A Software Engineer at Staples (Canada) serves as a vital contributor to the digital transformation of a leading retail and business services provider. In this role, you are responsible for building, maintaining, and scaling the software solutions that power the company’s internal operations and customer-facing digital experiences. Your work directly impacts how the business manages inventory, processes orders, and serves its wide range of Canadian clients.
This position is particularly interesting due to the blend of legacy system integration and modern development practices. You will often find yourself navigating complex business requirements while collaborating with cross-functional teams, including product managers, developers, and operations leads. If you are a candidate who enjoys technical problem-solving within a high-impact, enterprise retail environment, this role offers a unique opportunity to influence the infrastructure that keeps Staples (Canada) competitive.
Initial Screening Call
reportedThe title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.
What to demonstrate
- Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
- Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
- Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year
How to prepare
- Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
- Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
- Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter
Technical Assessments
reportedInput bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.
What to demonstrate
- Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
- Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
- Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
- Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply
How to prepare
- For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
- For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
- Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last
Panel Interviews
reportedWhere the day includes a partner from product, design or data, that conversation is weighted like the technical ones and prepared for least. They are deciding one thing: whether having you in the room makes their decisions cheaper. That means options with costs attached, not implementation detail and not "it depends". An estimate someone can plan against — a range, the assumption that would push it to the high end, and what you would drop to hit the low one — is worth more than a confident single number, which everyone present already knows is wrong.
What to demonstrate
- Whether an estimate comes as a range with the assumption most likely to break it, and states what a specific scope cut would actually buy
- Whether a technical constraint is handed over as a choice with consequences on their side, rather than as a verdict they have no standing to argue with
- Whether you establish what decision is on the table before proposing anything
- Whether risk is raised while it can still change the plan, with the trigger that would confirm it, instead of reported afterwards as a slip
How to prepare
- Take a project that shipped late and write the two-sentence warning you could have given three weeks earlier, naming what you would have needed decided at that point
- Rehearse one estimate out loud until it arrives in three parts: the range, the single assumption that would blow it, and the smallest thing you would cut to protect the date
- Rewrite an objection you have actually made — the "we can't do that" version — as two options with their costs, so the choice ends up with the person who owns it
PracHub editorial advice for the preparation topics above.
Treating a timed-out write as a failed write
A timeout says the response did not arrive, not that the work did not happen; the server may well have committed and then lost the connection. Retrying a non-idempotent create after a timeout is the standard way to end up with two of something, and those duplicates land precisely when the system is already degraded and least able to absorb them. The discipline is to treat a timeout as unknown: either the write carries an idempotency key so the retry is safe by construction, or the client re-reads authoritative state before deciding what to do, and the interface says unknown rather than showing a failure that invites a second click.
Holding money in a floating-point type, or rounding it more than once
Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.
Check-then-act on shared state
Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.
Answering a debugging question with a guess instead of a bisection
Give a procedure that halves the search space at each step: confirm the symptom reproduces, establish the last known-good version, input or timestamp, then bisect over commits, over the data, or over the layers of the request path. A plausible cause with no way to confirm it is the same move whether it happens to be right or wrong, which is why it scores nothing.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Hold a tenant to a trailing sixty-second request limit
The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.
Approach
- Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while
front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request. - Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
- Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate,
prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact. - Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (
tokens,last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual. - Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
- Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.
Worked solution 25 min
- Implement the deque version and instrument the per-request pop count, then confirm total pops equal total pushes over a run.
- Generate a burst that places R requests in the last 100 ms of one minute and R more in the first 100 ms of the next.
- Run that burst through the exact deque, a fixed 60-second counter, and the weighted two-bucket estimate, recording admissions in the trailing 60 seconds at every instant.
- Size the memory as R x active tenants x 8 bytes at R = 1,000 and 200,000 tenants, and compare it against what a token bucket would need.
Follow-up
- One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
- Quotas rather than rate limits: the check is
select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes. - How do you return an accurate
Retry-Afterfrom the exact algorithm without a second scan?
Fold a deduplicated usage stream into hourly rollups
You are given one day of usage_event rows, up to 250 million, each carrying event_id, tenant_id, workspace_id, environment, sku, quantity numeric(20,6), idempotency_key, occurred_at and ingested_at. Produce usage_rollup_hourly cells keyed (tenant_id, workspace_id, sku, hour_start) with quantity_sum, event_count and source_max_ingested_at. An event counts once per (tenant_id, idempotency_key). The rollup grain has no environment column, so state your filter. One pass. Give your time and space bounds, and say what the deduplication actually costs in memory.
Approach
- Bucket on
occurred_at, neveringested_at:hour_start = date_trunc('hour', occurred_at at time zone 'UTC'). The two columns answer different questions.occurred_atsays which hour the customer is billed for;ingested_atsays how current the fold is. Using the second for the first makes late data invisible instead of correctable. - The fold is trivial and the deduplication is the entire cost, so price it before designing anything clever. An exact set over
(tenant_id, idempotency_key)at 250M entries, stored as a 16-byte 128-bit hash in an open-addressed table at 0.7 load factor, needs about 357M slots at 16 bytes each, roughly 5.7 GB. The fix is partitioning byhash(tenant_id) % Pso each shard holds 1/P of the set and no tenant's keys straddle shards. - Rule out a Bloom filter as a replacement, in the right direction: a false positive reports 'already seen' for an event never seen, so you drop a real event and lose revenue with no error raised. It is usable only as a negative pre-filter in front of the exact set, where a miss is conclusive and a hit must fall through to the real lookup.
- Accumulate in scaled integers, not binary floating point.
numeric(20,6)admits values below 10^14, so one event scaled to micro-units can reach 10^20, past int64's 9.22 x 10^18; use a 128-bit or arbitrary-precision accumulator unless you first bound the per-event maximum. binary64 represents integers exactly only to 2^53, about 9.01 x 10^15, and cannot represent 0.1 at all, so two runs that sum in different orders disagree. - Carry
source_max_ingested_at = max(ingested_at)over the events folded into each cell, and countevent_countover accepted, post-dedup events. Without that watermark there is no way to prove later what a number did and did not include, which is the first question any reconciliation asks. - State the environment filter explicitly, because the rollup grain cannot record it. A fold that quietly includes
stagingbills non-production traffic; one that quietly excludes it loses a cost signal. Production-only is the billing answer, and either way it belongs in the job name and the output metadata. Complexity: O(n) time, O(distinct dedup keys) space, dominated by the dedup set rather than by the cells.
Follow-up
- A producer retries at 23:59:59 and the retry lands at 00:00:01. The unique index on the daily-partitioned table must include the partition key. What gets double-counted, and what is the smallest change that fixes it?
- The consumer acknowledges its batch before committing the fold. Which failure loses revenue now, and which arrangement duplicates instead?
- What makes a re-run over the same day produce byte-identical rollups?
Order a job dependency graph and find its critical path
A workspace defines up to 50,000 jobs with up to 200,000 dependency edges and an estimated duration_seconds per job. Given the edge list, reject the graph if it contains a cycle and name one cycle's nodes; otherwise return a valid execution order, the earliest possible completion time with unlimited workers, and the set of jobs whose slack is zero. Then say which single job to shorten in order to cut the completion time, and by exactly how much. State the complexity of each part.
Approach
- Kahn's algorithm for the order: compute indegrees, seed a queue with zero-indegree nodes, emit and decrement. O(V + E), which at 50,000 and 200,000 is milliseconds. If fewer than V nodes are emitted, the graph contains a cycle.
- Kahn detects a cycle but cannot name one. The nodes left with indegree above zero contain every cycle, so run one DFS restricted to that residual subgraph with three-colour marking and report the stack slice from the grey node the back edge points at. That is the difference between a usable error message and 'dependency cycle detected'.
- Earliest completion with unlimited workers is the longest path, which is NP-hard on a general graph and linear on a DAG. State the precondition, then relax in topological order:
earliest_finish[v] = duration[v] + max(earliest_finish[u] for u in preds(v)), taking the max over an empty predecessor set as zero. The makespan T is the maximum over all nodes. O(V + E). - Second pass in reverse topological order for
latest_finish, thenslack[v] = latest_finish[v] - earliest_finish[v]. Zero-slack nodes form the critical path, and there can be several disjoint critical paths, so return the set rather than one chain.slack[v] = 0is exactly the statement that some longest path runs through v; equivalently, the longest path through v has lengthT - slack[v]. - The speed-up bound is the point of the question, and the obvious form of it is wrong. Shortening a zero-slack job v by d, with 0 <= d <= duration[v], cuts the makespan by
min(d, T - L_avoid(v)), whereL_avoid(v)is the longest path in the graph with v deleted: the longest path that avoids v, not the second-longest path overall. The two coincide only when the runner-up path misses v. Counterexample: A of 10 s feeds both B of 5 s and C of 4 s, so T = 15 s and the second-longest path is 14 s, yet shortening A by 10 s leaves a makespan of 5 s. The realised gain is the full 10 s, because both paths ran through A and shrank together, whilemin(10, 15 - 14)predicts 1 s. The reason is structural: shortening v reduces every path through v by d and leaves every other path alone, so the new makespan ismax(T - d, L_avoid(v)). - Compute
L_avoid(v)the direct way: delete v and re-run the same forward relaxation, O(V + E) per candidate. The cheaper equivalent skips the deletion, sinceL_avoid(v)only ever matters through that max: setduration[v] := 0, recompute the makespan asT0(v) = max(T - duration[v], L_avoid(v)), and the gain ismin(d, T - T0(v)), which is identical for every d <= duration[v]. Only zero-slack jobs are candidates, because shortening a job with positive slack changes the completion time not at all. One relaxation is milliseconds at this size, so ranking a critical set in the hundreds costs O(k(V + E)) and is worth doing exactly; a critical set in the tens of thousands is not, and there you evaluate a shortlist, longest jobs first, and say that the answer is the best of that shortlist rather than the optimum.
Follow-up
- Only m workers are available. What happens to your answer, and what can you still promise about the schedule you produce?
- Edges arrive incrementally as the customer edits the pipeline. How do you detect a cycle at insert time without re-running Kahn over 250,000 elements?
- Durations are estimates. How would you express completion time as a distribution, and what breaks about the critical path once you do?
Paginate a tenant's delivery export without skipping rows
A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.
Approach
- Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
- Replace the position with a value predicate over a stable, unique, indexed ordering:
where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window. - Index
(tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit. - Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
- Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.
Worked solution 20 min
- Load about 50k deliveries for one tenant, then walk them with the OFFSET query while a writer inserts 10 rows/second, collecting every returned delivery_id.
- Compare the distinct ids collected against the set of ids that existed when the walk started, and record the shortfall.
- Repeat the walk with the keyset query and confirm every pre-existing id is returned exactly once.
- Run
explain (analyze, buffers)on page 1 and page 500 of each form and compare shared buffer hits.
Follow-up
- The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added
created_at <= $snapshotbound, and a materialised export table. - A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
- The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?
Find the join that inflates every invoice total
invoice_line_item holds line_id, invoice_id, tenant_id, sku, rate_tier, quantity, unit_price_micros, amount_minor (bigint), currency, kind, voided_at. invoice_payment_attempt holds attempt_id, invoice_id, tenant_id, amount_minor, status (succeeded, failed, pending), created_at, and an invoice has many attempts. A finance report runs select i.invoice_id, sum(l.amount_minor), count(p.attempt_id) from invoice i join invoice_line_item l using (invoice_id) join invoice_payment_attempt p using (invoice_id) group by 1 and the totals are wrong. Say precisely what the sum now equals, and write a version that is also correct for invoices with zero attempts.
Approach
- Compute what the query actually returns before fixing it. The two joins form a Cartesian product per invoice, so each line row repeats once per attempt row:
sum(l.amount_minor)is the true total multiplied by the attempt count, andcount(p.attempt_id)is attempts times lines. Three lines and two attempts report double the money and six attempts. - Reject the reflex repair.
count(distinct p.attempt_id)does fix the count, because attempt_id is unique.sum(distinct l.amount_minor)does not fix the sum, because two legitimate lines with equal amounts collapse into one. DISTINCT inside an aggregate deduplicates values, not rows, and the difference stays invisible until two lines happen to match. - Aggregate each branch to invoice grain before joining: one CTE summing lines by invoice_id, one counting attempts by invoice_id, then join the two results. A LATERAL subquery per invoice is equivalent and sometimes plans better when the outer set is small. Either way every aggregate stays at the grain it was defined at.
- Keep invoices with no attempts by making the attempt branch a LEFT JOIN with
coalesce(attempt_count, 0). An inner join here silently drops every unpaid invoice, which is usually the exact population finance is asking about. - Push each filter to its own grain:
where l.voided_at is nullbelongs inside the line CTE, not the outer query, or it would also filter the attempt branch through the join. Put the tenant predicate on both branches, since the denormalised tenant_id is what stops a wrong join crossing tenants. - Leave yourself a standing check: an invoice total is a function of its non-voided lines and of nothing about payments, so if changing the payment filter moves the money figure, the fan-out is back.
Follow-up
- Add a third branch for credit notes applied to the invoice. Does the CTE shape still hold, and when would a single pass with
filter (where ...)be better? - Over 500k invoices this report takes minutes. Which grain would you materialise, and how do you keep it correct when a line is voided?
- The same report is needed per tenant per month. What index makes the line CTE cheap?
Design a machine-readable error contract for the gateway
The edge gateway serves roughly 30k requests/second to SDKs and CI pipelines that retry automatically. Today every failure returns 500 with a prose message that clients string-match on. Design the error contract: the response body fields, and the status code for a malformed body, a revoked credential, a scope the credential lacks, a row belonging to another tenant, a reused idempotency key sent with a different body, an exceeded rate limit, and an unreachable dependency. For each, state whether the client may retry and on what schedule. Deliverable: the envelope schema plus the status-to-retry table.
Approach
- Split the envelope by audience: a stable
codestring for programs, amessagedocumented as human-only and free to change, arequest_idthat joins to gateway logs, and adetailsarray for per-field problems. The code list is an enum that only ever grows. - Assign status by who has to change something: 400/422 for the caller's bytes, 401 for a credential that no longer authenticates, 403 for a scope or entitlement, 404 rather than 403 for a row in another tenant because 403 confirms the identifier exists, 409 for an idempotency conflict, 429 for a limit, 503 for a dependency.
- Derive retryability from the method and the idempotency key rather than from the status: a 5xx or a timeout is an unknown outcome, not a failure, so GET/PUT/DELETE may be retried under HTTP semantics and POST only when it carries an idempotency key.
- Put the schedule in the response: Retry-After on 429 and 503 overrides the client's own backoff; otherwise capped exponential backoff with full jitter, sleeping uniformly in [0, min(cap, base * 2^attempt)], bounded by a total attempt budget so retries expire before the caller's deadline.
- Write the negative rules into the published contract: clients must never parse
message, must tolerate unknowncodevalues by falling back to the status class, and a code's meaning is never redefined once shipped.
Follow-up
- A customer reports they retried a 500 from POST /v1/runs and ended up with two sandboxes billed. Whose bug is it, and what in your contract permits their reading?
- You need to add a new error code next quarter without a version bump. What did the v1 contract have to say for that to be non-breaking?
Webhook fan-out with per-endpoint isolation and backoff
One domain event fans out to every matching subscription, producing a webhook_delivery row per (subscription_id, event_id, redelivery_seq). Peak unique event rate is 20k/second; attempts run five to ten times that once fan-out and retries are counted. One customer endpoint has returned 503 for six hours and its backlog holds days of events; every other customer must be unaffected. Design the delivery system: how a worker claims work, the backoff schedule, the per-endpoint circuit breaker, the queue partitioning, and whether you offer ordering per subscription. State the delivery guarantee in one sentence.
Approach
- State the guarantee first, because it determines the rest: at-least-once with a stable event_id, and the consumer documented as responsible for idempotency. Exactly-once over HTTP is not deliverable - the 200 can be lost after the customer has already committed - so any design that promises it is either lying or is really offering at-most-once.
- Partition work per subscription rather than into one global pool, with a concurrency cap per subscription. With a shared pool, the endpoint that has been dead for six hours consumes workers on retries that will fail, and every other customer's delivery latency rises: head-of-line blocking across tenants is the exact failure being designed against here.
- Claim by compare-and-set with a fencing token: UPDATE webhook_delivery SET status = 'in_flight', lease_token = $new, leased_until = now() + interval '60 seconds' WHERE delivery_id = $1 AND status IN ('pending','failed_retryable') AND (leased_until IS NULL OR leased_until < now()), and make the terminal write carry AND lease_token = $new so a paused worker's late write is rejected rather than overwriting a newer attempt. Find due work through the partial index on next_attempt_at WHERE status IN ('pending','failed_retryable'), so the scan is proportional to live rows rather than to the terminal rows that outnumber them by orders of magnitude.
- Use full jitter: sleep uniformly in [0, min(cap, base x 2^(attempt-1))]. Plain exponential backoff hands a recovering endpoint its entire backlog as one synchronised herd and knocks it over again; full jitter de-correlates it. Then check the schedule actually spans the retention you promise - with base 1 s and a 3,600 s cap, twenty attempts have an expected total elapsed time of only about 4.6 hours, so a twenty-four-hour promise needs roughly fifty-nine attempts or a larger cap.
- Trip a circuit per endpoint on consecutive failures or a failure ratio over a rolling window: stop dispatching, push next_attempt_at out or mark new deliveries dropped_circuit_open, and half-open with exactly one probe rather than a batch. Bound the backlog explicitly with a per-subscription cap or retention, and decide in advance whether a recovered endpoint receives six hours of events at full rate or a pointer telling it to fetch what it missed.
- Offer ordering only as an opt-in mode of one in-flight attempt per subscription, and price it honestly: with parallel attempts a retried event overtakes a newer one, so ordering requires serialisation, and serialisation means one slow endpoint blocks its own queue entirely. That converts a shared problem into that customer's own problem, which is the right place for it, but it is still a real cost.
Worked solution 35 min
- Compute the attempt rate: 20k unique events/second x mean fan-out x retry multiplier, and size worker pools and the per-subscription concurrency cap from it.
- Write the claim statement and the terminal write, and point at the clause that rejects a resumed worker's stale write.
- Tabulate the backoff for attempts 1 to 20 with base 1 s and cap 3,600 s, take the expected value of each full-jitter sleep as half its ceiling, and sum to get total expected elapsed coverage.
- Decide the policy for a subscription down six hours: events buffered, bytes held, and what the customer actually receives when it returns.
Follow-up
- The endpoint recovers. Does it receive six hours of events at full rate, and what does that do to it?
- Trace the exact code path by which an event belonging to one tenant could be signed and sent to another tenant's endpoint.
- A customer insists they never received an event your row marks delivered. What evidence do you have, and what does payload_digest let you prove?
Hourly rollups merge one hour and lose another
Reconciliation flags one tenant on one day. Summing usage_event.quantity by hour of occurred_at gives 24 non-empty hours, but usage_rollup_hourly holds 23 rows for that tenant, workspace and SKU, one of which carries roughly the sum of two adjacent hours. Other days reconcile exactly, and the affected date matches a civil-time transition. hour_start is documented as truncated to the hour in UTC. You have both tables, the rollup job source, and its runtime environment. Give an ordered checklist, the mechanism, and the correction path for a day that may already be sealed.
Approach
- Bisect by dimension until one cell explains the whole difference: tenant, then day, then SKU, then hour. A defect confined to a single transition date already rules out deduplication and late arrival, both of which are indifferent to which hour an event lands in.
- Read the truncation with its precondition stated: date_trunc on a timestamptz value is evaluated in the session TimeZone, not in UTC. If the job connects without pinning that setting, it inherits the server or container default.
- Follow that to the collision: in a zone that observes daylight saving, two distinct UTC hours map to the same local wall-clock label at the autumn transition, so both fold into one key under the unique constraint on (tenant_id, workspace_id, sku, hour_start) and their quantities sum into one row. At the spring transition a label never occurs and the row is simply absent.
- Confirm from data rather than from reading code: run the same aggregate twice, once with the session pinned to UTC and once with the job host zone, and check that the second reproduces the stored rollup exactly.
- Fix at the source by pinning the connection to UTC explicitly, or by truncating on occurred_at AT TIME ZONE 'UTC', rather than relying on a default that differs between a developer machine, CI and production.
- Correct according to status, not convenience: an open hour is recomputed with revision incremented, a sealed hour is frozen and the difference becomes an adjustment line on the next invoice with voided_by_line_id pointing at the line it reverses.
Follow-up
- The same job also emits a daily figure for a dashboard. Why can a correct hourly rollup still produce a wrong day, and what does the tenant's billing timezone have to do with it?
- How would you detect this class automatically rather than waiting for reconciliation, given that it only manifests twice a year per zone?
For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Rebuild the primitives by implementing them
- Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
- Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
- For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.
Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Arrays under an invariant: two pointers, sliding window, binary search
- Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
- Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
- Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.
Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.
Practice prompt ↗Practice prompt ↗03Sorting, heaps, and the greedy argument that has to be proved
- Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
- Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
- Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.
Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.
Practice prompt ↗Practice prompt ↗04Recursion, memoisation, and the step to a table
- Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
- Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
- Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.
Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Graphs, where most of the work is choosing the traversal
- Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
- Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
- Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.
Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.
Practice prompt ↗Practice prompt ↗06One day for everything that is not an algorithm
- Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
- Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
- Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.
Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.
Practice prompt ↗07Solve out loud, under time
- Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
- Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
- Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.
Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
A slipped date is only a bad story if you sat on it. What matters is what you believed when you gave the number, the signal that told you it was wrong, how many days passed before you said so, and what you cut rather than asking for more time. Scope you defended counts as much as scope you dropped.
Can you explain your key achievements in a project you worked on previ…
Can you explain your key achievements in a project you worked on previously?
Approach
- Give the blast radius: what could have broken, and what you measured.
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
How do you handle disagreements within a technical team?
How do you handle disagreements within a technical team?
Approach
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
Describe a significant challenge you faced in a project and the steps …
Describe a significant challenge you faced in a project and the steps you took to resolve it.
Approach
- Give the blast radius: what could have broken, and what you measured.
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
Why do you want to leave your previous job?
Why do you want to leave your previous job?
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Name the disagreement and how you resolved it with evidence.
- Give the blast radius: what could have broken, and what you measured.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
- 01
Can you explain your key achievements in a project you worked on previously?
- 02
How do you handle disagreements within a technical team?
- 03
Describe a significant challenge you faced in a project and the steps you took to resolve it.
- 04
Why do you want to leave your previous job?
Is this an official Staples (Canada) interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Staples (Canada). Rounds and questions reflect what candidates have reported, not a process Staples (Canada) has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long does the interview process typically take?
The timeline can vary, but generally, it involves several rounds spread over a few weeks. It is best to stay in regular contact with your recruiter to understand the expected schedule for your specific role.
PracHub interview research ↗What is the best way to prepare for the behavioral portion of the interview?
Use the STAR method (Situation, Task, Action, Result) to structure your answers. This ensures your responses are concise, focused, and highlight your specific contributions to the projects you discuss.
PracHub interview research ↗Is the technical interview focused on whiteboard coding or system design?
Expect a mix of both depending on the seniority level. For entry-level roles, there is often a greater emphasis on foundational problem-solving, whereas more senior roles will delve deeper into system design and architectural trade-offs.
PracHub interview research ↗What can I do to stand out as a candidate?
Show genuine interest in the business side of Staples (Canada). Researching the company’s current digital initiatives and preparing thoughtful questions about the team’s roadmap will demonstrate that you are a serious, engaged candidate.
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 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-22
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