilinged by the concurrency cap: at the fleet average of 0.0375 deliveries/second per destination (1.5k spread over 40k) a 10-second timeout is 0.375 slots. A destination that has queued retries behind it is a different regime - every slot refills the instant an attempt expires, so it sits pinned at its cap of 4 - and 150 of those hold 600 slots, more than a pool sized for healthy traffic, entirely consumed by endpoints that will never answer. The per-destination cap bounds one endpoint and says nothing about the aggregate, which is exactly why it alone is not containment.
- Contain with bulkheads and an admission bound rather than a larger pool. Cap total in-flight per pool and shard destinations across pools by a hash of destination id, so a correlated group - one provider, one region - cannot exceed its pool's share. A delivery refused admission and re-queued with backoff is strictly better than one holding a slot on behalf of a receiver that is not listening.
- Treat the timeout as two timeouts, and be exact about what shortening one buys. Connect and read are separate failures and both must be shorter than the budget of whatever is waiting. Occupancy is min(cap, arrival rate x timeout), so dropping the read ceiling from 10 seconds to 3 cuts a merely slow destination's occupancy proportionally, 0.375 slots to 0.11 at the fleet average. It does not cut the 4 slots held by one of the 150: a destination with a retry backlog arrives far above cap/timeout - 0.4/second at a 10-second timeout, 1.33/second at 3 - so it stays pinned at the cap either way and only the slot-seconds per attempt fall. What that does buy is detection rate: 3.3x more failures observed per second on the same four slots, which is how fast the circuit reaches its threshold. Pick the value from the measured latency distribution of successful deliveries, with their high percentile as the floor, not from a round number.
- Add a circuit per destination, counting a timeout as a failure. Once open, fail fast without taking a slot - that is the whole point, converting 4 held slots into zero. Half-open on a schedule with exactly one probe and close only if the probe succeeds, so a permanently dead endpoint costs one request per interval instead of a growing retry queue.
- Make retries safe and non-synchronising. Back off with full jitter, sleeping a random value in [0, min(cap, base x 2^attempt)], because a fixed delay re-synchronises every failed delivery to one destination into a simultaneous burst. Delivery is at-least-once, so the payload carries the event id under the signature and the receiver deduplicates on it; record the attempt against (destination, event id) rather than a bare success flag, so a lost response does not become a second business effect on the other side.
Worked solution 25 min
- Compute healthy in-flight from rate times latency, then slots held by 150 destinations at the cap and the full timeout, and compare both against one pool size.
- Write the pool sharding rule and the admission bound, and state what a refused delivery does next.
- Pick connect and read timeouts from the success-latency distribution, then compute min(cap, arrival rate x timeout) for an average destination and for one with a retry backlog, and say which of the two the shorter timeout actually moves.
- Write the circuit's state machine with its open threshold, probe interval and close condition, and the backoff formula with full jitter.
Follow-up
- The destination is not dead - it answers in 9.5 seconds with a 200. Does a failure-rate circuit open? Should anything shed that traffic, and on what signal?
- One destination requires deliveries in order. What does a per-destination concurrency of 4 do to that guarantee, and what would you change to offer it?
- A destination has been parked six hours with 900k undelivered events. What does resuming look like, and is delivering the whole backlog the right call?
One log partition stops advancing while the others drain
Search results for a subset of tenants are hours stale; the rest are current. The projection consumer reports lag of zero on 15 of 16 partitions and 400,000 on one. Its error rate is flat and its CPU is idle. outbox_event has no pending rows older than a second, so the relay has published everything it holds. Identify the mechanism, give the ordered checks, and state what you do in the first ten minutes versus what you change permanently.
Approach
- Read the lag distribution first. A slow consumer lags everywhere; zero on fifteen partitions and 400,000 on one is not throughput. Idle CPU on the stuck partition means the consumer is not advancing its offset at all, which points at one message it cannot get past rather than at a rate problem.
- Exonerate the producer before touching the consumer. No pending outbox rows older than a second means the relay published, so the event exists in the log. This separates never sent from sent and never applied, which are different code paths and usually different owners.
- Read the message at the stuck offset and the handler's log lines for its event_id. A flat error rate with no progress has two explanations and you must distinguish them: the handler is throwing and the retry loop is swallowing it, or the handler is blocking on something and never returning. Idle CPU with no error lines favours the second.
- Mitigate before diagnosing further. Move the offending event to a dead-letter store and commit the offset past it. Adding consumers does nothing here, because a partition is consumed by exactly one member of the group, and the blast radius is every aggregate hashed to that partition, not only the aggregate that produced the bad event.
- Fix permanently by bounding handler attempts and dead-lettering on exhaustion, so no single message can stop a partition. Then replay the dead-lettered event once the handler is fixed: it carries aggregate_id and aggregate_version, so a consumer that discards versions it has already applied can absorb the replay, and resource_revision is the fallback if the event itself is unusable.
Follow-up
- The dead-lettered event carried aggregate_version 7 and the projection had applied 6. What must the replay do differently if 8 and 9 landed in the meantime?
- How do you show staleness to the user while the partition is behind, given the API already returns the projection's watermark?
- What changes if the message is poison because a previous deploy wrote a payload shape the current code cannot parse?
For a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Numbers before diagrams
- Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
- Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
- Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.
Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.
Practice prompt ↗Practice prompt ↗Practice prompt ↗Worked solution ↗02One system, from requirements to schema
- Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
- Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
- Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.
Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.
Practice prompt ↗Practice prompt ↗03The consistency you are actually buying
- Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
- Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
- Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.
Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.
Practice prompt ↗Practice prompt ↗04Failure is the design
- For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
- Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
- Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
- Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.
Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Scaling the hot path
- Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
- Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
- Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
- Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.
Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.
Practice prompt ↗Practice prompt ↗06Keep the coding hand in, at the bar that applies to you
- Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
- Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
- Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.
Deliverable: One problem hardened to review standard, with its test list and one written review comment.
Practice prompt ↗Practice prompt ↗07Defend it while being interrupted
- Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
- Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
- Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.
Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.
Practice prompt ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Team size, service count and tickets closed say very little. Seniority shows in the decision you owned: what you chose not to build, which constraint you traded away, whose objection you had to resolve before anything could move. A large project where you executed someone else's plan is a small story.
Tell us about a time you had a technical disagreement with a teammate …
Tell us about a time you had a technical disagreement with a teammate and how you resolved it.
Approach
- Give the blast radius: what could have broken, and what you measured.
- Close with what you would do differently, concretely.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
Describe a time you had to make a technical trade-off between speed an…
Describe a time you had to make a technical trade-off between speed and quality.
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?
- What did you decide not to do, and why?
How do you stay updated with emerging technologies and integrate them …
How do you stay updated with emerging technologies and integrate them into your work?
Approach
- Close with what you would do differently, concretely.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
- 01
Tell us about a time you had a technical disagreement with a teammate and how you resolved it.
- 02
Describe a time you had to make a technical trade-off between speed and quality.
- 03
How do you stay updated with emerging technologies and integrate them into your work?
Is this an official Technext Technosoft interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Technext Technosoft. Rounds and questions reflect what candidates have reported, not a process Technext Technosoft has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How much time should I spend preparing?
Plan for at least 2–4 weeks of dedicated preparation. Focus on brushing up on your core language knowledge and practicing system design scenarios.
PracHub interview research ↗Is the technical interview focused on whiteboard coding?
Candidates report that the coding scenarios are realistic and relevant. Be prepared to write code that you would actually use in a production environment, with a focus on readability and error handling.
PracHub interview research ↗What is the culture like at Technext Technosoft?
Candidates report that the company values innovation, collaboration, and direct communication. Engineers are encouraged to take ownership of their projects and contribute ideas that improve the company's processes.
PracHub interview research ↗How long does the process take from start to finish?
The timeline can vary, but the process generally moves efficiently. You should receive updates on your status promptly after each stage.
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