F5Networks Software Engineer Interview Questions & Guide 2026

The 60-second brief

This guide covers what a Software Engineer at F5Networks is expected to do and how to prepare for the interview.

Find out what you will be typing into. A shared plain-text editor with no autocomplete, compiler or test runner changes what you have to hold in your head, and practising inside your own configured environment hides exactly that gap.

Focus areas

• Evolve APIs without breaking pinned SDK clients
• Build at-least-once pipelines with explicit deduplication horizons
• Make every write idempotent under client retries

The F5Networks interview process

F5Networks candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Phone Screening

The person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.

What to demonstrate

• Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
• Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
• Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural

How to prepare

• Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
• Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
• Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one

Onsite Interview

Nobody in the room with you decides this. Interviewers typically write their rounds up separately, often before seeing anyone else's, and the outcome is settled later from those write-ups. A split panel gets resolved by whichever note carries specific evidence, so what you want out of each room is one concrete thing that person could write down: a bug you caught yourself, a trade-off you named, a decision you owned. The rest is arithmetic. The project you describe in a behavioural conversation is often the same system you sketched an hour earlier, and the two accounts have to agree.

What to demonstrate

• Whether the scale, team size and timeline you attach to a project hold steady when that project resurfaces in a different round
• Whether each interviewer leaves with a specific thing to cite rather than a general impression of competence
• Whether a trade-off you defended in one round survives a challenge in another, instead of being quietly swapped for the answer the new interviewer seemed to want
• Whether a question you have already answered earlier in the day gets the same answer at the same depth, without visible impatience

How to prepare

• Write a one-page sheet per project fixing the figures you will quote — request volume, data size, team size, elapsed time, what broke — and say them aloud from the sheet until they come out identical every time
• For each round on the schedule, decide in advance the one sentence you want in that person's notes, then check in a mock that you said it outright instead of leaving it to be inferred
• Have someone ask you the same project question twice, an hour apart, and diff the two answers for numbers that moved or a trade-off that reversed

Where candidates lose points

Serialising a tenant's writes through select ... for update on a single counter row

It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.

Holding money in a floating-point type, or rounding it more than once

Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.

Abandoning working code to chase the optimal solution

Get the straightforward version correct, state its complexity, and only then optimise, keeping the working version until the faster one passes the same cases. A correct quadratic solution with a stated path to linear beats a half-written optimal one that never ran.

Sorting when the problem never required a total order

Match the algorithm to the guarantee actually needed: the top k comes from a size-k heap in O(n log k) time and O(k) space, distinctness needs a set rather than an ordering, and a small bounded integer key range admits a linear counting pass. A full O(n log n) sort is the right default only when you genuinely need everything in order.

Questions & practice

Given a singly linked list and a pointer to an arbitrary node, delete that node in O(1) time without traversing from the head.

Medium · data structures and algorithms

Approach

1. Walk one small example through your approach before writing the whole thing.
2. Name the brute-force solution and its complexity before improving on it.
3. Restate the input: its shape, its size, and what is guaranteed about it.

Follow-up

• What is the worst case, and how likely is it on real data?
• How does this change if the input no longer fits in memory?

Given an array of integers where one element is missing, write a function to find the missing integer in O(N) time.

Medium · data structures and algorithms

Approach

1. Walk one small example through your approach before writing the whole thing.
2. Choose the data structure from the access pattern, not from familiarity.
3. Restate the input: its shape, its size, and what is guaranteed about it.

Follow-up

• How does this change if the input no longer fits in memory?
• Which test case would catch an off-by-one here?

Find the contiguous subarray within a one-dimensional array of numbers that yields the maximum sum.

Medium · data structures and algorithms

Approach

1. Walk one small example through your approach before writing the whole thing.
2. Choose the data structure from the access pattern, not from familiarity.
3. State the target complexity and say which constraint rules the naive version out.

Follow-up

• Which test case would catch an off-by-one here?
• How does this change if the input no longer fits in memory?

Determine whether a given binary tree is height-balanced (where the height difference between left and right subtrees never exceeds 1).

Medium · data structures and algorithms

Approach

1. Restate the input: its shape, its size, and what is guaranteed about it.
2. State the target complexity and say which constraint rules the naive version out.
3. Name the brute-force solution and its complexity before improving on it.

Follow-up

• Which test case would catch an off-by-one here?
• How does this change if the input no longer fits in memory?

What is the output of sizeof on an empty C++ class, and why does the compiler enforce this memory allocation rule?

Medium · languages, concurrency and fundamentals

Approach

1. Name what is shared across threads and what owns each piece of state.
2. Identify the window where an invariant is briefly untrue.
3. Reach for the cheapest primitive that closes the race, not the broadest lock.

Follow-up

• Where could this allocate more than you expect?
• What happens if two callers reach this at the same time?

Explain the memory layout of a compiled C/C++ program (Stack, Heap, BSS, Data, Text segments) and how dynamic heap allocation functions.

Medium · languages, concurrency and fundamentals

Approach

1. Identify the window where an invariant is briefly untrue.
2. Distinguish a value from a reference to it, and say which one you handed out.
3. Say what the runtime actually does before reasoning about the code.

Follow-up

• How would you prove the race exists rather than suspect it?
• Where could this allocate more than you expect?

Locate a billing reconciliation gap without rescanning ninety million events

Hard · reconciliation · dimensional-bisection · watermarks · hypothesis-testing

A tenant's sealed invoice total is 0.4% below the sum of its raw usage_event rows for the period. That tenant has 90 million events over 30 days in a table partitioned daily on ingested_at, and its rollups carry source_max_ingested_at, revision and sealed_at. Recomputing all 30 days from raw is correct, and you are not going to do it. Give the procedure that locates the divergent (workspace, sku, hour) cell, the cost of each probe, and the one query you run before any of it.

Approach

1. Run the free query first. Sum raw quantity for the period restricted to ingested_at <= source_max_ingested_at of the sealed rollups, and compare that against the unrestricted sum. The rollup stores the watermark precisely so this can be answered without a scan. If the whole 0.4% sits above the watermark, nothing is broken: it is late data, it becomes an adjustment line, and the investigation ends in one query.
2. Only if the gap survives that test do you bisect, and you bisect by dimension rather than by rows. Compare 30 per-day totals, then inside the offending day compare the 6 SKUs, then the workspaces, then the 24 hours. That is roughly 30 + 6 + W + 24 grouped probes, each an indexed range scan over one daily partition for one tenant, against O(N) per attempt for the naive re-fold.
3. Quantify why naive is not merely slow but unusable mid-incident: at a generous 200,000 rows/second sequential, 90 million rows is about 7.5 minutes per attempt, you will want ten attempts, and every one competes for I/O on the same partitions live ingest is writing. The diagnostic worsens the backlog it is diagnosing.
4. Before fetching each comparison, state what it would look like under each hypothesis. Two adjacent hours off by equal and opposite amounts is occurred_at versus ingested_at bucketing. A whole day offset by exactly N hours is a timezone applied at the wrong layer. A gap confined to one SKU in one workspace is an environment filter. The same (tenant_id, idempotency_key) present in two ingested_day partitions is the dedup horizon losing a retry that crossed midnight.
5. Make the next bisection cheap by storing the aggregate you keep recomputing. A per-(tenant_id, ingested_day) count and quantity checksum turns step two from thirty probes into one read, and it is the same number the reconciliation job already produces.
6. Whatever you find, the sealed period does not change value. The correction is an adjustment line pointing at the line it reverses, carrying its own source_rollup_watermark, because the original invoice is the evidence of what the customer was charged.

Worked solution (35 min)

1. Reproduce the shape locally: generate 2 million events for one synthetic tenant over 5 days, fold them, then inject three defects, namely 0.2% of events bucketed by ingested_at, a handful of duplicate idempotency_key values whose retries cross midnight, and a block of events ingested after the seal.
2. Write the watermark-bounded query first and record how much of the gap it explains on its own.
3. Bisect by day, then SKU, then hour, recording the probe count and the rows each probe touches.
4. For each located cell, write down the predicted signature before querying it, then check whether the data matches the prediction.
5. Time a full re-fold of the 2 million rows and extrapolate to 90 million.

Expected result: The watermark-bounded query accounts for the post-seal block entirely and removes it from the investigation. The ingested_at bucketing appears as adjacent hours off by equal and opposite amounts. The midnight-crossing duplicates appear as one (tenant_id, idempotency_key) present in two ingested_day partitions. Bisection reaches each cell in fewer than 70 probes.

Verify it yourself

• The three injected defects sum to the original gap with no unexplained remainder.
• Each probe's row count is bounded by one day-partition for one tenant, not by the period.
• Removing the source_max_ingested_at bound from the first query makes a gap appear, and grow, while ingest runs, even after every defect is removed.

Follow-up

• The gap is 0.4% in one direction on one day and 0.4% the other way the next day. What does that shape rule in, and what does it rule out?
• How do you distinguish a duplicate from a restatement, given revision and recomputed_at on the rollup?
• Ingest is still running while you investigate. What makes your two numbers comparable at all?

Decide which facts an invoice line copies instead of joining

Medium · denormalisation · immutable records · foreign keys · rounding

invoice_line_item already denormalises tenant_id, which is reachable through invoice_id, and stores amount_minor even though quantity times unit_price_micros would recompute it. A reviewer asks you to normalise both away, and separately asks whether the tenant's legal name and billing address should be copied onto the invoice header. Decide each case. For every field you keep denormalised, name the read pattern or the invariant that justifies it, the anomaly the copy can develop, and the mechanism that prevents that anomaly here.

Approach

1. Split the question into two kinds of copy, because they fail differently. A copy of a currently mutable fact is a cache: it drifts and needs invalidation. A copy of a fact frozen at write time is not a cache at all, it is the record of what happened, and normalising it away destroys information the source no longer holds.
2. Keep tenant_id on the line. It costs 8 bytes, it leads every index on the table so no read is ever accidentally cross-tenant, and it turns a wrong join into an empty result rather than another tenant's money. Prevent the drift structurally: a unique constraint on invoice (invoice_id, tenant_id) plus a composite foreign key from the line on (invoice_id, tenant_id) makes a mismatched pair impossible, so the database enforces agreement instead of a code review.
3. Keep amount_minor. Rounding must happen exactly once, at a named site, with a stated mode (half-even here). If readers recompute from quantity and unit_price_micros, every reader owns a rounding decision, and half-up and half-even diverge systematically across thousands of lines rather than cancelling out. A check constraint can bound the stored value but deliberately cannot re-derive it.
4. Copy the legal name and billing address onto the invoice header, written once and never updated. The statement must show what was true when it was sealed, and the tenant record will change afterwards. This is a snapshot for the same reason source_rollup_watermark is stored per line: without it, nobody can reconstruct what the customer was told.
5. Name the read pattern that pays for all of it. Rendering, dispute response and export are per-tenant, per-period reads over thousands of lines that would otherwise join back to slowly changing dimensions that no longer hold the historical value. The write side is a once-per-period batch, so the extra columns cost nothing that matters.
6. Concede the case where the reviewer is right: a mutable operational attribute such as the tenant's current plan name has no business on a line. If a report wants it, join. If a statement needs the plan as of the period, that is another snapshot and it belongs on the header with the rest.

Follow-up

• Write the composite foreign key and the unique constraint it requires on the parent. What does it cost on every line insert, and what does it do to a bulk load?
• A tenant is renamed after being invoiced. Which rows change, and what does the customer see on last quarter's PDF?
• Where does currency live, and what breaks if a tenant's billing currency changes between two periods?

Paginate a tenant's delivery export without skipping rows

Easy · keyset pagination · indexing · multi-tenancy

A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.

Approach

1. Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
2. Replace the position with a value predicate over a stable, unique, indexed ordering: where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window.
3. Index (tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit.
4. Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
5. Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.

Worked solution (20 min)

1. Load about 50k deliveries for one tenant, then walk them with the OFFSET query while a writer inserts 10 rows/second, collecting every returned delivery_id.
2. Compare the distinct ids collected against the set of ids that existed when the walk started, and record the shortfall.
3. Repeat the walk with the keyset query and confirm every pre-existing id is returned exactly once.
4. Run explain (analyze, buffers) on page 1 and page 500 of each form and compare shared buffer hits.

Expected result: A keyset query using the row comparison (created_at, delivery_id) < ($2, $3) with order by created_at desc, delivery_id desc limit 100, backed by (tenant_id, created_at, delivery_id) scanned backwards; plus the statement that OFFSET drops rows under concurrent insert and costs time proportional to the offset, and that updated_at is disqualified because it mutates on retry.

Verify it yourself

• Buffer counts for page 500 are the same order of magnitude as page 1.
• Every id present before the walk is returned exactly once when only inserts run concurrently.
• The query plan shows no Sort node above the index scan.
• Two rows sharing a created_at value to the microsecond are both returned across the page boundary.

Follow-up

• The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added created_at <= $snapshot bound, and a materialised export table.
• A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
• The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?

Explain the step-by-step process of the TCP three-way handshake (SYN, SYN-ACK, ACK) and describe state transitions on client and server sockets.

Medium · system design and architecture

Approach

1. State the consistency you need, and where you are willing to be stale.
2. Name the read and write paths separately; they rarely have the same bottleneck.
3. Name the failure you are designing for, then the recovery path.

Follow-up

• What would you drop to keep the system up under load?
• What breaks first when traffic grows ten times?

How does SSL/TLS termination operate at the load balancer level, and what steps occur during an encrypted handshake?

Medium · system design and architecture

Approach

1. Name the failure you are designing for, then the recovery path.
2. State the consistency you need, and where you are willing to be stale.
3. Fix the scope first: who calls this, how often, and what they do when it fails.

Follow-up

• What would you drop to keep the system up under load?
• What breaks first when traffic grows ten times?

Publish limit semantics that clients can back off against

Medium · rate limiting · quotas · degraded mode

The gateway enforces three separate limits: a per-tenant token bucket, a monthly plan quota, and a cap on concurrent in-flight requests. Tenants hold several credentials and call from three regions, and today they see undifferentiated 429s. Design the client-facing contract: which headers carry which limit, the status codes that distinguish slow down from out of plan allowance from too many in flight, and what a well-behaved SDK does for each. Also state what the gateway returns when the shared counter store is unreachable, and what bound that choice implies.

Approach

1. Separate the three objects before naming a header. A bucket refills continuously, a quota does not refill until the period rolls, and a concurrency cap clears when an in-flight request finishes. They have different remedies, so they cannot share a status code.
2. Map them accordingly: 429 with Retry-After for the bucket, where waiting works; 429 with a distinct code and a reset hint for concurrency, where waiting works only if something else completes; and 402 or 403 with a plan code for quota, where waiting never works and the SDK must surface it rather than sleep on it.
3. Emit RateLimit-Limit, RateLimit-Remaining and RateLimit-Reset for the bucket only, and document Remaining as advisory. It comes from a shared counter that other requests are changing concurrently, so a client treating it as a reservation has built a race.
4. State the aggregation in the docs: the limit applies per tenant across every credential and every region. The cheap per-pod bucket of rate/N is correct only when traffic spreads evenly, and a tenant whose connections land on a few pods is throttled well under its published limit while a widely spread one exceeds it.
5. Decide the degraded mode in advance and publish it. Failing open serves unmetered traffic during a counter outage; failing closed converts a counter outage into a total outage. Pick one, bound it, for example a local fallback bucket at a fraction of the limit for the duration, and say so in the contract.
6. Specify SDK behaviour: honour Retry-After over local backoff, apply full jitter otherwise, never retry 402 or 403, and cap total attempts so retries expire before the caller's own deadline.

Worked solution (25 min)

1. Write a three-row table of limit, refill behaviour, status code, headers, and the client's correct action.
2. Write the exact header set for one throttled response and one quota-exhausted response, showing that they differ.
3. Write the documentation sentence that states aggregation across credentials and regions, and the sentence marking Remaining advisory.
4. Choose and justify the counter-store failure behaviour, then state the numeric bound it puts on overage or on availability.
5. Write the SDK's decision function: given status and headers, return sleep duration or surface to the caller.

Expected result: Three distinguishable responses, with Retry-After authoritative on the two retryable ones, quota exhaustion never retried, Remaining documented as advisory, and a named degraded-mode behaviour carrying an explicit overage or availability bound.

Verify it yourself

• A client can tell the three conditions apart without parsing prose.
• Exactly one of the three conditions is documented as never worth retrying.
• The aggregation statement mentions credentials and regions, not just the tenant.
• The degraded-mode answer names a number, not only a direction.

Follow-up

• Two services under the same tenant each read Remaining = 50 and each send 50 requests. What did the contract promise, and what actually happens?
• How would you keep a burst from a staging workspace out of the production workspace's share of the same tenant's bucket?

Invoice detail latency triples after an ORM relationship refactor

Easy · n+1 query · orm · latency regression

An invoice detail endpoint returned in 40 ms at p99 last week. After a refactor replaced a hand-written join with ORM relationship access it returns in 1.4 s, and the regression grows with the number of invoice_line_item rows on the invoice. Database CPU rose, but no statement in the slow-query log exceeds 3 ms. You have request traces with per-span SQL, the ORM statement log, and a staging copy of the data. Produce an ordered diagnostic checklist, the measurement that confirms the cause before any code change, and the fix.

Approach

1. Count statements per request before reading any statement duration. A slow-query log hides this class by construction, because every individual query is fast and only their number is wrong; take one trace and count SQL spans.
2. Establish proportionality rather than asserting it: sample invoices with 5, 20, 60 and 200 line items and plot statements per request against line count. A straight line of slope 1 through an intercept of one or two identifies a lazy relationship load, and no index or cache would move that line.
3. Locate the emitting attribute access in the refactored code and check whether the same shape repeats one level deeper, for instance a tax or adjustment collection hanging off each line, which turns the cost quadratic.
4. Fix with a bounded statement count: either one join that fetches invoice and lines together, or two statements where the second is WHERE invoice_id = $1 AND tenant_id = $2. Keep tenant_id in the predicate so the read stays tenant-scoped even though invoice_id already implies it.
5. Choose between the two deliberately: the join duplicates the wide parent row across N children on the wire, the two-statement form avoids that for one extra round trip. Prefer the join for narrow parents and the split for wide ones.
6. Pin it with a per-request statement-count assertion in a test that varies line count, because a latency assertion passes on a small fixture and would not have caught this.

Follow-up

• The endpoint now also needs per-line tax rows. Show the shape that keeps statement count constant instead of reintroducing the same defect one level down.
• How does this change if a transaction-pooling proxy sits between the service and the database, so each statement may land on a different backend session?
• The same page paginates invoices with LIMIT and OFFSET. Why is that a second, independent defect, and what replaces it?

Suggested practice checklist

For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.

Day 1 — Rebuild the primitives by implementing them

• Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
• Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
• For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.

Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.

Day 2 — Arrays under an invariant: two pointers, sliding window, binary search

• Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
• Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
• Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.

Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.

Day 3 — Sorting, heaps, and the greedy argument that has to be proved

• Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
• Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
• Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.

Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.

Day 4 — Recursion, memoisation, and the step to a table

• Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
• Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
• Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.

Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.

Day 5 — Graphs, where most of the work is choosing the traversal

• Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
• Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
• Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.

Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.

Day 6 — One day for everything that is not an algorithm

• Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
• Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
• Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.

Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.

Day 7 — Solve out loud, under time

• Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
• Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
• Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.

Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.

Outage answers are graded on instruments

Nobody is scoring your stamina at three in the morning. What carries weight is which signal told you something was wrong, what you measured before touching anything, what you rolled back versus what you fixed forward, and why you picked one. 'We restarted it and it went away' is a story about not knowing.

Describe a project where you had to quickly learn an unfamiliar technology stack or programming language to deliver a critical feature.

Medium · behavioural and engineering judgement

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. 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?

Argue against failing open when the control plane is unreachable

Hard · revocation · cache staleness · fail-open vs fail-closed · influence

The gateway caches credential-to-context decisions with a sixty-second TTL. A design proposal says that when control-plane reads fail, pods should keep serving from expired entries indefinitely so a control-plane outage never becomes a product outage. You believe that converts every revocation into an unbounded one. Describe a design you argued against while it was still a live proposal: what you measured or modelled to make the case, what you conceded, who decided, and what happened afterwards. Say what would have changed your mind before the decision, not after it.

Approach

1. Reframe it from a values argument into a bounded-staleness argument. Both sides already accept the cache; the disagreement is only about the ceiling on how long a revoked credential keeps authorising. Put a number on the table — serve stale for up to fifteen minutes, then fail closed — and make the other side argue against a number rather than against a principle.
2. Bring arithmetic rather than adjectives: the rate of revocations with revoked_reason in ('suspected_leak','auth_version_bump'), the observed distribution of control-plane unavailability, and the product of the two, which is expected requests served by revoked credentials per outage-hour. At 30k requests/second the unbounded version is not a subtle exposure and the number says so.
3. Concede the strong half of the opposing case first, because that is what buys you the room: failing closed turns one service's outage into a total outage across three regions, and a control plane doing tens of writes per second is not engineered to the gateway's availability target. A proposal you have not steelmanned reads as reflex.
4. Propose the asymmetry that usually resolves this: stale entitlements cost bounded money (a quota fifteen minutes out of date over-serves by a computable amount), while a stale revocation costs unbounded access. Split the cached decision by what it authorises, give the two halves different staleness ceilings, and let the entitlement half fail open while the revocation half fails closed.
5. State the propagation dependency plainly, since it is the part that is missed: validity is also derived from the principal's auth_version, so password reset and sign-out-everywhere flow through this same cache. A design that bounds staleness for explicit revocation and not for auth_version bumps has only fixed half of it.
6. Say who decided, and what you did afterwards in either outcome: write the decision down with its number and a review date, and instrument the exposure you were worried about so the next round of the argument is settled by data instead of by seniority.

Follow-up

• Publish-subscribe invalidation is lossy under a partition, and a TTL is the only hard bound. What TTL do you pick, and what does it cost you at 30k requests/second?
• The key was revoked because it was found in a public repository. Does your answer change, and where does that urgency live in the design?
• You lost the argument and six weeks later the failure you predicted happens. What do you say in the review, and what do you not say?

Resolve a review disagreement over a quota check

Easy · code review · isolation levels · write skew · disagreement

A colleague's pull request enforces a per-tenant quota by selecting the current count and then inserting when it is under the limit. You flag it as a race. They reply that the transaction already runs at repeatable read, so the snapshot makes it safe, and the tests pass. Walk through taking that disagreement to a resolution: what you write in the review, what you demonstrate rather than assert, which fix you propose and why, and what you do if they still disagree after all of it.

Approach

1. Answer the claim precisely instead of restating your objection, because they have made a specific technical argument. In PostgreSQL, repeatable read is snapshot isolation; this is write skew, which snapshot isolation permits by design. Both transactions read a count that is stable within their own snapshot, insert disjoint rows that the other cannot see, and both commit, so the limit is exceeded by exactly the concurrency.
2. Demonstrate rather than cite. Two psql sessions, both BEGIN ISOLATION LEVEL REPEATABLE READ, both select the count, both insert, both commit: it succeeds. Repeat at SERIALIZABLE and the second commit fails with serialization_failure, SQLSTATE 40001. That takes two minutes, ends the argument without anyone conceding a position, and leaves an artefact for the next reviewer.
3. Offer the options with their costs rather than a verdict. Serialisable plus a retry loop on 40001 is correct but obliges every caller to retry and degrades under contention. An increment-and-compare on a counter row — update tenant_quota set used = used + 1 where tenant_id = $1 and used < limit returning used — is safe even at read committed, because a blocked updater re-evaluates the WHERE clause against the row version it finally locks, and zero rows returned means full. A unique or exclusion constraint that makes the surplus write fail is the third.
4. Name the plausible non-fix explicitly, since it is what usually gets merged instead: folding the count into the insert as insert ... select ... where (select count(*) ...) < limit is still racy under read committed, because the subquery cannot see the other transaction's uncommitted rows. It looks atomic and is not.
5. Say what you do if they still disagree: escalate the decision rather than the disagreement. Attach the reproduction, hand it to the service owner or a third reviewer, and state that you will not block the merge if the owner accepts the risk knowingly — and that you want that acceptance written down.
6. Close with the general lesson worth leaving in the review thread: a passing suite is weak evidence for a concurrency claim because it runs one request at a time. Ask for a test that runs two.

Follow-up

• Write the counter-row version. Does your answer change if the quota counts child rows rather than a column?
• Under serialisable, who performs the retry, and what does the API client see if the retry also fails?
• This is the third disagreement with the same reviewer this month. What changes in how you review?