A Software Engineer at SOLV Energy—specifically within the context of HV SCADA Engineer II roles—is a critical bridge between complex energy infrastructure and digital control systems. You are not just writing code; you are building the software solutions that monitor and manage high-voltage energy assets, ensuring the stability, efficiency, and reliability of renewable energy projects.
This role is highly impactful because your work directly influences how energy data is captured, analyzed, and acted upon in real-time. You will be expected to thrive at the intersection of software development and industrial automation. It is a position for engineers who enjoy solving tangible, physical-world problems and who want to see their code contribute to the larger mission of sustainable energy production.
Phone Screening
reportedBefore anything technical happens, someone has to decide which rung of the ladder your loop is calibrated to, and that decision sets the bar for every round after it. It comes from how you describe scope, not from your title, because titles do not convert cleanly between companies. The weak version of the answer is team size and years. The strong version names the largest change you shipped where nobody reviewed the design, what would have broken if you had been wrong, and what you were paged for. Get the level said out loud on this call, because the range and the loop both follow from it.
What to demonstrate
- Whether the scope in your own account maps onto a level the team actually has an opening at, so a mismatch ends the process cheaply rather than after four interviewers have spent a day
- Whether your title needs re-mapping: the same word describes very different amounts of independent decision-making at a twenty-person company and a ten-thousand-person one
- Whether your compensation expectation can be filled at that level in the structure the role pays in, which is why the number gets asked for before any engineer is scheduled
How to prepare
- Write down two changes from the last two years: the largest one you designed with nobody reviewing the design, and the largest one where someone more senior did. Lead with the first when scope comes up, and be ready to say which parts of the second were yours
- Ask which level the loop is calibrated to and what changes at the level above it, then plan your weeks from that answer rather than from the posting
- Settle a total-compensation range beforehand with the split named, base against bonus against equity and its vesting period, so a question about numbers gets a number instead of the word market
Technical Interview
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
Team-Focused Interview
reportedBecause the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.
What to demonstrate
- Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
- Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
- Whether you can be redirected in the middle of an answer without restarting it from the beginning
- Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it
How to prepare
- Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
- Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
- Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing
PracHub editorial advice for the preparation topics above.
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.
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.
Saying 'eventually consistent' without naming the anomaly a user would see
Describe the concrete symptom you are choosing to accept: the author reloads and their own comment is missing for two seconds, or two devices show different balances for a minute. The class of consistency model is a technical label; the tolerable anomaly is the actual product decision.
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.
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.
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?
Size dock doors from overlapping appointment windows
For one node and one local calendar day you have up to 200,000 planned dock appointments from shipment_leg: leg_id, planned_arrive_at, planned_depart_at (both TIMESTAMPTZ), plus the node's IANA zone. A door is occupied over [arrive, depart). Return the minimum number of doors that lets every appointment start on time, and the maximal interval over which that peak is sustained. Target O(n log n). Say how you derive the day's boundaries and how you break ties between an arrival and a departure at the same instant.
Approach
- Name the result before computing it: with one interchangeable door type, the minimum door count equals the maximum number of simultaneously occupied intervals. That equality is not a heuristic — interval graphs are perfect, so their chromatic number equals their clique number, and greedy left-to-right assignment achieves it. Say this, because it is what licenses solving a scheduling question with a counter.
- Emit 2n endpoints,
(t, +1)at each arrival and(t, -1)at each departure, sort byt, and at equaltorder the-1before the+1. Half-open occupancy means a trailer leaving at 10:00 frees the door for one arriving at 10:00; ordering arrivals first inflates the answer by exactly the number of back-to-back handoffs, which on a well-packed schedule is most of them. - Scan once with a running counter, tracking the maximum and the endpoint index where it was first reached. The peak is attained on a half-open interval between two consecutive endpoints,
[t_i, t_{i+1}), not at an instant — report it that way or the operations team cannot act on it. Extend the interval while the counter stays at the maximum. - Complexity: O(n log n) dominated by the sort, O(n) space. If rows already arrive ordered by
planned_arrive_atfrom an index, use the min-heap variant instead — push each departure, pop all departures at or before the current arrival, and the heap size is the current occupancy. Same time bound, O(peak) space rather than O(n), which matters when peak occupancy is 40 and n is 200,000. - Derive the day's boundaries from the zone, not from arithmetic on instants. Convert local midnight and the next local midnight to instants in the node's IANA zone; on a transition day that span is 23 or 25 hours, so adding 86,400 seconds silently drops or duplicates an hour of appointments. Decide explicitly what a zero-length dwell means — under half-open semantics it occupies nothing — and state it.
Worked solution 25 min
- Write the endpoint tuple with the tie-break baked into the sort key — for example
(t, kind)where departure sorts as 0 and arrival as 1 — rather than handling ties in the scan body. - Implement the scan, returning both the maximum and the
[t_i, t_{i+1})span at which it holds. - Build a fixture of six appointments where three are strictly back-to-back at the same minute, and hand-compute the answer before running.
- Add an appointment spanning the local DST transition and derive the day window from the zone, asserting the appointment count inside the window.
- Re-run with the input shuffled to confirm the result depends only on the set of intervals.
Follow-up
- Doors are typed: some break pallets, some are parcel-only. Does max-overlap still equal the door count?
- Appointments have a tolerance — a trailer may start up to 20 minutes late without penalty. How does that change the objective, and is it still solvable by a sweep?
- Half the appointments are actuals and half are plans. Which do you sweep for tomorrow's staffing, and which for last week's utilisation report?
Find the busiest ten-minute window per reader after deduplication
One day of observation_event from fixed readers: source_id (reader), subject_id (container), occurred_at, received_at, dedupe_key. 300 million rows, partner and device replays resend whole windows, and some events arrive six hours after they occurred. For each reader, return the 10-minute event-time window containing the most distinct containers, with the window's start. Target linear work per reader after sorting. State the window semantics you chose, and what the job does with an event that lands after its window was already reported.
Approach
- Deduplicate before windowing, on
dedupe_key, and keep the two concerns separate. A replayed batch is a no-op only if the duplicate is removed first; deduplicating inside the window logic counts a resend as traffic and reports a congestion peak that never happened. Dedupe is a hash set per reader-day or a sort-unique on the key — the same sort you need anyway. - Partition by
source_idand sort each partition byoccurred_at, never byreceived_at. Arrival order is an artefact of device connectivity and partner batch cadence; a reader that went offline for six hours would otherwise show its entire backlog as one impossible burst. Sorting dominates: O(p log p) per reader of size p, O(n log n) overall. - Two pointers over the sorted partition with a hash multiset
container -> countand a separatedistinctinteger. Advance the right pointer, incrementing the count and bumpingdistinctonly on a 0-to-1 transition; advance the left pointer whileoccurred_at[left] <= occurred_at[right] - 600s, decrementing and loweringdistinctonly on a 1-to-0 transition, and erasing the key from the map so the map's size anddistinctnever diverge. Each element enters and leaves once: O(p) amortised after the sort, O(window occupancy) space. - Justify the candidate set rather than scanning every possible start. With half-open windows
[t, t+600), sliding a window right until its left edge coincides with an event can only add events and never removes one, so the maximum is attained at some event'soccurred_at; checking the p windows anchored at events is sufficient and exhaustive. State the half-open convention explicitly — an event exactly att+600belongs to the next window, and flip-flopping on that produces off-by-one disagreements between this job and whatever reads its output. - Handle lateness by choosing a policy, not by hoping. Either run over a closed event-time batch with a stated lateness allowance — the day plus six hours, so the six-hour tail is inside the batch and the day-late tail is not — or emit under a watermark and restate the affected windows when a late event lands. Both are defensible; what is not defensible is silently dropping the late event, because that is how the busiest window quietly becomes the one with the best connectivity.
Follow-up
- Two events for one container arrive from two readers a second apart. Is that congestion, a misread, or one container passing two portals?
- You need this continuously rather than as a daily batch. What changes, and where does the memory go?
- A device's clock is three hours fast. What does that do to this result, and how would you detect it from
clock_offset_ms?
Add a required unit column and uniqueness to a live ledger
inventory_movement holds 4.2 billion rows on PostgreSQL 15 and takes about 2,000 inserts per second around the clock. Two changes are required with no downtime: add uom_code TEXT NOT NULL, whose truthful value must come from the conversion factors in force at each row's occurred_at rather than from today's item master; and make idempotency_key UNIQUE, although duplicates predating the retry fix already exist. Give the ordered steps, the lock each one takes, its expected duration and its rollback.
Approach
- ADD COLUMN with a non-volatile DEFAULT has been metadata-only since PostgreSQL 11, so no rewrite, but it still takes ACCESS EXCLUSIVE for an instant. That lock queues behind any long-running reader while blocking every statement behind it, so one 10-minute analytics scan turns a millisecond change into a 10-minute outage. Set lock_timeout to 2s and retry in a loop rather than waiting.
- Reach NOT NULL without a scan under an exclusive lock: ADD CONSTRAINT ... CHECK (uom_code IS NOT NULL) NOT VALID, which is instant, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE and runs concurrently with inserts, then SET NOT NULL, which PostgreSQL 12 and later proves from the validated check and so skips its own scan. Drop the redundant check afterwards.
- The truthful backfill is the long pole and must read the versioned conversion factor as of each row's occurred_at, not the current item master, or a supplier moving a case from 12 units to 10 retroactively reinterprets last quarter. Batch by primary-key range, a few thousand rows per transaction, with a sleep between batches and a resumable cursor in a control table. At 5,000 rows a second, 4.2 billion rows is roughly ten days of throttled work: that is a schedule, not a failure, and it must not block the deploy.
- The UNIQUE index cannot be built until the duplicates are gone. CREATE UNIQUE INDEX CONCURRENTLY aborts on the first violation and leaves an INVALID index behind that is ignored by queries but still carries update overhead on every insert, and it can only be removed with DROP INDEX CONCURRENTLY. So: find the duplicates, resolve each one as a compensating reversal rather than a DELETE, since every duplicate key is a double-applied movement and deleting it fixes the constraint while leaving the position wrong.
- Then CREATE UNIQUE INDEX CONCURRENTLY, which makes two passes over the table and waits out concurrent transactions, so budget hours here, and note it cannot run inside a transaction block: a migration framework that wraps every step in BEGIN will fail on this one. Promote it with ALTER TABLE ... ADD CONSTRAINT ... UNIQUE USING INDEX, which is a brief ACCESS EXCLUSIVE.
- Question whether the constraint should cover all history. At roughly 40 bytes per key plus overhead, a btree over 4.2 billion rows is in the hundreds of gigabytes, while a retry never arrives a year later. A partial unique index bounded by a fixed occurred_at literal, rotated forward on a schedule, gives the same protection for a fraction of the size, and the predicate must be a literal because now() is not immutable.
Worked solution 45 min
- Write the migration as numbered steps with the lock mode, expected duration and rollback beside each, including the lock_timeout retry wrapper on every ACCESS EXCLUSIVE step.
- On a copy at reduced scale, run ADD COLUMN while a long SELECT is open and observe writers queueing behind the lock request, then repeat with lock_timeout set and confirm the migration backs off instead.
- Run the batched backfill against a versioned conversion table, kill it mid-run and restart it, confirming the control-table cursor resumes without redoing or skipping a batch.
- Seed duplicate idempotency keys, attempt CREATE UNIQUE INDEX CONCURRENTLY, and confirm the INVALID index left behind, then drop it concurrently, resolve the duplicates as reversals and rebuild.
- Compare the full unique index size against the partial one bounded by a fixed occurred_at literal.
Follow-up
- Sequence the application deploy against these steps. When exactly may the code start writing uom_code, and when may it start requiring it?
- The backfill is on day six and the conversion-factor table is itself corrected. What happens to the rows already written?
- How do you roll back after SET NOT NULL is in place and half the traffic depends on the column?
Write the reconciliation query that finds ledger and position variance
inventory_movement(movement_id, item_id, node_id, lot_id, state, delta_qty) is the ledger; inventory_position(item_id, node_id, lot_id, state, qty, version, last_movement_id, reconciled_through_movement_id) is its running sum, written in the same transaction as each movement. Write one query that reports every key whose stored qty disagrees with the sum of its movements, including keys present in only one of the two tables, bounded by a watermark movement_id you pass in. State what the job does with each variance, and how you avoid re-summing four billion rows every night.
Approach
- Aggregate once and join once: a CTE doing SELECT item_id, node_id, lot_id, state, SUM(delta_qty) AS ledger_qty FROM inventory_movement WHERE movement_id <= :w GROUP BY 1,2,3,4, then FULL OUTER JOIN inventory_position USING those four columns. One pass plus one hash aggregate, O(n) in movements scanned; a correlated subquery per position row instead costs one index scan per key and is orders of magnitude slower at 40 million keys.
- Use FULL OUTER JOIN, not LEFT. A key with movements and no position row is the signature of a lost position update, and a position row with no movements is the signature of a write that bypassed the ledger. Both disappear under a LEFT JOIN from either side.
- Compare with IS DISTINCT FROM and COALESCE the join keys out of both sides, so a missing row reads as a variance rather than as NULL = NULL evaluating to unknown and being filtered away.
- Run both sides in one snapshot: either a single statement, or a REPEATABLE READ transaction. Reading positions in one transaction and movements in another manufactures variances for every key written in between.
- Make it incremental with reconciled_through_movement_id: sum only movements in (reconciled_through, :w] and add that delta to the previously verified total. The catch is that sequence values are handed out before commit, so a movement whose id is below :w can become visible after the scan and would then never be re-summed. Set :w to a max(movement_id) observed a few minutes earlier, beyond the longest write transaction, and full-scan one partition per night on a rotation as a backstop.
- A variance is reported, not repaired. Rewriting qty to match the ledger hides the double-applied retry that caused it; only a physical count may produce a cycle_count_adjust movement, and the location should be flagged unreliable to planning until it is counted.
Follow-up
- The job finds 14 variances, all on one node, all negative. What do you look at first, and what would make you suspect the reconciliation job itself rather than the data?
- How do you run this without the nightly scan competing with the ATP read path for buffer cache and I/O?
- What does the query return for a key whose only movements are a receipt and its exact reversal, and is that row worth reporting?
How do you ensure high availability and data integrity in your softwar…
How do you ensure high availability and data integrity in your software designs?
Approach
- Say what you would check first and why it is the highest-information step.
- State your assumptions explicitly before working the problem.
- Work from the requirement backwards to the design.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Can you explain how you would approach troubleshooting a data communic…
Can you explain how you would approach troubleshooting a data communication issue in a SCADA environment?
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
- What assumption would you test first?
- How would you know your answer was wrong?
Available-to-promise reads for a cart across candidate nodes
The checkout page asks for availability of up to 20 lines against up to 30 candidate nodes, 3,000 times per second at peak, inside a 25 ms p99 budget. Availability at a node is that node's on_hand position minus the outstanding held and committed allocations against it. Design the read path: what is cached, keyed how, invalidated by what, and what the service returns when the inventory primary is unreachable. State which direction the error must go and why.
Approach
- Size the fan-out before choosing anything: 20 lines by 30 nodes is 600 (item, node) pairs per call and 1.8 million pair reads per second. Per-pair round trips are ruled out by tail amplification alone. If each lookup exceeds 20 ms one time in a hundred, the chance at least one of 600 does is 1 - 0.99^600, about 99.8%, so the call's p99 becomes the slowest of 600. One batched multi-get, or one query with a composite IN list, per call.
- Fix the cache entry: key (item_id, node_id), value {on_hand_qty, outstanding_alloc_qty, position_version}. Availability is computed at read time from those two numbers, so a movement and an allocation status change invalidate the same key rather than two different ones.
- Invalidate by version, not by TTL alone. The Availability Service publishes (key, version) on every committed write; the cache applies it as a compare-and-set that discards a version lower than the one held. That makes out-of-order invalidations safe, and a lost invalidation message costs staleness bounded by a 1-2 second backstop TTL instead of permanent divergence.
- State the contract out loud: this path is advisory. The authoritative check is the conditional allocation write against the position row. Cache staleness therefore sets the rate of promise-then-retry, not the rate of overselling, and that is the whole reason a cache is permissible in front of inventory at all.
- Degrade deliberately. With the primary unreachable, serve last-known values minus a safety buffer and collapse the response to a coarse bucket (in stock / low / out) so no surface can quote an exact count from stale data. Err toward under-promising: the cost is a lost sale, against a short ship, a re-source and a second shipment's freight on the other side.
Worked solution 25 min
- Compute the per-call fan-out and the resulting pair-read rate, then decide batched versus per-pair from the tail-amplification arithmetic rather than from taste.
- Write the cache entry layout and key, and name the exactly two events that must invalidate it: any movement on that position, and any allocation entering or leaving held/committed.
- Write the invalidation message shape and the compare-and-set rule that makes an out-of-order invalidation harmless.
- Write the degraded response shape, the safety-buffer rule, and one sentence stating the direction of error and its cost on each side.
Follow-up
- A promoted item invalidates its 30 node keys every 50 ms. What stops the invalidation stream from evicting entries faster than they can be refilled, and what does a read do on a miss for that key?
- The candidate node set spans two regions. Does the read cross the region boundary, and what does that do to the 25 ms budget?
- How do you measure whether the safety buffer is set correctly, rather than guessing it?
Duplicate bookings and second trailers after a partner slowdown
A partner's booking API ran at 45 s p99 for ninety minutes. Afterwards 214 shipment_leg rows carry two booking references, 41 stops received two trailers, and finance holds duplicate invoices. booking_idempotency_key is generated and persisted on the leg before the call and is sent to the partner, whose documented dedupe window is 60 seconds. The HTTP client retries timeouts three times, and a sweeper re-tenders any leg still in status 'planned' after ten minutes. Explain how duplicates happened despite the key, and give the corrected flow.
Approach
- Establish that the duplicates are timeouts rather than double-clicks. Join the gateway request log to shipment_leg for the 214 legs and check whether each duplicate pair carries the same idempotency key and how far apart the calls landed. Same key, calls minutes apart, first attempt ending in a client timeout is the ambiguous-outcome signature; different keys would mean the key is being regenerated and is a different bug.
- Name why the key did not help. The key was correct and persisted, but deduplication was delegated to the partner, whose window is 60 seconds. A retry at 45 s may land inside it; the third retry does not, and the ten-minute sweeper re-tender certainly does not, so the partner sees a new request and books a second truck. A short partner window plus a long retry horizon is an at-least-once system wearing an at-most-once label.
- Identify the missing state. The leg goes planned to tendered only on a 2xx, so a timeout leaves it exactly as if nothing was sent, which is false: the partner may have committed before the connection dropped. The state machine needs an explicit in-doubt status between requested and confirmed, and nothing may issue a new tender for a leg in that state.
- Replace re-issuing with asking. On any ambiguous outcome, move the leg to in-doubt and let a sweeper resolve it by querying the partner for that key or for bookings against our shipment reference, and only tender again when the query definitively returns nothing. Where the partner offers no query, the manifest or the status feed is the authority and resolution is slower, so the in-doubt state must be allowed to persist rather than being timed out into a retry.
- Move retry ownership out of the process. A retry decorator around an HTTP client disappears on deploy or crash, so the persisted state is the only thing that can drive recovery: a sweeper reading legs in requested or in-doubt, with bounded attempts, backoff and an alert on in-doubt depth. The call site should not retry an irreversible effect at all.
- Deal with the 214 that already exist, since code cannot undo them. Reconcile against the partner's manifest, cancel the duplicate where the cancellation window still allows it, and for legs already executed record the second cost as a compensating entry and dispute the invoice. State plainly that this is a business process with a cost, which is exactly why the guard belongs before the call.
Follow-up
- The partner offers no lookup by your key and no search by reference. What is your resolution path and how long may a leg stay in doubt?
- Which other effects in this system have the same shape and the same missing state?
- How do you test this without a degraded partner, and what does the fault injection have to simulate that a plain timeout does not?
For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Rebuild the primitives by implementing them
- Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
- Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
- For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.
Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Arrays under an invariant: two pointers, sliding window, binary search
- Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
- Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
- Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.
Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.
Practice prompt ↗Practice prompt ↗03Sorting, heaps, and the greedy argument that has to be proved
- Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
- Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
- Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.
Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.
Practice prompt ↗Practice prompt ↗04Recursion, memoisation, and the step to a table
- Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
- Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
- Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.
Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Graphs, where most of the work is choosing the traversal
- Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
- Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
- Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.
Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.
Practice prompt ↗Practice prompt ↗06One day for everything that is not an algorithm
- Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
- Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
- Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.
Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.
Practice prompt ↗Practice prompt ↗07Solve out loud, under time
- Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
- Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
- Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.
Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Nobody is scoring your stamina at three in the morning. What carries weight is which signal told you something was wrong, what you measured before touching anything, what you rolled back versus what you fixed forward, and why you picked one. 'We restarted it and it went away' is a story about not knowing.
Please, tell me about yourself.
Please, tell me about yourself.
Approach
- 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.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
What do you consider to be your greatest accomplishment?
What do you consider to be your greatest accomplishment?
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
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
Can you describe a time you had to manage a complex task under a tight…
Can you describe a time you had to manage a complex task under a tight deadline?
Approach
- Name the disagreement and how you resolved it with evidence.
- 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?
How do you handle working in a team?
How do you handle working in a 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
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
- 01
Please, tell me about yourself.
- 02
What do you consider to be your greatest accomplishment?
- 03
Can you describe a time you had to manage a complex task under a tight deadline?
- 04
How do you handle working in a team?
Is this an official SOLV Energy interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at SOLV Energy. Rounds and questions reflect what candidates have reported, not a process SOLV Energy 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 technical interviews are considered moderate. They focus on practical application rather than theoretical riddles, so being able to explain your past work is more important than memorizing complex algorithms.
PracHub interview research ↗What is the typical timeline from the first screen to an offer?
The process is designed to be efficient. From the initial phone screen to the final team interview, the process often concludes within a few weeks, provided you are responsive and scheduling aligns.
PracHub interview research ↗Is this a remote-friendly role?
Yes, the current HV SCADA Engineer II postings indicate that the position is remote, allowing for flexibility while still requiring high-level collaboration with your distributed team.
PracHub interview research ↗What differentiates successful candidates?
Successful candidates are those who can clearly articulate how their technical skills solved a real-world problem. They show a genuine interest in the energy sector and communicate their thought process clearly during technical discussions.
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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