A Machine Learning Engineer at TD operates at the critical intersection of financial services, data science, and scalable engineering. You are not just building models; you are architecting robust, production-grade AI solutions that drive decision-making, enhance customer personalization, and fortify the security of one of North America’s largest financial institutions. Your work directly impacts how millions of customers interact with banking services, from fraud detection systems to predictive analytics for wealth management.
This role requires a unique balance of theoretical rigor and pragmatic software engineering. You will be expected to navigate the complexities of large-scale, high-velocity financial data while adhering to the stringent regulatory and ethical standards of a major bank. It is a position of significant influence, where the ability to translate abstract business problems into performant, reliable machine learning pipelines is the primary measure of success.
Given the scale of data at TD, emphasize your experience with productionizing models and maintaining performance in high-stakes environments during your interviews.
Initial Screening Call
reportedBefore anything technical happens, someone has to decide which rung of the ladder your loop is calibrated to, and that decision sets the bar for every round after it. It comes from how you describe scope, not from your title, because titles do not convert cleanly between companies. The weak version of the answer is team size and years. The strong version names the largest change you shipped where nobody reviewed the design, what would have broken if you had been wrong, and what you were paged for. Get the level said out loud on this call, because the range and the loop both follow from it.
What to demonstrate
- Whether the scope in your own account maps onto a level the team actually has an opening at, so a mismatch ends the process cheaply rather than after four interviewers have spent a day
- Whether your title needs re-mapping: the same word describes very different amounts of independent decision-making at a twenty-person company and a ten-thousand-person one
- Whether your compensation expectation can be filled at that level in the structure the role pays in, which is why the number gets asked for before any engineer is scheduled
How to prepare
- Write down two changes from the last two years: the largest one you designed with nobody reviewing the design, and the largest one where someone more senior did. Lead with the first when scope comes up, and be ready to say which parts of the second were yours
- Ask which level the loop is calibrated to and what changes at the level above it, then plan your weeks from that answer rather than from the posting
- Settle a total-compensation range beforehand with the split named, base against bonus against equity and its vesting period, so a question about numbers gets a number instead of the word market
Technical Assessment
reportedMost 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.
Technical Interviews
reportedMost of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.
What to demonstrate
- Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
- Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
- Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
- Whether a failing case is isolated and explained before any edit is made to the code
How to prepare
- From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
- Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
- Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.
PracHub editorial advice for the preparation topics above.
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.
Finishing a solution without stating its complexity
Give time and space in the same breath as the code, and define n explicitly when there are two sizes, since n nodes and m edges are not interchangeable. Space is the half that gets skipped: count the auxiliary structures you allocate and the recursion stack at its deepest, not only the answer you hand back.
Reading the constraints as preamble rather than as part of the problem
The bounds are usually there to eliminate the obvious approach: n up to 10^5 makes an O(n^2) scan roughly 10^10 operations, far outside any per-test time budget, and an input larger than memory rules out loading it at all. When a bound is not given, ask for it, then say out loud which approach it kills.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
How would you evaluate the performance of a model beyond simple accura…
How would you evaluate the performance of a model beyond simple accuracy?
Approach
- Pick the metric from the cost of each error type, not from habit.
- Name the simplest model that could work and what would make you move past it.
- Say how you would validate it, and where leakage could enter the split.
Follow-up
- Where could label leakage enter this setup?
- How would you know the model is overfitting?
Explain the trade-offs between bias and variance in a model.
Explain the trade-offs between bias and variance in a model.
Approach
- Name the simplest model that could work and what would make you move past it.
- Say how you would validate it, and where leakage could enter the split.
- 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?
Explain the difference between bagging and boosting techniques.
Explain the difference between bagging and boosting techniques.
Approach
- Pick the metric from the cost of each error type, not from habit.
- Name the simplest model that could work and what would make you move past it.
- State the learning problem: the label, the unit of prediction and how the model is used.
Follow-up
- What changes if the classes are heavily imbalanced?
- Where could label leakage enter this setup?
Implement a basic clustering algorithm from scratch.
Implement a basic clustering algorithm from scratch.
Approach
- State the learning problem: the label, the unit of prediction and how the model is used.
- Pick the metric from the cost of each error type, not from habit.
- Say how you would validate it, and where leakage could enter the split.
Follow-up
- What changes if the classes are heavily imbalanced?
- Where could label leakage enter this setup?
How would you optimize the memory usage of a data processing script?
How would you optimize the memory usage of a data processing script?
Approach
- Walk one small example through your approach before writing the whole thing.
- Restate the input: its shape, its size, and what is guaranteed about it.
- Choose the data structure from the access pattern, not from familiarity.
Follow-up
- How does this change if the input no longer fits in memory?
- What is the worst case, and how likely is it on real data?
Given an array of integers, find the longest subarray with a sum equal…
Given an array of integers, find the longest subarray with a sum equal to k.
Approach
- Name the brute-force solution and its complexity before improving on it.
- Choose the data structure from the access pattern, not from familiarity.
- Walk one small example through your approach before writing the whole thing.
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?
Merge partitioned event streams into one ordered feed with bounded lateness
The read-model service consumes 64 log partitions carrying about 4,000 events per second in total. Each partition is ordered within itself, but partitions drift by up to 30 seconds, and the activity feed must present a tenant's events in occurred_at order. Produce the merge. State its complexity, the buffer it requires in events and in bytes, what happens when one partition is idle, and what you do with an event that arrives after you have already emitted its position. Payloads average 1 KB.
Approach
- Merge with a min-heap over the 64 partition heads keyed on (occurred_at, event_id): O(log P) per event and O(n log P) overall. The tie-break on event_id is what makes the output deterministic when two partitions carry the same millisecond, which matters because the feed is paginated and a non-deterministic order reorders pages under the reader.
- Emitting the heap head is only correct once every partition has produced everything up to that timestamp, so the emit condition is a watermark: the minimum across partitions of the highest occurred_at seen, less the allowed lateness. Events are held until the watermark passes them, which is what turns individually ordered streams into a jointly ordered one.
- Size the buffer from the lateness rather than guessing: 4,000 events per second times 30 seconds is 120,000 buffered events, and at 1 KB each about 120 MB of heap. That number is the real price of the ordering guarantee and belongs in front of whoever asked for it.
- Handle the idle partition explicitly, because it fails the feed rather than corrupting it: a partition with no traffic never advances its own maximum, so the watermark freezes and output stops entirely. Either every partition emits a periodic idle marker carrying the broker's current time, or the watermark falls back to wall clock for a partition silent beyond a threshold.
- Choose the late-event policy from what the projection is keyed on. The projection upserts on (aggregate_id, aggregate_version) and discards a version it has already applied, so a late event is safe to apply out of order and correctness never depended on the merge at all. Apply it, recompute the affected feed page, and count lateness so the 30-second budget can be re-derived from data rather than folklore.
- Say what the merge does not buy: ordering is guaranteed within one aggregate by the log's partitioning, and no watermark makes the cross-aggregate order authoritative. Two events from different aggregates in the same millisecond have no true order, so the feed's order is a presentation choice that must be stable rather than correct.
Worked solution 35 min
- Write the heap comparator on (occurred_at, event_id) and the per-partition head refill.
- Write the watermark computation and the emit-loop condition, then list which buffered events are held at a chosen instant.
- Compute the buffer at 4,000 events per second, 30 seconds and 1 KB per event, and state what fraction of a worker's heap that represents.
- Add the idle-partition marker and trace the watermark with one silent partition, both with and without the marker.
- Write the late-event path and name the key that makes applying it safe.
Follow-up
- The lateness budget is raised to five minutes. What is the new buffer, and what besides memory changes?
- The consumer restarts. Where does it resume from, and what does the feed look like for the first 30 seconds?
- One partition is ten minutes behind because its producer is slow. Do you stall the feed or emit without it?
Stop tag and share joins from fanning out a page
resource_tag is (resource_id, tag_id) with PK (resource_id, tag_id); resource_share is (resource_id, shared_with_user_id, permission). The tagged-and-shared listing inner-joins resource to both, filters tenant_id, tag_id = ANY($2) and shared_with_user_id = $3, orders by updated_at DESC and takes 50. Pages come back with fewer than 50 distinct resources and the total in the header is far too high. Explain the row multiplication, rewrite both the page query and the count query so each is correct, and name the index each one needs. PostgreSQL 16.
Approach
- Do the arithmetic against the predicates that are actually there. An inner join emits one row per matching child row, and both joins are filtered: tag_id = ANY($2) admits only the requested tags, shared_with_user_id = $3 admits one user's share rows. So a resource holding three of the requested tags and shared with $3 once yields three rows, not one — the multiplier is its count of matching tags times its share rows for that single user, and that second factor is 1 unless the table admits duplicate (resource_id, shared_with_user_id) pairs. LIMIT 50 then limits rows rather than resources, and COUNT(*) counts pairs — the header is the product, not the population.
- Reject DISTINCT as the fix. It deduplicates after the product has been built, so the planner must materialise and sort the fanned-out set before the LIMIT can apply, and it leaves any SUM or AVG in the same select list wrong.
- Rewrite both filters as semi-joins, keeping resource as the only row source: AND EXISTS (SELECT 1 FROM resource_tag rt WHERE rt.resource_id = r.resource_id AND rt.tag_id = ANY($2)) and the same shape against resource_share. A semi-join stops at the first match per resource and preserves the driving index order, so ORDER BY updated_at DESC, resource_id DESC LIMIT 50 still stops after 50 rows.
- Count with the same predicates and no join at all: SELECT count(*) FROM resource r WHERE r.tenant_id = $1 AND r.status = 'active' AND EXISTS (...) AND EXISTS (...). Nothing multiplies a resource, so the number is the population.
- Attach the tags for display after the page has been cut — LEFT JOIN LATERAL (SELECT array_agg(rt.tag_id) FROM resource_tag rt WHERE rt.resource_id = p.resource_id) ON TRUE over the 50 returned rows. Aggregate over the page, never over the tenant.
- Index both directions and say which query each serves: PK (resource_id, tag_id) serves the lateral lookup, (tag_id, resource_id) serves the EXISTS probe by tag, and resource_share needs (shared_with_user_id, resource_id) for the same reason. An index covering one direction only leaves the other as a scan.
Worked solution 30 min
- Build a tenant where each resource carries 0-5 tags from a 20-tag vocabulary and is shared with 0-4 distinct users, then bind $2 to three tags and $3 to a user holding shares on about half the resources. Run the joined query and compare its row count to the distinct resource count on page one.
- Run COUNT(*) on the joined shape and on the EXISTS shape and compare both to a ground truth computed from distinct ids; then give $3 a second permission row on 10% of resources and record which of the two counts moves.
- EXPLAIN both page queries and compare rows-read plus the presence of a Sort or HashAggregate node above the join.
- Add (tag_id, resource_id), re-run the EXISTS probe, and record the plan change on the inner side.
Follow-up
- The filter changes from 'any of these tags' to 'all of these tags'. Rewrite it and state what it costs relative to the ANY form.
- A resource can be shared with the same user twice under different permissions. Does your count change, and should it?
- Where does the correct total come from when the tenant holds 4M resources and the header must not cost 200 ms?
Hold a per-tenant active cap against concurrent creates
A tenant on the standard plan may hold at most 50 resources with status='active'. The create handler runs SELECT count(*) FROM resource WHERE tenant_id = $1 AND status = 'active', compares to 50, then inserts. Two creates arrive 3 ms apart on different instances and the tenant lands at 51. Name the anomaly, say whether PostgreSQL 16 READ COMMITTED or REPEATABLE READ prevents it and why, then give an implementation that holds the cap at READ COMMITTED with the exact statements. Finally, say what changes when the cap is 'at most one running export per tenant' on job_run.
Approach
- Name it: write skew. The two transactions read an overlapping set and write disjoint rows, so there is no row-level conflict for the engine to detect and each commit is individually legal.
- Rule out the levels precisely. READ COMMITTED takes a fresh snapshot per statement and takes no lock on the counted rows, so both see 49. PostgreSQL's REPEATABLE READ is snapshot isolation: it removes non-repeatable reads and phantoms within the snapshot but still admits write skew, because the anomaly is not a re-read of a changed row, it is a read of a set that a concurrent transaction invalidates. Only SERIALIZABLE closes it, by tracking the read dependency and aborting one transaction with SQLSTATE 40001 — a guarantee that exists only if the application re-runs the whole transaction from the read.
- Convert the set predicate into a single-row conflict: keep tenant.active_resource_count and run UPDATE tenant SET active_resource_count = active_resource_count + 1 WHERE tenant_id = $1 AND active_resource_count < 50 in the same transaction as the INSERT. Zero affected rows is the cap, returned as 409. The row lock serialises the decision at any isolation level, and contention is bounded to one tenant's row — which is also the fair-scheduling unit, unlike a global counter that would convoy every tenant behind one row.
- State the cost you just took on: a counter is a second source of truth that can drift, so every path that changes status must adjust it inside the same transaction, and a periodic reconciliation has to exist, with resource_revision as the authority for what the count should have been.
- For the job case the invariant is expressible per row, so let the database hold it: a partial unique index on job_run (tenant_id, job_type) WHERE status IN ('queued','running') makes a second running export unwritable and the loser takes 23505, mapped to 409. That is strictly better than a counter — no drift, no reconciliation — and it is available only because the cap is one rather than fifty.
- Add the retry discipline each route demands: under SERIALIZABLE both 40001 and deadlock 40P01 are retryable and the retry must re-execute the read, while under READ COMMITTED with the counter nothing retries, because the conflict is reported to the caller rather than raised as an error.
Follow-up
- A resource moves from archived back to active. Which statements change, and what breaks if the counter update and the status change land in different transactions?
- The cap becomes plan-dependent and a plan can change mid-month. Where does the number 50 live, and who reads it?
- How do you detect after the fact that the counter drifted, without locking the table?
Choose what to break when replication lag reaches forty seconds
Reads are served from two replicas: 14k requests/second, about 85% absorbed by cache, so roughly 2.1k reads/second reach the database. Writes go to the primary at 1.2k/second. A tenant's backfill drives replication lag from under 100 ms to 40 seconds and it is still climbing. Sessions that have just written are pinned to the primary. Decide, endpoint class by endpoint class, whether to serve stale, fail, or route to the primary, and justify each choice with the load it adds to the primary. Then state what you would have built beforehand.
Approach
- Establish blast radius before cause, because mitigation and diagnosis have different deadlines. The decisive arithmetic is what happens if the database reads move to the primary: 2.1k reads/second on top of 1.2k writes/second roughly triples its operation count, on the node already absorbing the backfill that caused this. Reads and writes are not equal in cost, so treat that as an argument against a blanket move rather than as a capacity model - but it is enough to rule out routing everything to the primary.
- Classify endpoints by what staleness costs, not by how important they feel. Reads whose staleness is invisible - listings, search, counters - stay on the replica and return the watermark so the client can tell. Reads that immediately follow that same session's write keep their primary pin, which is a small bounded slice of traffic rather than the whole 2.1k/second. Reads that feed a decision with a side effect - authorisation, quota, the read half of a read-modify-write - must not be stale at all, because a 40-second-old permission row is the stale-permission failure wearing a different costume; those go to the primary or fail.
- Shed instead of queueing. If the must-be-fresh class alone exceeds the primary's headroom, refuse its lowest-value slice with 503 and a retry-after. A request queued behind a saturated primary holds a connection for a client that has already given up, and the retry storm that follows is what turns degradation into an outage. Bound the connection pool per role so the read fallback cannot consume the write path's connections - that bulkhead is the single decision that determines whether writes survive the next ten minutes.
- Attack the cause in parallel, since it is the one thing that can be stopped. The backfill is the load generator. A backfill that reads replication lag as its throttle signal and pauses above a threshold would have made this a non-event, with batch sizes small enough that each batch's write volume is a fraction of what a replica can apply per second. That is most of the answer to what should have existed beforehand.
- Name the mechanism you would prefer over session pinning. Capture the write position at commit and require the read path to be at or past it: compare the primary's pg_current_wal_lsn() at commit time against the replica's pg_last_wal_replay_lsn(), and fall back to the primary only for the specific request that is ahead of the replica. Session pinning is the cheap approximation and it over-pins - every read in the window goes to the primary whether or not it needed to, which is a share of the cost being paid right now.
Follow-up
- Lag returns to normal in nine minutes. Which mitigation do you remove first, and which one stays permanently?
- A user reports their change did not save, and the write committed. Trace the path that produces that report and name the signal that would have shown it before the report arrived.
- The replica is 40 seconds behind but otherwise healthy. Do you take it out of rotation? What does that do to the other replica's lag?
Convert the listing endpoint from offset pages to stable cursors
GET /v1/resources returns a tenant's resources newest-updated first, today with page and per_page, backed by index (tenant_id, status, updated_at DESC, resource_id DESC). Callers are a browser feed and a nightly sync job that walks every page. Users report items appearing twice or vanishing between pages, and page 400 is slow. Design the cursor contract: what the cursor contains and how it is encoded, the exact WHERE and ORDER BY, what happens when a row's updated_at changes mid-walk, how a client detects the end, and what the sync job does when a cursor is rejected.
Approach
- Separate the two defects, because they have different fixes. Cost: OFFSET n makes the engine produce and discard n rows, so price grows with page depth rather than page size and page 400 pays for 400 pages of work. Correctness: while the set shifts, rows cross the offset boundary and are skipped or repeated, and nothing in the response lets the client detect it.
- Write the seek: WHERE tenant_id = $1 AND status = $2 AND (updated_at, resource_id) < ($k, $id) ORDER BY updated_at DESC, resource_id DESC LIMIT n. The tie-break is not decoration - updated_at is not unique, and two rows sharing a timestamp across a page boundary reintroduce exactly the skip this was adopted to remove. PostgreSQL seeks the composite index on the row-value comparison directly; on an engine that does not optimise a row constructor, expand it into the equivalent OR form or the plan quietly degrades to a scan.
- Encode the cursor as an opaque token carrying the sort key, the id, and a fingerprint of the filter and sort order, signed or at minimum validated. A cursor replayed against a different sort or filter must be a 400 with its own code, not a silently wrong page - the sync job cannot notice the difference otherwise.
- State the guarantee precisely rather than generously. Keyset is stable against inserts and deletes elsewhere in the set, because the position is a value and not a count. It is not a snapshot: updated_at is mutable, so a row that is edited during the walk re-sorts and may be seen twice or not at all. If the sync job needs exactly-once coverage, order on an immutable key such as (created_at, resource_id), or walk resource_revision by revision_id and treat updated_at as data.
- Define termination and limits in the response, not in the client's inference: fetch n+1 rows, return n, and emit next_cursor only when the extra row existed. Absence of next_cursor is the sole end signal, because a page can legitimately come back short when rows are filtered after retrieval. Cap n and document the cap rather than honouring per_page=10000.
- Migrate without a flag day: keep page and per_page working, add the cursor, count usage per credential, and remove the offset path only once the sync job's traffic on it is zero.
Worked solution 25 min
- Write the old and new queries side by side and state the rows examined for page 400 under each.
- Define the cursor payload field by field, including the filter and sort fingerprint, and choose the encoding.
- Write the end-of-results rule and the cap, then the 400 response for a cursor that does not match the current query.
- Construct the mid-walk edit case: a row updated between page two and page three, and say exactly what the client sees.
- Write the deprecation plan for page and per_page, including the signal that says removal is safe.
Follow-up
- The client wants a total count and the ability to jump to page 400. What can you honestly offer instead, and what does each option cost?
- How do you paginate backwards, and what does that require of the index?
One log partition stops advancing while the others drain
Search results for a subset of tenants are hours stale; the rest are current. The projection consumer reports lag of zero on 15 of 16 partitions and 400,000 on one. Its error rate is flat and its CPU is idle. outbox_event has no pending rows older than a second, so the relay has published everything it holds. Identify the mechanism, give the ordered checks, and state what you do in the first ten minutes versus what you change permanently.
Approach
- Read the lag distribution first. A slow consumer lags everywhere; zero on fifteen partitions and 400,000 on one is not throughput. Idle CPU on the stuck partition means the consumer is not advancing its offset at all, which points at one message it cannot get past rather than at a rate problem.
- Exonerate the producer before touching the consumer. No pending outbox rows older than a second means the relay published, so the event exists in the log. This separates never sent from sent and never applied, which are different code paths and usually different owners.
- Read the message at the stuck offset and the handler's log lines for its event_id. A flat error rate with no progress has two explanations and you must distinguish them: the handler is throwing and the retry loop is swallowing it, or the handler is blocking on something and never returning. Idle CPU with no error lines favours the second.
- Mitigate before diagnosing further. Move the offending event to a dead-letter store and commit the offset past it. Adding consumers does nothing here, because a partition is consumed by exactly one member of the group, and the blast radius is every aggregate hashed to that partition, not only the aggregate that produced the bad event.
- Fix permanently by bounding handler attempts and dead-lettering on exhaustion, so no single message can stop a partition. Then replay the dead-lettered event once the handler is fixed: it carries aggregate_id and aggregate_version, so a consumer that discards versions it has already applied can absorb the replay, and resource_revision is the fallback if the event itself is unusable.
Follow-up
- The dead-lettered event carried aggregate_version 7 and the projection had applied 6. What must the replay do differently if 8 and 9 landed in the meantime?
- How do you show staleness to the user while the partition is behind, given the API already returns the projection's watermark?
- What changes if the message is poison because a previous deploy wrote a payload shape the current code cannot parse?
For a candidate senior enough that the loop turns on design and judgement rather than on whether the coding round gets finished. Five days build one system properly and then stress it; coding gets a single maintenance day, on the assumption that the risk at this level is an unexamined tradeoff rather than a missed algorithm.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Numbers before diagrams
- Build your own reference card of the figures you will re-derive all week: bytes for a realistic record, requests per second implied by a given daily active count, and the storage that a year at a given write rate produces. Derive each one rather than copying it, because the derivation is what survives a follow-up.
- Turn one product statement into capacity requirements. From ten million daily users at four writes and forty reads each, state the peak-to-average factor you are assuming and why, then produce peak write QPS, peak read QPS and a year of storage.
- Write the two numbers whose order of magnitude changes the design, the read-to-write ratio and the working-set size against memory per node, and state the threshold at which each one flips your answer.
Deliverable: A one-page numbers card and one worked capacity estimate with every assumption written down.
Practice prompt ↗Practice prompt ↗Practice prompt ↗Worked solution ↗02One system, from requirements to schema
- Spend the first ten minutes producing only functional requirements, non-functional targets with numbers attached, a p99 latency, a durability expectation, a consistency requirement, and an explicit out-of-scope list.
- Define the interface before the boxes: the three or four endpoints, their parameters, what each returns, and which of them are idempotent.
- Write the data model, then write the single access pattern that justifies it, and state what the schema would have to become if the dominant access pattern were the other one.
Deliverable: One design carried to endpoint-and-schema depth, with non-functional targets expressed as numbers and a written out-of-scope list.
Practice prompt ↗Practice prompt ↗03The consistency you are actually buying
- Write out what a client sees under asynchronous replication when its write commits on the leader and its next read is served by a lagging follower, then write the two fixes, pinning that session's reads to the leader for a bounded window or carrying a version token the replica must reach, and the cost of each.
- Work the quorum arithmetic on paper for N of three with W and R of two, and separate what R + W > N does guarantee, that any read set intersects any write set, from what it does not: on its own it is not linearizability, and a sloppy quorum that accepts writes on nodes outside the preference list breaks even the intersection.
- Take two storage choices with different defaults, a single-leader relational store committing synchronously and a quorum-replicated store that converges eventually, and write the specific product behaviour that would be wrong under each, rather than a general statement about which is stronger.
Deliverable: A page separating what quorum overlap guarantees from what it does not, with one concrete product misbehaviour attached to each gap.
Practice prompt ↗Practice prompt ↗04Failure is the design
- For one write path, work through the case where the client times out after the server has already committed, then design the idempotency key: who generates it, how long it is retained, and what the duplicate request returns.
- Express the retry policy as parameters rather than as a word: maximum attempts, base delay, backoff factor, jitter, and which error classes are retried at all. Then state why retrying a non-idempotent write without a key is a correctness bug and not merely waste.
- Compute the fan-out effect on tail latency. If a request waits on ten backends and each independently exceeds its p99 one percent of the time, the chance at least one is slow is 1 - 0.99^10, about ten percent. Then write why independence is the optimistic assumption and what correlates them in practice.
- Name the backpressure mechanism for one queue or one dependency in the design, a bounded queue with shedding or a concurrency limit, and write what the caller is told when it engages.
Deliverable: One write path with an idempotency design, a parameterised retry policy, and a written tail-latency calculation with its assumption named.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Scaling the hot path
- Choose cache-aside or write-through for one read path and write the staleness window each produces, then name the invalidation event and what the system does when that event is lost.
- Design against the stampede: either coalesce requests so only one recomputes a missing key, or refresh early with jittered expiry, and write why identical TTLs on keys populated in the same moment produce a synchronised expiry and a thundering herd.
- Shard one table by a key you choose, then answer the two questions that break the choice: which queries now require a scatter-gather, and what happens to the distribution when one tenant is a hundred times larger than the median.
- Write the cost of adding a node under plain modulo placement, where nearly every key moves, against consistent hashing, where roughly one key in n+1 moves, and state what virtual nodes are for.
Deliverable: A caching and sharding decision for one path, each with its failure mode and its rebalancing cost written beside it.
Practice prompt ↗Practice prompt ↗06Keep the coding hand in, at the bar that applies to you
- Solve one medium problem in thirty minutes, then spend twenty more making it production-shaped: named invariants, validation at the boundary, and errors that distinguish a caller mistake from an internal fault.
- Write the tests you would require of a colleague's version of that function: one for empty input, one for the boundary, and one for the case the implementation is most likely to get wrong.
- Read a piece of your own code from six months ago and write the change you would ask for, phrased as you would actually phrase it in review.
Deliverable: One problem hardened to review standard, with its test list and one written review comment.
Practice prompt ↗Practice prompt ↗07Defend it while being interrupted
- Run a forty-five-minute design mock with an interviewer briefed to change a requirement halfway, a tenfold traffic increase or a new strict consistency requirement, and to push on one number you estimated.
- Rehearse the two sentences a senior loop is listening for: naming the tradeoff you are choosing against and why, and saying what you would measure to learn that the choice was wrong.
- Prepare the design you regret: a real decision, the constraint that produced it, what it cost, and what you changed afterwards.
Deliverable: Mock notes recording how the design changed under the new requirement, plus a written account of one regretted decision.
Practice prompt ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Conflict answers where you were right and everyone came round are the weakest ones. Stronger: the evidence you went and collected, what would have changed your mind, and what you did in the weeks after the call went against you. Implementing a design you argued against, properly, is a specific and checkable behaviour.
Describe a time you worked with a cross-functional team to solve a bus…
Describe a time you worked with a cross-functional team to solve a business problem.
Approach
- Close with what you would do differently, concretely.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
How do you handle imbalanced datasets in a fraud detection context?
How do you handle imbalanced datasets in a fraud detection context?
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Pick a story where you made the decision, not one where you watched it.
- 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?
Ship under a deadline and bound the debt you chose
You have four days to ship a tenant-facing listing endpoint. The version you would defend uses keyset pagination over (tenant_id, status, updated_at DESC, resource_id DESC); the version you can finish uses LIMIT/OFFSET with no matching index. Describe a deadline call you actually made of this shape: what you shipped, what you knowingly deferred, how you bounded the damage with a mechanism rather than an intention, and the specific numeric condition that would force the follow-up. Name who you told and where you wrote it down.
Approach
- Name the deferred failure precisely instead of calling it slow. OFFSET n makes the database produce and discard n rows, so cost grows with page depth; without an index matching the sort, every matching row is read and sorted before the limit applies; and rows inserted between two page fetches shift across the boundary so items are skipped or repeated with nothing in the response to signal it.
- Bound the blast radius with something mechanical rather than a promise: cap maximum page depth, cap page size, restrict the endpoint to one internal caller, or keep it behind a flag. State which failure each cap removes and which it leaves standing.
- Attach a number to the trigger and wire it to an alarm: the first tenant crossing N resources, or the endpoint's p99 crossing its share of the 400 ms budget, so the debt announces itself instead of waiting to be remembered.
- Write it where the next engineer looks, which is the code and the ticket, not a chat message: what was deferred, why, the cap, and the trigger.
- Report what actually happened in your real example, including the case where the trigger never fired and the debt was correctly never repaid.
Follow-up
- At what page depth does the offset version breach your latency budget, given your page size and row counts?
- What breaks first when you switch to keyset pagination later, and what does a client holding an old page token see?
- Who would have overruled you if you had asked for two more days, and did you ask?
- 01
Describe a time you worked with a cross-functional team to solve a business problem.
- 02
How do you handle imbalanced datasets in a fraud detection context?
- 03
You have four days to ship a tenant-facing listing endpoint. The version you would defend uses keyset pagination over (tenant_id, status, updated_at DESC, resource_id DESC); the version you can finish uses LIMIT/OFFSET with no matching index. Describe a deadline call you actually made of this shape: what you shipped, what you knowingly deferred, how you bounded the damage with a mechanism rather than an intention, and the specific numeric condition that would force the follow-up. Name who you told and where you wrote it down.
Is this an official TD interview guide?
No. It is PracHub's own research and practice material for the Machine Learning Engineer role at TD. Rounds and questions reflect what candidates have reported, not a process TD has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long does the hiring process typically take?
While it can vary by team and role level, candidates should generally expect the process to span several weeks from the initial screening to a final decision.
PracHub interview research ↗Is the technical assessment purely coding-based?
No, the technical assessment typically combines LeetCode-style coding challenges with conceptual questions about machine learning to ensure a well-rounded skill set.
PracHub interview research ↗What is the most important thing I can do to prepare?
Be intimately familiar with every project on your resume. You should be able to explain the technical hurdles, your specific design choices, and the business impact of every model you have worked on.
PracHub interview research ↗Does TD value academic research or industry experience more?
TD values the ability to solve real-world problems. While academic research is respected, you must be able to demonstrate how your skills apply to the practical constraints of a production environment.
PracHub interview research ↗Sources & methodology 3 sources ↗
No official company page is cited. Rounds and questions come from candidate reports and PracHub editorial material; each source shows the date it was read.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
PracHub page · Accessed 2026-09-30 - 02PracHub Machine Learning Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-30 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-30