Interview guides · Katomaran Technologies Private
Katomaran Technologies Private Software Engineer Interview Questions & Guide 2026
API design practice for Katomaran Technologies Private software engineer candidates: compatibility, versioning, and changes you cannot take back.
The 60-second brief
A Software Engineer at Katomaran Technologies Private serves as a core builder of the company’s digital infrastructure. Whether you are working on web development, full-stack applications, or specialized machine learning initiatives, your primary objective is to translate complex technical requirements into functional, high-quality code. This role is pivotal in maintaining the company’s competitive edge by ensuring that products are both scalable and user-centric.
Prepare in one language you know well enough to debug in rather than the one you think reads best. Under a clock an unfamiliar language costs you standard-library lookups and iteration mechanics, and that time comes out of your thinking budget, not your typing budget.
Focus areas
- Make every write idempotent under client retries
- Bound blast radius with per-tenant concurrency limits
- Evolve APIs without breaking pinned SDK clients
Overview
The position offers significant exposure to the full development lifecycle, from initial design and architecture to deployment and refinement. You will work within collaborative teams where your ability to solve technical hurdles—such as model optimization or front-end responsiveness—directly impacts the company's success. It is an environment that values practical application and a hands-on approach to problem-solving, making it an excellent space for engineers who enjoy taking ownership of their work.
The company often values practical, portfolio-based demonstrations of skill, especially for those entering the industry as freshers or early-career professionals.
The Katomaran Technologies Private interview process
Katomaran Technologies Private candidates report 4 rounds · ≈ 3-5 weeks. The stages below are what candidates describe, not a published process.
Resume Screening
The 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
HR Discussion
When a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.
What to demonstrate
- Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
- Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
- Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
- Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened
How to prepare
- Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
- Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
- Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure
Technical Assessment
What this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.
What to demonstrate
- Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
- Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
- Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly
How to prepare
- Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
- Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
- Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live
Final Rounds
Coding rounds mostly set a floor. They decide whether you clear the bar, not where you land on the ladder. Level tends to come out of the design discussion and the ownership stories, so the question worth auditing beforehand is whether the scope you describe matches the scope of the job. Work that stops at your own service, or a story whose hard part was writing the code rather than getting several people to agree on an interface, reads a level below where you think you are interviewing, and that gap is usually resolved downwards.
What to demonstrate
- Whether the largest thing you describe owning ran end to end — the decision, the migration path, the rollout, and what you did when it went wrong — or stopped at the change you merged
- Whether design answers include what you would not build, what you would defer, and what you would measure before committing, rather than only what the boxes are
- Whether a disagreement in a story was settled with something checkable — a benchmark, a prototype, a written proposal — instead of by seniority or by waiting it out
- Whether you can say which calls you made alone and which you escalated, and why the line sat where it did
How to prepare
- Write your largest piece of owned work as a timeline of decisions — who decided what, when, and what you did when the plan broke — then delete every sentence whose subject is "we" and see how much survives
- Take one system you know well and drill the migration answer: how old and new paths run side by side under live traffic, how you compare their outputs, what the rollback is once writes are going to both, and which step you would not automate
- Map each line of the ladder in the job posting to a specific thing you have done, find the line you cannot support, and prepare the closest evidence you have plus an honest account of the gap
Where candidates lose points
Paginating a growing table with limit and offset
Two unrelated defects share the idiom. Correctness: rows inserted or deleted between page requests shift the window, so a consumer walking an export skips rows and sees others twice, which for a customer-facing sync is silent data loss rather than an error anyone notices. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N times the page size and a deep page on a large table degrades from milliseconds to seconds. Keyset pagination over a stable, unique, indexed ordering -- where (created_at, id) < ($1, $2) order by created_at desc, id desc limit $3 -- is constant-cost per page and immune to shifting, on the precondition that the cursor columns never change value for a row, which disqualifies updated_at as a cursor.
One shared connection pool for every tenant and every query class
A single tenant with a large table and a missing index can occupy every connection with slow queries, and every other tenant then waits in connection acquisition -- a queue invisible in database metrics, because the database itself looks healthy while the application starves. Containment is bulkheads: separate pools or per-tenant concurrency caps for interactive requests, background jobs and exports, a statement timeout low enough that a pathological query dies before it accumulates, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks indefinitely. One caveat worth knowing in advance: if a transaction-pooling proxy sits in front of the database, session-scoped behaviour changes, so session-level advisory locks and settings applied outside a transaction do not survive the way they do on a direct connection.
Assuming the bug is in the framework
Suspect your own code first: read the stack trace top to bottom, check which versions are actually installed rather than which ones you believe are, and reproduce in isolation before blaming a library that thousands of people run daily. When the fault really is upstream, you need that minimal reproduction to say so credibly anyway.
Finishing a solution without stating its complexity
Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.
Questions & practice
Schedule ordered webhook retries with a heap of subscription queues
Medium · heap · scheduling · backoff · head-of-line-blocking
Design the in-memory scheduler for webhook delivery. Up to 20 million rows sit in status pending or failed_retryable across 200,000 subscriptions, each row carrying next_attempt_at and attempt_count, and each endpoint having a circuit breaker. Deliveries for one subscription must be attempted in order, so at most one attempt per subscription may be in flight. Support due(now), complete(delivery, outcome) and insert(delivery) in O(log S), where S is the subscription count rather than the delivery count. Give the backoff formula you schedule retries with.
Approach
- Key the global heap by subscription, not by delivery. Each subscription owns a FIFO of its due deliveries in event order; the heap holds one entry per eligible subscription, keyed by its head's
next_attempt_at. That is 200,000 heap entries instead of 20 million, and it makes the one-in-flight rule structural rather than a check somebody can forget. due(now): peek the minimum. If its key is in the future, sleep until then instead of spinning. Otherwise pop it, move the subscription into an in-flight set, and do not re-push it. A subscription absent from the heap cannot be dispatched twice, which is precisely how ordering is preserved.complete: on success, drop the head and re-push the subscription keyed by its new head, or leave it out when the queue empties. On a retryable failure, incrementattempt_countand setnext_attempt_at = now + uniform(0, min(cap, base * 2^attempt)), sampled uniformly across the whole interval. That is full jitter; deterministic backoff re-synchronises the herd you just created.- Circuit breaker: park the subscription in a second heap keyed by its half-open time, so an endpoint dead for six hours costs one heap entry and zero attempts rather than consuming worker slots. Admit exactly one probe at half-open and close the breaker only on its success.
- Say the price of the ordering guarantee out loud. One in-flight attempt per subscription means an endpoint answering in 10 seconds drains at 0.1 deliveries/second however many workers you run, and its backlog grows until it recovers. If the customer does not need order, allow k in flight and document delivery as unordered; that is the trade, and it is a product decision.
- All three operations are O(log S) with O(S) resident heap memory and the queues themselves backed by the store. The database-backed equivalent is a partial index on
(subscription_id, next_attempt_at) where status in ('pending','failed_retryable')claimed withFOR UPDATE SKIP LOCKED, and the write-back must be fenced onlease_tokenso a worker that stalled and resumed cannot overwrite a newer attempt.
Follow-up
- One subscription has 4 million queued deliveries. What stops it from starving the other 199,999, and what does your heap look like under that load?
- A customer requests redelivery of last Tuesday's events. Where do those rows enter your structure, and what keeps them from reordering live traffic?
- The process restarts. How much state do you rebuild, and what stops every subscription from being attempted in the same second?
Seal an hour under late data with bounded memory
Hard · watermarks · late-data · quantile-sketch · conditional-write
Metering ingest reads 256 partitions at 10,000 to 40,000 events/second. Events carry occurred_at and ingested_at, and during a producer replay the gap between them is hours. Seal each UTC hour once no more than 50 parts per million of that hour's eventual quantity can still arrive, using memory that does not grow with the size of the replay. Define the watermark, the lateness parameter and how you measure it, the structure holding open hours, and the write that performs the seal. State what an idle partition does to your watermark.
Approach
- Two clocks, two jobs. Bucket by
occurred_at, because that is the hour the customer is billed for, and advance the watermark oningested_at, because that is what the fold has consumed and whatsource_max_ingested_atrecords. Conflating them is what makes late data invisible. - The global watermark is the min over partitions of each partition's committed
ingested_at, not the max: the fold is trustworthy only as far as the slowest partition. The consequence is that one idle partition pins the watermark forever and nothing seals, so an idle partition must promote its watermark to wall clock after a stated idle timeout, and that timeout becomes a correctness parameter, because a partition that is slow rather than idle gets sealed past. - Choose the lateness L from the measured distribution of
ingested_at - occurred_at, weighted by quantity rather than by event count. The target is 50 ppm of the hour's quantity, and a replay is rare in events while carrying disproportionate mass, so an event-weighted quantile picks an L that is comfortably wrong at exactly the moment it matters. - Measure that quantile in bounded memory. A Greenwald-Khanna summary gives epsilon-approximate quantiles in O((1/epsilon) log(epsilon n)) space; a t-digest costs more per merge but has relative error that tightens at the tails, which is the half of the distribution you are reading at p99.99. Keep a separate summary per tenant class, because one tenant's batch importer is not the population.
- Hold open hours in a min-heap keyed by
hour_start. When the watermark advances, pop every hour withhour_end + L < Wand seal it: O(log H_open) per advance and O(1) amortised per event to touch its bucket. Memory is open hours multiplied by distinct(tenant, workspace, sku)keys, so cap the number of simultaneously open hours and spill the oldest intousage_rollup_hourlyasstatus='open'with arevisionbump. While an hour is open the row is upsertable, so the store is your overflow. - The seal itself is a conditional write:
update ... set status='sealed', sealed_at=now() where status='open' returning .... Two sealers race on every restart, and the loser must see zero rows and stop rather than write a second value. After the seal, an event for that hour is not an upsert but an adjustment, andsource_max_ingested_atis what proves it arrived afterwards.
Follow-up
- A replay starts during the sealing window for a period you are about to close. What do you do, and what is the customer-visible consequence of each option?
- Your measured quantity-weighted p99.99 lateness is six hours and the invoice must be issued at 02:00 UTC on the first. How do you reconcile those two numbers?
- How would you detect that L has drifted before it costs you an hour's quantity?
Hold a tenant to a trailing sixty-second request limit
Medium · sliding-window · two-pointer · rate-limiting · tenant-skew
The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.
Approach
- Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while
front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request. - Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
- Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate,
prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact. - Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (
tokens,last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual. - Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
- Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.
Worked solution (25 min)
- Implement the deque version and instrument the per-request pop count, then confirm total pops equal total pushes over a run.
- Generate a burst that places R requests in the last 100 ms of one minute and R more in the first 100 ms of the next.
- Run that burst through the exact deque, a fixed 60-second counter, and the weighted two-bucket estimate, recording admissions in the trailing 60 seconds at every instant.
- Size the memory as R x active tenants x 8 bytes at R = 1,000 and 200,000 tenants, and compare it against what a token bucket would need.
Expected result: The exact deque never admits more than R in any trailing 60-second window. The fixed counter admits close to 2R across the boundary. The weighted estimate lands between the two on this burst and approaches 2R once the previous window's requests are packed at its end.
Verify it yourself
- Total pops equal total pushes, which is what makes the amortised O(1) claim true rather than merely asserted.
- Sliding the whole burst by 30 seconds changes the fixed counter's result and leaves the deque's unchanged.
- Running two processes each limiting to R/2, against a tenant that sends everything to one of them, shows that tenant throttled at half its contractual limit.
Follow-up
- One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
- Quotas rather than rate limits: the check is
select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes. - How do you return an accurate
Retry-Afterfrom the exact algorithm without a second scan?
Decide which facts an invoice line copies instead of joining
Medium · denormalisation · immutable records · foreign keys · rounding
invoice_line_item already denormalises tenant_id, which is reachable through invoice_id, and stores amount_minor even though quantity times unit_price_micros would recompute it. A reviewer asks you to normalise both away, and separately asks whether the tenant's legal name and billing address should be copied onto the invoice header. Decide each case. For every field you keep denormalised, name the read pattern or the invariant that justifies it, the anomaly the copy can develop, and the mechanism that prevents that anomaly here.
Approach
- Split the question into two kinds of copy, because they fail differently. A copy of a currently mutable fact is a cache: it drifts and needs invalidation. A copy of a fact frozen at write time is not a cache at all, it is the record of what happened, and normalising it away destroys information the source no longer holds.
- Keep tenant_id on the line. It costs 8 bytes, it leads every index on the table so no read is ever accidentally cross-tenant, and it turns a wrong join into an empty result rather than another tenant's money. Prevent the drift structurally: a unique constraint on invoice (invoice_id, tenant_id) plus a composite foreign key from the line on (invoice_id, tenant_id) makes a mismatched pair impossible, so the database enforces agreement instead of a code review.
- Keep amount_minor. Rounding must happen exactly once, at a named site, with a stated mode (half-even here). If readers recompute from quantity and unit_price_micros, every reader owns a rounding decision, and half-up and half-even diverge systematically across thousands of lines rather than cancelling out. A check constraint can bound the stored value but deliberately cannot re-derive it.
- Copy the legal name and billing address onto the invoice header, written once and never updated. The statement must show what was true when it was sealed, and the tenant record will change afterwards. This is a snapshot for the same reason
source_rollup_watermarkis stored per line: without it, nobody can reconstruct what the customer was told. - Name the read pattern that pays for all of it. Rendering, dispute response and export are per-tenant, per-period reads over thousands of lines that would otherwise join back to slowly changing dimensions that no longer hold the historical value. The write side is a once-per-period batch, so the extra columns cost nothing that matters.
- Concede the case where the reviewer is right: a mutable operational attribute such as the tenant's current plan name has no business on a line. If a report wants it, join. If a statement needs the plan as of the period, that is another snapshot and it belongs on the header with the rest.
Worked solution (25 min)
- Write the DDL: unique (invoice_id, tenant_id) on invoice, the composite FK from the line, and a comment on each denormalised column saying whether it is a snapshot or a cache.
- Attempt to insert a line whose tenant_id differs from its invoice's and confirm the foreign key rejects it.
- Rename a tenant, re-render a sealed invoice, and confirm the rendered name is the one stored on the header.
- Recompute amount_minor from quantity times unit_price_micros for a thousand synthetic lines rounding half-up, sum both ways, and record the divergence from the stored half-even values.
Expected result: tenant_id kept and enforced by a composite foreign key to (invoice_id, tenant_id); amount_minor kept because rounding happens once with a stated mode; legal name and billing address snapshotted onto the invoice header and never updated; mutable operational attributes left normalised. Each decision paired with its read pattern, its anomaly, and the mechanism that prevents the anomaly.
Verify it yourself
- A line cannot be attached to an invoice belonging to another tenant, and the rejection comes from the database rather than the application.
- Re-rendering a sealed invoice after editing the tenant record produces identical output.
- Rounding per line and rounding once at the end differ measurably over a thousand lines, which is the argument for storing the value.
- No column on the line is recomputed by any reader.
Follow-up
- Write the composite foreign key and the unique constraint it requires on the parent. What does it cost on every line insert, and what does it do to a bulk load?
- A tenant is renamed after being invoiced. Which rows change, and what does the customer see on last quarter's PDF?
- Where does currency live, and what breaks if a tenant's billing currency changes between two periods?
Find the join that inflates every invoice total
Medium · join fan-out · aggregation grain · sql correctness
invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.
Approach
- Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row:
sum(l.amount_minor)is the true total multiplied by the attempt count, andcount(p.attempt_id)is attempts times lines. Three lines and two attempts report double the money and six attempts. - Reject the reflex repair.
count(distinct p.attempt_id)does fix the count, because attempt_id is unique.sum(distinct l.amount_minor)does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match. - Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
- Keep invoices with no attempts by making the attempt branch a LEFT JOIN with
coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about. - Push each filter to its own grain:
where l.voided_at is nullbelongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants. - Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.
Follow-up
- Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with
filter (where ...)be better? - Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
- The same report is needed per tenant per month. What index makes the line CTE cheap?
Explain the core differences between HTML, CSS, and JavaScript in a production environment.
Medium
Approach
- Clarify what is being asked and what a complete answer contains.
- State your assumptions explicitly before working the problem.
- Say what you would check first and why it is the highest-information step.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
How did you handle the constraints provided in your technical assignment?
Medium
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?
Why did you choose specific technologies for your project architecture?
Medium
Approach
- Work from the requirement backwards to the design.
- Say what you would check first and why it is the highest-information step.
- State your assumptions explicitly before working the problem.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
A resumable usage export that never skips a row
Easy · pagination · indexing · tenant scoping · visibility
Customers pull their own rows from usage_event through GET /v1/usage to reconcile against their own systems. The table is append-only and partitioned daily on ingested_at; a large tenant adds millions of rows a day while the export is being walked, and a client may pause for hours and resume. Specify the cursor, the index it requires, the tenant scoping, the ordering guarantee you can honestly offer, the per-page cost as the walk deepens, and what the client must do to avoid missing rows.
Approach
- Rule out LIMIT/OFFSET on two independent grounds and say both, because fixing only one leaves the other. Correctness: rows inserted between page requests shift the window, so a walking client skips rows and repeats others, which for a reconciliation consumer is silent data loss rather than an error anyone sees. Cost: the database still produces and discards the skipped rows, so page N costs time proportional to N x page_size and a deep page degrades from milliseconds to seconds.
- Use keyset pagination over a stable, unique, indexed ordering: WHERE tenant_id = $1 AND (ingested_at, event_id) > ($2, $3) ORDER BY ingested_at, event_id LIMIT $4, carrying the last row's pair as the cursor. The row-value comparison navigates a composite btree directly, so each page is O(log n + page_size) and stays constant as the walk deepens. The precondition is that the cursor columns never change value for a row, which ingested_at satisfies and updated_at would not.
- Lead the index with tenant_id - (tenant_id, ingested_at, event_id) - which is simultaneously the correctness guard and the plan choice. An index on (ingested_at) alone forces a filter across every tenant's rows, and on a table where one tenant holds most of them that is fine only for that tenant and terrible for everyone else. Because ingested_at is also the partition key, a resumed cursor prunes to the partitions from the cursor forward.
- Name the visibility hazard rather than assuming it away: in PostgreSQL now() is transaction start time, so a transaction that starts at T, inserts, and commits at T+8 s writes a row whose ingested_at is T but which becomes visible only at T+8 s. A walker that has already passed T never returns it. The gap equals the writer's longest transaction, so the mitigation is either a safety lag - serve only rows older than now() minus the longest permitted transaction - or a client that re-walks a trailing overlap window and deduplicates on event_id, which is stable and unique.
- Offer the guarantee you can actually keep: ordering by (ingested_at, event_id) with no claim whatsoever about occurred_at order, and completeness only behind the safety lag or with the documented overlap-and-dedup obligation on the client. Saying this in the API reference is part of the design, because the client's reconciliation logic is what has to absorb it.
Worked solution (20 min)
- Write the keyset query and the exact index it needs, then read the query plan and confirm there is no Sort node.
- Insert rows concurrently while walking with a LIMIT/OFFSET pager and count the distinct rows returned against the rows that exist; repeat with the keyset pager and compare.
- Construct the out-of-order commit case by hand: open a transaction, insert, hold it open while the walker passes that timestamp, then commit, and check whether the walker ever returns that row.
- Choose the mitigation - safety lag or client overlap plus event_id dedup - and write down the number it depends on.
Expected result: Keyset pagination on (tenant_id, ingested_at, event_id) with a matching composite index, constant per-page cost and no sort; the guarantee limited to ingested_at ordering with nothing claimed about occurred_at; completeness delivered by a safety lag equal to the longest permitted write transaction, or by a documented client overlap window with deduplication on event_id.
Verify it yourself
- The plan for a page should be an index scan with no Sort node. A Sort means the index column order does not match the ORDER BY and the constant-cost claim is false.
- Under concurrent inserts the offset pager and the keyset pager return different sets. Count the difference rather than reasoning about it.
- Your safety lag must be at least the longest transaction the writing service permits. If you cannot name that number, the completeness guarantee is not enforceable and should not be documented.
Follow-up
- The customer wants to reconcile by occurred_at instead of ingested_at. What breaks, and what would you offer them in its place?
- One tenant starts a full-history export. How do you keep it from occupying every connection in the pool?
- A customer reports a missing row. What do you check first, and what would each answer tell you?
One tenant's counter writes stall the whole connection pool
Medium · lock contention · connection pool · multi-tenancy
A change that made a per-tenant usage counter correct now produces site-wide latency whenever one large tenant writes: unrelated endpoints time out waiting for a connection while database CPU stays low and no statement is slow. The change wraps the counter update in a transaction that takes SELECT ... FOR UPDATE on one row, calls an external pricing service, then updates and commits. Give an ordered checklist, the arithmetic that bounds that tenant's write rate, and three repairs with the cost each one accepts.
Approach
- Separate waiting from working. Low database CPU alongside high application latency points at a queue, so instrument connection-acquisition wait separately from query execution time; that queue forms in the application and is invisible in database metrics, which is why the database looks healthy throughout.
- Confirm the lock rather than assuming it: sample waiting sessions and group by wait event, relation and tuple. Contention concentrated on one tuple belonging to one tenant is the signature; a deadlock would instead show the database aborting transactions after its detection timeout, which is not happening here.
- Do the arithmetic out loud. Throughput on a serialised row is one divided by the lock hold time, and the hold spans the external call, so a 20 ms pricing call caps that tenant near 50 writes per second no matter how many pods run. Every waiter also holds a pooled connection while it queues, so the shared pool drains and unrelated tenants fail at acquisition.
- Repair one: shrink the critical section to a single statement with the price resolved before the transaction opens. Cost is a stale price for the duration of one request and a second round trip; benefit is a hold time measured in the database's own execution time.
- Repairs two and three change where the contention lives rather than how long it is held. Sharding the counter into per-(tenant, bucket) rows and summing on read multiplies write throughput by the shard count, at the cost of an aggregate on every read and a shard count you must size against the largest tenant rather than the median. Accumulating in memory and flushing periodically removes the per-write round trip entirely, paid for with a bounded loss window on crash, which is acceptable for a rate limiter and not for a billing counter.
- Contain independently of which repair wins: a separate pool or per-tenant concurrency cap for this write class, a statement timeout low enough that a pathological query dies before it accumulates waiters, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks.
Follow-up
- What would a genuine deadlock look like here, which two code paths would produce one, and how does the database's response differ from what you observed?
- If a transaction-pooling proxy sits in front of the database, which of your three repairs changes behaviour, and what stops working that would have worked on a direct connection?
- The counter also enforces a quota. Why is SELECT the count and then INSERT still wrong after you have fixed the contention?
Suggested practice checklist
Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.
Day 1 — Diagnostic, scored before you study anything
- Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
- Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
- Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.
Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.
Day 2 — Largest gap: find the boundary rather than the subject
- Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
- Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
- Re-attempt one of them from blank four hours later with nothing open.
Deliverable: A sub-skill map with the two blocking sub-skills circled.
Day 3 — Drill the blocking sub-skill by repeating the shape
- Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
- State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
- Have someone else read your one-sentence rule and find the precondition you left out.
Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.
Day 4 — Second gap, plus maintenance on the strongest area
- Run the same sub-skill decomposition on the second-largest gap in half the time.
- Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
- Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.
Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.
Day 5 — The gap that is not a skill
- Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
- Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
- Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.
Deliverable: Two recordings with a counted reduction in time-to-first-question.
Day 6 — Retest under day-one conditions
- Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
- For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
- Write down which single block you would still lose the offer on.
Deliverable: A second scored rubric placed beside the first, with one named remaining risk.
Day 7 — Full loop under interview conditions
- Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
- Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
- Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.
Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.
Bounding a vague ticket before writing code
When the requirements were thin, the interesting part is how you fenced the problem off: the assumption you wrote down, who you got to confirm it, the narrow version you shipped first so the rest stayed cheap to change. Guessing and being right is luck. Guessing in writing, where someone could correct you, is method.
How do you approach learning new technologies when faced with an unfamiliar task?
Medium · behavioural and engineering judgement
Approach
- Close with what you would do differently, concretely.
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
Tell me about a time you had to troubleshoot a bug in your code.
Medium · behavioural and engineering judgement
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
Why are you interested in joining Katomaran Technologies Private?
Medium · behavioural and engineering judgement
Approach
- Close with what you would do differently, concretely.
- Name the disagreement and how you resolved it with evidence.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?