A Software Engineer at Pragmatic Techsoft delivers high-performance solutions for the company's clients. As a team member, you are responsible for translating complex business requirements into scalable code, keeping Pragmatic Techsoft's software architecture efficient and maintainable. Your work directly influences the success of the company's product deployments for its clients.
The environment at Pragmatic Techsoft is one of continuous technical growth and rigorous problem-solving. You will be expected to balance immediate project deliverables with long-term architectural health, often working with enterprise-grade databases and sophisticated application frameworks. This role is ideal for engineers who thrive on technical depth and take pride in crafting elegant solutions to intricate logic challenges.
Pragmatic Techsoft places a high premium on database proficiency. Whether you are working on Odoo development or custom applications, your ability to model data and write efficient queries will be a central theme of your daily work.
Online Assessment
reportedInput bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.
What to demonstrate
- Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
- Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
- Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
- Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply
How to prepare
- For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
- For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
- Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last
Technical Rounds
reportedWhat this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.
What to demonstrate
- Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
- Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
- Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly
How to prepare
- Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
- Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
- Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live
Advanced System Design
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
PracHub editorial advice for the preparation topics above.
Treating a timed-out write as a failed write
A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.
Serialising a tenant's writes through select ... for update on a single counter row
It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.
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.
Never running a concrete value through the code
Trace one small input and one edge input by hand, index by index, out loud. Re-reading your own code catches design mistakes; walking a real value through it catches the off-by-one, the uninitialised accumulator and the loop that never advances.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Seal an hour under late data with bounded memory
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?
Find peak concurrent sandbox usage from run intervals
Given up to 5 million job_run rows for one tenant over one day, with run_id, started_at, finished_at, status and wall_clock_limit_seconds, report the maximum number of sandboxes running at once, the earliest instant that maximum is reached, and the first run_id that would breach a per-tenant cap of C. started_at is null while a run is queued; finished_at is null both for runs still executing and for runs in status lost. Treat a run as occupying [started_at, finished_at). Give the complexity and state how you handle each null.
Approach
- Turn each run into two sweep events,
(started_at, +1)and(end, -1), then sort the 2n events by timestamp with-1ordered before+1at equal timestamps. That tie-break is what makes the interval half-open, so a run finishing at 10:00:00 and one starting at 10:00:00 never overlap. - Decide each null out loud before sweeping, because each choice moves the answer. A null
started_atmeans queued and contributes nothing. A nullfinished_atwith statusrunningorleasedis clipped to the window end. Statuslosthas no observed end at all, so clip it atstarted_at + wall_clock_limit_secondson the grounds that the supervisor owns the timeout, and record that you did. The table'scheck (finished_at is null or started_at is not null)guarantees you never see an end without a start. - Sweep once, maintaining a running counter, the maximum, and the timestamp at which the maximum was first attained (update
peak_atonly on a strict increase, or you will report the last such instant instead of the earliest). Capture the firstrun_idwhose+1takes the counter to C+1 during the same sweep rather than in a second pass. - Complexity: O(n log n) dominated by the sort, O(n) space. If rows already arrive ordered by
started_at, a min-heap of end times gives O(n log k) time and O(k) space with k the peak concurrency, which is the better shape when the rows come from an index scan on(tenant_id, started_at). - If second resolution is acceptable, counting-sort the endpoints into an 86,400-slot delta array and prefix-sum it: O(n + T) time and O(T) space, which beats the comparison sort at 5 million rows. It answers only at second granularity, so state which resolution the cap is defined in.
Worked solution 20 min
- Write the null policy as three lines of prose first, one per case, and keep them beside the output.
- Emit 2n endpoint tuples
(timestamp, delta, run_id)and sort on the key(timestamp, delta)so-1precedes+1. - Sweep, tracking
cur,peak,peak_atupdated only on a strict increase, and the firstrun_idwhose+1takescurto C+1. - Build a fixture with two runs where one ends exactly when the next starts, three genuinely overlapping runs, one run with a null
finished_atand statusrunning, and one with statuslostand a 300-secondwall_clock_limit_seconds. - Re-run with every timestamp shifted by a constant and confirm the peak is unchanged while
peak_atshifts by the same constant.
Follow-up
- Now report peak concurrency per tenant for 10,000 tenants from one globally sorted stream. What changes about memory and about the sort?
- The cap has to be enforced at dispatch rather than reported afterwards. What does the admission check look like, and where does it race?
- How would you answer 'peak concurrency within any 5-minute window' without re-sorting?
Order a job dependency graph and find its critical path
A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.
Approach
- Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
- Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
- Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order:
earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E). - Second pass in reverse topological order for
latest_finish, thenslack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain.slack[v] = 0is exactly the statement that some longest path runs through v; equivalently, the longest path through v has lengthT - slack[v]. - The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by
min(d, T - L_avoid(v)), whereL_avoid(v)is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, whilemin(10, 15 - 14)predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan ismax(T - d, L_avoid(v)). - Compute
L_avoid(v)the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, sinceL_avoid(v)only ever matters through that max: setduration[v] := 0, recompute the makespan asT0(v) = max(T - duration[v], L_avoid(v)), and the gain ismin(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.
Follow-up
- Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
- Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
- Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?
Find the join that inflates every invoice total
invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.
Approach
- Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row:
sum(l.amount_minor)is the true total multiplied by the attempt count, andcount(p.attempt_id)is attempts times lines. Three lines and two attempts report double the money and six attempts. - Reject the reflex repair.
count(distinct p.attempt_id)does fix the count, because attempt_id is unique.sum(distinct l.amount_minor)does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match. - Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
- Keep invoices with no attempts by making the attempt branch a LEFT JOIN with
coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about. - Push each filter to its own grain:
where l.voided_at is nullbelongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants. - Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.
Worked solution 25 min
- Create one invoice with three lines of 1000, 1000 and 500 minor units and two payment attempts, then run the original query.
- Confirm it reports 5000 and 6 rather than 2500 and 2.
- Apply
sum(distinct l.amount_minor)and confirm the total becomes 1500, which is worse rather than better. - Write the two-CTE version with a LEFT JOIN and coalesce, and confirm 2500 and 2.
- Add a second invoice with lines and no attempts and confirm it still appears.
Follow-up
- Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with
filter (where ...)be better? - Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
- The same report is needed per tenant per month. What index makes the line CTE cheap?
Rebuild an hourly rollup with deduplication and late-arrival accounting
From usage_event (event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at, ingested_at), produce the values usage_rollup_hourly should hold for one tenant over one day: per (workspace_id, sku, hour_start) the deduplicated quantity_sum, event_count and source_max_ingested_at, bucketed by occurred_at. Duplicates share (tenant_id, idempotency_key). Also report, per hour, the running total across the day and the share of quantity that arrived more than two hours after the hour began. Write the query, and state which duplicates a daily unique index cannot catch.
Approach
- Deduplicate in its own CTE before any aggregation, because a SUM cannot be un-summed:
row_number() over (partition by tenant_id, idempotency_key order by ingested_at, event_id) = 1. Include the tiebreaker. Without it the surviving row is non-deterministic when two duplicates share an ingested_at, and a rollup described as deterministically recomputable then disagrees with itself between runs. - Bucket on occurred_at and nothing else, and pin the timezone explicitly.
date_trunc('hour', timestamptz)truncates in the session's TimeZone setting, so the same query run by a session set to a non-UTC zone buckets differently; use the three-argumentdate_trunc('hour', occurred_at, 'UTC')on PostgreSQL 16 or later, ordate_trunc('hour', occurred_at at time zone 'UTC') at time zone 'UTC'before that. Filterenvironment = 'production'explicitly, since metering covers three environments and billing covers one. - Aggregate to the grain with
sum(quantity),count(*)andmax(ingested_at). The last is not decoration: it is the watermark the row consumed up to, and without it there is no way to prove afterwards what a number did and did not include. - Compute the late share inside the dedup-and-aggregate step as a conditional aggregate,
sum(quantity) filter (where ingested_at > hour_start + interval '2 hours'), then divide by the hour's total. Compute the running total as a window over the already aggregated rows:sum(quantity_sum) over (partition by workspace_id, sku order by hour_start rows between unbounded preceding and current row). Running either over raw rows puts the duplicates back. - Answer the index question exactly. The unique constraint is on (ingested_day, tenant_id, idempotency_key), because a unique index on a partitioned table must contain the partition key. It therefore deduplicates only within one ingest day and admits a duplicate whose retry crosses midnight or whose replay runs a week later. That is why this CTE dedups across the whole window being recomputed, and why the dedup horizon is a correctness parameter rather than a retention cost.
- Keep the numeric type all the way through. quantity is numeric so the sums are exact; a cast to double precision anywhere in this pipeline reintroduces drift that surfaces only as a few unreconcilable cents per tenant per month, long after the query is out of anyone's mind.
Follow-up
- A dispute forces the same recompute over 40 days for one tenant. What changes about the dedup CTE's memory use and the chosen plan, and what would you do about it?
- Two runs a minute apart return different quantity_sum values for an hour that is already closed. Give two mechanisms that produce that, and the single query that distinguishes them.
- Express the same rollup incrementally so it does not re-scan the day each time the watermark advances. What does the incremental version stop being able to answer?
Explain the concept of abstraction and provide a code example.
Explain the concept of abstraction and provide a code example.
Approach
- Work from the requirement backwards to the design.
- 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
- What assumption would you test first?
- How would you know your answer was wrong?
Describe the core differences between various Java collections.
Describe the core differences between various Java collections.
Approach
- Say what you would check first and why it is the highest-information step.
- Work from the requirement backwards to the design.
- 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?
How do you swap two variables without using a third variable?
How do you swap two variables without using a third variable?
Approach
- Work from the requirement backwards to the design.
- 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
- What assumption would you test first?
- How would you know your answer was wrong?
Write a program to find the second largest number from an array.
Write a program to find the second largest number from an array.
Approach
- State your assumptions explicitly before working the problem.
- Work from the requirement backwards to the design.
- 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?
Publish limit semantics that clients can back off against
The gateway enforces three separate limits: a per-tenant token bucket, a monthly plan quota, and a cap on concurrent in-flight requests. Tenants hold several credentials and call from three regions, and today they see undifferentiated 429s. Design the client-facing contract: which headers carry which limit, the status codes that distinguish slow down from out of plan allowance from too many in flight, and what a well-behaved SDK does for each. Also state what the gateway returns when the shared counter store is unreachable, and what bound that choice implies.
Approach
- Separate the three objects before naming a header. A bucket refills continuously, a quota does not refill until the period rolls, and a concurrency cap clears when an in-flight request finishes. They have different remedies, so they cannot share a status code.
- Map them accordingly: 429 with Retry-After for the bucket, where waiting works; 429 with a distinct code and a reset hint for concurrency, where waiting works only if something else completes; and 402 or 403 with a plan code for quota, where waiting never works and the SDK must surface it rather than sleep on it.
- Emit RateLimit-Limit, RateLimit-Remaining and RateLimit-Reset for the bucket only, and document Remaining as advisory. It comes from a shared counter that other requests are changing concurrently, so a client treating it as a reservation has built a race.
- State the aggregation in the docs: the limit applies per tenant across every credential and every region. The cheap per-pod bucket of rate/N is correct only when traffic spreads evenly, and a tenant whose connections land on a few pods is throttled well under its published limit while a widely spread one exceeds it.
- Decide the degraded mode in advance and publish it. Failing open serves unmetered traffic during a counter outage; failing closed converts a counter outage into a total outage. Pick one, bound it, for example a local fallback bucket at a fraction of the limit for the duration, and say so in the contract.
- Specify SDK behaviour: honour Retry-After over local backoff, apply full jitter otherwise, never retry 402 or 403, and cap total attempts so retries expire before the caller's own deadline.
Worked solution 25 min
- Write a three-row table of limit, refill behaviour, status code, headers, and the client's correct action.
- Write the exact header set for one throttled response and one quota-exhausted response, showing that they differ.
- Write the documentation sentence that states aggregation across credentials and regions, and the sentence marking Remaining advisory.
- Choose and justify the counter-store failure behaviour, then state the numeric bound it puts on overage or on availability.
- Write the SDK's decision function: given status and headers, return sleep duration or surface to the caller.
Follow-up
- Two services under the same tenant each read Remaining = 50 and each send 50 requests. What did the contract promise, and what actually happens?
- How would you keep a burst from a staging workspace out of the production workspace's share of the same tenant's bucket?
Regional error rate explodes after a dependency merely slows
A control-plane read replica in one region degrades from 4 ms to 120 ms. Within ninety seconds that region's gateway error rate rises from 0.01% to 40% and its p99 becomes bimodal, one mode near the old p99 and one at the client timeout. The other two regions are unaffected. The gateway retries control-plane reads three times with exponential backoff and no jitter. Give an ordered checklist that separates trigger from amplifier, the offered-load arithmetic, and the controls that break the loop.
Approach
- Split the incident into three questions before touching a control: what started it, what amplified it, and what would make recovery slow. Here they are the replica slowdown, the retry policy interacting with queueing, and a synchronised unjittered herd at recovery. They are different mechanisms and each needs its own control.
- Read the distribution rather than the mean. A bimodal p99 with one mode pinned at the client timeout is two populations, not one degraded path; split latency by cache hit and miss and confirm the fast mode is hits and the timeout mode is misses that reached the replica.
- Do the load arithmetic. Three retries turn one client request into up to four upstream requests, so offered load reaches roughly 4x on a dependency that is already slower, and it arrives at the worst moment. With utilisation approaching one, queueing delay grows superlinearly, which is why a 30x latency increase upstream does not produce a 30x increase downstream, it produces timeouts.
- Break the loop with controls that bound offered load rather than with more attempts: a concurrency limit on the control-plane client so at most N calls are in flight and the remainder fail fast, a circuit breaker scoped per dependency and region, and a retry budget capping retries at a small fraction of base traffic so amplification has a ceiling that does not depend on how many clients are retrying.
- Add full jitter to whatever retries survive, sleeping uniformly in [0, min(cap, base x 2^attempt)], so attempts de-correlate instead of arriving in waves aligned to the moment of failure.
- Decide the unreachable-dependency behaviour in advance, because it is the actual product decision underneath: serving from an expired credential cache keeps the product available while extending a revoked key's life past the stated bound, and failing closed converts a dependency degradation into a total outage. State the mode and the staleness number rather than letting the timeout choose.
Follow-up
- The replica recovers. Describe what happens in the first ten seconds with your controls in place versus without them.
- Which single metric would have paged before the error rate moved, and why is upstream latency by itself not it?
- Requests that fail fast under the concurrency limit still need an answer. What does the gateway return, and what does it do to the usage event it would otherwise have emitted?
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 ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.
Unblock an engineer on a job run that finished twice
An engineer two weeks into the team brings you a job_run row showing status succeeded with an exit_code written by a worker declared dead ten minutes earlier; the retry attempt also shows succeeded. They have spent a day adding logging and are no closer. You have twenty minutes and you do not want to take the keyboard. Describe how you unblock someone: the question you ask first, what you let them find themselves, the concept you name and when, and how you check the next day that they own the fix rather than having watched you produce it.
Approach
- Ask what they expect rather than what they see: which statement set status to succeeded, and what did it check before writing? That question points directly at the update's WHERE clause, which is where the answer lives, and it costs them nothing to answer, so it does not read as a test.
- Let them build the timeline themselves from the row: queued_at, started_at, leased_until, finished_at and worker_id, on both the original run and the retry. Two different worker_ids with a lease expiry between them tells the whole story, and they will see it before you say it.
- Name the concept once the evidence has earned it. A lease bounds time; it does not prevent a write. The store has to reject a stale writer, which means the update carries a fencing token the row compares — update job_run set status = 'succeeded' where run_id = $1 and lease_token = $2 and status = 'running' — and a long garbage-collection pause or a brief partition is enough to produce what they are looking at.
- Point at the second, less obvious half and let them decide it: 'lost' exists in the status enum precisely so a run whose worker vanished is not recorded as failed, because failed asserts an outcome nobody observed and the system then bills and retries on that assertion. Ask them what these two rows should have said.
- Leave them with the next step rather than the patch — a test that kills the first worker after the sandbox exits and before the row is written — and say when you are available again, so the offer is real rather than polite.
- Check ownership the next day by what they produced, not by asking if it went well: a test that reproduces the window proves they understood it; a test that only asserts the new WHERE clause proves they copied it. Ask them to explain it to a third person and listen for whether the explanation is theirs.
Follow-up
- They propose a longer lease instead of a token. What do you say, and what breaks when legitimate runs last thirty minutes?
- How can you tell whether your explanation landed or they simply deferred to you?
- The same engineer hits a variant of this next month. What did you fail to teach the first time?
Own the incident where invoices undercounted metered usage
A metering consumer acknowledged each batch before committing the fold into usage_rollup_hourly. A rolling deploy restarted consumers mid-batch for two hours; roughly 1.4M usage_event rows were acknowledged and never folded, and 61 invoices sealed against the resulting rollups before anyone noticed. Take the owner's role. Describe an incident of comparable blast radius you owned: how it surfaced, the query that sized the loss, what you stopped first, and how the money was corrected. Give a wall-clock timeline and one thing you got wrong while it was still live.
Approach
- Open with the invariant that broke and the direction of the error, because they determine everything else: acknowledging before committing makes the consumer at-most-once, so this loses events rather than duplicating them, and loss raises no error anywhere. A listener who hears 'we lost revenue silently' knows immediately why detection took two hours.
- Size it with a stated reconciliation rather than an adjective: sum(quantity) from usage_event grouped by (tenant_id, sku, hour of occurred_at) over the window, against usage_rollup_hourly.quantity_sum on the same keys, filtered to environment='production' because staging and sandbox are metered but not billed. Then bisect by hour and tenant until single cells explain the gap. Say how long that ran and whether a replica could serve it while the incident was live.
- Separate mitigation from fix and say which came first. Mitigation is holding the sealing job, because a sealed row is frozen by design and every minute of sealing converts a recoverable rollup into an invoice correction. The fix is moving the acknowledgement after the commit, which re-introduces duplicates that the dedup check on (tenant_id, idempotency_key) must now absorb.
- State the correction path in the domain's own terms: sealed periods are never edited, so each affected tenant gets an adjustment line on the next invoice with kind='adjustment' and voided_by_line_id pointing at the line it reverses, priced against the same rate tier and carrying the watermark it priced against. That is four separate numbers — tenants affected, minor units, the cycle the adjustment lands in, and when customers were told.
- Close on one prevention control with its cost, not five: a per-hour reconciliation comparing raw sum to rollup sum that pages above a threshold. Name the threshold and the false-page rate you accepted, because a detector nobody will keep staffed is not prevention.
- Name a mistake you made inside the response window — the wrong first hypothesis, a mitigation that made it worse — rather than a design mistake from six months earlier. That is the part candidates rehearse away and interviewers weight heavily.
Follow-up
- Your fix moves the acknowledgement after the commit. What breaks now, and what absorbs it?
- One undercharged tenant has since churned. Do you bill them, and who decides?
- How would you have caught this in ten minutes instead of two hours, and what would that detector cost you in pages per week?
Estimate a tenant-leading index migration you have never run
Someone needs a date. usage_event carries an index on (occurred_at) and needs (tenant_id, occurred_at); the largest tenant holds roughly a hundred times the median tenant's rows, the table is partitioned daily with years of retention, and you have never run a migration on a table this large. Give an estimate you would defend: how you decompose the work, the two or three numbers you would go and measure first, the range and confidence you state, and what you commit to when the person asking needs a single date today.
Approach
- Refuse the bare number and then give one anyway, in the form that is actually useful: a range plus the measurement that collapses it. 'Four to eleven days; one afternoon building this index on a restored copy of the largest partition takes that to within a day' is an answer, while 'it depends' is not.
- Decompose by failure mode rather than into equal chunks, because that is where estimates go wrong. On a partitioned parent you create the index ON ONLY the parent, build each partition's index with CREATE INDEX CONCURRENTLY, then ALTER INDEX ... ATTACH PARTITION, at which point the parent index becomes valid. CONCURRENTLY does not block writes but scans each partition twice, waits out older transactions, cannot run inside a transaction block, and on failure leaves an invalid index you must drop concurrently and retry.
- Name the two unknowns that dominate and price them: build time on one restored partition of realistic size, and whether the planner actually chooses the new index for the skewed tenant, since selectivity for a tenant holding most of the rows is a different question from selectivity for the median tenant. Both are half-day measurements against a replica, and both are cheaper than being wrong by a week.
- State the assumptions the range is conditional on, because that is what makes a slip a re-estimate instead of a credibility event: no partition above a stated row count, one concurrent build at a time so it does not compete with ingest for I/O, and an ingest backlog that can absorb the added write amplification while both indexes exist.
- Budget the step nobody budgets: verification and the old index's removal. Dropping the old index is fast, but deciding it is safe to drop means confirming no plan still uses it, and that confirmation waits on real traffic across a full weekly cycle rather than on your patience.
- Answer the single-date request honestly. Commit to a date for the first checkpoint — the measured build number from the replica — and to re-estimating on that date, and say plainly what you are not committing to yet. A date with a scheduled re-estimate is worth more to the asker than a confident wrong one, and you should say why in those words.
Follow-up
- The concurrent build fails half way through the largest partition. What is the state of the database and what do you do next?
- Your estimate slips by sixty percent. Which assumption broke, and at what point would you have known?
- The person asking needs the date for a customer commitment. Does your answer change?
- 01
An engineer two weeks into the team brings you a job_run row showing status succeeded with an exit_code written by a worker declared dead ten minutes earlier; the retry attempt also shows succeeded. They have spent a day adding logging and are no closer. You have twenty minutes and you do not want to take the keyboard. Describe how you unblock someone: the question you ask first, what you let them find themselves, the concept you name and when, and how you check the next day that they own the fix rather than having watched you produce it.
- 02
A metering consumer acknowledged each batch before committing the fold into usage_rollup_hourly. A rolling deploy restarted consumers mid-batch for two hours; roughly 1.4M usage_event rows were acknowledged and never folded, and 61 invoices sealed against the resulting rollups before anyone noticed. Take the owner's role. Describe an incident of comparable blast radius you owned: how it surfaced, the query that sized the loss, what you stopped first, and how the money was corrected. Give a wall-clock timeline and one thing you got wrong while it was still live.
- 03
Someone needs a date. usage_event carries an index on (occurred_at) and needs (tenant_id, occurred_at); the largest tenant holds roughly a hundred times the median tenant's rows, the table is partitioned daily with years of retention, and you have never run a migration on a table this large. Give an estimate you would defend: how you decompose the work, the two or three numbers you would go and measure first, the range and confidence you state, and what you commit to when the person asking needs a single date today.
Is this an official Pragmatic Techsoft interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Pragmatic Techsoft. Rounds and questions reflect what candidates have reported, not a process Pragmatic Techsoft 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 interview?
Dedicate at least two to four weeks to reviewing core CS fundamentals, specifically focusing on SQL queries and Java programming, as these are the most common areas of assessment.
PracHub interview research ↗What differentiates a successful candidate from others?
Successful candidates don't just provide the correct answer; they explain their methodology, discuss potential trade-offs, and demonstrate a deep understanding of why their solution is the most efficient.
PracHub interview research ↗What is the culture like at Pragmatic Techsoft?
Candidates describe Pragmatic Techsoft as a result-oriented company that emphasizes technical excellence and professional growth. Engineers there are expected to be self-starters who take pride in the quality of their work.
PracHub interview research ↗Can I expect to meet with senior leadership?
Yes, it is common for the final technical rounds to be conducted by senior developers or even the CEO, especially when the role requires a high level of technical responsibility.
PracHub interview research ↗Sources & methodology 3 sources ↗
No official company page is cited. Rounds and questions come from candidate reports and PracHub editorial material; each source shows the date it was read.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
PracHub page · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-22
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