ALLEN Digital Software Engineer Interview Questions & Guide 2026

The 60-second brief

As a Software Engineer at ALLEN Digital, you are at the forefront of transforming the education landscape through technology. This role is pivotal in building scalable, high-performance systems that support millions of students and educators. You will contribute to products that handle complex educational workflows, ranging from doubt-clearing systems to sophisticated billing and scheduling platforms.

The shape of the workload matters more for your prep than the industry label does. Read-heavy serving, write-heavy ingestion and scheduled batch processing have different binding constraints and fail in different places, so find out which one the team lives in before picking design topics.

Focus areas

• Bound blast radius with per-tenant concurrency limits
• Scope every query and cache key by tenant
• Build at-least-once pipelines with explicit deduplication horizons

Overview

The work at ALLEN Digital requires a balance of rigorous engineering practices and a user-centric mindset. You will be expected to design modular, testable, and extensible code while tackling real-world problems that directly impact the scale and reliability of the platform. This position is ideal for engineers who thrive in fast-paced environments and are passionate about building robust software architectures that can evolve alongside the company's rapid growth.

The ALLEN Digital interview process

ALLEN Digital candidates report 5 rounds · ≈ 4-6 weeks. The stages below are what candidates describe, not a published process.

Initial Technical Assessment

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

Hands-on Coding

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.

Architectural Design

You cannot drill a format you do not know, so put the preparation into material that travels. Three pieces of your own work, each rehearsed until you can take a follow-up you did not anticipate, will carry a conversation or a code walkthrough equally well. Specificity is what separates that from filler. A number needs its definition before it means anything: a p99 is over some window and measured at some hop, and a server-side figure excludes the queueing and network time a client would see. The number you cannot qualify is the one to leave out.

What to demonstrate

• Whether your examples carry detail only someone who did the work would hold, such as what the binding constraint actually was, which alternative you rejected and why it was worse, and what you measured on each side of the change
• Whether a number survives one follow-up, meaning you can say what it was measured over and whether it moved because of your change or merely alongside it
• Whether a failure is described with the specific change that followed it, rather than a lesson stated in general terms
• Whether your part in a team effort is stated accurately, including what other people did

How to prepare

• Write a page on each of three projects covering the constraint, the option you rejected, the measurement before and after, and what went wrong. Cut any line you cannot take a follow-up on, since you are writing the parts you will be pressed on rather than a summary.
• Recover the real figures while you still have access: request volume, data size, latency with its percentile and window, team size, timeline. Note where each came from, whether a dashboard, a design document or memory, and mark the estimates so you can say which they are out loud.
• Take your weakest project story to someone who works in a different area and have them ask why four times in succession. The point where you run out of answer is the part to go and re-read before the round.

Behavioral Discussions

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

Final Managerial Conversation

Coding rounds mostly set a floor. They decide whether you clear the bar, not where you land on the ladder. Level tends to come out of the design discussion and the ownership stories, so the question worth auditing beforehand is whether the scope you describe matches the scope of the job. Work that stops at your own service, or a story whose hard part was writing the code rather than getting several people to agree on an interface, reads a level below where you think you are interviewing, and that gap is usually resolved downwards.

What to demonstrate

• Whether the largest thing you describe owning ran end to end — the decision, the migration path, the rollout, and what you did when it went wrong — or stopped at the change you merged
• Whether design answers include what you would not build, what you would defer, and what you would measure before committing, rather than only what the boxes are
• Whether a disagreement in a story was settled with something checkable — a benchmark, a prototype, a written proposal — instead of by seniority or by waiting it out
• Whether you can say which calls you made alone and which you escalated, and why the line sat where it did

How to prepare

• Write your largest piece of owned work as a timeline of decisions — who decided what, when, and what you did when the plan broke — then delete every sentence whose subject is "we" and see how much survives
• Take one system you know well and drill the migration answer: how old and new paths run side by side under live traffic, how you compare their outputs, what the rollback is once writes are going to both, and which step you would not automate
• Map each line of the ladder in the job posting to a specific thing you have done, find the line you cannot support, and prepare the closest evidence you have plus an honest account of the gap

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.

Treating a timed-out write as a failed write

A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.

Not asking what the system looks like if it dies halfway through

For any multi-step write, say what state remains if the process stops between step two and step three, and what brings it back: a single transaction, a saga with compensating actions, an outbox, or a reconciliation job. Partial failure is routine at any real call volume, so 'that shouldn't happen' is an answer with nothing behind it.

Trusting input because it came from your own front end

Anything crossing a trust boundary is hostile: parameterise queries instead of building SQL by concatenation, validate against an allow-list rather than a deny-list, and bound the size of anything you allocate from a request. Raising this unprompted in an API or design question is a cheap and unusually strong signal.

Questions & practice

Order a job dependency graph and find its critical path

Medium · topological-sort · dag-longest-path · cycle-detection · critical-path

A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.

Approach

1. Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
2. Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
3. Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order: earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E).
4. Second pass in reverse topological order for latest_finish, then slack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain. slack[v] = 0 is exactly the statement that some longest path runs through v; equivalently, the longest path through v has length T - slack[v].
5. The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by min(d, T - L_avoid(v)), where L_avoid(v) is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, while min(10, 15 - 14) predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan is max(T - d, L_avoid(v)).
6. Compute L_avoid(v) the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, since L_avoid(v) only ever matters through that max: set duration[v] := 0, recompute the makespan as T0(v) = max(T - duration[v], L_avoid(v)), and the gain is min(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.

Follow-up

• Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
• Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
• Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?

Parse and verify a timestamped multi-signature webhook header

Easy · parsing · hmac · constant-time-compare · replay-protection

An inbound webhook carries a signature header of at most 1 KiB shaped t=<unix seconds>,v1=<64 hex chars>, with up to five v1 values during secret rotation and possibly unknown scheme keys. You hold the raw request body bytes and the currently active signing secrets. Write the parser and the verifier: accept when any active secret reproduces a signature and the timestamp is within a five-minute tolerance in either direction, reject otherwise. Single left-to-right pass over the header, no regular expression. State what is inside the MAC and why.

Approach

1. Parse in one scan: split on ,, then on the first = only, since a value may itself contain = under a future scheme. Accept t exactly once and treat a second t as a reject rather than last-wins. Push every v1 onto a short list and ignore any other key, so a v2 can be introduced later without breaking this verifier.
2. Say what is signed: HMAC-SHA256 over the exact byte string <t>.<raw body bytes>, yielding 32 bytes or 64 hex characters. The timestamp sits inside the MAC because otherwise an attacker replays yesterday's body with its still-valid signature and only has to edit the header timestamp.
3. Hash the bytes as received. Verifying against a re-serialised JSON body is the usual defect: key order, whitespace and number formatting all change the bytes while the parsed objects compare equal, so signatures fail for honest senders and the popular 'fix' is to stop checking.
4. Compare in constant time over fixed-length digests. Decode the hex to 32 bytes, accumulate acc |= a[i] ^ b[i] across the whole length, and test acc == 0 at the end. Evaluate every candidate without an early exit; at five candidates that is five HMACs over the body, linear in body size and negligible beside the network.
5. Apply the tolerance as a two-sided bound, rejecting when |now - t| > 300 seconds. A sender whose clock runs ahead of yours is an ordinary case, and an unbounded future timestamp is a free replay window.
6. Complexity: O(L) over the header producing k candidates, plus k HMACs at O(|body|) each. Space is O(k) beyond the body itself. Do the cheap rejections, including the tolerance check, before any cryptography runs.

Worked solution (15 min)

1. Write the grammar on one line before coding: header := field (',' field)*, field := key '=' value, split on the first = only.
2. Implement the parser to return {t: int, v1: [hex, ...]}, rejecting a missing t, a duplicate t, any v1 that is not 64 hex characters, and a header over 1 KiB, all before any cryptography runs.
3. Implement the verifier: for each active secret compute HMAC-SHA256(secret, f'{t}.'.encode() + raw_body), compare it in constant time against each parsed v1, and OR the results with no early exit.
4. Test with a valid signature; the same body with t moved 400 seconds into the past; the same body with t 400 seconds into the future; a header carrying an unknown v2= alongside a valid v1; and a body re-serialised with different JSON key order.

Expected result: The valid case accepts. Both out-of-tolerance cases reject, including the future one. The unknown v2 field is ignored and the v1 still verifies. The re-serialised body fails, which is correct and is exactly why the raw bytes must be retained.

Verify it yourself

• Timing the comparison over 10,000 mismatching signatures that differ at the first character versus the last shows no separation you can measure.
• Removing t from the MAC input makes an old captured body with its original signature pass. Write that test, watch it pass, then put t back.
• Every rejection except the signature mismatch itself returns before an HMAC is computed.

Follow-up

• The body is 40 MB. What changes about where you verify, and what can you do before the whole body has arrived?
• A customer reports that signatures fail for exactly the requests whose body contains a non-ASCII character. What is your first hypothesis?
• How do you rotate the signing secret with no failed deliveries, and how long do both secrets stay live?

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.

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?

Model credential revocation so history survives the delete

Easy · soft delete · partial indexes · check constraints · audit history

tenant_api_key stores key_id, tenant_id, workspace_id, name, key_prefix, secret_hash, scopes text[], status (active, revoked, expired, compromised), auth_version, created_at, expires_at, last_used_at, revoked_at, revoked_reason. Rotation inserts a new row and revocation never deletes, because an incident review asks which credential served a request last quarter. Write the constraints that enforce: a label is unique only among a tenant's live keys, revoked_at and status can never disagree, and scopes is never empty. Then write the authentication lookup predicate, and name one column in this table that must stay out of it.

Approach

1. Reach for a partial unique index rather than a plain UNIQUE: create unique index on tenant_api_key (tenant_id, name) where revoked_at is null. Any number of revoked rows may share a label, the live namespace stays unique per tenant, and the revoked majority is not in the index at all, so it stays small on a table that only grows.
2. Tie the nullable timestamp to the enum so the two cannot drift: check ((revoked_at is not null) = (status in ('revoked','compromised'))) and check ((revoked_at is null) = (revoked_reason is null)). A revocation that records no reason is the one an incident review cannot use.
3. Write the emptiness check as check (cardinality(scopes) > 0), not array_length(scopes, 1) > 0. array_length returns NULL for an empty array, a CHECK constraint passes when its expression is NULL, so the array_length version accepts exactly the value it was written to reject.
4. Make the lookup a single index probe with every liveness condition inside it: where secret_hash = $1 and revoked_at is null and (expires_at is null or expires_at > now()) and auth_version = $2, backed by a unique index on secret_hash. Nothing is filtered in application code, so there is no path that forgets a clause.
5. Keep last_used_at out of that predicate. It is written asynchronously and is allowed to lag by a minute, so it is a usage signal; feeding it into an authorisation decision makes the decision depend on a write that may be late, batched away or lost.
6. Flag the modelling smell while you are here: expired is derivable from expires_at < now(), so storing it as a status obliges a job to keep it true and guarantees the column is wrong between the expiry instant and that job's next run. Derive it in the predicate; keep the stored status for states that are decisions rather than clock readings.

Follow-up

• Rotation issues a replacement while the old key stays live for a 30-day overlap. What does the uniqueness rule become, and what does the UI show to tell two same-named keys apart?
• A password reset bumps the principal's auth_version. No row in this table changed. How does the next request fail, and what query counts how many keys that bump just killed?
• A key turns up in a public repository. Which columns let you find it, and what do you write to the row?

Find the join that inflates every invoice total

Medium · join fan-out · aggregation grain · sql correctness

invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.

Approach

1. Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row: sum(l.amount_minor) is the true total multiplied by the attempt count, and count(p.attempt_id) is attempts times lines. Three lines and two attempts report double the money and six attempts.
2. Reject the reflex repair. count(distinct p.attempt_id) does fix the count, because attempt_id is unique. sum(distinct l.amount_minor) does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match.
3. Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
4. Keep invoices with no attempts by making the attempt branch a LEFT JOIN with coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about.
5. Push each filter to its own grain: where l.voided_at is null belongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants.
6. Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.

Worked solution (25 min)

1. Create one invoice with three lines of 1000, 1000 and 500 minor units and two payment attempts, then run the original query.
2. Confirm it reports 5000 and 6 rather than 2500 and 2.
3. Apply sum(distinct l.amount_minor) and confirm the total becomes 1500, which is worse rather than better.
4. Write the two-CTE version with a LEFT JOIN and coalesce, and confirm 2500 and 2.
5. Add a second invoice with lines and no attempts and confirm it still appears.

Expected result: The original sum equals the true line total multiplied by the number of payment attempts and the count equals attempts times lines; the corrected query pre-aggregates each branch to invoice grain in separate CTEs and LEFT JOINs them with coalesce, with voided lines excluded inside the line CTE and the tenant predicate applied to both branches.

Verify it yourself

• The money total is unchanged by adding or removing payment attempts.
• An invoice with zero attempts appears with its total and a zero count.
• Two lines with identical amounts both contribute to the total.
• Voiding a line changes the money figure and leaves the attempt count alone.

Follow-up

• Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with filter (where ...) be better?
• Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
• The same report is needed per tenant per month. What index makes the line CTE cheap?

How would you structure a system to handle concurrency?

Medium · system design and architecture

Approach

1. Name the read and write paths separately; they rarely have the same bottleneck.
2. State the consistency you need, and where you are willing to be stale.
3. Choose a partition key and say what query it makes expensive.

Follow-up

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

Discuss your approach to modularizing service and controller layers.

Medium · system design and architecture

Approach

1. Choose a partition key and say what query it makes expensive.
2. Name the failure you are designing for, then the recovery path.
3. State the consistency you need, and where you are willing to be stale.

Follow-up

• How does this behave when that dependency is down for an hour?
• What breaks first when traffic grows ten times?

Build a Cab Billing System (managing drivers, vehicles, and trip calculations).

Medium · system design and architecture

Approach

1. Choose a partition key and say what query it makes expensive.
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

• How does this behave when that dependency is down for an hour?
• What would you drop to keep the system up under load?

Design a Slot booking review system for online doubt clarification.

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?

Walk me through a problem involving train scheduling.

Medium

Approach

1. State your assumptions explicitly before working the problem.
2. Clarify what is being asked and what a complete answer contains.
3. Say what you would check first and why it is the highest-information step.

Follow-up

• What assumption would you test first?
• How would you know your answer was wrong?

How do React hooks function, and specifically, what is the purpose of the useState hook?

Medium

Approach

1. Say what you would check first and why it is the highest-information step.
2. Clarify what is being asked and what a complete answer contains.
3. State your assumptions explicitly before working the problem.

Follow-up

• What assumption would you test first?
• How would you know your answer was wrong?

Per-tenant rate limiting that holds across one hundred pods

Medium · rate limiting · hot keys · atomicity · write skew

The same gateway enforces each tenant's plan rate limit in requests/second and its monthly quota. Traffic is heavily skewed: a handful of tenants exceed all others combined, while a small tenant's requests may all land on one of the 120 pods. Both limits must hold in aggregate rather than per pod, per credential or per region, and the check may add at most 2 ms at p99. Specify the algorithm, where the state lives, the exact operation performed per request, and the behaviour when the counter store is unreachable - separately for the rate limit and the quota.

Approach

1. Do the arithmetic on per-pod buckets before discarding them, and be precise about which half fails. Giving each pod rate/N keeps the sustained aggregate correct - 120 buckets of 5 rps sum to the 600 rps limit - but a tenant whose traffic all lands on one pod is throttled at 5 rps, a hundred and twentieth of what it pays for. The over-admission comes from burst depth, not sustained rate: if each pod carries a full bucket depth B, the aggregate burst is 120B against a limit that intended to allow B.
2. Move the rate limit to a shared counter accessed in exactly one atomic round trip - a token bucket or GCRA as a single server-side script returning allow/deny plus a retry-after - never a read followed by a write. Keep the key tenant-scoped and region-local; a globally shared counter costs a cross-region round trip that alone exceeds the 2 ms budget, so shard the global limit per region in proportion to observed regional share and accept that a tenant whose traffic shifts regions is briefly throttled until the shards rebalance.
3. Cut round trips for the largest tenants with leased tokens: a pod leases 50 at a time and spends them locally, reducing counter round trips by 50x at the cost of bounded over-admission of pods x lease_size, which is 6,000 requests in the worst case here. Shrink the lease for tenants with small limits, where that bound would swamp the limit itself.
4. Treat the quota as a different problem, not a longer window. It is a count that other in-flight requests are changing, so select-the-usage-then-insert is write skew: read-committed permits it and repeatable read permits it too, because snapshot isolation is exactly what allows two transactions to read a consistent count and both write. Enforce it inside one statement - an atomic increment whose returned value is compared, or UPDATE ... SET used = used + $1 WHERE used + $1 <= limit RETURNING - or with a constraint that makes the surplus write fail.
5. Write the unavailability policy per limit rather than one policy for both. The rate limit fails open to a conservative local bucket, because it protects capacity and failing closed turns a counter blip into a total outage. The quota fails open for tenants whose last known usage was well below the cap and closed for those at or above it, since that is the only decision with revenue attached, and reconciles from the usage ledger on recovery - which is why the monthly counter must be re-seedable from that ledger rather than being the only copy.

Worked solution (30 min)

1. For a 600 rps limit across 120 pods, compute the sustained rate and the aggregate burst under local buckets of depth B, then the throughput of a tenant whose traffic all lands on one pod.
2. Write the shared-counter operation as a single atomic script: its inputs, its return value, and the number of network round trips per request.
3. Size a lease of 50 tokens: compute the worst-case over-admission as pods x lease_size and the reduction in round trips, then decide the smallest limit at which the lease is still acceptable.
4. Write the two unavailability policies side by side and the reconciliation step that runs when the counter store returns.

Expected result: A per-region shared token bucket keyed (tenant, region, window) executed as one atomic script at roughly one sub-millisecond intra-region round trip; the monthly quota enforced by a single atomic increment-and-compare that is re-seedable from the usage ledger; leases only for high-limit tenants with over-admission bounded at 120 x 50 = 6,000 requests; the rate limit failing open to a conservative local bucket and the quota failing closed only above the last known cap.

Verify it yourself

• Separate the two halves of the local-bucket claim: sustained aggregate is correct at rate/N per pod, burst is not. If your answer says local buckets let a tenant exceed its sustained rate, the arithmetic does not support it.
• Re-read your quota check and confirm no branch reads the counter and then writes it. If the predicate is not inside the write or enforced by a constraint, the race is still there.
• Compare worst-case lease over-admission against the smallest tenant limit you apply leases to. If 6,000 exceeds that tenant's per-minute allowance, the lease size is wrong for that tier.

Follow-up

• One tenant is 40% of all traffic and its counter key lands on a single shard. What do you change, and what does that cost in accuracy?
• Clients receive 429s and retry. What stops the well-behaved ones from synchronising into a herd?
• A tenant reports being throttled below its limit. Which evidence do you produce, and does your design emit it today?

Regional error rate explodes after a dependency merely slows

Hard · retry storm · circuit breaker · partial outage

A control-plane read replica in one region degrades from 4 ms to 120 ms. Within ninety seconds that region's gateway error rate rises from 0.01% to 40% and its p99 becomes bimodal, one mode near the old p99 and one at the client timeout. The other two regions are unaffected. The gateway retries control-plane reads three times with exponential backoff and no jitter. Give an ordered checklist that separates trigger from amplifier, the offered-load arithmetic, and the controls that break the loop.

Approach

1. Split the incident into three questions before touching a control: what started it, what amplified it, and what would make recovery slow. Here they are the replica slowdown, the retry policy interacting with queueing, and a synchronised unjittered herd at recovery. They are different mechanisms and each needs its own control.
2. Read the distribution rather than the mean. A bimodal p99 with one mode pinned at the client timeout is two populations, not one degraded path; split latency by cache hit and miss and confirm the fast mode is hits and the timeout mode is misses that reached the replica.
3. Do the load arithmetic. Three retries turn one client request into up to four upstream requests, so offered load reaches roughly 4x on a dependency that is already slower, and it arrives at the worst moment. With utilisation approaching one, queueing delay grows superlinearly, which is why a 30x latency increase upstream does not produce a 30x increase downstream, it produces timeouts.
4. Break the loop with controls that bound offered load rather than with more attempts: a concurrency limit on the control-plane client so at most N calls are in flight and the remainder fail fast, a circuit breaker scoped per dependency and region, and a retry budget capping retries at a small fraction of base traffic so amplification has a ceiling that does not depend on how many clients are retrying.
5. Add full jitter to whatever retries survive, sleeping uniformly in [0, min(cap, base x 2^attempt)], so attempts de-correlate instead of arriving in waves aligned to the moment of failure.
6. Decide the unreachable-dependency behaviour in advance, because it is the actual product decision underneath: serving from an expired credential cache keeps the product available while extending a revoked key's life past the stated bound, and failing closed converts a dependency degradation into a total outage. State the mode and the staleness number rather than letting the timeout choose.

Follow-up

• The replica recovers. Describe what happens in the first ten seconds with your controls in place versus without them.
• Which single metric would have paged before the error rate moved, and why is upstream latency by itself not it?
• Requests that fail fast under the concurrency limit still need an answer. What does the gateway return, and what does it do to the usage event it would otherwise have emitted?

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.

Risky changes and the reversal path

For anything that touched live traffic, be ready to say how you would have undone it: a flag, a staged rollout, dual writes with the old path still authoritative. Once the old column is dropped or the source rows are overwritten there is no reverse, so name what you kept a copy of and for how long.

Walk me through your past work experience and the technical challenges you solved.

Medium · behavioural and engineering judgement

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. 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 was the biggest hurdle you faced in your work, and how did you overcome it?

Medium · behavioural and engineering judgement

Approach

1. Give the blast radius: what could have broken, and what you measured.
2. State the situation in two sentences and spend the rest on the reasoning.
3. Pick a story where you made the decision, not one where you watched it.

Follow-up

• What did you decide not to do, and why?
• What would you do differently if you ran that again?

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?