So Energy · Software Engineer
Updated · 2026-09-24

So Energy Software Engineer
Interview Questions & Guide 2026

THE 60-SECOND BRIEF

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.

Most rounds test how you handle an under-specified problem more than what you recall. A wrong first approach you correct out loud is recoverable; a constraint you assumed silently and were never given is the expensive mistake.

PracHub has no confirmed round sequence for So Energy. Treat the sections below as preparation areas and confirm the format with your recruiter.

Make every external side effect safe to retryKeep a site operating during a network partitionReserve inventory without overselling under concurrent allocation

34 min read

Browse Software Engineer questions

See the practice prompts

13Practice promptsAcross five skill areas
3With worked solutionsIncluded in the practice prompts

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.

01

Preparation focus

editorial

No 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 interview preparation framework ↗

PracHub editorial advice for the preparation topics above.

01

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.

02

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.

03

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.

04

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.

9 technical prompts3 include a worked solution

Find the hottest write keys under a fixed memory budget

medium
top-kheapstreaming sketches

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
  1. 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).
  2. 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.
  3. 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.
  4. 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.
  5. 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

hard
set coverbitmask dpapproximation

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
  1. 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.
  2. 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 in rem, with dp[0] = (0, 0) and the value a (node_count, total_distance) pair compared lexicographically. Let b be the lowest set bit of rem; any cover must contain some node that stocks line b, 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.
  3. Quote the memory for what you actually store, not for a count array alone. Over 2^25 masks you need three dense arrays: uint8 node count (33.5 MB), uint32 total distance (134.2 MB, integer metres so the tie-break compares exactly and the array stays fixed-width), and uint8 chosen 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.
  4. Reconstruction is free in this formulation and needs no parent-mask array: because every transition eliminates the lowest uncovered line, the predecessor of rem under the stored choice v is exactly rem & ~cover(v), so walk rem from the full mask down to 0 emitting choice[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 of ceil(popcount(rem) / max_v popcount(cover(v) & rem)), which prunes long before the theoretical bound bites.
  5. 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.
  6. 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

easyWorked solution
parsingunit-of-measuretime zones

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
  1. 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.
  2. Parse the decimal as an integer mantissa plus a scale. 4.500 becomes (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.
  3. Convert by integer arithmetic and demand exactness: numerator = mantissa * factor; emit numerator / 10^scale only when numerator % 10^scale == 0, otherwise reject the line with a reason code. Worked: 4.500 CS at 12 eaches per case gives 54000/1000 = 54, exact. 0.125 PL at 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.
  4. 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.
  5. 20260317 carries 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
  1. Write the integer conversion as a pure function (mantissa, scale, factor) -> Result<int64> and unit-test it before touching the file reader.
  2. Table-drive the cases: exact conversion, non-dividing conversion, a mantissa near int64 overflow, a zero quantity, and a unit not in the ladder.
  3. Wire the scanner, emitting one rejection row per bad line with the raw segment attached so the partner can be shown what was sent.
  4. 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.
  5. Re-run the whole file and confirm identical output, including identical rejections.
EXPECTED RESULT`4.500 CS` at factor 12 yields 54 eaches; `0.125 PL` at factor 36 is rejected as non-exact rather than stored as 4 or 5; every emitted row carries both the eaches integer and the source unit; the date-only field resolves through the node's zone, not the server's.
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 +0100 on some dates. Does that change your storage decision for a future appointment as well as a past event?

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.

Small steps. Visible outcomes.0 / 7 completed
ONE WEEK · YOUR PACE

Prepare, practise & reflect

One practical outcome each day. Spend longer where you need it.

0 / 7 done
01Measure 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…

medium
behavioural and engineering judgement

How do you handle specific technical challenges when working under pressure?

Approach
  1. State the situation in two sentences and spend the rest on the reasoning.
  2. Name the disagreement and how you resolved it with evidence.
  3. 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…

medium
behavioural and engineering judgement

Can you describe a time you took the initiative to improve a process or system?

Approach
  1. Pick a story where you made the decision, not one where you watched it.
  2. Name the disagreement and how you resolved it with evidence.
  3. 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?

medium
behavioural and engineering judgement

Can you explain your experience with microservices architecture?

Approach
  1. Pick a story where you made the decision, not one where you watched it.
  2. State the situation in two sentences and spend the rest on the reasoning.
  3. 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?

medium
behavioural and engineering judgement

What is your experience working with Java or Kotlin?

Approach
  1. Name the disagreement and how you resolved it with evidence.
  2. Close with what you would do differently, concretely.
  3. 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?

PracHub interview preparation framework ↗
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.