As a Software Engineer at PDF Solutions, you play a pivotal role at the intersection of advanced analytics and semiconductor manufacturing. The company provides critical software and hardware solutions that enable semiconductor manufacturers to increase yield and reduce time-to-market. Your work directly impacts how data is transformed into actionable intelligence, influencing the efficiency of complex production environments.
This role is both technically demanding and strategically significant. You are not merely writing code; you are building systems that solve high-stakes engineering problems, often involving large-scale data processing, hardware-software integration, or specialized domain-specific applications like e-beam optics or semiconductor process control. Candidates who thrive here are those who enjoy bridging the gap between theoretical software architecture and the physical constraints of manufacturing hardware.
You will likely work in multidisciplinary teams, collaborating with domain experts, data scientists, and hardware engineers. Because operates in a highly technical niche, your ability to understand the "why" behind the code—specifically the physics and statistical foundations of the semiconductor industry—is what differentiates a good engineer from a great one. Expect a fast-paced environment where your contributions have immediate, visible impacts on product performance and manufacturing reliability.
Screening Call
reportedThe title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.
What to demonstrate
- Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
- Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
- Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year
How to prepare
- Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
- Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
- Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter
Technical Deep-Dives
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
Stakeholder Interaction
reportedWhen 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
Final Decision-Making
reportedNobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.
What to demonstrate
- Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
- Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
- Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
- Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience
How to prepare
- Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
- For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
- Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed
PracHub editorial advice for the preparation topics above.
Long-lived shared credentials for client systems
One static key used by the connector runtime, a debugging script and three engineers cannot be attributed to a person in an audit, cannot be expired at engagement close, and cannot be rotated without breaking an unknown number of consumers at once. The expensive part of the eventual incident is not the leak; it is that rotation requires finding every holder under time pressure, and nothing recorded who they were. Issue per-principal, per-target, short-lived grants from a broker so rotation is a no-op, attribution is automatic, and closure is a TTL rather than a search.
Tenant context leaking across a connection pool
Setting the tenant on a connection and relying on it for the rest of the request looks correct in every test that runs one request at a time. Under transaction pooling the connection goes back to the pool with that session state still attached, and the next checkout, serving a different client, inherits it; row-level security then enforces the previous tenant's policy flawlessly and returns the wrong rows. It reproduces only under concurrency, which is the load profile your integration tests do not have. Use a transaction-scoped setting, and assert that the setting matches the request's tenant immediately before the first query rather than trusting that it was set correctly upstream.
Abandoning working code to chase the optimal solution
Get the straightforward version correct, state its complexity, and only then optimise, keeping the working version until the faster one passes the same cases. A correct quadratic solution with a stated path to linear beats a half-written optimal one that never ran.
Treating a network call as though it were a local function call
A remote call can be slow, fail, or return after you stopped waiting, so name the timeout, the retry policy, and what the caller sees while the dependency is down. A call with no timeout turns one slow dependency into an exhausted thread or connection pool in every service upstream of it.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Assign connector bindings to rollout waves and name the cycle
Connector bindings declare ordering dependencies: an edge u -> v means v must not run in a wave earlier than one after u has succeeded. You have up to 50,000 bindings and 200,000 edges, and a misconfigured dependency may have created a cycle. Assign each binding the earliest wave number such that every predecessor sits in a strictly earlier wave, return the total wave count, and if a cycle exists return the bindings on one concrete cycle rather than a boolean. The implementation must be iterative: a dependency chain can be 50,000 deep.
Approach
- Build an adjacency list and an indegree array, then run Kahn's algorithm from every indegree-zero node. Bindings with no predecessors get wave 0. O(V+E) time, O(V+E) space.
- Relax waves as you decrement: when popping u, set wave[v] = max(wave[v], wave[u]+1) for each successor before enqueueing v at indegree zero. Because v is only enqueued after its last predecessor is popped, wave[v] is 1 + max over predecessors, which is the earliest legal wave. Total wave count is max(wave)+1, the longest path in nodes.
- If Kahn emits fewer than V nodes, the residual subgraph is exactly the nodes lying on or downstream of a cycle. Every residual node still has an in-edge inside the residual, otherwise it would have been popped, so walking backwards along in-edges cannot dead-end and must revisit a node within |residual| steps; the segment between the two visits is a concrete cycle. O(V) with a visited-position map.
- Use an explicit stack or a queue rather than recursion. Recursive DFS on a 50,000-node chain exceeds the default stack in most runtimes, and the failure looks like a crash rather than a cycle.
- At this size, recompute the whole plan whenever an edge changes. Incremental topological order maintenance costs far more code than the O(V+E) rebuild it saves.
Follow-up
- Two bindings land in the same wave but hit the same client endpoint, whose rate budget is shared. How does the wave plan have to change, and what problem class does that turn into?
- One dependency holds only for overlapping watermark ranges rather than globally. Is the dependency graph still a DAG over bindings, and what is the right vertex if not?
- Report the critical path so an operator can see which chain sets the wave count, not just that it is five.
Find peak window utilisation and the first rejected request
One connector binding's requests against a client endpoint arrive as timestamps t[0..n-1] in milliseconds, sorted non-decreasing, n up to 1,000,000. The endpoint's budget is R requests per rolling window of W milliseconds, and a request at time t is admitted only if strictly fewer than R already-admitted requests fall in the half-open window (t-W, t]. Return (a) the maximum number of requests falling in any W-millisecond window, counting every request regardless of admission, and (b) the index of the first request the limiter would reject. Both parts in O(n) time.
Approach
- Part (a) is a two-pointer scan: for each right index, advance left while t[right] - t[left] >= W, which leaves the window exactly (t[right]-W, t[right]], and track max(right-left+1). Left is monotone non-decreasing, so total work is O(n) amortised with O(1) extra space. The maximum over all windows of width W is attained at a window whose right edge is a request timestamp, so scanning right edges is sufficient.
- Part (b) is not the same scan. A rejected request consumes no budget, so the window must count admitted requests only. Keep a FIFO deque of admitted timestamps: pop from the front while front <= t - W, admit if the deque holds fewer than R, and return the first index where it does not. O(n) total pushes and pops, O(R) memory.
- Pin the boundary convention before writing either loop. With a half-open (t-W, t] window a request exactly W milliseconds after an earlier one does not see it; switching to a closed window changes the answer at every boundary and is where most disagreements with the client's own limiter come from.
- Note what fixed buckets of width W buy and cost: O(1) memory, but they admit up to 2R inside a span shorter than W, R at the end of one bucket and R at the start of the next. If buckets are unavoidable, halve the bucket width or use a weighted estimate across two adjacent buckets.
- If part (a) already exceeds R, the binding cannot run at this concurrency under any limiter shape. That is a scheduling problem, not a retry-policy problem, and no backoff tuning fixes it.
Worked solution 25 min
- Take W = 10,000 ms, R = 3, and timestamps in milliseconds: [0, 1000, 2000, 3000, 11000, 11500, 12000, 12500].
- Part (a): at right=3 (t=3000) the window (-7000, 3000] holds 4 requests. At right=7 (t=12500) advance left past 0, 1000 and 2000 to index 3, since 12500-3000 = 9500 < 10000; the window holds indices 3..7, which is 5 requests.
- Confirm no other right edge beats 5 by inspecting each, then record the peak.
- Part (b): admit index 0, 1 and 2 (deque grows to 3 entries, none expiring since t-W is negative). At index 3, t=3000: nothing to pop, deque size is 3, which is not strictly fewer than R=3, so reject.
- Check what happens after the rejection if rejected timestamps were wrongly enqueued: indices 4 onwards would also be refused despite the deque having expired entries by t=11000.
Follow-up
- Three bindings and four workers share this one endpoint's quota. Where does the counter live, and what is the correct behaviour when that store is unreachable?
- Add full-jitter backoff to the retry path and state precisely what it prevents that plain exponential backoff does not.
- The client documents the limit as 'about 100 per minute'. How would you measure the real one without tripping it in their production?
Measure credential exposure past engagement close without double counting
For one engagement you have credential_grant rows: (grant_id, principal_id, target_system_id, issued_at, not_after, revoked_at which may be NULL). The cutoff instant T is the engagement's ends_on plus access_grace_days. A grant is live over the half-open interval [issued_at, min(not_after, revoked_at)). There are up to 200,000 grants across target systems. For each target_system_id, return the total wall-clock time after T during which at least one grant was live, plus the disjoint segments that make it up. Grants that overlap must be counted once.
Approach
- Clip each grant to [max(issued_at, T), min(not_after, coalesce(revoked_at, +inf))) and discard any interval whose start is not strictly before its end. O(n), and it removes every grant that already expired inside the engagement window.
- Bucket the survivors by target_system_id and sort each bucket by start. O(n log n) overall, and sorting dominates the whole algorithm.
- Sweep each bucket once, holding one open segment [s,e): if the next start is greater than e, emit [s,e) and open a new segment, otherwise set e = max(e, next_end). O(n) after the sort, O(1) working state, O(number of emitted segments) output.
- Total exposure is the sum of emitted segment lengths, which is the measure of the union. Summing per-grant durations instead answers a different question and is unbounded above by the elapsed wall clock.
- Produce a second, conservative figure that ignores revoked_at and uses not_after alone. revoked_at records that you asked for revocation; whether the client's identity provider honoured it is not something this table knows, and a self-contained token validated offline stays valid to its own expiry regardless.
- If a bucket does not fit in memory, sort externally and sweep the stream, or push end timestamps into a min-heap keyed by end and pop those below the current start. Same O(n log n), memory proportional to the maximum number of concurrently live grants.
Follow-up
- Three grants overlap and one of them belongs to an offboarded contractor. What must the sweep emit so exposure can be attributed per principal?
- Which of your two numbers is the real bound on access if the client system validates tokens offline, and what does that imply about issuing TTLs in the first place?
- Run this across 20,000 engagements as a nightly job with a fixed memory budget. What changes in the shape of the computation?
Fix a billing report that double-counts through a join
A monthly report joins engagement to work_record (engagement_id, principal_id, work_date, minutes, is_billable, status) and to invoice_line (engagement_id, period_start, period_end, amount_minor, line_type, status) in a single FROM clause, then aggregates sum(minutes), count(DISTINCT principal_id) and sum(amount_minor) grouped by engagement. Finance reports the minutes are roughly triple the timesheets while the headcount column looks correct. Explain the arithmetic, then write the corrected query for one client and one period, returning zero rather than NULL for engagements with no work.
Approach
- Name the arithmetic before touching SQL: joining two independent one-to-many children of the same parent produces the Cartesian product per engagement, |W| x |L| rows. sum(minutes) is therefore multiplied by the invoice-line count and sum(amount_minor) by the work-record count. Three lines per engagement — fees, expense, credit note — is exactly the factor of three finance reported.
- Explain why the headcount column looked fine: count(DISTINCT principal_id) collapses the duplication, so it is correct by accident. That is precisely why the bug survived review, and it is the reason a column agreeing with expectations is not evidence that a join is at the right grain.
- Pre-aggregate each branch to the join grain first: one subquery per fact table, grouped by engagement_id, each producing at most one row per engagement. Join those results, which are now one-to-one, so no multiplication is possible by construction rather than by care.
- Drive the outer query from engagement with LEFT JOINs and wrap each aggregate in COALESCE, because sum over zero rows is NULL rather than 0. For counts, count the key column and not count(*), which returns 1 for a non-matching LEFT JOIN row.
- Keep credit notes inside the amount sum deliberately: amount_minor is negative for line_type = 'credit_note', so filtering them out overstates what was billed. Exclude status = 'void' instead, and state that choice in the query as a comment because the next reader will assume the opposite.
- Compare cost: the fan-out plan materialises |W| x |L| intermediate rows before aggregating, while pre-aggregation reads each table once under its own index and hash-joins two small results. The base-table I/O is identical; the intermediate work differs by orders of magnitude on a large engagement.
Worked solution 25 min
- Build a fixture: one engagement with two work records of 60 and 90 minutes and three invoice lines.
- Run the broken join and confirm sum(minutes) returns 450 rather than 150, and that count(DISTINCT principal_id) is unaffected.
- Rewrite with one aggregating CTE per fact table and LEFT JOIN both onto engagement.
- Add an engagement with invoice lines but no work records and confirm it returns 0 rather than NULL or a missing row.
Follow-up
- The report now needs a per-principal breakdown alongside the per-engagement totals. Does your shape survive, or do you need a different grain entirely?
- How do you stop the next analyst reintroducing this — a view, a naming convention, or a test that fails on a fixture with two children?
- Why does sum(DISTINCT minutes) not fix it, and what does it actually compute?
Return the latest attempt per connector idempotency key
connector_run stores one row per attempt: run_id, connector_id, idempotency_key, attempt_no SMALLINT, status (queued, running, succeeded, failed, ambiguous, cancelled, abandoned), failure_class, watermark_from, watermark_to, started_at, finished_at, unique on (connector_id, idempotency_key, attempt_no). For one connector_id, write the query returning the highest-numbered attempt per idempotency_key, keeping only those keys whose latest attempt is not succeeded, ordered by watermark_from. Give the index that serves it, and say what the plan does without that index.
Approach
- Use DISTINCT ON (idempotency_key) with ORDER BY idempotency_key, attempt_no DESC. PostgreSQL requires the ORDER BY to lead with exactly the DISTINCT ON expressions, and that requirement is what makes the row picked deterministic rather than whichever the scan happened to reach first.
- Apply the not-succeeded filter outside, in a wrapping SELECT. Putting it in the inner WHERE changes the question from 'keys whose latest attempt failed' to 'keys that have any non-succeeded attempt', which reports a key that failed twice and then succeeded, and is the single most common wrong answer here.
- Know the portable alternative: row_number() OVER (PARTITION BY idempotency_key ORDER BY attempt_no DESC) in a subquery, filtered = 1 outside. A window function cannot be filtered in WHERE because WHERE is evaluated before the window, and PostgreSQL has no QUALIFY clause, so the subquery is mandatory rather than stylistic.
- Index (connector_id, idempotency_key, attempt_no DESC). With connector_id fixed by equality, the index supplies the exact order DISTINCT ON needs and the Sort node disappears. It does not reduce rows read: through PostgreSQL 17 there is no loose index scan, so the executor still walks every attempt in range rather than skipping to each next distinct key.
- Give the complexity honestly: without the index the plan is a scan plus an O(n log n) sort over the connector's attempts; with it, O(n) ordered reads and constant working memory, where n is attempts in range, not distinct keys. If the distinct keys are a tiny fraction of attempts, a recursive skip-scan is the next step.
Follow-up
- attempt_no is not monotonic in time because a retry was queued out of order. Does your query still pick the right row, and which column should it actually order by?
- How would you return the latest attempt per key across all connectors in one environment without the query degrading to a full sort?
- An ambiguous latest attempt means the target may or may not have been written. What does this query feed, and should ambiguous be treated as failed here?
Can you discuss your background in e-beam optics?
Can you discuss your background in e-beam optics?
Approach
- State your assumptions explicitly before working the problem.
- Clarify what is being asked and what a complete answer contains.
- Say what you would check first and why it is the highest-information step.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
You are given a specific problem description; walk me through your ste…
You are given a specific problem description; walk me through your step-by-step solution.
Approach
- Work from the requirement backwards to the design.
- Say what you would check first and why it is the highest-information step.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Share one client's rate budget across every job you run
A connector_binding carries an endpoint, an auth method and a rate budget. Several bindings, an ad-hoc backfill and an operator's debugging script all hit the same client-owned system, whose quota is per client account and shared across every one of them. Your workers are horizontally scaled and stateless. The client returns 429 with Retry-After, and sometimes 503 with nothing. Design the internal contract governing outbound calls: where the limit is enforced, what a worker asks for and gets back, how a 429 propagates, and what a run does when it cannot get budget.
Approach
- Key the limiter on the resource actually being consumed: (client_id, target_system_id), never on worker, process or binding. A per-worker limit multiplies by the worker count, which is exactly wrong when the quota is one shared number, and it loosens silently the moment someone scales the deployment.
- Make budget an explicit acquisition with a deadline rather than an implicit sleep. A worker calls acquire(key, cost, deadline) against a shared token bucket — a Redis Lua script keeps the read-modify-write atomic — and receives granted, a wait duration, or refused. Return the refusal rather than blocking a worker for minutes; a blocked worker is a worker not doing another client's work.
- Add a concurrency bound alongside the rate bound, because they catch different failures. A token bucket happily permits a burst of fifty simultaneous in-flight requests that satisfies the per-second rate and still exhausts the client's connection pool. Hold a bounded lease per (client_id, target_system_id) with a TTL so a crashed worker's lease expires instead of wedging the binding.
- Treat the client's own signal as authoritative and shared. A 429 with Retry-After must pause every caller on that key, not just the run that received it, so the limiter takes a cooldown on the key itself. A 503 with no Retry-After gets exponential backoff with full jitter — sleep = random(0, min(cap, base·2^attempt)) — because synchronised retries from a dozen bindings after a client outage are a self-inflicted herd against the quota you are about to need.
- Adapt downward on evidence and upward only slowly. Cut the effective rate on repeated 429s, recover gradually, and record the measured sustained rate per binding: the documented limit and the enforced limit differ often enough that treating the client's documentation as fact is the specific error here.
- Define what a refused run is. It is not a failure: record it as queued or abandoned with its own reason, leave the watermark untouched, and let the scheduler re-claim the same range later. Counting budget exhaustion as a failure pollutes the per-binding failure rate you alert on and buries the client-owned problems underneath it.
Worked solution 30 min
- Write the limiter's key and list every caller that shares it; name one that is easy to forget.
- Write acquire()'s signature and its three outcomes, and say what the worker does with each.
- Compute the worst-case in-flight request count under a rate limit alone at your burst size, then add the concurrency bound that fixes it.
- Trace a 429 arriving at one of five concurrent runs on the same key and say what the other four do next.
- Choose the run status and failure_class for budget exhaustion, then check your alerting query still isolates client-owned failures.
Follow-up
- An operator's debugging script drains the whole budget at 09:00 and the scheduled runs starve. What in your design prevents that, and what priority scheme would you accept?
- How do you measure the client's real limit without tripping it during their business hours?
Connector workers are OOM-killed every day and a half
Connector runtime workers are OOM-killed after 18 to 40 hours and a restart clears it. RSS grows monotonically at about 1.9 MB per completed run, flat regardless of how many records that run moved, and workers serving more distinct bindings do not grow faster than workers serving one. Live heap measured immediately after a forced full collection rises with the same slope as RSS. Traffic is unchanged week over week. Give the ordered diagnosis, the specific object you expect at the root, and the fix.
Approach
- Prove retention before hunting for it. RSS and the post-collection live set rising with the same slope means the objects are reachable, which is a leak in the ordinary sense. RSS rising while the live set stays flat would be allocator fragmentation or native buffers, and that is a different investigation with different tools.
- Find the unit the growth is keyed on by division rather than by guessing. 1.9 MB per completed run, flat across record counts and flat across distinct bindings, points at something retained exactly once per run. Per-record growth would implicate an accumulating buffer; per-binding growth would implicate a legitimate cache that is merely unbounded.
- Diff two heap snapshots an hour apart by retained size and walk the dominator tree to a single root. Expect a module-level dictionary whose entry count equals the number of runs completed since process start, which is the observation that confirms step two.
- Read the key and the value separately, because they are two different bugs. The key contains something run-scoped, an idempotency key or a correlation id, which is why the hit rate is zero and the map only ever grows. The value is megabytes rather than kilobytes because the compiled schema validator closes over the first parsed page, so every entry pins a client payload as well.
- Fix both halves. Re-key on (connector_id, source_schema_version), whose cardinality is in the hundreds and stable. Cap the map with a fixed-size LRU so an unforeseen key can never make it unbounded again. Stop the closure capturing the page, so an entry costs a compiled schema rather than a schema plus a payload.
- Make the class of bug visible next time: export entry count and hit rate for every process-level cache. A zero hit rate is the unmistakable signature of a run-scoped key and appears within minutes, whereas this one took 18 hours to present itself as an OOM kill.
Follow-up
- Those retained pages are client payloads sitting in a long-lived shared process. What does that mean beyond memory?
- A worker killed mid-run leaves connector_run rows in 'running'. Who reconciles them, and how do you avoid applying that range twice?
- Same symptom, but the post-collection live set is flat and only RSS climbs. What are you looking at instead, and which tool do you reach for?
For someone fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Measure before reasoning
- Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
- Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
- Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.
Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.
Practice prompt ↗Practice prompt ↗Worked solution ↗02References, copies, and the bugs they produce
- Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
- Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
- Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.
Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.
Practice prompt ↗Practice prompt ↗03Types, once, in a language that checks them
- Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
- Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
- Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.
Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.
Practice prompt ↗Practice prompt ↗04Concurrency, starting with what actually runs at the same time
- Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
- Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
- Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.
Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Debugging as a procedure rather than an instinct
- Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
- Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
- Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.
Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.
Practice prompt ↗Practice prompt ↗06Tests that catch the bug you are about to write
- Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
- Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
- Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.
Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.
Practice prompt ↗Practice prompt ↗07Debug something broken, out loud
- Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
- Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
- Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.
Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
A migration is a cost you chose to pay, not an achievement. The story is what the old system made expensive, what you measured before committing, what kept serving traffic during the cutover, and what you would have done if the numbers had come back flat. Without those, a rewrite reads as taste.
How do you manage conflict within a technical team?
How do you manage conflict within a technical team?
Approach
- Give the blast radius: what could have broken, and what you measured.
- 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 would you do differently if you ran that again?
- How did you know your change caused the improvement?
Can you describe a time you handled a difficult situation with a colle…
Can you describe a time you handled a difficult situation with a colleague?
Approach
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
What is your experience with LabView and its application architectures…
What is your experience with LabView and its application architectures?
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
How would you explain your experience with semiconductor processes?
How would you explain your experience with semiconductor processes?
Approach
- Close with what you would do differently, concretely.
- Pick a story where you made the decision, not one where you watched it.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
- 01
How do you manage conflict within a technical team?
- 02
Can you describe a time you handled a difficult situation with a colleague?
- 03
What is your experience with LabView and its application architectures?
- 04
How would you explain your experience with semiconductor processes?
Is this an official PDF Solutions interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at PDF Solutions. Rounds and questions reflect what candidates have reported, not a process PDF Solutions has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How difficult are the technical interviews?
They are designed to be challenging but fair. They focus on practical problem-solving rather than rote memorization, so prioritize understanding concepts over memorizing syntax.
PracHub interview research ↗What is the typical timeline for the hiring process?
The process can take anywhere from a few weeks to a month. You can ask your recruiter for an updated timeline during your initial screen.
PracHub interview research ↗How should I prepare for the behavioral questions?
Use the STAR method (Situation, Task, Action, Result) to structure your answers. Ensure your examples are professional and highlight your specific contributions to team success.
PracHub interview research ↗Is there a preference for specific technical backgrounds?
General software engineering skill matters, but experience with hardware-software integration, statistical analysis, or semiconductor processes will set you apart.
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