At iTech Robotics and Automation, the Software Engineer role is the bridge between conceptual industrial automation and high-performance execution. You are not just writing code; you are architecting the logic that drives complex assembly lines, battery manufacturing, and electric vehicle (EV) infrastructure. Your contributions directly affect the efficiency and precision of large-scale robotic systems and support the company's mission to advance automated manufacturing.
This role requires a unique blend of software proficiency and hardware awareness. You will work within interdisciplinary teams, collaborating with mechanical and controls engineers to solve high-stakes challenges in real-time environments. Whether you are optimizing a PLC program or designing software architectures for battery assembly, the work is intellectually demanding and offers the opportunity to see your code manifest in physical, industrial-scale robotics.
Technical Screening
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
System Design Interview
reportedAn unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.
What to demonstrate
- Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
- Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
- Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating
How to prepare
- Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
- Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
- Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes
Past Project Experiences
reportedYou cannot drill a format you do not know, so put the preparation into material that travels. Three pieces of your own work, each rehearsed until you can take a follow-up you did not anticipate, will carry a conversation or a code walkthrough equally well. Specificity is what separates that from filler. A number needs its definition before it means anything: a p99 is over some window and measured at some hop, and a server-side figure excludes the queueing and network time a client would see. The number you cannot qualify is the one to leave out.
What to demonstrate
- Whether your examples carry detail only someone who did the work would hold, such as what the binding constraint actually was, which alternative you rejected and why it was worse, and what you measured on each side of the change
- Whether a number survives one follow-up, meaning you can say what it was measured over and whether it moved because of your change or merely alongside it
- Whether a failure is described with the specific change that followed it, rather than a lesson stated in general terms
- Whether your part in a team effort is stated accurately, including what other people did
How to prepare
- Write a page on each of three projects covering the constraint, the option you rejected, the measurement before and after, and what went wrong. Cut any line you cannot take a follow-up on, since you are writing the parts you will be pressed on rather than a summary.
- Recover the real figures while you still have access: request volume, data size, latency with its percentile and window, team size, timeline. Note where each came from, whether a dashboard, a design document or memory, and mark the estimates so you can say which they are out loud.
- Take your weakest project story to someone who works in a different area and have them ask why four times in succession. The point where you run out of answer is the part to go and re-read before the round.
Culture Alignment
reportedThis round is deciding whether a change you make without supervision can be allowed to reach production. It is scored on what you knew at the moment you decided, not on how it turned out, so a story that opens with the result and works backwards reads as luck retold as judgement. Say what the options were, what you did not know, what you did to shrink the unknown before committing, and what you accepted as the worst plausible case. The detail that separates answers is a bound: how many users, how much data, and for how long, if you had been wrong.
What to demonstrate
- Whether the reasoning you give was available at the time you decided rather than after the result came in, since a story whose deciding evidence arrived later describes an outcome and not a judgement
- Whether you can put units on the exposure (users, rows, minutes of degraded service) and whether the containment you chose actually bounded it: a canary bounds the request path it fronts, while a background job writing to a shared table reaches every user regardless of which version served their requests
- Whether the reversal path existed before you shipped or was improvised during the incident, and whether it restores state or only stops further damage
How to prepare
- For your three largest changes, write down the one thing you would have had to be wrong about for it to fail, and what your best estimate of it was on the day you shipped. If you never held an estimate, that is the gap the follow-up questions will find
- Write the undo procedure for one of those changes as it existed at the time, then mark which steps restore data and which only stop new damage. Turning a flag off or reverting a deploy ends the new writes; rows already written come back only from a copy you kept, and a dropped column comes back empty unless something outside the schema holds the values
- Rehearse one story from the decision point forward and stop before the outcome, then have someone ask what you would do next. If the story only works with the ending attached, it is an anecdote rather than a decision you can defend
PracHub editorial advice for the preparation topics above.
Floating-point quantities and implicit unit-of-measure conversions
Binary floating point cannot represent most decimal fractions exactly, so a chain of pallet-to-case-to-each conversions accumulates a residue that a later rounding turns into a unit that was created or destroyed, breaking conservation with no concurrency involved at all. The bug is slow and non-local: it surfaces as a cycle-count variance weeks later, at a node nobody changed, and it is untraceable because the arithmetic looked correct at every individual step. The fix is to hold quantities as integers in the smallest transacting unit, carry the UoM explicitly on every row that carries a number, store conversion factors as integers that are versioned in time, and reject a conversion that does not divide exactly instead of rounding it. The same discipline applies to money on the freight side, where an accessorial in floating-point dollars produces invoice reconciliation differences of a cent that cost more to investigate than the freight.
Holding a database transaction open across a physical or third-party operation
It is natural to open a transaction, lock the position, call the rating or tendering API, and commit on the response, and it works perfectly until the partner's p99 goes from 200 milliseconds to thirty seconds. At that point every request holding a lock on a hot item-node pair queues behind it, the connection pool fills with transactions that are waiting on the network rather than on the database, and an unrelated service sharing the pool fails at the same moment. In this domain the effect is amplified because demand concentrates on a few hot rows during a promotion or a seasonal peak, exactly when partner latency is also degraded. The structural fix is to keep transactions short and local - commit the state change together with an outbox row, let a relay perform the external call, and reconcile asynchronously - accepting at-least-once delivery and making the effect idempotent rather than trying to stretch a transaction over something the database cannot roll back.
Assuming fixed-width integer arithmetic cannot overflow
In languages with fixed-width integers, including C, C++, Java, Go and Rust, computing a midpoint as (lo + hi) / 2 overflows once the sum passes the type's maximum, so write lo + (hi - lo) / 2 instead. Say which language you are in: arbitrary-precision integers, as in Python or Ruby, remove this specific hazard and none of the others.
Reading the constraints as preamble rather than as part of the problem
The bounds are usually there to eliminate the obvious approach: n up to 10^5 makes an O(n^2) scan roughly 10^10 operations, far outside any per-test time budget, and an input larger than memory rules out loading it at all. When a bound is not given, ask for it, then say out loud which approach it kills.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Trace a suspect lot forward to every shipped order line
lot_edge(parent_lot_id, child_lot_id) has 400 million rows with btree indexes on both (parent_lot_id, child_lot_id) and (child_lot_id, parent_lot_id). lot_shipment(lot_id, order_line_id) maps lots to shipped lines. Given one suspect received lot, return every downstream order_line_id. The reachable subgraph can reach 2 million lots and 12 levels; the full edge table does not fit in memory. Give the traversal, its complexity in terms of the reachable subgraph, and what you do if the data contains a cycle.
Approach
- Genealogy is a directed acyclic graph, not a tree: a production run consumes several lots and one lot splits across many shipments, so a node is reached by several paths and a tree traversal revisits it combinatorially. Breadth-first from the suspect lot over the
(parent_lot_id, ...)index, with a hash set of visited lot ids, is O(V' + E') where V' and E' count only the reachable subgraph — 2 million 8-byte ids is 16 MB of payload and a few tens of MB with open-addressing overhead, which is why the visited set is affordable and the edge table's 400 million rows never enter memory. - Batch each level into one indexed read. Issue
WHERE parent_lot_id = ANY($1)with a few thousand parents per call instead of a query per node: at 12 levels and 2 million nodes, per-node round trips dominate the wall clock entirely, and the recall's deadline is a real-world response time, not a query-plan aesthetic. The batch size is the tuning knob between round-trip count and per-statement planning cost. - Keep the visited set even though the schema says DAG. Rework and repack loops — material returned into an earlier lot — do occur, and an unguarded traversal on a cyclic edge set does not terminate; a set membership test is O(1) and turns a hang into a correct answer plus a data-quality alert. Also cap the traversal by level and by node count, and fail loudly on the cap rather than returning a partial set silently.
- Join to
lot_shipmentonce at the end over every visited lot id, not over the leaves and not per node. A lot is routinely both shipped and consumed: half a lot goes out on an order line while the other half is repacked into a child lot, so it has alot_shipmentrow and outgoing edges at the same time and is not a leaf. Restricting the join to nodes with no outgoing edges drops exactly those order lines, which is under-recall — the one failure mode this trace is not allowed to have. One bulk join against an index onlot_idover the whole visited set is a single pass; per-node lookups multiply the round-trip problem by the number of visited lots. - State the semantics honestly: reachability answers which shipped lines could contain material from the suspect lot, and without propagating edge quantities it is a superset of physical containment. For a recall that is the correct direction of error — over-recall, never under-recall — and saying so is the point, because a trace that returns quickly and incompletely will be believed. The upstream question, which received lots are in a suspect shipped unit, is the same traversal over the reverse index, which is why both indexes exist.
Worked solution 30 min
- Draw a six-lot fixture containing a merge (two parents into one child), a split (one parent into three children), and one interior lot that is partly shipped on an order line and partly consumed into a child lot, then write down the expected reachable set and the expected order lines by hand.
- Implement level-batched BFS with an explicit frontier list, a visited hash set, and a per-level batch read.
- Add a cycle to the fixture and confirm the traversal terminates and flags it rather than hanging.
- Bulk-join
lot_shipmentonce over every visited lot id — the whole visited set, not the leaves — and count the queries issued, confirming it is one per level plus one. - Re-run with the frontier order reversed to confirm the output set is unchanged.
Follow-up
- A recursive CTE would express this in ten lines. What is its cost here if the index on the traversal column is missing, and how would you notice?
- You need the quantity of suspect material in each downstream line, not just reachability. What has to be on the edge, and what does that do to the traversal?
- The traversal must run while a production run is writing new edges. What does a consistent answer mean here, and what do you cut it at?
Fold late out-of-order status events into a monotonic leg state
Leg statuses form the chain planned < tendered < accepted < picked_up < in_transit < delivered, with exception and cancelled outside it. You ingest 300 million observation_event rows a day carrying subject_id (leg), observed_status, occurred_at, received_at, clock_offset_ms and dedupe_key; five percent arrive over an hour late and a tail arrives a day late. Write the merge that produces each leg's current status, prove the result depends only on the set of events and not their order, and state the per-event cost.
Approach
- Define the per-leg state as a tuple ordered totally:
(status_rank, occurred_at, source_priority, event_id), and definemerge(a, b)as the maximum under that order. Rank leads;occurred_atand the rest only break ties. The proof is then one line: maximum over a total order is idempotent, commutative and associative, so the state space withmergeis a join-semilattice and the fold over any permutation of the event set yields the same value. That is also why no global sort is needed — the O(n log n) disappears and the fold is O(1) per event, O(distinct legs) state. - Rank before time, deliberately.
occurred_atis device-reported and sometimes wrong; if it led the comparison, the leg's status would be handed to whichever device has the most badly skewed clock. Ranking first means a bad clock can only reorder events within one rank. Correctoccurred_atbyclock_offset_mswhere it was measured, and where the offset exceeds a threshold keep the observation but bar it from winning a tie — the event is still evidence, it just stops being authority. - Make the lattice total, because
exceptionandcancelledare not points on the chain andmaxover a partial order is undefined for incomparable pairs. Modelcancelledas an absorbing top element, and carryexceptionas a boolean flag merged by OR alongside the rank rather than as a rank of its own — otherwise an exception at rank 4 either blocksdeliveredat rank 6 or is erased by it, and both are wrong. A product of semilattices is a semilattice, so the order-insensitivity argument still holds over the tuple. - Deduplicate on
dedupe_keyeven though the merge is idempotent. Idempotence covers a byte-identical redelivery; a partner replaying a window with a freshevent_idis not identical, and the tie-break would let the replay displace the original for no reason. Dedupe on the stable key, then fold — both halves are needed, and conflating them is the usual defect. - Partition the stream by
subject_idso all events for one leg land on one partition: broker ordering guarantees are per-partition only, and while a commutative merge makes interleaving harmless, a concurrent read-modify-write on the same leg row is still a lost update. Where that cannot be guaranteed, make the write a compare-and-set on a version column, or a conditional update that only raises the state. Because the state is mergeable, shard maps combine cleanly and replaying the whole day converges to the same answer — which is what makes reprocessing safe. - Say where genuine regressions live: a delivered shipment that is refused and returns is a new leg, not a rank decrease. Monotonicity is only sound if real reversals are modelled as new objects, and keeping every superseded observation is what lets you answer the dispute later, so the fold discards nothing — it marks losers
superseded.
Follow-up
- An event arrives a week late for a leg already closed and invoiced. Does it change the status, the on-time metric, both, or neither?
- Two events for one leg have the same rank and the same
occurred_atfrom different sources. What decides, and is that decision stable across a replay? - How would you test order-insensitivity rather than assert it?
Find the hottest write keys under a fixed memory budget
One day of writes to inventory_movement is 200 million rows; each carries (item_id, node_id). Distinct pairs reach 60 million. Return the 200 pairs with the most writes so they can be moved behind a single writer before the next peak. First give the exact answer assuming you may hold 60 million counters, then give an answer under a 200 MB budget, and state the error guarantee that second answer carries. Both must be one pass over the day's rows unless you argue for a second.
Approach
- Exact version: one hash pass to build
(item_id, node_id) -> int64, then a single scan of the map maintaining a min-heap of size K = 200, pushing and popping when the incoming count exceeds the heap root. O(n + d log K) with n = 2e8 and d = 6e7, versus O(d log d) if you sort all 60 million entries — a factor of roughly log d / log K, about 3.4x, for no benefit. The heap must be a min-heap: the root is the weakest survivor, so eviction is O(log K). - Cost the map honestly. 60 million entries at a 16-byte key, an int64 count and per-entry overhead is 2-3 GB in most runtimes, which is the whole reason for the second version rather than a theoretical aside.
- Bounded version: Misra-Gries with m counters. Keep at most m keys with counts; on a key already present increment it; on a new key when fewer than m are held, insert at 1; otherwise decrement every held counter and drop those reaching zero. With m = 1,000,000 at roughly 40 bytes per counter, that is about 40 MB, comfortably inside 200 MB alongside the heap.
- State the guarantee as a number, not as 'approximate'. Misra-Gries returns an estimate satisfying f - n/(m+1) <= f_hat <= f, so with n = 2e8 and m = 1e6 every count is an undercount by at most 200. If the 200th hottest pair has hundreds of thousands of writes, a 200-count error cannot reorder the top of the list — and if the counts near rank 200 are within 200 of each other, the ranking there was never meaningful for the decision being made. The sketch also merges: run one per shard, combine, prune back to m, and the bound holds with n the combined length.
- If exactness is genuinely required, argue for two passes rather than more memory: the Misra-Gries pass yields a candidate superset of size m, and a second pass counts only those candidates exactly in O(m) memory. That is the standard way to get an exact top-K under a memory bound, and the cost is one extra read of the day's rows.
Follow-up
- The decision downstream is which keys get a single writer. Does a rank error at position 190 change that decision, and what does the answer tell you about how much accuracy to buy?
- Write volume on one key is bursty — a promotion concentrates it into twenty minutes. Does a daily top-K find it, and what window would?
- How do you keep the sketch across a worker restart without replaying the day?
Stop two allocators promising the same unit of stock
inventory_position is keyed (item_id, node_id, lot_id, state) and holds qty BIGINT, which is on-hand stock and is moved only by the movement ledger; available_qty BIGINT, maintained on the same row; and version. The declared invariant is available_qty = qty - SUM(allocation.qty) over that key in status held or committed, enforced by CHECK (available_qty >= 0) and CHECK (available_qty <= qty). allocation holds allocation_id, order_line_id, item_id, node_id, lot_id, qty, tier, status (held, committed, consumed, released, expired) and expires_at. Twelve allocator threads try to claim the last eight units of one item at one node inside the same 50 ms. Give the exact statements that let at most eight units be claimed, then say what changes if identical code runs under PostgreSQL REPEATABLE READ versus InnoDB REPEATABLE READ, and what a retried allocate for the same order line must do.
Approach
- Collapse the check and the write into one statement whose WHERE clause is the business rule: UPDATE inventory_position SET available_qty = available_qty - :q, version = version + 1 WHERE item_id = :i AND node_id = :n AND lot_id = :l AND state = 'on_hand' AND available_qty >= :q. qty is deliberately untouched: a hold is a promise, not a physical move, so only a movement may change on-hand and the ledger-equals-position invariant survives. Rowcount zero is a refusal, not an error, and the allocation row is inserted in the same transaction only when the rowcount is one.
- If the deployment carries no available_qty column and availability must be derived, the alternative is SELECT ... FOR UPDATE on the position row followed by the SUM over held and committed allocations inside the same transaction. That is correct but holds the row lock for two statements instead of one, which matters on a key taking hundreds of writes a second.
- Isolation is not a substitute for the atomic step, and it changes the error contract. Under PostgreSQL READ COMMITTED the losing UPDATE blocks, then re-evaluates its predicate against the newly committed row version and returns rowcount 0, which the handler already handles. Under PostgreSQL REPEATABLE READ the same statement instead raises SQLSTATE 40001 and the whole transaction must be retried, so a handler that only tests rowcount fails closed on an exception it never expected. InnoDB REPEATABLE READ does neither: a locking read or UPDATE performs a current read of the latest committed row, so the loser blocks and then sees rowcount 0, with a deadlock rather than a serialization failure as its exceptional case.
- Make the retry a no-op with a partial unique index: UNIQUE (order_line_id, node_id, lot_id) WHERE status IN ('held','committed'). Then a retried allocate conflicts instead of double-claiming, and the handler reads the existing allocation and returns it.
- Never source the availability number from a replica or a cache for this path. Replication lag is largest during the write burst that made the key hot, so the guarantee fails precisely when it is load-bearing.
- For the hottest item-node pairs, shed load without overselling: route allocations for that key through a single writer with a bounded queue and refuse with a conservative not-available when the queue is full. Splitting the position into N sub-rows cuts contention N-fold but strands the remainder across sub-rows and produces false out-of-stock when the last units are spread thin.
Follow-up
- One order line needs two units from two different nodes, atomically. What do you do when those positions live in different shards and no distributed transaction is available?
- Soft-tier allocations expire. Show the sweeper statement, and say what stops it from releasing an allocation that is being committed at that instant.
- Under sustained contention your CAS retries. What is the retry budget, and what does the caller see when it is exhausted?
Attribute multi-stop freight cost without fanning it out
shipment_leg(leg_id, shipment_id, sequence_no, linehaul_cost_cents, accessorial_cost_cents, actual_arrive_at) joins to fulfilment_order_line through leg_order_line(leg_id, order_line_id, allocated_weight_g). One multi-stop truckload leg carries 40 order lines. The current report SUMs linehaul_cost_cents after joining all three tables and returns roughly forty times the real freight spend. Write the query that returns cost per order for one month, explain the fan-out in terms of the join's row count, and make the per-order amounts sum exactly to the leg totals in integer cents.
Approach
- Name the cardinality: joining shipment_leg to leg_order_line produces one row per (leg, line) pair, so the result has SUM over legs of lines-per-leg rows rather than one row per leg. SUM(linehaul_cost_cents) over that result counts each leg's cost once per line it carries, which is why 40 lines gives 40 times the spend. The bug is not the SUM, it is aggregating a parent measure across a child-expanded row set.
- Compute a share before aggregating anything. In one pass: leg.linehaul_cost_cents * b.allocated_weight_g / SUM(b.allocated_weight_g) OVER (PARTITION BY b.leg_id), which keeps the denominator local to the leg and needs no self-join.
- Do the division in integer cents and settle the remainder deliberately. Floor each share, then distribute the leftover cents one each by descending remainder, ordered by a stable tiebreak such as order_line_id. Floating-point shares leave a residue that makes the freight accrual miss by a cent per leg, which costs more to investigate than the freight.
- Aggregate to the order only after the share exists: GROUP BY order_id over the per-line allocated cents. Restrict the month on shipment_leg.actual_arrive_at, not on the order date, so a leg is attributed in the period it actually ran.
- Index for the shape of the query: leg_order_line needs (leg_id) for the window partition and (order_line_id) for the reverse lookup, and shipment_leg needs an index on actual_arrive_at to bound the month without a full scan.
- Apply the same treatment to accessorial_cost_cents separately. Accessorials such as a detention charge often belong to one stop rather than to the whole leg, so weight-sharing them across every line on the load misattributes cost even though the total ties out.
Worked solution 30 min
- Build a fixture with one leg at 100,003 cents carrying three lines weighted 1000 g, 1000 g and 1 g.
- Write the naive join and observe the total reported as three times the leg cost.
- Rewrite using the SUM(...) OVER (PARTITION BY leg_id) share, floor to integer cents, and add a largest-remainder pass ordered by remainder then order_line_id.
- Aggregate to order level and compare the grand total against SELECT SUM(linehaul_cost_cents + accessorial_cost_cents) FROM shipment_leg for the same window.
Follow-up
- A line is cancelled after the leg ran. Does its share of the freight disappear, move to the remaining lines, or stay as a cost with no revenue?
- Weight is the wrong basis for a light bulky pallet. What would you use instead, and does the reconciliation check still hold?
- The carrier invoice arrives three weeks later and differs from the rated cost. Where does the correction land, and does it re-run this allocation?
These questions evaluate how you structure complex systems to ensure s…
These questions evaluate how you structure complex systems to ensure scalability, reliability, and maintainability.
Approach
- Choose a partition key and say what query it makes expensive.
- Name the read and write paths separately; they rarely have the same bottleneck.
- Fix the scope first: who calls this, how often, and what they do when it fails.
Follow-up
- What breaks first when traffic grows ten times?
- What would you drop to keep the system up under load?
How do you balance modularity against performance constraints in your …
How do you balance modularity against performance constraints in your code?
Approach
- Name the read and write paths separately; they rarely have the same bottleneck.
- Name the failure you are designing for, then the recovery path.
- Choose a partition key and say what query it makes expensive.
Follow-up
- What breaks first when traffic grows ten times?
- What would you drop to keep the system up under load?
How would you design the software architecture for an EV battery assem…
How would you design the software architecture for an EV battery assembly line?
Approach
- Choose a partition key and say what query it makes expensive.
- Name the read and write paths separately; they rarely have the same bottleneck.
- Fix the scope first: who calls this, how often, and what they do when it fails.
Follow-up
- What breaks first when traffic grows ten times?
- How does this behave when that dependency is down for an hour?
How do you manage version control and deployment in a production facil…
How do you manage version control and deployment in a production facility?
Approach
- Name the read and write paths separately; they rarely have the same bottleneck.
- Name the failure you are designing for, then the recovery path.
- Fix the scope first: who calls this, how often, and what they do when it fails.
Follow-up
- What would you drop to keep the system up under load?
- What breaks first when traffic grows ten times?
Explain the process of troubleshooting a robotic system that has stall…
Explain the process of troubleshooting a robotic system that has stalled mid-sequence.
Approach
- State your assumptions explicitly before working the problem.
- Clarify what is being asked and what a complete answer contains.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
This category tests your foundational knowledge of industrial automati…
This category tests your foundational knowledge of industrial automation, control systems, and software lifecycle management within a manufacturing context.
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
- How would you know your answer was wrong?
- What assumption would you test first?
Gateway that books carriers at most once through timeouts
The Carrier Integration Gateway buys bookings and labels from 30 partners. Peak is 300 booking calls per second, partner p99 is 3 seconds, and stalls of several minutes occur. Each partner admits roughly 10 requests per second and rejects the rest. A booking is physically irreversible: a duplicate is a second trailer and a second invoice. Design the call path and the retry path so one shipment leg is booked at most once, including what is recorded before the call and how an ambiguous timeout is resolved.
Approach
- Generate booking_idempotency_key and persist it on shipment_leg with status requested before the socket is opened. The durable row is the memory, not the process, which is what makes a crash mid-call recoverable. Send the key to the partner where their contract supports one, but never depend on their deduplication: their window is short, their key is often derived from fields you may legitimately change, and older integrations have no such concept.
- Add an explicit unknown status between requested and confirmed, entered on any timeout, connection reset or 5xx. Nothing on the request path may re-issue from unknown. A sweeper reads rows in unknown and asks the partner about that key - a lookup by reference, or the next manifest - and re-issues with the same key only when the partner affirmatively has no such booking.
- Make invocation asynchronous from the caller. The orchestrator commits its saga step and an outbox row in one local transaction; a relay publishes; the gateway consumes. That is at-least-once publication, not exactly-once, so the gateway consumer must itself deduplicate on (leg_id, booking_idempotency_key) - the outbox moves the problem, it does not remove it.
- Size the queues from the numbers. By Little's law, 300 calls per second held 3 seconds is 900 concurrent calls, so connection and worker budgets are per-partner and total 900 plus headroom. A partner receiving 100 per second of demand against a 10 per second limit accumulates 90 per second of backlog: 5,400 queued after a minute, which is a 9-minute drain at their limit. So every partner gets a bounded queue with a token bucket, and overflow sheds by class - rate quotes dropped, tenders deferred against their cut-off, and a tender that will miss its physical departure escalated to an alternate carrier instead of waiting.
- Separate retryable from terminal. A partner rejection - embargoed lane, undeliverable address, expired account - is not a retry candidate; it goes to a dead-letter queue keyed by leg with the partner's raw error preserved, and the orchestrator sees a terminal failure it can re-source rather than a call that quietly never completes.
Worked solution 30 min
- Draw the leg's booking state machine - planned, requested, unknown, confirmed, failed - and label each edge with the exact event that causes it.
- Write what is committed before the call and what after, identify the crash window between them, and show the sweeper resolving it without a distributed transaction.
- Compute in-flight concurrency from Little's law and per-partner backlog growth at the stated limit, then set queue bounds and shed classes from those numbers.
- Work the three ambiguous outcomes - client timeout, reset after the request was sent, 500 after a partner-side commit - and state the sweeper's action for each.
Follow-up
- A partner has no idempotency concept and no lookup by reference. What is left to you, and what do you hand to operations?
- A stall lasts 40 minutes and 60,000 legs are queued. What do you cancel, what do you keep, and how do you avoid a thundering retry the moment the partner recovers?
- Where is the line between the gateway retrying and the orchestrator re-sourcing, and who owns the cut-off decision?
Wave release screen slows linearly with the line count
A wave-release screen lists open fulfilment_order_line rows for one node and, per line, its held allocation rows and the on_hand inventory_position quantity for that (item_id, node_id, lot_id, state). p95 is 40 ms for a 25-line wave and 3.4 s for an 800-line wave. The database reports 1,601 statements for the 800-line request, each under 1 ms, and no individual query is slow. Diagnose the cause, give the fix, and state the statement count and p95 you expect afterwards.
Approach
- Read the shape out of the numbers before forming a theory. 1,601 statements for 800 rows is 1 + 2N: one driver query plus two per line. Every statement being sub-millisecond rules out a bad plan, so the time is round trips, not work.
- Check that the arithmetic accounts for the whole regression. 1,600 round trips at roughly 2 ms each is about 3.2 s against a 40 ms baseline, which covers the observed 3.4 s. If it had only covered half, there would be a second defect and the N+1 fix alone would disappoint.
- Identify the two per-row statements by their normalised text: a single-row allocation lookup keyed by order_line_id and a single-row inventory_position lookup keyed by the four-column primary key. Confirm they are lazy relationship loads by removing those two fields from the response and watching the count fall to 1.
- Replace them with two batch statements. Collect the order_line_ids and item_ids from the driver query, then SELECT ... FROM allocation WHERE order_line_id = ANY($1) AND status IN ('held','committed') and SELECT ... FROM inventory_position WHERE node_id = $1 AND item_id = ANY($2) AND state = 'on_hand', and join them into maps in application memory. Three statements per request, independent of wave size.
- Verify the access paths rather than assuming them. The partial unique index on allocation (order_line_id, node_id, lot_id) WHERE status IN ('held','committed') leads with order_line_id, so it serves the first batch. inventory_position's primary key is (item_id, node_id, lot_id, state), so the second batch scans on the item_id, node_id prefix and filters on state; lot_id sits between them and is unconstrained, which is fine for a few hundred keys but is the reason this is a prefix scan rather than a point lookup.
Follow-up
- The screen adds a per-line column showing the last movement's reason_code. How do you fetch that for 800 lines without reintroducing the fan-out?
- A wave can now reach 12,000 lines. At what point does ANY($1) stop being the right tool, and what replaces it?
- Nothing here was a slow query. What monitor would have caught this before an operator did?
Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Coding, one pass at shallow depth
- Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
- For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
- Fix nothing today. The value of the pass is the unfixed record.
Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.
Practice prompt ↗Practice prompt ↗Practice prompt ↗Worked solution ↗02Design, one pass at shallow depth
- Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
- After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
- Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.
Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.
Practice prompt ↗Practice prompt ↗Practice prompt ↗03Fundamentals and the practical rounds
- Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
- Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
- Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.
Deliverable: Eight scored short answers and one written reading of unfamiliar code.
Practice prompt ↗Practice prompt ↗04The rounds that are about you, and the map
- Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
- Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
- Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.
Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.
Practice prompt ↗Practice prompt ↗Worked solution ↗05First chosen area, to the depth you skipped
- Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
- After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
- Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.
Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.
Practice prompt ↗Practice prompt ↗06Second chosen area, where the gap is coverage rather than speed
- Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
- Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
- Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.
Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.
Practice prompt ↗Practice prompt ↗07Reassemble the loop
- Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
- Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
- Reduce the week to one page holding only the rules you can state without reading them.
Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.
Practice prompt ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
A migration is a cost you chose to pay, not an achievement. The story is what the old system made expensive, what you measured before committing, what kept serving traffic during the cutover, and what you would have done if the numbers had come back flat. Without those, a rewrite reads as taste.
Describe a situation where you encountered a critical system failure; …
Describe a situation where you encountered a critical system failure; how did you identify the root cause?
Approach
- Name the disagreement and how you resolved it with evidence.
- Give the blast radius: what could have broken, and what you measured.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
How do you handle competing project priorities in a fast-paced enginee…
How do you handle competing project priorities in a fast-paced engineering environment?
Approach
- Close with what you would do differently, concretely.
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
Resolve a review disagreement over clamping a position
A pull request changes the pick path from a plain decrement to qty = GREATEST(qty - :n, 0) on inventory_position, because the CHECK (qty >= 0) constraint has been failing a few times a day and paging the on-call. The author argues the page is noise and a negative on-hand is meaningless anyway. You disagree. Describe a code review disagreement you had: the argument you made, how you kept it about the system rather than the person, what you conceded, and how it ended. State what the clamp destroys and what you proposed be done with the failures instead.
Approach
- Name the information the clamp destroys, in one sentence: the constraint failure is the only signal that the ledger and the shelf disagree, and GREATEST turns a detectable variance into a silent one that surfaces later as a picker standing at an empty location for an order the system promised.
- Concede the real half of the author's case immediately, because it is real: paging a human on a data condition that a human cannot fix at 3am is bad design. The disagreement is not whether the page should stop but whether stopping the page should also stop the evidence.
- Propose the version that satisfies both: catch the constraint violation at the application boundary, write a cycle_count_adjust movement with its own reason code and reference so the correction is a first-class row rather than an invisible edit, fail the pick task with a specific error the floor can act on, and move the alert from a page to a variance report per item-node pair.
- Anchor the argument in the append-only model rather than in taste. Positions are derived from inventory_movement, so a clamp makes the position stop equalling the sum of its movements and the reconciliation job now reports a permanent, unexplainable variance on that key.
- Say how you handled the disagreement mechanically: the comment thread, then a short synchronous conversation once it became clear the thread was circling, and a written decision so the next person reading the clamp knows why it is not there.
- State the concession you actually made, including if you lost. Losing a review and writing the detector anyway is a legitimate and common outcome worth describing plainly.
Follow-up
- The clamp has already been in production for a month. How do you find out how much stock it silently absorbed?
- Where should the adjustment movement get its reference, given that no count sheet exists at pick time?
- The author is more senior than you and unconvinced. What is your next move?
- 01
Describe a situation where you encountered a critical system failure; how did you identify the root cause?
- 02
How do you handle competing project priorities in a fast-paced engineering environment?
- 03
A pull request changes the pick path from a plain decrement to qty = GREATEST(qty - :n, 0) on inventory_position, because the CHECK (qty >= 0) constraint has been failing a few times a day and paging the on-call. The author argues the page is noise and a negative on-hand is meaningless anyway. You disagree. Describe a code review disagreement you had: the argument you made, how you kept it about the system rather than the person, what you conceded, and how it ended. State what the clamp destroys and what you proposed be done with the failures instead.
Is this an official iTech Robotics and Automation interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at iTech Robotics and Automation. Rounds and questions reflect what candidates have reported, not a process iTech Robotics and Automation 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?
The interviews are designed to be challenging but fair. They focus on practical, real-world scenarios rather than abstract puzzles, so focus on your hands-on experience and problem-solving process.
PracHub interview research ↗What is the typical timeline for the hiring process?
While it can vary based on business needs, the process typically moves from the initial screen to a final decision within 3–5 weeks. Candidates report that the process is kept efficient.
PracHub interview research ↗How much preparation time do you recommend?
Plan for at least 1–2 weeks of focused preparation, specifically reviewing your past projects and refreshing your knowledge of the core technical domains relevant to the position.
PracHub interview research ↗What differentiates a successful candidate?
Successful candidates demonstrate a blend of technical mastery and a pragmatic, collaborative mindset. Being able to explain "why" you made a design choice is often more important than the specific tool you used.
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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