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Altec Industries Software Engineer Interview Questions & Guide 2026

System design practice for Altec Industries software engineer candidates: reach a data model and an API contract, not a box diagram.

The 60-second brief

A Software Engineer at Altec Industries plays a pivotal role in bridging the gap between cutting-edge software solutions and the heavy equipment that powers utility and infrastructure projects globally. You are not just writing code; you are building the digital intelligence that ensures safety, efficiency, and reliability in the field. Whether you are developing internal management tools, optimizing data pipelines, or working on systems that interface with physical machinery, your work directly affects how Altec Industries' teams deliver critical services.

Prepare in one language you know well enough to debug in rather than the one you think reads best. Under a clock an unfamiliar language costs you standard-library lookups and iteration mechanics, and that time comes out of your thinking budget, not your typing budget.

Focus areas

Detect concurrent edits instead of losing writes
Trace a symptom to a mechanism under load
Paginate large result sets with keyset cursors

Overview

This role requires a balance of technical precision and a practical, problem-solving mindset. Altec Industries operates in a space where software must be as robust and dependable as the equipment the company manufactures. You will work in a collaborative environment alongside mechanical, electrical, and systems engineers, making communication and cross-functional teamwork essential. If you are looking to apply your technical skills to real-world, tangible outcomes that keep the lights on and the infrastructure moving, this position offers a unique combination of scale and impact.

The Altec Industries interview process

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

HR Screening

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

What to demonstrate

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

How to prepare

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

Technical Interviews

Most of the time lost in this format is not lost to thinking. It goes to a standard-library call you half-remember, an off-by-one in a loop bound, and a debugging loop that mutates code at random until something passes. When output is wrong, stop re-reading the whole function: take the smallest input that reproduces it and walk the state through by hand, printing intermediates if the environment allows. Guessing at a fix without a failing case you understand is how a five-minute bug becomes twenty, and the clock does not pause while you do it.

What to demonstrate

Whether you reach the right structure without a detour, and can write it from memory rather than only recall that one exists
Whether overflow is considered where the language has fixed-width integers, since a signed 32-bit value stops at 2,147,483,647 and then wraps in Java, is undefined behaviour in C++, and does not arise in Python, whose integers grow instead
Whether recursion depth is treated as a constraint on large inputs, given that CPython's default limit is 1000 frames and a deep recursion can exhaust the stack in any language where an iterative version would not
Whether a failing case is isolated and explained before any edit is made to the code

How to prepare

From an empty file and with no references open, implement the pieces you lean on most: a heap push and pop, an iterative DFS with an explicit stack, and a binary search whose midpoint is written lo + (hi - lo) / 2, which avoids the overflow that (lo + hi) / 2 can hit in a fixed-width integer type
Time yourself on the ten library calls you look up most, such as sorting with a custom comparator, splitting and joining strings, and finding the next key at or above a value in an ordered map, until the lookup is gone
Take a solution you know is broken and, before touching it, write one sentence naming the input, the expected value and the actual value. Repeat until you do it without deciding to.

Onsite Interviews

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

What to demonstrate

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

How to prepare

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

Where candidates lose points

Shipping a migration and the code that depends on it as a single change

During any rolling deploy, and for as long as a rollback remains possible, old and new code execute against the same schema at the same time. A migration that drops or renames a column breaks every instance that has not restarted yet, and code that requires a column the migration has not applied breaks every instance that restarted early. The discipline is expand then contract: add the new column nullable, write both shapes, backfill in batches, move reads across once the backfill is verified, and only then stop writing the old shape and drop it - four deploys, usually spread over days. It feels disproportionate until the first rollback, at which point it is the only reason the previous version still runs.

Running a schema change as though the lock lasts as long as the statement

In PostgreSQL an ALTER TABLE that needs an ACCESS EXCLUSIVE lock must first wait for every open transaction touching that table, and while it waits, later queries needing a conflicting lock queue behind it rather than overtaking it. A DDL statement that would execute in milliseconds, issued while a thirty-second analytics query is open, therefore stalls all traffic on that table for thirty seconds: the outage length is set by the longest open transaction, not by the change. The defences are specific and worth knowing by name - set lock_timeout low and retry rather than queue, add columns without a volatile default so no table rewrite occurs (from version 11 a non-volatile default is a metadata-only change), build indexes with CREATE INDEX CONCURRENTLY while accepting that it cannot run inside a transaction block and leaves an invalid index behind if it fails, and add constraints as NOT VALID followed by a separate VALIDATE CONSTRAINT, which takes a weaker lock.

Trusting input because it came from your own front end

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

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.

Questions & practice

Find overlapping job attempts and peak concurrency from lease records

Medium · sweep line · intervals · leases

A day of job_run history yields about 50,000,000 attempt records: (job_run_id, job_type, attempt, started_at, finished_at which is NULL when the worker died, lease_expires_at). Leases expire on a clock, so a job that outran its lease ran twice. Produce (a) every job_run_id whose attempts overlapped in wall-clock time and (b) the peak number of simultaneously running attempts per job_type with the minute it occurred. Target O(n log n). State how you treat a NULL finished_at and what clock skew does to your answer.

Approach

Define the interval before sorting anything: an attempt occupies [started_at, COALESCE(finished_at, lease_expires_at)). finished_at is observed and lease_expires_at is only a promise, so every attempt without a finish contributes an estimate and the whole result is a lower bound on overlap rather than an exact count.
For peak concurrency, sweep: emit 2n endpoints, sort by (timestamp, kind) with ends ordered before starts at equal timestamps, then walk the sequence maintaining a counter per job_type and record each type's maximum with its timestamp. O(n log n) dominated by the sort, O(n) space, or O(1) extra if the sort is external and the walk streams.
For overlap detection, do not compare attempts pairwise. A single global sort by (job_run_id, started_at) gives both the grouping and the order; within a group, keep the maximum end seen so far and report an overlap exactly when the next start is less than that running maximum, which is one linear pass after the sort.
Half-open intervals matter and are easy to get wrong: with closed intervals an attempt ending at the same millisecond another begins reads as concurrency two, and across 50,000,000 records that artefact swamps the real signal.
State the clock caveat: started_at and finished_at are written by different workers, so under skew of a few hundred milliseconds an apparent overlap shorter than that bound is not evidence. Filter reported overlaps by a minimum duration, or prefer timestamps written by whichever component heartbeats the lease.
Scale the sort rather than assuming it fits: the sweep emits two endpoints per attempt, so 50,000,000 records become 100,000,000 endpoints, and at roughly 24 bytes each, an 8-byte timestamp plus a 4-byte job_type plus a kind flag padded to alignment, that is about 2.4 GB of sort keys before any scratch space. Either push the ordering into the database behind an index on (job_type, started_at) or run an external merge sort in chunks; the overlap pass sorts n records rather than 2n, so it is the cheaper of the two.

Worked solution (30 min)

Write the interval derivation with the COALESCE and state in one line which of the two end sources is observed and which is assumed.
Write the concurrency sweep: the endpoint tuples, the sort key including the end-before-start tie-break, and the per-job_type counter.
Hand-trace four attempts of one job, two disjoint and two overlapping by three seconds, and confirm the overlap detector fires exactly once.
Add the skew filter as a minimum overlap duration, state the value you chose, and justify it from how the timestamps are written.

Expected result: One global sort by (job_run_id, started_at) driving a running-maximum-end check for overlaps, plus a sweep over 2n endpoints with ends ordered before starts for per-job_type peak concurrency, both O(n log n), with half-open intervals, NULL finished_at falling back to lease_expires_at, and overlaps shorter than the clock-skew bound excluded.

Verify it yourself

Two attempts touching at exactly the same instant report concurrency 1, not 2.
A job with a single attempt never appears in the overlap output regardless of its lease.
The reported peak timestamp for a job_type lies inside at least one of that type's intervals.

Follow-up

A handler is not idempotent and you have found 400 overlapping jobs. Which of them actually caused damage, and what would you query to find out?
Peak concurrency for one job_type is 4 against a configured cap of 4. Is the cap working, or is the data hiding attempts that never started?
How would you compute both answers incrementally as records arrive rather than in a daily batch?

Identify the heaviest tenants in a five-minute window under memory pressure

Medium · top-k · heavy hitters · streaming

The edge service handles about 3,000 requests per second across roughly 50,000 tenants, peaking near 9,000. Expose the 50 heaviest tenants by request count over the trailing five minutes so limits can be tightened before one tenant's backfill starves the fleet. You may not retain five minutes of raw records. Give the exact solution and its memory, then the bounded-memory approximation with its error stated as a formula, and say which you would ship and at what tenant cardinality that choice changes.

Approach

Do the exact version first, because it is affordable at this cardinality: a ring of 300 one-second counters per tenant, advanced lazily, is 1,200 bytes of counters per tenant and roughly 60 to 90 MB for 50,000 tenants with overhead. Carry a running total and subtract the bucket you overwrite so a window read is O(1) rather than 300 adds.
Extract the top 50 with a size-k min-heap over the tenant sums: O(d log k) for d tenants, against O(d log d) to sort them all. Maintaining the heap continuously instead of on query requires a tenant-to-heap-index map, because incrementing a count already inside the heap means sifting from a known position, and without that map you rebuild the heap on every request.
State the approximation precisely rather than gesturing at sketches. Misra-Gries with m counters retains every item whose true count exceeds N/(m+1), and each retained count underestimates the truth by at most N/(m+1). With m = 1,000 and N = 900,000 requests in the window the error is roughly 900 requests, which is fine for spotting a tenant sending 50,000 and useless for ranking two tenants 200 apart.
Say what breaks when the window slides: Misra-Gries and Space-Saving are insert-only and cannot be decremented as records age out. The workable construction is one summary per sub-window, say ten seconds, with 30 summaries merged at query time, and the merged error is the sum of the per-summary errors, so the bound degrades linearly in the number of sub-windows.
Choose and defend it: at 50,000 tenants the exact rings cost under 100 MB in a process that already holds more, so ship exact. Keep the sketch for the case that actually motivates it, a per-principal or per-IP key where cardinality runs to millions and is not bounded by anything you control.
Raise the fleet problem before it is asked: each of 20 to 40 instances sees only its share, and the top 50 of one shard is not the top 50 of the fleet. Either aggregate counts centrally or accept that a per-instance threshold multiplied by instance count is the limit you are really enforcing.

Follow-up

The heaviest tenant is heavy because of one export job rather than user traffic. Should the limiter treat those as the same tenant?
Two tenants sit tied at the boundary of the top 50. Does your answer flap, and does the flapping matter?
You switch to per-principal keys and cardinality goes to 10 million. Walk through what changes.

Diff a projection against the primary without per-row point reads

Hard · reconciliation · range hashing · throttling

The listing projection has drifted and some rows show a stale version. The primary holds 40,000,000 resource rows across 12,000 tenants while serving 1,200 writes and 14,000 reads per second. The obvious repair, reading each resource row and comparing its version against the projection, is correct and would eventually finish. Explain precisely why it is unacceptable here, then give a diff that finds the differing rows, state its complexity, and make it safe to run against a live primary. Replication lag is usually under 100 ms and is not bounded.

Approach

Quantify the naive cost rather than calling it slow: 40,000,000 point reads at even 0.5 ms each is over five hours serialised, and the only lever is concurrency, which is exactly what you cannot spend. The primary's pool is sized for the write path, and 40,000,000 random reads evict the buffer cache that sustains the 85 percent cache hit rate, so the audit degrades the system it is auditing.
Replace random access with one ordered pass per side. Both sides can be read in (tenant_id, resource_id) order, which is a sequential scan on each and a merge join in O(n) time and O(1) memory. For a dense diff that is the whole answer, and it reads the primary once instead of 40,000,000 times.
For the expected sparse case, compare range hashes instead of rows: partition the key space, compute per range an order-independent aggregate over hash(resource_id, version), compare aggregates, and descend only into ranges that differ. With d differing rows and branching factor B, at most d ranges mismatch per level, so the drill-down examines O(d log_B(n/d)) ranges and reads full rows only in mismatching leaves.
Aggregate with a sum modulo 2^64 or a multiset hash, never XOR. XOR is order-independent but self-cancelling, so two rows wrong in the same way, or a row duplicated on one side, leave the range aggregate matching and the range is declared clean.
Pin the comparison to a point in time or it reports lag as drift: consider only rows whose updated_at is older than now minus a lag margin, and re-check each candidate mismatch individually before repairing. At 1,200 writes per second a diff without this reports thousands of false positives, and an unattended repairer would then overwrite live rows with stale values.
Make the run resumable and throttled: batch by range key, persist the last completed range, and watch a signal such as replica lag or primary CPU, pausing rather than pressing on. A reconciliation that cannot be stopped and resumed gets killed halfway and restarted from zero, which is how a repair becomes an incident.

Follow-up

The diff reports 900 stale rows. How do you decide between patching those rows and rebuilding the projection from resource_revision?
Same job, but the projection lives in a search index that cannot be scanned in key order. What changes?
How would you run this continuously at low cost instead of only as incident response?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

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

Approach

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

Follow-up

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

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

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

Approach

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

Worked solution (25 min)

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

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

Verify it yourself

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

Follow-up

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

Describe a typical day in your current or most recent role.

Medium

Approach

State your assumptions explicitly before working the problem.
Clarify what is being asked and what a complete answer contains.
Work from the requirement backwards to the design.

Follow-up

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

What makes you the best candidate for this position?

Medium

Approach

Say what you would check first and why it is the highest-information step.
State your assumptions explicitly before working the problem.
Work from the requirement backwards to the design.

Follow-up

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

Shard by tenant when one tenant outgrows a single shard

Hard · sharding · hot partition · routing · online migration

One primary holds resource, resource_revision, outbox_event and idempotency_key for every tenant and is at its write ceiling at 1.2k writes/second. tenant_id leads every index. Shard across eight primaries. One tenant holds 22% of all rows and by itself exceeds a single shard's write capacity. Design the routing, the split of that tenant, and the online move of a tenant between shards with writes continuing. State what breaks for queries that are tenant-scoped today, and exactly what a write must do when it arrives at the old shard after the cutover.

Approach

Route on a unit smaller than a tenant from the start. Make the routing key (tenant_id, bucket) with a fixed bucket count - 64 over eight shards - and keep a directory mapping each (tenant_id, bucket) to a shard, carrying a version and cached in every service. An ordinary tenant has all 64 buckets pointing at one shard and behaves exactly as it does today; only the hot tenant has its buckets spread. Hashing tenant_id alone spreads tenants evenly, gives you no way to move one, and has no answer at all for a tenant larger than a node. The bucket count is the part you cannot change later without rehashing rows, so pick it well above the shard count and rebalance by moving buckets, not by re-bucketing.
For the tenant that exceeds one node, its buckets must land on different shards - that is the whole point of bucketing it, and buckets confined to its own shard would rename the rows while leaving every write on the node whose ceiling it already exceeds. Size it from measured numbers rather than from its row share: 22% of rows says nothing about write rate. The current primary tops out near 1.2k writes/second on this hardware and workload, so a tenant peaking at W writes/second needs its buckets spread over at least ceil(W / headroom-per-shard) shards, where the divisor is the share of each shard's ceiling you are willing to give it while that shard still serves other tenants - not the full 1.2k. Size on its peak, not its mean.
Fix the co-location invariant at the right grain. What must commit in one transaction is a resource, its resource_revision row and its outbox_event row, so the bucket is a property of the resource: derive it once at creation and stamp it into resource_id, and every later revision and event routes with its parent for free. Per-tenant co-location was never the requirement, and mistaking it for one is what makes a tenant look unsplittable. What genuinely breaks is an invariant spanning two resources of one tenant - a per-tenant counter, uniqueness across its resources - which now needs either a home-shard table or two-phase commit, and 2PC at this write rate is not a serious option.
Keep the idempotency constraint arbitrating, because it is now enforced per shard. PRIMARY KEY (tenant_id, idempotency_key) only continues to reject a retry if the same key always lands on the same shard, so derive the create-path bucket from hash(tenant_id, idempotency_key) and mint the new resource_id inside that bucket, which also puts the key row and the resource it guards in one transaction. A bucket chosen from anything that differs between a request and its retry - a timestamp, the worker id, a client-supplied resource id - splits one key across two shards, both inserts succeed, and the write endpoint's retry safety is silently gone.
State what the split costs the hot tenant's reads. Its listing, one 21-entry index scan today, becomes a scatter-gather: every bucket-shard returns 21 rows, a coordinator merges and discards the surplus, latency becomes the slowest shard's rather than the median's, and the keyset cursor has to carry a position per bucket instead of one (updated_at, resource_id) pair. Counts over that tenant fan out the same way. The relay becomes one leader per shard; consumers are unaffected because their ordering guarantee was always per aggregate and a resource's events never leave its bucket.
Move one bucket at a time, reversibly, and fence the straggler at the shard rather than at the caller. Copy from a snapshot while the bucket stays read-write, tail changes until the remaining delta is a few seconds of writes, fence writes for that (tenant_id, bucket) alone with a retryable status, apply the final delta, bump the routing version. Scoping the fence to a bucket is what makes a seconds-long freeze affordable. Then have each shard store the routing epoch it believes it holds for each (tenant_id, bucket) and reject any write carrying an older one: without that token, a service on a stale map commits successfully to a database nothing will ever read again, and the loss stays invisible for days. Outside the data path, anything that aggregated across tenants in one query - admin reporting, the A-Z index, global counters - becomes a fan-out across eight shards with a merge, and per-tenant uniqueness survives only on tables that stay whole on the tenant's home shard.

Worked solution (40 min)

Write the routing lookup keyed by (tenant_id, bucket), its version field, where it is cached and invalidated, and the request-path cost.
From the tenant's measured peak write rate and the per-shard headroom you will grant it, compute how many shards its buckets must span, assign them, and show no single shard carries its whole write rate.
Trace one create end to end: which value picks the bucket, where resource_id gets it stamped, and why the revision, outbox and idempotency rows land on the same shard.
Write the move steps for one bucket, then the epoch check the shard performs on every write, and trace a stale-map write through it.

Expected result: A versioned (tenant_id, bucket) to shard directory cached in every service, with ordinary tenants pointing all their buckets at one shard. The hot tenant's buckets are deliberately assigned to different shards, enough of them that its peak write rate divided across them fits the headroom granted on each, because no single node can carry it. The bucket is derived once and stamped into resource_id, so a resource, its revisions and its outbox rows commit together on one shard while the tenant spans several; only cross-resource invariants within the tenant lose their single-shard transaction. The create-path bucket comes from hash(tenant_id, idempotency_key) so the per-shard unique constraint still arbitrates retries. The tenant's listing becomes a scatter-gather with a per-bucket cursor. Migration moves one bucket at a time - snapshot copy, tail, seconds-long fence for that bucket, final delta, routing-version bump - and a routing epoch per (tenant_id, bucket) is checked by the shard so a stale caller's write is rejected rather than committed where nothing will read it.

Verify it yourself

The hot tenant's buckets resolve to more than one shard, and its peak write rate split across them is inside the headroom each shard has after its other tenants.
A resource's rows, its revisions, its outbox rows and its idempotency key row always resolve to a single shard, so no one write spans two.
The same (tenant_id, idempotency_key) routes to the same bucket on every retry, so the per-shard unique constraint still rejects the duplicate.
A write carrying an old routing epoch is rejected by the shard rather than merely misrouted by the caller.

Follow-up

The fence lasts 90 seconds instead of 4 because the final delta keeps growing. What is happening, and what do you do while the tenant is fenced?
Two tenants must merge into one account. What does that cost under this scheme, and which step is not reversible?
A shard is lost entirely. Which tenants are affected, and what is the source of truth for rebuilding them?

Listing latency scales with page size, not with filters

Easy · n+1 · latency · query batching

The tenant listing endpoint reads resource filtered by tenant_id and status, ordered by updated_at DESC, and returns each row plus the owner's display name from app_user and the actor of that resource's latest resource_revision. p99 is 55 ms at 10 rows per page and 1.4 s at 200. Database telemetry shows 401 statements per request, each under 1 ms, and nothing in the slow-query log. Diagnose the cause and give the fix, stating the statement count per request and the p99 you expect afterwards.

Approach

Read the counters before forming a theory. 401 statements for 200 rows is one driver query plus two per row, and sub-millisecond execution with an empty slow-query log rules out a bad plan. The time is round trips, which is why it is invisible in every per-query metric and scales with rows returned rather than with filter selectivity.
Name the two per-row statements from their normalised text: a single-row app_user lookup by user_id, and a resource_revision lookup by resource_id ordered by version DESC LIMIT 1. Confirm by dropping those two response fields and watching the statement count fall to one. That locates the calls in the serialisation layer, not the repository.
Check that the arithmetic accounts for the whole gap. Measure one round trip to the replica in isolation; 400 trips at roughly 3 ms of network plus 0.2 ms of execution is about 1.3 s on top of a 55 ms baseline, which matches. If the multiplication had fallen short, the N+1 would only be part of the story and you would keep looking.
Batch both lookups. Collect owner_user_ids and resource_ids from the driver query, then issue WHERE tenant_id = $1 AND user_id = ANY($2) for the users, and PostgreSQL's SELECT DISTINCT ON (resource_id) ... WHERE resource_id = ANY($2) ORDER BY resource_id, version DESC for the latest revision, which the UNIQUE (resource_id, version) index serves directly. On an engine without DISTINCT ON, use a lateral join or a row_number window. Three statements per request at any page size.
Keep the tenant predicate in the batched query. The per-row version was implicitly scoped because its ids came from tenant-scoped rows; a batched user_id = ANY(...) with no tenant_id is an unscoped read that behaves correctly only as long as the id list is trustworthy.
Re-measure at 10, 50 and 200 rows and confirm the statement count is constant. Latency should now track bytes returned.

Follow-up

The page size is capped at 200 today. What breaks first if it is raised to 2,000, and is it still this bug?
How do you stop the next N+1 from reaching production, given that no individual query is slow and the endpoint's tests pass?
The latest-revision actor is only used to render an avatar. Make the case for denormalising it onto resource, and name the write anomaly that introduces.

Suggested practice checklist

Roughly ninety minutes on weeknights with one longer weekend block. The plan cuts scope rather than compressing everything, on the assumption that one thing finished per night beats four half-started.

Day 1 — Fix the scope and take a cold baseline

Read the role description and write the three things the loop will almost certainly test, then write an explicit not-doing list and keep it visible all week.
Take one twenty-five-minute coding problem and one fifteen-minute design prompt cold, and write the single sentence naming what blocked each, because those two sentences decide where the remaining evenings go.
Set the week's rule: one thing finished every night, including the night you only have forty minutes.

Deliverable: A one-page scope with a not-doing list and two cold attempts, each carrying one sentence on what blocked it.

Day 2 — One pattern, written three times from blank

Choose the single pattern most likely to appear in your loop and write it three times from an empty file rather than editing the previous attempt.
On the third pass, write the invariant as a comment before the loop body and the complexity before the first line of code.
Stop at ninety minutes even if the third version is imperfect, and write the one thing you would fix given another hour.

Deliverable: Three independent implementations of the same pattern plus a note on what changed between them.

Day 3 — One design, only to the depth you can defend

Take one system shape and go only as far as requirements, interface and data model, refusing to draw a box you could not survive a follow-up about.
Attach one number to each non-functional requirement, deriving it rather than asserting it, and write the assumption the number rests on.
Write the one tradeoff you are choosing against and the observation that would make you reverse it.

Deliverable: One design at interface-and-schema depth with derived numbers and one written reversible tradeoff.

Day 4 — Only the fundamentals you will have to defend

Write, in under two hundred words each, the answers to the two questions that follow almost any implementation: why this structure and not the obvious alternative, and what happens to this code at a hundred times the input.
Write what an index actually costs: faster lookups on the indexed columns against a write that now maintains a second structure, plus the cases where the planner declines to use it anyway, low selectivity, or a predicate wrapping the column in a function.
Delete any answer you cannot deliver aloud in under a minute, since an answer that needs reading is not an answer you have.

Deliverable: Three written answers, each under two hundred words and each timed aloud.

Day 5 — Your own work, timed

Write a ninety-second and a four-minute version of your main project and time both aloud rather than reading them.
Prepare the two follow-ups that always come: what you would do differently, and how you knew it worked.
Put one number in the first sentence and be ready to say exactly where it came from and what it excludes.

Deliverable: Two timed narratives with one defensible number in the opening line.

Day 6 — The one full rehearsal, in the weekend block

Run a sixty-minute mock covering a coding round and a design round in one sitting with no break, because sustained attention is the thing evenings have not trained.
Immediately afterwards, and before hearing any feedback, write the three moments you lost the thread.
Spend the rest of the block only on those three moments, and on nothing you merely feel shaky about.

Deliverable: Mock notes naming three failure moments with a specific fix written under each.

Day 7 — Taper

Write the twenty-minute warm-up you will actually do on the morning: one problem you can already solve from a blank file, one design you can narrate, and nothing you have never seen.
Re-read only your own notes from this week and open no new material.
Write the logistics down: the editor or shared document you will be working in, whether execution and lookups are permitted, and the sentence you will use when you do not know something.

Deliverable: A one-page card holding the design structure, the project numbers, and the logistics.

Outage answers are graded on instruments

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

Tell me about a time you had to complete a task under significant time pressure.

Medium · behavioural and engineering judgement

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 would you do differently if you ran that again?
How did you know your change caused the improvement?

What do you hope to get out of your experience with us?

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Close with what you would do differently, concretely.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

Tell me about your experience and the projects you have worked on.

Medium · behavioural and engineering judgement

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.
Name the disagreement and how you resolved it with evidence.

Follow-up

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

What are your personal values, and how do they influence your work?

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Give the blast radius: what could have broken, and what you measured.
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?