Lendable (UK) · Software Engineer
Updated · 2026-09-24

Lendable (UK) Software Engineer
Interview Questions & Guide 2026

THE 60-SECOND BRIEF

A Software Engineer at Lendable (UK) is a critical contributor to a high-velocity, fintech-focused engineering team. You will not just be writing features; you will be building and maintaining the core infrastructure that powers financial products, ensuring reliability, scalability, and precision in every line of code. The work is deeply technical, frequently involving complex domain modeling and the orchestration of distributed systems.

Seniority moves the scope further than the words in the title do. An earlier-career loop mostly checks that you implement something correctly and can reason about its cost, while a senior loop checks that you can pick between two defensible designs and say what you gave up.

Lendable (UK) candidates report 4 rounds · ≈ 3-5 weeks. The stages below are what candidates describe, not a published process.

Store money as integer minor unitsReconcile the ledger against processor settlement filesMake every money-moving endpoint idempotent by key

39 min read

Browse Software Engineer questions

See the practice prompts

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

A Software Engineer at Lendable (UK) is a critical contributor to a high-velocity, fintech-focused engineering team. You will not just be writing features; you will be building and maintaining the core infrastructure that powers financial products, ensuring reliability, scalability, and precision in every line of code. The work is deeply technical, frequently involving complex domain modeling and the orchestration of distributed systems.

The role demands a high degree of craftsmanship. Because you are working in a sensitive financial domain, your decisions directly impact user trust and business outcomes. You will likely work with modern stacks (often including PHP/Symfony or Kotlin) and be expected to apply rigorous architectural standards—such as Domain-Driven Design (DDD), Hexagonal Architecture, and SOLID principles—to ensure that the platform remains maintainable as it scales.

Expect an environment where "good enough" is rarely the standard. You will be challenged to think about edge cases, financial precision, and the long-term extensibility of your solutions. This is an ideal role for engineers who enjoy deep, structured problem-solving and who view software development as a disciplined, deliberate practice rather than just a task-completion exercise.

01

Initial Screening Call

reported

Half of this call is the part candidates treat as small talk: start date, notice period, work authorisation and its timing, location and time zone, on-call, and the number. Those are what kill offers late, after several engineers have each spent a day. Surfacing a hard constraint now costs you nothing and occasionally buys you something, since a loop compressed to fit a competing deadline can usually only be arranged if it is asked for early. The common failure is deflecting the compensation question twice, then discovering at offer stage that the band never reached your number.

What to demonstrate

  • Whether your hard constraints are compatible with the role before a loop gets booked: earliest start, notice period, what authorisation you hold and when it needs action, days on site, willingness to carry a pager
  • Whether you give a compensation range with something behind it, such as current total compensation or a competing timeline, rather than leaving the band untested
  • Whether your stated timeline is real, since a competing deadline raised now is something scheduling can sometimes work around and the same deadline raised at offer stage usually is not

How to prepare

  • Write each constraint down in one line before the call and state them as facts rather than negotiating them live under a question you were not expecting
  • Set your range from two or three current data points for that level and location, and name the structure you are quoting in, so the number is comparable to the one they are holding
  • If another process is running, say where it stands and by when, and ask directly whether this loop can be scheduled inside that window
PracHub interview research ↗
02

Technical Exercise

reported

The same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.

What to demonstrate

  • In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
  • In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
  • Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes

How to prepare

  • Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
  • Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
  • For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first
PracHub interview research ↗
03

Technical Deep-Dive

reported

Most of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.

What to demonstrate

  • Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
  • Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
  • Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
  • Whether a failing case is isolated and explained before any edit is made to the code

How to prepare

  • From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
  • Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
  • Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.
PracHub interview research ↗
04

Behavioral Fit Interview

reported

Many of these questions are about something that went wrong, and the grading sits mostly in the hours after you knew. Who found out first, whether that was you or an alert or a user, how long it took you to say it out loud, and whether the people who needed the news got it while they could still act on it. Engineers under-tell this part because it feels like confessing. The pattern it is looking for is the opposite: the quiet fix, an incident absorbed without telling anyone, after which nothing changed and the same failure is still available.

What to demonstrate

  • How the problem was found, and whether that route was one you had built or one that happened to you, since a user reporting it first means your instrumentation did not cover that failure
  • Whether time-to-detect and time-to-tell are separate numbers in your account and whether you know both, because a fast fix that nobody heard about until the retro is a different answer from a slow one that was announced immediately
  • Whether the resolution left something durable behind, a check that fires or a default that changed, rather than depending on people remembering to be careful
  • Whether you can say what the failure cost without either inflating it or waving it away

How to prepare

  • Reconstruct one incident you were part of as a timeline with clock times: first bad request, first signal, first person who knew, first message outside the team, mitigation, permanent fix. The gaps between those entries are what gets asked about
  • Look up the configuration of the signal that caught it, including its evaluation window and threshold. An alert defined on a five-minute aggregate cannot fire until the condition holds across that window, which puts a floor under time-to-detect that has nothing to do with how severe the failure was. Be able to say what that floor was and whether anyone had chosen it deliberately
  • Prepare one story where you escalated early and the severity turned out to be smaller than you thought, including what it cost the people you pulled in. Without it, every answer you give about raising alarms is unfalsifiable
PracHub interview research ↗

PracHub editorial advice for the preparation topics above.

01

Treating money as a decimal with two places

ISO 4217 exponents are 0 for currencies such as JPY and KRW, 2 for most, and 3 for BHD, KWD, JOD, OMR and TND, so a hard-coded multiply-by-100 is off by a factor of 100 or 10 depending on the currency, in opposite directions. Floating point is worse: IEEE 754 binary64 cannot represent 0.1 exactly, so repeated accrual accumulates drift that appears as a handful of minor units in the daily reconciliation and then gets 'fixed' by widening the match tolerance, which is how a genuine break becomes invisible. The only forms that survive a reconciliation are integer minor units with the exponent carried alongside the currency code, or a fixed-scale decimal type with exactly one documented rounding point.

02

Retrying a charge after a timeout

A timeout is not a failure; it is an unknown outcome, and the request may have been processed in full with only the response lost. Re-sending it without an idempotency key that the processor itself honours produces a duplicate charge, which is a customer-visible incident and usually a dispute. The correct handling is to treat the state as unknown, query the processor for that key or client reference, and only then decide. The mechanism also depends on the key being generated once by the caller and reused across every attempt — generating a fresh key per retry turns the whole scheme into a no-op while leaving all the code that appears to implement it in place.

03

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.

04

Never running a concrete value through the code

Trace one small input and one edge input by hand, index by index, out loud. Re-reading your own code catches design mistakes; walking a real value through it catches the off-by-one, the uninitialised accumulator and the loop that never advances.

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

Answer as-of balance queries over an append-only entry log

hard
prefix sumsoffline queriesappend-only

Given 400 million ledger_entry rows (entry_id, account_id, direction, amount_minor, currency, business_date) and 2 million queries of (account_id, currency, as_of_date) asking for the balance at the end of that business date, produce every answer. The obvious solution — per query, sum that account's entries with business_date <= as_of_date — is correct. Say precisely why it will not finish, then give one that will, with time and space complexity. Corrections are posted as new entries carrying their own business_date.

Approach
  1. Cost the naive version in numbers before rejecting it. Spread uniformly over 20 million accounts, each query touches about 20 rows behind a per-account index and 2 million queries is 4e7 row touches — perfectly fine. The problem is skew: one pooled clearing or merchant settlement account holding 3e7 entries, taking 10% of the queries, is 6e12 row touches. Name the skew; 'n is large' is not the reason.
  2. The structural fact that buys a cheap answer: entries are append-only and never updated, so a prefix sum over an account's entries ordered by (business_date, entry_id) is stable — nothing behind position i can change. No mutable-balance design offers that, and it is why the storage is worth paying for.
  3. Offline sweep, when all queries are known up front: externally sort entries by (account_id, currency, business_date, entry_id) and queries by (account_id, currency, as_of_date), then merge-walk both with a running sum, emitting each query's answer as the sweep passes its date. O((n + q) log(n + q)) dominated by the sort, O(1) beyond sort buffers, one sequential pass over each input instead of 2 million random seeks.
  4. Online alternative: materialise end-of-day snapshots — one row per (account_id, currency, business_date) that had activity, holding the cumulative total. A query becomes one index seek for the latest snapshot at or before as_of_date, O(log n) per query, over far fewer rows than n. Use snapshots when queries arrive singly and the sweep when they arrive as a batch.
  5. Corrections are the subtlety: an entry posted today but dated back changes historical answers, so every snapshot for that account from that date forward is stale. Either keep a Fenwick tree over dates per account (O(log D) update and prefix query) or recompute that account's snapshots from the corrected date onward. Then be precise about what reproducibility means — yesterday's statement is reproducible as of a stated snapshot time, not identical forever.
  6. Bound the resources: int64 sums throughout, no float; 400 million rows at roughly 48 bytes of the columns you actually need is about 19 GB, so the sort is external and its fan-out is chosen from the sort buffer, not from the row count.
Follow-up
  • One account holds 30% of all entries. What does the external sort do with it, and what would you do for that one key instead?
  • Queries now arrive online at 500 per second. Which design survives, and what does keeping the other one warm cost?
  • A correction lands with a business_date 90 days back. Which snapshots are now wrong, and how does a reader find out?

Compute peak held exposure from overlapping authorisation holds

medium
sweep lineintervalsauthorisation holds

An account has up to 2 million authorisations on one business date: (auth_id, amount_minor, created_at, expires_at) with the hold live over the half-open interval [created_at, expires_at), plus capture events (auth_id, captured_minor, captured_at) that reduce the hold at captured_at, and explicit reversals that drop the remainder to zero. Timestamps are microsecond-precision timestamptz. Return the maximum total held amount across the day and the earliest instant it is reached, with complexity. Then say what changes if the deliverable is the peak per minute instead.

Approach
  1. Expand each authorisation into signed delta events rather than reasoning about intervals: +amount at created_at, -remaining at expires_at, -captured_minor at each captured_at, -remaining at reversed_at. The problem collapses to a running sum over a sorted event list.
  2. Sort the 2n to 4n events by (timestamp, sign) with negative deltas ordered first on a tie. The half-open convention forces that: at exactly expires_at the hold is already gone, so a - must land before a + at the same instant or you report a one-microsecond peak that never existed. O(n log n) time, O(n) space.
  3. Sweep once, tracking running, best and best_at, taking the first instant that attains the maximum. Say out loud which tie rule you are using — 'the peak' is ambiguous when the same level is reached twice, and the caller needs to know which instant they are being handed.
  4. If timestamps are bucketed (1,440 minute buckets for the per-minute variant), drop the sort for a difference array: add the delta at the start bucket, subtract at the end bucket, prefix-sum once. O(n + B) time and O(B) space, strictly better, at the cost of answering only at bucket resolution.
  5. Assert the invariant during the sweep: running must never go negative. A negative total means a capture exceeded its authorisation, which is an invariant violation upstream rather than a sweep bug — fail loudly instead of clamping at zero and reporting a plausible number.
  6. Handle carry-in: a hold created before the window contributes its remaining amount as the sweep's initial value, not as a + event inside the window. Omitting that is the off-by-a-day that makes the first minute of every day look artificially quiet.
Follow-up
  • An incremental authorisation raises an existing hold after the fact. Where does that event go, and does it disturb the tie rule?
  • You now need the peak for 10 million accounts inside a nightly window. What changes, and what must the partitioning key be?
  • The peak sizes a funding transfer. Does the business date or the timestamp decide which day that transfer lands on?

Detect duplicate-charge bursts in an out-of-order authorisation stream

hardWorked solution
sliding windowevent timewatermarksdeduplication

Authorisations arrive as (instrument_token_id, amount_minor, currency, event_time, arrival_time) at roughly 3,000 per second, up to 60 seconds late and out of order. Flag any token with three or more authorisations of identical (amount_minor, currency) inside any 10-minute window of event time. Report each flag once, as early as correctness allows. State memory per key and in total, how late an event you will accept, and what you do with one that arrives after you have already reported — or already declined to report — that window.

Approach
  1. Key state by (instrument_token_id, amount_minor, currency), not by token: the predicate is about identical amounts, so the window belongs to the triple. Each key holds its own event-time-ordered deque.
  2. In-order, this is two pointers: on insert, pop from the front while front <= new - 10 min, then flag if the deque reaches length 3. Amortised O(1) per event, memory proportional to that key's window occupancy.
  3. Out-of-order arrival breaks append-only monotonicity, so insert in position instead. With lateness bounded at 60 seconds the insertion point is always near the tail, so a short sorted vector or a 600-bucket per-second ring keeps it O(w) with tiny w; a balanced tree per key is correct but over-built for a one-minute reorder.
  4. Drive decisions off a watermark of max(event_time seen) - 60 s, never off wall clock. Anything older than the watermark is too late to change an answer and is counted in a late_dropped metric. Without an explicit watermark you have still chosen a lateness policy — you just cannot state it or test it.
  5. Report-once needs its own state: per key, a set of already-flagged 10-minute window ids. A later event inside an already-flagged window must not re-flag, and a late event that completes a window you never flagged must flag — which is why state is retired at window close plus the lateness bound, not at window close.
  6. Size it: state lives 660 seconds, so at 3,000 events per second there are about 1.98 million in flight, plus one flagged-window set per active key. Bound total memory explicitly and shed by key age, and say what shedding costs — a shed key can miss a flag, making the threshold a product decision rather than a tuning knob.
Worked solution 40 min
  1. Write the key, the window, the watermark and the report-once rule in four lines before any code; every later bug is one of these left implicit.
  2. Implement per-key state as a sorted deque of event times plus a set of flagged window ids, both retired at watermark - 660 s.
  3. Fixture 1, in order: identical amounts at t, t+1 min and t+9 min produce one flag. A fourth at t+11 min evicts t+1 exactly under the half-open rule pop while front <= new - 10 min, leaving two events and no second flag; change that comparison to strict < and the same input flags twice. Pick one and write it in the spec.
  4. Fixture 2, out of order: deliver the same three events as t+9, t, t+1 and assert the flag fires on the third arrival with the same window id as fixture 1.
  5. Fixture 3, too late: the t event arrives 90 seconds after the watermark passed it, so no flag fires and late_dropped increments by one.
  6. Replay all three fixtures under twenty random arrival orders and assert the flag set is identical whenever every event is inside the lateness bound.
EXPECTED RESULTFixture 1 flags exactly once, at the t+9 event; the t+11 event does not flag under the half-open rule and does under the closed one, which is the boundary the fixture exists to pin. Fixture 2 yields the same single flag with the same window id. Fixture 3 yields no flag and `late_dropped == 1`. All twenty shuffles produce an identical flag set.
Follow-up
  • Make the threshold and window configurable without rebuilding all in-flight state on every change. What does that constrain in the data structure?
  • Two stream partitions hold events for the same token. What does that force the partitioning key to be?
  • An event arrives three hours late. Does any answer change, and who is told?

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.

Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.

Talk me through a time you had to work on a complex problem and how yo…

medium
behavioural and engineering judgement

Talk me through a time you had to work on a complex problem and how you solved it.

Approach
  1. Name the disagreement and how you resolved it with evidence.
  2. Close with what you would do differently, concretely.
  3. Give the blast radius: what could have broken, and what you measured.
Follow-up
  • What did you decide not to do, and why?
  • What would you do differently if you ran that again?

Tell me about a big problem you have solved and the trade-offs you mad…

medium
behavioural and engineering judgement

Tell me about a big problem you have solved and the trade-offs you made.

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 would you do differently if you ran that again?
  • How did you know your change caused the improvement?

When are you the happiest in your work?

medium
behavioural and engineering judgement

When are you the happiest in your work?

Approach
  1. Pick a story where you made the decision, not one where you watched it.
  2. Close with what you would do differently, concretely.
  3. Name the disagreement and how you resolved it with evidence.
Follow-up
  • How did you know your change caused the improvement?
  • What did you decide not to do, and why?

Describe a time you had conflict on a team and how you dealt with it.

medium
behavioural and engineering judgement

Describe a time you had conflict on a team and how you dealt with it.

Approach
  1. Pick a story where you made the decision, not one where you watched it.
  2. Give the blast radius: what could have broken, and what you measured.
  3. Close with what you would do differently, concretely.
Follow-up
  • How did you know your change caused the improvement?
  • What did you decide not to do, and why?
  • 01

    Talk me through a time you had to work on a complex problem and how you solved it.

  • 02

    Tell me about a big problem you have solved and the trade-offs you made.

  • 03

    When are you the happiest in your work?

  • 04

    Describe a time you had conflict on a team and how you dealt with it.

PracHub interview preparation framework ↗
Is this an official Lendable (UK) interview guide?

No. It is PracHub's own research and practice material for the Software Engineer role at Lendable (UK). Rounds and questions reflect what candidates have reported, not a process Lendable (UK) has published, and they change over time. Confirm the current format and scope with your recruiter.

PracHub interview research ↗
How long should I spend on the take-home test?

While the suggested time is often 2–4 hours, do not rush the design. Focus on quality, clean abstractions, and test coverage. It is better to build a smaller feature perfectly than a large feature poorly.

PracHub interview research ↗
What is the most common reason for rejection in the technical stage?

Candidates often fail by ignoring basic engineering standards, such as using floating-point numbers for money, missing test coverage, or failing to decouple business logic from the framework.

PracHub interview research ↗
Is the culture at Lendable (UK) collaborative?

Yes, the team values transparency and professional growth. They are looking for engineers who are eager to learn, contribute to technical discussions, and uphold high standards.

PracHub interview research ↗
How should I prepare for the "culture" interview?

Be prepared to discuss your past experiences with honesty. They value self-reflection; be ready to share a time you received constructive feedback or a time you failed and what you learned from it.

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.