cswork / libraryOpen practice workspace

Interview guides · OpenText

OpenText Machine Learning Engineer Interview Questions & Guide 2026

OpenText Machine Learning Engineer interview guide: reported rounds, practice questions with worked solutions and a 7-day prep plan.

The 60-second brief

As a Machine Learning Engineer at OpenText, you are at the forefront of transforming how the world's largest enterprises manage, secure, and extract value from their information. OpenText is a global leader in Information Management, and AI is increasingly embedded into the core of its product suite, from intelligent document processing to enterprise search and predictive analytics. In this role, you will build and deploy models that process massive volumes of unstructured and structured data, directly impacting products like OpenText Aviator and Magellan.

The title spans product, platform and infrastructure work, and which of those the seat actually is decides whether design or algorithms carries more weight in your preparation. The posting rarely settles it; what the team is on call for usually does.

Focus areas

  • Bound every outbound call with a timeout
  • Trace a symptom to a mechanism under load
  • Detect concurrent edits instead of losing writes

Overview

Your work will bridge the gap between theoretical data science and production-grade engineering. You will collaborate closely with data scientists, backend engineers, and product managers to design scalable machine learning pipelines. Because OpenText serves highly regulated and complex industries, your solutions must be not only accurate but also highly secure, robust, and capable of operating at enterprise scale.

This role is critical, challenging, and highly rewarding. You will tackle complex problems involving Natural Language Processing (NLP), computer vision for document analysis, and large-scale recommendation systems. If you are passionate about building AI solutions that solve real-world business problems and drive automation for millions of users, this is the perfect environment to showcase your expertise.

The OpenText interview process

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

HR 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

Technical Evaluation

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

Take-Home Assignment

Nobody is in the room when this is read, so the submission stands in for what an ordinary pull request from you would cost to review. That makes navigability the first thing being judged. A reader should reach the code that does the actual work within a minute of cloning, without tracing three layers of indirection to get there. What most separates a strong submission from a weak one is separation itself: the logic that implements the prompt kept apart from file reading, argument parsing and printing, so each piece can be read, and tested, on its own.

What to demonstrate

  • Whether the entry point makes the shape of the program obvious: what is read, what is computed, what is written, in that order
  • Whether the core logic can be called without touching the filesystem, a socket or stdin, since that is what makes it testable at all
  • Whether names hold up on a second reading; a function called parse that also writes, or a total that means something different from the README, costs the reader more than a longer name would have
  • Whether the commit history reads as a sequence of steps someone could follow, rather than one commit containing everything

How to prepare

  • Clone your own finished submission into an empty directory and read it in the order a stranger would, README then entry point then downward, noting every place you rely on something the code never told you, and fix those places
  • Take an exercise you have already solved and restructure it so the module holding the rules imports no file, CLI or network library, leaving input parsing and output as a thin shell around it
  • Build the history while you work, committing at points where the tests pass, so the log shows the order you actually solved the problem in

Technical and Managerial Rounds

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

Where candidates lose points

Paginating with LIMIT/OFFSET over a set that changes while the client is reading it

OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.

Assuming an isolation level prevents the anomaly you actually have

Isolation levels are named by the SQL standard but implemented differently, so any claim about one is only true of a named engine. PostgreSQL defaults to READ COMMITTED, where every statement takes a fresh snapshot, so two statements inside one transaction can legitimately disagree about the same row. Its REPEATABLE READ is snapshot isolation: it removes non-repeatable and phantom reads but permits write skew, where two transactions each read a set, each conclude their own write is safe, both commit, and the combined result violates a constraint that no single row expresses. Only SERIALIZABLE closes that, and it closes it by aborting a transaction with a serialization failure (SQLSTATE 40001), which means the guarantee is theoretical unless the application has a retry loop. InnoDB's REPEATABLE READ is a different mechanism again - plain SELECTs read a consistent snapshot while locking reads and writes see the latest committed row - so a read-modify-write inside one transaction can act on a value that the transaction's own earlier SELECT never returned.

A queue or buffer with no bound

Every producer-consumer boundary needs a capacity and a policy for reaching it: block the producer, shed load, or drop the oldest entry. Unbounded buffering converts a temporary slowdown into memory exhaustion and hides the backpressure signal that would have revealed the consumer was falling behind.

Comparing floating-point values for equality, or holding money in them

Binary floating point cannot represent 0.1 exactly, so repeated addition drifts and an equality check fails on values that are mathematically equal. Store currency as integer minor units or a decimal type, and compare floats against a tolerance you chose for a stated reason.

Questions & practice

Explain how Gradient Boosting works and compare it to Random Forests.

Medium · machine learning fundamentals

Approach

  1. Say how you would validate it, and where leakage could enter the split.
  2. Name the simplest model that could work and what would make you move past it.
  3. State the learning problem: the label, the unit of prediction and how the model is used.

Follow-up

  • How would you know the model is overfitting?
  • Where could label leakage enter this setup?

What evaluation metrics would you use for a highly imbalanced classification problem, and why?

Medium · machine learning fundamentals

Approach

  1. Name the simplest model that could work and what would make you move past it.
  2. State the learning problem: the label, the unit of prediction and how the model is used.
  3. Say how you would validate it, and where leakage could enter the split.

Follow-up

  • How would you know the model is overfitting?
  • What changes if the classes are heavily imbalanced?

Explain the bias-variance tradeoff and how it impacts model performance.

Medium · machine learning fundamentals

Approach

  1. Name the simplest model that could work and what would make you move past it.
  2. Pick the metric from the cost of each error type, not from habit.
  3. Say how you would validate it, and where leakage could enter the split.

Follow-up

  • How would you know the model is overfitting?
  • What changes if the classes are heavily imbalanced?

What is the difference between Generative and Discriminative models?

Medium · machine learning fundamentals

Approach

  1. State the learning problem: the label, the unit of prediction and how the model is used.
  2. Name the simplest model that could work and what would make you move past it.
  3. Pick the metric from the cost of each error type, not from habit.

Follow-up

  • What changes if the classes are heavily imbalanced?
  • Where could label leakage enter this setup?

Given an array of integers, return the indices of the two numbers that add up to a specific target.

Medium · coding and algorithms

Approach

  1. Walk one small example through your approach before writing the whole thing.
  2. Restate the input: its shape, its size, and what is guaranteed about it.
  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?
  • What is the worst case, and how likely is it on real data?

Implement a binary search tree and write a method to find the lowest common ancestor of two nodes.

Medium · coding and algorithms

Approach

  1. Choose the data structure from the access pattern, not from familiarity.
  2. Walk one small example through your approach before writing the whole thing.
  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?
  • Which test case would catch an off-by-one here?

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.

Approach

  1. Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
  2. Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
  3. Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
  4. Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
  5. Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
  6. Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.

Worked solution (30 min)

  1. Write the edge-loading query with the tenant predicate on both endpoints and state what it does with a cross-tenant edge.
  2. Implement iterative Tarjan with an explicit stack and confirm on a three-node cycle that it emits one component of size three.
  3. Build the transpose restricted to the visited set and mark every node with an in-edge from outside it as refused, carrying the referrer id.
  4. Run Kahn over the condensation and verify the emitted order against the referrer-before-referenced rule.
  5. Size the CSR arrays for 2,000,000 nodes and 8,000,000 edges and compare against a boxed adjacency map.

Expected result: An iterative O(V+E) traversal over a tenant-scoped CSR subgraph, SCC condensation so cycles archive atomically as one component, a transpose-based refusal list naming the external referrer for each blocked resource, and a Kahn topological order over the condensation, with recursion replaced by an explicit stack because of graph depth rather than style.

Verify it yourself

  • A two-node cycle with no external referrers archives as one unit, and adding a single external referrer moves both nodes to the refusal list.
  • No resource appears in the archive order before something that references it, except within its own component.
  • A counter on edge visits matches the loaded edge count exactly, confirming each edge is traversed once.

Follow-up

  • The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
  • The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
  • An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.

Approach

  1. Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
  2. For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
  3. Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
  4. Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.
  5. Answer the two negatives explicitly. owner_user_id alone is not a left prefix of the new index, so it degrades to a full scan of the index at best. Ordered by updated_at, the query still seeks on the (tenant, owner) pair but must sort, because only created_at is ordered within that pair.
  6. Measure both sides with EXPLAIN (ANALYZE, BUFFERS) and compare estimated against actual rows at the lowest node — a 2M-versus-200 misestimate there is usually what chose the plan, and adding an index will not fix a statistics problem.

Worked solution (25 min)

  1. Load 2M resource rows across 5k owners in one tenant, run the query under EXPLAIN (ANALYZE, BUFFERS), and record the node reading the most rows plus the Sort Method line.
  2. Create (tenant_id, owner_user_id, created_at DESC) concurrently and re-run, confirming the Sort node is gone and actual rows fall to about 20.
  3. Run the two negative cases and capture the plan for each.
  4. Re-run the original tenant listing query to confirm the new index has not displaced the index that query depends on.

Expected result: With (tenant_id, owner_user_id, created_at DESC) the plan is an Index Scan with no Sort, roughly 20 rows and a handful of buffers, sub-millisecond. owner_user_id alone still scans; ORDER BY updated_at seeks on the pair and then sorts that owner's rows.

Verify it yourself

  • Actual rows at the leaf node is on the order of the LIMIT, not the tenant's row count.
  • No Sort node appears above the index scan.
  • Buffers read stays constant as one owner's row count grows; if it grows with it, the ORDER BY is not being satisfied by the index.
  • pg_stat_user_indexes shows idx_scan climbing on the new index and not on the listing index for this endpoint.

Follow-up

  • 90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
  • A dashboard runs this for 40 owners in one page load. What changes about the design?
  • How do you roll this index out on a table taking 1.2k writes/second, and what does it cost on every insert from then on?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

resource carries version INT NOT NULL DEFAULT 1. resource_revision holds revision_id, resource_id, version, actor_user_id, change_kind, patch JSONB, request_id, created_at with UNIQUE (resource_id, version). outbox_event holds aggregate_type, aggregate_id, aggregate_version, event_type, payload, status. A PUT carries the version the client read. Write the exact statements for the single transaction that applies the edit, records the revision and enqueues 'resource.updated', and give the handler's branch on zero affected rows. Then say what PostgreSQL 16 does under READ COMMITTED when two of these updates hit one row at once.

Approach

  1. One transaction, three writes, no network call inside it: UPDATE resource SET title = $3, version = version + 1, updated_at = now() WHERE resource_id = $1 AND tenant_id = $4 AND version = $2; then INSERT the resource_revision row at version $2 + 1; then INSERT the outbox_event row at the same aggregate_version. The event goes to a table rather than a broker because no transaction spans both.
  2. Branch on the affected-row count before doing anything else. Zero has three causes — stale version, wrong tenant, row gone — so re-read once and map to 409 carrying the current version, or 404 for an id outside the caller's tenant, which also stops the endpoint confirming that another tenant's id exists.
  3. State the engine behaviour instead of assuming it. Under READ COMMITTED the second UPDATE blocks on the row lock, and when the first commits PostgreSQL re-evaluates the WHERE clause against the newly committed row, so the version predicate now fails and the statement reports zero rows. Under REPEATABLE READ the identical collision raises SQLSTATE 40001 instead, so the handler must fold both shapes into one conflict response.
  4. Keep UNIQUE (resource_id, version) even though the predicate already serialises writers. It is what makes a lost update unwritable if any other path ever reaches the revision table, and it converts a logic bug into 23505 rather than into a silently missing history row.
  5. Refuse to auto-retry the whole PUT. A retry re-reads the winner's state and reapplies an intent formed against data that no longer exists — the silent overwrite the version token was added to detect. Return the conflict; merge field-wise only if the patches are provably disjoint.
  6. Note that now() is the transaction timestamp in PostgreSQL, so resource.updated_at, the revision's created_at and the outbox row share one instant, which is what later makes reconciliation between the three tables unambiguous.

Follow-up

  • A client sends the version it read ten minutes ago and the resource has moved three versions. What is in your 409 so it can resolve the conflict without a full re-fetch?
  • Two editors, two disjoint fields, no overlap. Does your answer still refuse the second write, and should it?
  • Every write now touches a second hot table. How do you keep the outbox insert and its partial index from becoming the write bottleneck at 1.2k writes/second?

Discuss the trade-offs between batch processing and real-time inference for a recommendation engine.

Medium · ml system design and production

Approach

  1. Separate the offline training path from the online serving path.
  2. Say where features come from at serving time and how they match training.
  3. Fix the product goal and the online metric before choosing any model.

Follow-up

  • How would you roll the new model out safely?
  • What happens when a feature is missing at serving time?

Walk me through how you would deploy a PyTorch model into a production environment.

Medium · ml system design and production

Approach

  1. Name what you would monitor after launch and what triggers a retrain.
  2. Say where features come from at serving time and how they match training.
  3. Separate the offline training path from the online serving path.

Follow-up

  • How would you detect drift before the metric drops?
  • How would you roll the new model out safely?

Design a scalable system to classify millions of incoming enterprise documents daily.

Medium · system design

Approach

  1. Name the failure you are designing for, then the recovery path.
  2. Name the read and write paths separately; they rarely have the same bottleneck.
  3. Choose a partition key and say what query it makes expensive.

Follow-up

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

Keep one unresponsive destination from stalling all webhook delivery

Medium · fan-out · circuit breaker · timeouts · bulkheads

Egress delivery sends about 1.5k webhooks/second to 40k destinations, with a per-destination concurrency cap of 4 and a 10-second connect-plus-read timeout. One destination begins accepting connections and never responding; within the hour 150 destinations behave the same way. Design the delivery path so unrelated destinations are unaffected: the pool structure, the timeouts, the retry policy, the per-destination circuit, and what is recorded so a retry is not a second effect at the receiver. State how many in-flight slots the degraded destinations hold and why that number decides the design.

Approach

  1. Start with the number, and with the law that produces it. In-flight work is arrival rate times time in service, so 1.5k/second against a healthy 200 ms response needs about 300 concurrent slots. Per destination the same product applies, ceilinged by the concurrency cap: at the fleet average of 0.0375 deliveries/second per destination (1.5k spread over 40k) a 10-second timeout is 0.375 slots. A destination that has queued retries behind it is a different regime - every slot refills the instant an attempt expires, so it sits pinned at its cap of 4 - and 150 of those hold 600 slots, more than a pool sized for healthy traffic, entirely consumed by endpoints that will never answer. The per-destination cap bounds one endpoint and says nothing about the aggregate, which is exactly why it alone is not containment.
  2. Contain with bulkheads and an admission bound rather than a larger pool. Cap total in-flight per pool and shard destinations across pools by a hash of destination id, so a correlated group - one provider, one region - cannot exceed its pool's share. A delivery refused admission and re-queued with backoff is strictly better than one holding a slot on behalf of a receiver that is not listening.
  3. Treat the timeout as two timeouts, and be exact about what shortening one buys. Connect and read are separate failures and both must be shorter than the budget of whatever is waiting. Occupancy is min(cap, arrival rate x timeout), so dropping the read ceiling from 10 seconds to 3 cuts a merely slow destination's occupancy proportionally, 0.375 slots to 0.11 at the fleet average. It does not cut the 4 slots held by one of the 150: a destination with a retry backlog arrives far above cap/timeout - 0.4/second at a 10-second timeout, 1.33/second at 3 - so it stays pinned at the cap either way and only the slot-seconds per attempt fall. What that does buy is detection rate: 3.3x more failures observed per second on the same four slots, which is how fast the circuit reaches its threshold. Pick the value from the measured latency distribution of successful deliveries, with their high percentile as the floor, not from a round number.
  4. Add a circuit per destination, counting a timeout as a failure. Once open, fail fast without taking a slot - that is the whole point, converting 4 held slots into zero. Half-open on a schedule with exactly one probe and close only if the probe succeeds, so a permanently dead endpoint costs one request per interval instead of a growing retry queue.
  5. Make retries safe and non-synchronising. Back off with full jitter, sleeping a random value in [0, min(cap, base x 2^attempt)], because a fixed delay re-synchronises every failed delivery to one destination into a simultaneous burst. Delivery is at-least-once, so the payload carries the event id under the signature and the receiver deduplicates on it; record the attempt against (destination, event id) rather than a bare success flag, so a lost response does not become a second business effect on the other side.

Worked solution (25 min)

  1. Compute healthy in-flight from rate times latency, then slots held by 150 destinations at the cap and the full timeout, and compare both against one pool size.
  2. Write the pool sharding rule and the admission bound, and state what a refused delivery does next.
  3. Pick connect and read timeouts from the success-latency distribution, then compute min(cap, arrival rate x timeout) for an average destination and for one with a retry backlog, and say which of the two the shorter timeout actually moves.
  4. Write the circuit's state machine with its open threshold, probe interval and close condition, and the backoff formula with full jitter.

Expected result: Destinations sharded across bounded pools with a total in-flight cap per pool, separate connect and read timeouts chosen from the success-latency distribution, a per-destination circuit that fails fast while open and probes once per interval, and full-jitter backoff on retry. A shorter read timeout lowers occupancy only for destinations arriving below cap/timeout; a dead destination with a retry backlog stays pinned at its cap of 4 whatever the timeout, and the circuit opening is what takes it to zero. Deliveries carry a signed event id the receiver deduplicates on, and attempts are recorded per (destination, event id). Degraded destinations hold at most their pool's share rather than 600 slots of a shared pool.

Verify it yourself

  • Slots held by degraded destinations are bounded by the pool shard, not by how many destinations degrade.
  • An open circuit reduces slots held by that destination to zero between probes.
  • Lowering the read timeout does not reduce successful deliveries, verified against the measured success-latency distribution, and does not by itself reduce the slots a backlogged dead destination holds.
  • A redelivered event is discarded by the receiver on its event id rather than applied twice.

Follow-up

  • The destination is not dead - it answers in 9.5 seconds with a 200. Does a failure-rate circuit open? Should anything shed that traffic, and on what signal?
  • One destination requires deliveries in order. What does a per-destination concurrency of 4 do to that guarantee, and what would you change to offer it?
  • A destination has been parked six hours with 900k undelivered events. What does resuming look like, and is delivering the whole backlog the right call?

p99 jumped on one listing filter while p50 stayed flat

Medium · query planning · composite index · percentiles · tail latency

After a release that added an owner_user_id filter to the resource listing, p99 rose from 90 ms to 1.9 s while p50 stayed at 40 ms. Traffic and row counts are unchanged. resource carries the index (tenant_id, status, updated_at DESC, resource_id DESC). The new query filters tenant_id and owner_user_id, orders by updated_at DESC, resource_id DESC, and takes 20 rows. On PostgreSQL, explain the shape of the regression, prove it from a query plan, and give the index you would add.

Approach

  1. Start from the shape. A flat p50 with a moved p99 means a subset of requests changed cost, not all of them, so the first job is naming the subset. Bucket the endpoint's latency by the tenant's row count; the natural hypothesis is that large tenants are a small share of requests and all of the tail.
  2. Get the plan for the new query on a large tenant with EXPLAIN (ANALYZE, BUFFERS). Expect an index scan over the tenant's range, a filter discarding most of it, then a Sort feeding the Limit, possibly reporting Sort Method: external merge Disk. Read actual rows on the scan node, not estimated.
  3. Explain why the existing index cannot serve it. A composite B-tree is seekable only as a left prefix, and with no equality predicate on status the scan cannot treat updated_at as an ordering, because rows in the tenant's range are ordered by status first. Everything matching must be read and sorted before LIMIT 20 can apply, so a tenant with 400,000 rows pays 400,000 rows to return 20.
  4. Add (tenant_id, owner_user_id, updated_at DESC, resource_id DESC). Equality on the first two columns leaves the index ordered by updated_at within that pair, so the plan becomes an index scan that stops after 20 rows with no Sort node. PostgreSQL can scan a B-tree backwards, so the DESC markers matter only if the two sort columns ever disagree in direction; keeping them explicit documents the order the keyset cursor depends on.
  5. Price the fix. This is a fourth index on a table taking about 1.2k writes/second, and every insert and version bump maintains it. Justify it against the query it serves, and check whether it makes an existing index redundant, which here it does not, since the original still serves the status-filtered default listing.
  6. Re-measure per tenant-size bucket rather than in aggregate. A fleet-wide p99 can improve while the largest tenant is still on the old plan.

Follow-up

  • The endpoint paginates with OFFSET. What does page 500 cost with your index, and what does the keyset version cost?
  • How would you have caught this before release, given that a 10,000-row seed database produces the same plan shape at an unnoticeable cost?
  • If a fourth index were unacceptable on write grounds, what else could serve this query?

Suggested practice checklist

Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.

Day 1 — Coding, one pass at shallow depth

  • Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
  • For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
  • Fix nothing today. The value of the pass is the unfixed record.

Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.

Day 2 — Design, one pass at shallow depth

  • Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
  • After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
  • Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.

Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.

Day 3 — Fundamentals and the practical rounds

  • Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
  • Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
  • Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.

Deliverable: Eight scored short answers and one written reading of unfamiliar code.

Day 4 — The rounds that are about you, and the map

  • Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
  • Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
  • Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.

Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.

Day 5 — First chosen area, to the depth you skipped

  • Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
  • After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
  • Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.

Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.

Day 6 — Second chosen area, where the gap is coverage rather than speed

  • Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
  • Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
  • Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.

Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.

Day 7 — Reassemble the loop

  • Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
  • Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
  • Reduce the week to one page holding only the rules you can state without reading them.

Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.

The failure that was yours

Keep one story where the bad call was yours rather than a dependency's or a manager's. Name the check that would have caught it, whether you added that check afterwards, and whether it has fired since. Answers that route blame outward end the conversation early; answers that end in a guardrail someone still relies on tend to open it up.

Tell me about a time you had to explain a complex machine learning concept to a non-technical stakeholder.

Medium · behavioural and collaboration

Approach

  1. Close with what you would do differently, concretely.
  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

  • What did you decide not to do, and why?
  • How did you know your change caused the improvement?

Reverse your own decision and price the reversal

Medium · reversibility · measurement · migrations

Describe a technical decision you made and later reversed. Pick one that cost something: a service you split and merged back, a cache you added and removed, an index you created that pushed the planner onto a worse plan, a projection you rebuilt from scratch. State what you believed when you decided, the measurement that changed your mind, how long the wrong version ran in production, and what the reversal cost in migrations, dual writes, and a deprecation window for callers you did not own.

Approach

  1. State the original rationale without irony, in the version you would still defend given what was known then. If it is not defensible, the story is about carelessness rather than judgement, and a different example serves you better.
  2. Give the measurement that moved with a before and after: the p99 that did not improve, the cache hit rate that sat at 40%, the plan that flipped to a sequential scan once the table passed a size you can name.
  3. Cost the reversal in steps, not adjectives: expand-and-contract deploys, the dual-write window, the callers who had to be notified, the rows already written in the wrong shape that had to be backfilled or abandoned.
  4. Distinguish reversal from rewrite by naming what you kept. Most good reversals preserve the schema or the interface and undo one decision inside it, which is also why they were affordable.
  5. Finish on the process change: the smallest experiment that would have produced the same measurement in a day, and why you did not run it the first time.

Follow-up

  • What in that decision was irreversible, and did you know it was irreversible when you made it?
  • How did you tell the people who had already built on top of the original decision?
  • What do you now measure before committing to a change of this size?

Argue against a design, lose, and commit anyway

Medium · disagreement · service boundaries · decision records

Describe a design you argued against and lost. State the failure you predicted as a named mechanism, not a feeling about complexity: two services that would need one transaction, a projection with no rebuild path, a write path with no idempotency key. Say what evidence you brought, what the decision maker weighed instead, and what you did after the decision was made: what you instrumented, what you wrote down, and whether the prediction came true. Five minutes.

Approach

  1. State the prediction in falsifiable form up front: the mechanism, the condition that triggers it, and the observable outcome. A prediction that cannot be checked also cannot be credited to you later.
  2. Show the evidence you had at the time and label each piece honestly as measured, analogous, or intuition. Keeping the intuition is fine; disguising it as data is the thing that erodes your standing in the next argument.
  3. Represent the opposing case at full strength, including the constraint you did not control: a fixed date, a team boundary, or the fact that the decision was cheap to reverse and yours was not.
  4. Make disagree-and-commit concrete. Name the artefact you left behind so the prediction could be settled without you: the alert and its threshold, the counter on the dashboard, the decision note that recorded the trade-off and the condition that would revisit it.
  5. Report the outcome without editing it. If the design held and your predicted mechanism never fired, say so and say what you had mis-weighted, which is more persuasive than a vindication story.

Follow-up

  • What threshold on that alert would have proved you right, and did anyone ever look at it?
  • If the same proposal arrived tomorrow with the same deadline, would you argue it the same way?
  • How did you behave toward the design once it shipped and started failing in a different way than you predicted?