As a Software Engineer at So Energy, you sit at the intersection of complex backend systems and the mission-critical goal of providing sustainable, efficient energy solutions. This role is not just about writing code; it is about building resilient, scalable architectures that support the energy transition. You will contribute to a microservices-based ecosystem, often working with technologies like Java and Kotlin, keeping So Energy's platforms robust while delivering a seamless experience for its customers.
The environment at So Energy is one of transparency and continuous improvement. The role suits engineers who are willing to challenge the status quo, mentor junior team members, and take ownership of technical debt. If you are a proactive problem solver who enjoys working in a collaborative, feedback-oriented culture where leadership is genuinely interested in your perspective on architectural and cultural improvements, this role offers a significant opportunity to drive meaningful change.
Preparation focus
editorialNo round sequence has been reported for this company, so work the categories below and confirm the format with your recruiter.
What to demonstrate
- Breadth across SQL, experimentation and product reasoning
- Ability to state assumptions before choosing a method
How to prepare
- Drill the practice exercises below and time yourself
- Prepare three quantified stories about decisions you drove
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.
Retrying an irreversible external call without an owned idempotency key
Timeouts and connection resets are ambiguous by construction - the partner may have committed before the response was lost - so a generic retry policy around an HTTP client is, in this domain, a machine for buying two labels and dispatching two trucks. Relying on the partner's deduplication is not a substitute, because their window is usually short, their key is often derived from fields you may legitimately change on retry, and many older integrations have no such concept at all. The workable pattern is to generate the key yourself, persist it with an explicit unknown status before the call, and resolve ambiguity by querying the partner for that key rather than re-issuing; the sweeper that does this is the component that has to be correct, not the call site. Teams get this right for payments and then forget that a carrier tender, a warehouse work release and an EDI shipping notice have the same shape.
Treating a network call as though it were a local function call
A remote call can be slow, fail, or return after you stopped waiting, so name the timeout, the retry policy, and what the caller sees while the dependency is down. A call with no timeout turns one slow dependency into an exhausted thread or connection pool in every service upstream of it.
Arguing past a hint
When the interviewer asks what happens for a particular input or floats a different data structure, stop and take it seriously; it is almost always a correction rather than idle curiosity. Talking over it converts a recoverable wrong turn into a data point about how you handle review.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need 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?
Split an order across the fewest nodes that can fill it
A fulfilment order has L lines, L up to 25. After filtering by capability, service time and stock, N candidate nodes remain; N is usually under 14 but can reach 200. Each node covers a known subset of the lines in full. Return a minimum-size set of nodes covering every line, tie-broken by total distance. Give an exact algorithm for the small case with its exact cost in operations and bytes, say precisely why enumerating node subsets stops working, and give what you ship when N is 200.
Approach
- Name it: this is minimum-cardinality set cover, which is NP-hard, so the useful move is choosing which parameter you are exponential in rather than hunting for a polynomial exact algorithm. Enumerating subsets of nodes is 2^N and correct — at N = 14 that is 16,384 subsets and finishes instantly, which is exactly why it survives review and then fails in production at N = 200, where 2^200 is about 1.6e60.
- Be exponential in L instead, because L is the parameter you bound. Index the DP by the set of lines still uncovered:
dp[rem]is the cheapest way to cover the lines inrem, withdp[0] = (0, 0)and the value a(node_count, total_distance)pair compared lexicographically. Letbbe the lowest set bit ofrem; any cover must contain some node that stocks lineb, so the transition ranges only over those nodes:dp[rem] = min over v with b in cover(v) of dp[rem & ~cover(v)] + (1, dist(v)). Worst case O(2^L * N) — at L = 25 and N = 14 that is 33,554,432 states times 14, about 4.7e8 transitions — and the lowest-bit restriction cuts the real branching factor to the nodes stocking one specific line. - Quote the memory for what you actually store, not for a count array alone. Over 2^25 masks you need three dense arrays:
uint8node count (33.5 MB),uint32total distance (134.2 MB, integer metres so the tie-break compares exactly and the array stays fixed-width), anduint8chosen node (33.5 MB, since N <= 200 fits a byte) — about 201 MB, roughly six times the 33.5 MB that a bare count array would suggest. The same layout at L = 20 is 6.3 MB, which is why this is comfortable up to about 20 lines and needs a deliberate decision at 25. - Reconstruction is free in this formulation and needs no parent-mask array: because every transition eliminates the lowest uncovered line, the predecessor of
remunder the stored choicevis exactlyrem & ~cover(v), so walkremfrom the full mask down to 0 emittingchoice[rem]. If the reachable state count is far below 2^L — coverage sets are usually highly correlated — a top-down memoized recursion over a hash map trades the 201 MB for roughly 40-60 bytes per visited state; measure the visited count first, because a hash map that ends up touching most masks is strictly worse than the dense arrays. Add a branch-and-bound cut ofceil(popcount(rem) / max_v popcount(cover(v) & rem)), which prunes long before the theoretical bound bites. - When exact is out of reach, ship greedy with a stated guarantee: repeatedly take the node covering the most currently-uncovered lines, using bitset popcounts, O(N * ceil(L/64)) word operations per pick and at most min(N, L) picks. Greedy is within H(L) <= ln L + 1 of optimal (Chvatal), and no polynomial algorithm achieves (1 - o(1)) ln L unless P = NP (Dinur and Steurer, 2014) — so effort belongs in the pre-filter and the tie-break, not in a cleverer heuristic. At L = 25 the bound is H(25), about 3.82x worst case, while the realistic gap on correlated coverage sets is zero or one node.
- Close on the objective, because fewest nodes is not cheapest. Two shipments from distant nodes routinely cost more than three from near ones, and a split also costs a second box, a second carrier pickup and a worse customer experience. If cost is the real objective it is weighted set cover — greedy by cost divided by newly-covered lines, same logarithmic bound — and if a node can cover a line only partially, the line mask is no longer sufficient state, since you must track remaining quantity per line; that is the point at which you stop hand-rolling and either restrict splits or hand an MILP to a solver with a time budget.
Follow-up
- The pre-filter is what keeps N small. What is in it, and what happens to your exact path the day someone loosens it?
- One node can cover a line only partially. Show where the bitmask formulation breaks and what state replaces it.
- You have 40 milliseconds inside a checkout call. Which of these runs there, and what runs asynchronously afterwards?
Parse partner quantity and date segments without floating point
A partner sends a delimited flat file whose segments look like QTY+17:4.500:CS and DTM+232:20260317. Quantities carry up to three decimal places in a named unit; each item has an integer pack factor to eaches that is versioned by effective date. Produce observation_event rows with an integer quantity in eaches, the unit recorded, and occurred_at resolved, rejecting any line whose conversion is not exact. Files reach 4 GB. Single pass, no floating point. State your bounds and what you do with the zoneless date.
Approach
- Scan with indices, not a regex. Find the segment terminator, then split the payload on
:by scanning forward and slicing — O(bytes) with no backtracking and no per-line allocation beyond the output row. A regex with nested quantifiers over 4 GB is where this job silently becomes CPU-bound. - Parse the decimal as an integer mantissa plus a scale.
4.500becomes(mantissa = 4500, scale = 3): accumulate digits into an int64, count digits after the point, and never call the runtime's float parser. Reject a mantissa that would overflow int64 when multiplied by the pack factor, checking before the multiply rather than after. - Convert by integer arithmetic and demand exactness:
numerator = mantissa * factor; emitnumerator / 10^scaleonly whennumerator % 10^scale == 0, otherwise reject the line with a reason code. Worked:4.500 CSat 12 eaches per case gives 54000/1000 = 54, exact.0.125 PLat 36 eaches per pallet gives 4500/1000 = 4.5, which is half a sealed unit and must fail at the boundary, not round three services away. - Select the pack factor by the file's effective date, not by the current item master. Factors are properties of a point in time — a supplier moving a case from 12 to 10 must not reinterpret last quarter's file — so the lookup is a range query on the factor's validity interval.
20260317carries no zone and no time. Attach the IANA zone of the node the segment refers to, captured at ingestion, because it cannot be recovered downstream; then pick and document one convention for a date-only field (start of local day is the usual choice) and store the raw local string beside the resolved instant so the assumption stays visible. Complexity: O(total bytes) time, O(longest line) working space, one pass.
Worked solution 20 min
- Write the integer conversion as a pure function
(mantissa, scale, factor) -> Result<int64>and unit-test it before touching the file reader. - Table-drive the cases: exact conversion, non-dividing conversion, a mantissa near int64 overflow, a zero quantity, and a unit not in the ladder.
- Wire the scanner, emitting one rejection row per bad line with the raw segment attached so the partner can be shown what was sent.
- Feed a fixture file containing a date on a day the destination zone changes offset, and assert the resolved instant against a hand-computed value.
- Re-run the whole file and confirm identical output, including identical rejections.
Follow-up
- The same file is re-sent in full tomorrow with three lines changed. What makes the second ingest a no-op for the unchanged lines?
- A line names a unit that is not in the ladder for that item. Reject the line, reject the file, or quarantine it — and who finds out?
- The partner starts sending an offset like
+0100on some dates. Does that change your storage decision for a future appointment as well as a past event?
Fold late, out-of-order scan events into one status per leg
observation_event holds event_id, subject_type, subject_id, observed_status, occurred_at, received_at, dedupe_key (UNIQUE) and processing_status. A lookup table maps observed_status to status_rank SMALLINT along a progress lattice. Write one query returning the current status of every shipment leg that received an event in the last hour, such that a picked_up event arriving after a delivered event never changes the answer. Then explain exactly which part of your ordering makes a shuffled replay of the same batch produce identical output.
Approach
- Separate the two jobs. Deduplication is the UNIQUE on dedupe_key at ingest, because partner feeds resend whole windows; ordering is the fold. Either alone leaves a defect: dedupe without a lattice still lets a stale event win, and a lattice without dedupe double-counts a replayed window in anything that aggregates.
- Select the affected legs in one CTE by received_at >= now() - interval '1 hour', then fold over every event for those legs, not only the recent ones. Restricting the fold's input to the recent window is the failure: a lone late picked_up would be the only row considered and the leg would regress.
- Fold as a maximum over the lattice: SELECT DISTINCT ON (subject_id) subject_id, observed_status FROM ... ORDER BY subject_id, status_rank DESC, occurred_at DESC, event_id DESC. The equivalent form is ROW_NUMBER() OVER (PARTITION BY subject_id ORDER BY status_rank DESC, occurred_at DESC, event_id DESC) = 1.
- The determinism comes from the final event_id tiebreak. Rank and occurred_at alone leave two equal-rank events at the same instant to be resolved by heap order, which changes when rows are re-inserted, so the shuffled replay differs in exactly the cases nobody tests. With a total order over the tuple the fold is a max, which is commutative and associative and therefore order-insensitive.
- Do not advance a leg on occurred_at alone. The timestamp is device-reported and can be wrong, which is why the lattice rank leads the ORDER BY and why clock_offset_ms is worth storing: a future-dated event with a low rank then cannot promote itself past a delivered.
- Keep superseded rows and mark processing_status = 'superseded' rather than deleting them. The discarded event is usually the one that explains a delivery dispute, and deleting it makes the trace unanswerable.
Follow-up
- An event arrives a week late for a leg that is already invoiced. Does the fold apply it, reject it as late, or apply it and emit a correction?
- The stream is partitioned by source_id rather than by subject_id. What ordering guarantee do you actually have, and what breaks?
- How would you materialise this fold so the customer-facing status read is not running a window function per request?
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?
How do you manage API versioning?
How do you manage API versioning?
Approach
- Work from the requirement backwards to the design.
- Say what you would check first and why it is the highest-information step.
- Clarify what is being asked and what a complete answer contains.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
How do you approach messaging patterns in distributed systems?
How do you approach messaging patterns in distributed systems?
Approach
- Work from the requirement backwards to the design.
- Clarify what is being asked and what a complete answer contains.
- State your assumptions explicitly before working the problem.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Fan shipment status out to five thousand customer webhooks
Customers subscribe to shipment status by webhook: 5,000 endpoints, some with an 8 second p99, some down for a day. The upstream is folded leg status, not raw observations. Requirements are per-shipment ordering at each subscriber, at-least-once delivery with a stable dedupe id, and no subscriber able to slow another. Assume 3 million legs a day, about six status transitions each, and a mean of three subscribers per shipment. Design the fan-out and quantify what folding first saves you.
Approach
- Fan out transitions, not observations, and do the arithmetic before designing delivery. Raw ingest averages 9,000 observations a second, about 780 million a day. Three million legs at six transitions each is 18 million notification sources a day, 208 per second, and three subscribers apiece is 54 million deliveries a day, 625 per second. Folding first is worth a factor of about 43 - 208 sources per second instead of 9,000, and 625 deliveries per second instead of roughly 27,000 - and that factor dominates every subsequent optimisation, so settle it before tuning delivery.
- Keep delivery state as a durable per-subscriber cursor over a per-subscriber queue rather than a row per attempt. At 54 million deliveries a day a row per attempt is roughly 19.7 billion rows a year before a single retry is counted, and retries only add to it; a cursor plus a bounded in-flight window gives retry and progress without any of them. Per-event rows earn their cost only where you must later prove a specific delivery in a dispute, so scope them to the subscribers who contract for that rather than to all 5,000.
- Preserve ordering deliberately: a bounded in-flight window greater than one can reorder. Give each subscriber's queue a small number of lanes keyed by hash(shipment_id), so order is strict within a lane and parallelism comes from lanes. Ordering is then a property of the key, not of a hopeful single-threaded consumer.
- Isolate subscribers with a per-subscriber concurrency cap drawn against a shared worker pool. By Little's law an 8 second endpoint receiving 20 notifications per second wants 160 workers in flight; without a cap it takes them from the pool and starves the other 4,999. The cap converts a slow subscriber into its own growing backlog instead of a shared outage.
- Handle the dead endpoint as a bounded liability: circuit-break after consecutive failures, stop attempting, let the cursor lag, and retain the queue to a fixed horizon such as 72 hours. At the horizon, advance the cursor, record a gap, and tell the subscriber to re-sync from the pull API. Unbounded retention for one dead customer is how the fan-out store becomes the next incident.
- Make at-least-once an explicit contract rather than an accident: every payload carries a stable event id and the leg's status_rank, so a customer can discard a duplicate or a stale retry on their side, and that requirement is documented rather than discovered.
Worked solution 30 min
- Derive the notification rate from legs per day, transitions per leg and subscribers per shipment, set it beside the raw observation rate, and state the reduction factor that justifies folding first.
- Choose between per-event delivery rows and a per-subscriber cursor, and show how the chosen shape still preserves per-shipment ordering under retry.
- Set a per-subscriber concurrency cap and compute an 8 second endpoint's in-flight occupancy at its share of traffic.
- Write the circuit-breaker states, the retention horizon and the re-sync contract for a subscriber that returns after a day offline.
Follow-up
- A customer demands exactly-once. What can you actually offer, and what must they implement to get the effect they want?
- One subscriber covers 40 percent of all shipments. What breaks first in a per-subscriber queue design, and what do you shard on instead?
- How do you replay a day of notifications for one subscriber after a bug on your side, without re-sending to the other 4,999?
One partner's leg statuses freeze while every other feed advances
Since 06:40 UTC no leg fed by one partner has changed status; every other feed is current. Ingest is healthy and CPU is flat, and observation_event rows from that source are still landing with current received_at and processing_status 'pending'. Consumer lag is growing on exactly one stream partition, whose head row has over 900 attempts and a byte-identical last_error on each one. Give the ordered diagnosis and the change that stops one unapplicable observation from stalling a partition again.
Approach
- Separate the partner not sending from us not applying, because both look identical to a customer. Rows landing with current received_at and processing_status 'pending' prove ingestion is fine and the fold is stuck; if received_at had gone stale instead, this would be a partner or transport incident and nothing in the consumer would be at fault.
- Read the blocked partition at its head, ordered by offset, and classify the error rather than counting it. An identical error string across 900 attempts is deterministic and implicates the payload; a varying set of connection resets, timeouts and 5xx implicates the downstream store. Only the deterministic case is a poison message.
- Reproduce outside the consumer by running the fold over that one event. The usual cause here is an observed_status the partner added that has no rank in the progress lattice, or a subject_id whose leg does not exist yet, so the fold throws before it can write.
- Decide the exit against the ordering contract, which is per subject and not per partition. Many legs hash into one partition, so blocking on one of them is a bug and not a guarantee. Park the row as processing_status 'rejected_malformed' with its offset in a dead-letter view, advance the offset, and block only the affected subject if the event is well-formed but not yet applicable.
- Make the retry policy classify instead of counting. Deserialisation failures, unknown enum values and constraint violations are terminal and must never be retried in place; IO errors and timeouts are retryable with bounded attempts and backoff. Alert on oldest-pending age per partition, which turns positive within seconds, rather than on lag, which drifts up slowly and looks like normal backlog.
Follow-up
- The parked event turns out to be a real delivered scan. How do you replay it after the code fix without double-applying, and what makes that safe?
- What should the customer-facing status show for legs that were queued behind it?
- The partition key is currently the leg id. What breaks if you key by partner instead, and what if you key randomly?
For someone fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Measure before reasoning
- Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
- Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
- Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.
Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.
Practice prompt ↗Practice prompt ↗Worked solution ↗02References, copies, and the bugs they produce
- Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
- Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
- Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.
Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.
Practice prompt ↗Practice prompt ↗03Types, once, in a language that checks them
- Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
- Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
- Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.
Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.
Practice prompt ↗Practice prompt ↗04Concurrency, starting with what actually runs at the same time
- Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
- Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
- Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.
Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Debugging as a procedure rather than an instinct
- Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
- Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
- Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.
Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.
Practice prompt ↗Practice prompt ↗06Tests that catch the bug you are about to write
- Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
- Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
- Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.
Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.
Practice prompt ↗Practice prompt ↗07Debug something broken, out loud
- Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
- Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
- Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.
Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Conflict answers where you were right and everyone came round are the weakest ones. Stronger: the evidence you went and collected, what would have changed your mind, and what you did in the weeks after the call went against you. Implementing a design you argued against, properly, is a specific and checkable behaviour.
How do you handle specific technical challenges when working under pre…
How do you handle specific technical challenges when working under pressure?
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
Can you describe a time you took the initiative to improve a process o…
Can you describe a time you took the initiative to improve a process or system?
Approach
- Pick a story where you made the decision, not one where you watched it.
- Name the disagreement and how you resolved it with evidence.
- Close with what you would do differently, concretely.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
Can you explain your experience with microservices architecture?
Can you explain your experience with microservices architecture?
Approach
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
What is your experience working with Java or Kotlin?
What is your experience working with Java or Kotlin?
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?
- What did you decide not to do, and why?
- 01
How do you handle specific technical challenges when working under pressure?
- 02
Can you describe a time you took the initiative to improve a process or system?
- 03
Can you explain your experience with microservices architecture?
- 04
What is your experience working with Java or Kotlin?
Is this an official So Energy interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at So Energy. Rounds and questions reflect what candidates have reported, not a process So Energy has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long does the entire interview process typically take?
The process is designed to be fast. From the initial recruiter screen to the final interview, the process keeps momentum high, often completing the stages within a few weeks.
PracHub interview research ↗How should I prepare for the pair programming session?
Focus on communicating your thought process aloud. Interviewers are less interested in perfect syntax and more interested in how you approach the problem and collaborate with your partner.
PracHub interview research ↗What is the culture like at So Energy?
So Energy's culture is defined by transparency and a desire to improve. The team is described as willing to admit mistakes and work together to fix them, which suits engineers who want to make a real impact.
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