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Peak Energy Software Engineer Interview Questions & Guide 2026

Peak Energy software engineer prep on operations: decide what you would need to see to know a release is going wrong.

The 60-second brief

The Software Engineer role at Peak Energy is central to the company's mission of transforming energy storage and grid infrastructure. As a member of Peak Energy's engineering team, you will develop the software layers that control the company's hardware systems, optimize energy throughput, and ensure the reliability of its products. This role is not just about writing code; it is about bridging the gap between high-level software architecture and physical, real-world energy systems.

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

Focus areas

  • Make every external side effect safe to retry
  • Fold late, out-of-order scan events by event time
  • Model units, lots and state machines precisely

Overview

You will contribute to high-impact projects that require a deep understanding of both software design and systems engineering. Whether you are working on manufacturing processes, energy storage systems (ESS), or reliability frameworks, your work will directly influence how Peak Energy builds and scales its technology. The role calls for engineers who are comfortable operating at the intersection of software and hardware, where precision and efficiency are paramount.

The Peak Energy interview process

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

Initial Conversation

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

What to demonstrate

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

How to prepare

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

Technical Deep-Dive

The same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.

What to demonstrate

  • In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
  • In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
  • Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes

How to prepare

  • Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
  • Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
  • For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first

Final Panel Presentation

Where 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

Where candidates lose points

Retrying an irreversible external call without an owned idempotency key

Timeouts and connection resets are ambiguous by construction - the partner may have committed before the response was lost - so a generic retry policy around an HTTP client is, in this domain, a machine for buying two labels and dispatching two trucks. Relying on the partner's deduplication is not a substitute, because their window is usually short, their key is often derived from fields you may legitimately change on retry, and many older integrations have no such concept at all. The workable pattern is to generate the key yourself, persist it with an explicit unknown status before the call, and resolve ambiguity by querying the partner for that key rather than re-issuing; the sweeper that does this is the component that has to be correct, not the call site. Teams get this right for payments and then forget that a carrier tender, a warehouse work release and an EDI shipping notice have the same shape.

Judging delivery windows and cut-offs in UTC

A delivery appointment, a receiving cut-off and a same-day promise are all statements about local wall-clock time at a specific stop, so evaluating them in UTC misclassifies on-time performance for any leg that crosses a zone, and misses cut-offs by an hour on the two days a year a region changes offset. The subtlety that catches experienced engineers is that converting a future local time to a UTC instant at write time bakes in the offset rules as they were known then: when a jurisdiction changes its rules, an appointment stored that way silently moves, which is why a future window is better stored as local wall time plus an IANA zone identifier and resolved at read time, while a past event is correctly stored as an instant. Duration arithmetic needs the same care, because a transit that spans a DST boundary is genuinely 23 or 25 hours long in local terms even though the elapsed instant count is unchanged. Storing only a date with no zone - common on partner feeds - is the version of this that cannot be fixed downstream, so the zone has to be captured at ingestion.

Assuming the bug is in the framework

Suspect your own code first: read the stack trace top to bottom, check which versions are actually installed rather than which ones you believe are, and reproduce in isolation before blaming a library that thousands of people run daily. When the fault really is upstream, you need that minimal reproduction to say so credibly anyway.

Optimising an axis nobody named

Ask which resource is actually scarce here: wall-clock latency, throughput, memory footprint, cost per request, or engineering time. Shaving a constant factor off an in-memory step is wasted effort when the same function makes a blocking remote call inside the loop.

Questions & practice

Find the busiest ten-minute window per reader after deduplication

Medium · sliding window · event time · deduplication

One day of observation_event from fixed readers: source_id (reader), subject_id (container), occurred_at, received_at, dedupe_key. 300 million rows, partner and device replays resend whole windows, and some events arrive six hours after they occurred. For each reader, return the 10-minute event-time window containing the most distinct containers, with the window's start. Target linear work per reader after sorting. State the window semantics you chose, and what the job does with an event that lands after its window was already reported.

Approach

  1. Deduplicate before windowing, on dedupe_key, and keep the two concerns separate. A replayed batch is a no-op only if the duplicate is removed first; deduplicating inside the window logic counts a resend as traffic and reports a congestion peak that never happened. Dedupe is a hash set per reader-day or a sort-unique on the key — the same sort you need anyway.
  2. Partition by source_id and sort each partition by occurred_at, never by received_at. Arrival order is an artefact of device connectivity and partner batch cadence; a reader that went offline for six hours would otherwise show its entire backlog as one impossible burst. Sorting dominates: O(p log p) per reader of size p, O(n log n) overall.
  3. Two pointers over the sorted partition with a hash multiset container -> count and a separate distinct integer. Advance the right pointer, incrementing the count and bumping distinct only on a 0-to-1 transition; advance the left pointer while occurred_at[left] <= occurred_at[right] - 600s, decrementing and lowering distinct only on a 1-to-0 transition, and erasing the key from the map so the map's size and distinct never diverge. Each element enters and leaves once: O(p) amortised after the sort, O(window occupancy) space.
  4. Justify the candidate set rather than scanning every possible start. With half-open windows [t, t+600), sliding a window right until its left edge coincides with an event can only add events and never removes one, so the maximum is attained at some event's occurred_at; checking the p windows anchored at events is sufficient and exhaustive. State the half-open convention explicitly — an event exactly at t+600 belongs to the next window, and flip-flopping on that produces off-by-one disagreements between this job and whatever reads its output.
  5. Handle lateness by choosing a policy, not by hoping. Either run over a closed event-time batch with a stated lateness allowance — the day plus six hours, so the six-hour tail is inside the batch and the day-late tail is not — or emit under a watermark and restate the affected windows when a late event lands. Both are defensible; what is not defensible is silently dropping the late event, because that is how the busiest window quietly becomes the one with the best connectivity.

Follow-up

  • Two events for one container arrive from two readers a second apart. Is that congestion, a misread, or one container passing two portals?
  • You need this continuously rather than as a daily batch. What changes, and where does the memory go?
  • A device's clock is three hours fast. What does that do to this result, and how would you detect it from clock_offset_ms?

Trace a suspect lot forward to every shipped order line

Medium · graph traversal · genealogy · indexing

lot_edge(parent_lot_id, child_lot_id) has 400 million rows with btree indexes on both (parent_lot_id, child_lot_id) and (child_lot_id, parent_lot_id). lot_shipment(lot_id, order_line_id) maps lots to shipped lines. Given one suspect received lot, return every downstream order_line_id. The reachable subgraph can reach 2 million lots and 12 levels; the full edge table does not fit in memory. Give the traversal, its complexity in terms of the reachable subgraph, and what you do if the data contains a cycle.

Approach

  1. Genealogy is a directed acyclic graph, not a tree: a production run consumes several lots and one lot splits across many shipments, so a node is reached by several paths and a tree traversal revisits it combinatorially. Breadth-first from the suspect lot over the (parent_lot_id, ...) index, with a hash set of visited lot ids, is O(V' + E') where V' and E' count only the reachable subgraph — 2 million 8-byte ids is 16 MB of payload and a few tens of MB with open-addressing overhead, which is why the visited set is affordable and the edge table's 400 million rows never enter memory.
  2. Batch each level into one indexed read. Issue WHERE parent_lot_id = ANY($1) with a few thousand parents per call instead of a query per node: at 12 levels and 2 million nodes, per-node round trips dominate the wall clock entirely, and the recall's deadline is a real-world response time, not a query-plan aesthetic. The batch size is the tuning knob between round-trip count and per-statement planning cost.
  3. Keep the visited set even though the schema says DAG. Rework and repack loops — material returned into an earlier lot — do occur, and an unguarded traversal on a cyclic edge set does not terminate; a set membership test is O(1) and turns a hang into a correct answer plus a data-quality alert. Also cap the traversal by level and by node count, and fail loudly on the cap rather than returning a partial set silently.
  4. Join to lot_shipment once at the end over every visited lot id, not over the leaves and not per node. A lot is routinely both shipped and consumed: half a lot goes out on an order line while the other half is repacked into a child lot, so it has a lot_shipment row and outgoing edges at the same time and is not a leaf. Restricting the join to nodes with no outgoing edges drops exactly those order lines, which is under-recall — the one failure mode this trace is not allowed to have. One bulk join against an index on lot_id over the whole visited set is a single pass; per-node lookups multiply the round-trip problem by the number of visited lots.
  5. State the semantics honestly: reachability answers which shipped lines could contain material from the suspect lot, and without propagating edge quantities it is a superset of physical containment. For a recall that is the correct direction of error — over-recall, never under-recall — and saying so is the point, because a trace that returns quickly and incompletely will be believed. The upstream question, which received lots are in a suspect shipped unit, is the same traversal over the reverse index, which is why both indexes exist.

Worked solution (30 min)

  1. Draw a six-lot fixture containing a merge (two parents into one child), a split (one parent into three children), and one interior lot that is partly shipped on an order line and partly consumed into a child lot, then write down the expected reachable set and the expected order lines by hand.
  2. Implement level-batched BFS with an explicit frontier list, a visited hash set, and a per-level batch read.
  3. Add a cycle to the fixture and confirm the traversal terminates and flags it rather than hanging.
  4. Bulk-join lot_shipment once over every visited lot id — the whole visited set, not the leaves — and count the queries issued, confirming it is one per level plus one.
  5. Re-run with the frontier order reversed to confirm the output set is unchanged.

Expected result: Every lot downstream of the suspect appears exactly once in the visited set; the merge node is visited once, not twice; the partly-shipped interior lot contributes its order line even though it has children; the cyclic fixture terminates with the same reachable set plus a flagged edge; the query count is levels + 1, not proportional to node count.

Verify it yourself

  • The reachable set is a superset of the set you computed by hand, and equal to it on the acyclic fixture.
  • Restricting the final join to lots with no outgoing edges drops the partly-shipped interior lot's order line — the failing test that pins the over-recall rule to code.
  • Running the same traversal over the reverse index from a leaf returns the suspect lot.
  • Removing the visited set makes the diamond fixture return duplicates — a failing test that documents why the set is there.

Follow-up

  • A recursive CTE would express this in ten lines. What is its cost here if the index on the traversal column is missing, and how would you notice?
  • You need the quantity of suspect material in each downstream line, not just reachability. What has to be on the edge, and what does that do to the traversal?
  • The traversal must run while a production run is writing new edges. What does a consistent answer mean here, and what do you cut it at?

Find the hottest write keys under a fixed memory budget

Medium · top-k · heap · streaming sketches

One day of writes to inventory_movement is 200 million rows; each carries (item_id, node_id). Distinct pairs reach 60 million. Return the 200 pairs with the most writes so they can be moved behind a single writer before the next peak. First give the exact answer assuming you may hold 60 million counters, then give an answer under a 200 MB budget, and state the error guarantee that second answer carries. Both must be one pass over the day's rows unless you argue for a second.

Approach

  1. Exact version: one hash pass to build (item_id, node_id) -> int64, then a single scan of the map maintaining a min-heap of size K = 200, pushing and popping when the incoming count exceeds the heap root. O(n + d log K) with n = 2e8 and d = 6e7, versus O(d log d) if you sort all 60 million entries — a factor of roughly log d / log K, about 3.4x, for no benefit. The heap must be a min-heap: the root is the weakest survivor, so eviction is O(log K).
  2. Cost the map honestly. 60 million entries at a 16-byte key, an int64 count and per-entry overhead is 2-3 GB in most runtimes, which is the whole reason for the second version rather than a theoretical aside.
  3. Bounded version: Misra-Gries with m counters. Keep at most m keys with counts; on a key already present increment it; on a new key when fewer than m are held, insert at 1; otherwise decrement every held counter and drop those reaching zero. With m = 1,000,000 at roughly 40 bytes per counter, that is about 40 MB, comfortably inside 200 MB alongside the heap.
  4. State the guarantee as a number, not as 'approximate'. Misra-Gries returns an estimate satisfying f - n/(m+1) <= f_hat <= f, so with n = 2e8 and m = 1e6 every count is an undercount by at most 200. If the 200th hottest pair has hundreds of thousands of writes, a 200-count error cannot reorder the top of the list — and if the counts near rank 200 are within 200 of each other, the ranking there was never meaningful for the decision being made. The sketch also merges: run one per shard, combine, prune back to m, and the bound holds with n the combined length.
  5. If exactness is genuinely required, argue for two passes rather than more memory: the Misra-Gries pass yields a candidate superset of size m, and a second pass counts only those candidates exactly in O(m) memory. That is the standard way to get an exact top-K under a memory bound, and the cost is one extra read of the day's rows.

Follow-up

  • The decision downstream is which keys get a single writer. Does a rank error at position 190 change that decision, and what does the answer tell you about how much accuracy to buy?
  • Write volume on one key is bursty — a promotion concentrates it into twenty minutes. Does a daily top-K find it, and what window would?
  • How do you keep the sketch across a worker restart without replaying the day?

Decide whether available-to-promise is derived or stored

Hard · denormalisation · read path · cache staleness

Available to promise at one node equals inventory_position.qty in state on_hand minus the sum of allocation.qty in status held or committed for that key. The read sits inside checkout: 4,000 reads per second at peak, p99 budget 20 ms. inventory_position holds 40 million rows; during a promotion a single hot item-node pair carries about 2,000 open allocations. Decide whether to compute this per read or to maintain available_qty on the position row. Give both paths, the failure mode of each, and what the browse read may do that the allocate read may not.

Approach

  1. Cost the derived read in tuples, not in plans. One index probe into the position is trivial; the aggregate over allocation touches one tuple per open claim on that key. Even if only a tenth of peak lands on the hot pair, 400 reads per second times 2,000 rows is 800,000 index tuples per second for one item at one node, and it gets worse exactly as the key gets hotter. A partial index on status IN ('held','committed') does not help, because the row count is the cost, not the selectivity.
  2. Maintaining available_qty on the position row is nearly free on the contention axis, which is the point most candidates miss. The allocator already takes a row lock or a compare-and-set on that same position row, so the extra column costs no lock that was not already held, and the read collapses to a single primary-key probe well inside 20 ms.
  3. It is not free on the correctness axis. available_qty becomes a second derived value that can drift, so it joins qty under the same reconciliation job with its own re-sum against the allocation table, and a drift alert on that comparison is the operational price of the read speed.
  4. Split the two reads rather than arguing about one. The browse read may serve from a replica or a short-TTL cache and must err low, showing less than is there. The allocate read must be on the primary and inside the same transaction as the conditional update, because replication lag peaks during the write burst that made the key hot. Conflating them is what makes a replica read look safe in a benchmark.
  5. Pick the TTL against the drain rate, with a number. If a promotion burns 2,000 units in a minute, a 30-second cache is a thousand units stale, so the displayed figure should be a banded signal such as in stock or low stock rather than an exact count that the allocate step will then contradict.
  6. Do not denormalise availability across nodes into a single network number. A yes-somewhere answer cannot be allocated against, so sourcing will disagree with the page, and the customer sees a promise retracted after checkout rather than an honest out-of-stock before it.

Worked solution (40 min)

  1. Load a position with 2,000 open allocations and measure p99 for the derived read at 400 requests per second against the same read from a maintained available_qty column.
  2. Add available_qty, backfill it per key under the same lock the allocator takes, and write the re-sum query that proves it equals qty minus held and committed.
  3. Run both read paths in shadow for a fixed window, logging every disagreement with its key and magnitude.
  4. Point the browse read at a replica with a 30-second TTL and the allocate read at the primary, then measure how often the browse figure is higher than the allocatable truth.

Expected result: The maintained column serves the read in a single primary-key probe inside the 20 ms budget at 4,000 reads per second, the derived path misses it on the hot key, and shadow comparison shows zero disagreements on the primary path while the replica path is stale in the low-availability direction only.

Verify it yourself

  • p99 for the maintained read is flat as open allocations on the key grow from 10 to 2,000; the derived read's is not.
  • The re-sum finds zero drift after a sustained allocate-and-release workload.
  • Every observed browse-versus-primary disagreement is an under-report, never an over-report.
  • Killing an allocator mid-transaction leaves available_qty and the allocation rows consistent after the expiry sweeper runs.

Follow-up

  • Roll this out on a live system. How do both paths run side by side, what do you compare, and what result gates the cut-over?
  • The reconciliation job finds available_qty drifted by 3 on one key. What do you do, and what must you not do?
  • Lot-tracked items multiply the position rows per item-node. Does your answer change, and where does the read now spend its time?

Fold late, out-of-order scan events into one status per leg

Medium · window functions · event time · monotonic fold

observation_event holds event_id, subject_type, subject_id, observed_status, occurred_at, received_at, dedupe_key (UNIQUE) and processing_status. A lookup table maps observed_status to status_rank SMALLINT along a progress lattice. Write one query returning the current status of every shipment leg that received an event in the last hour, such that a picked_up event arriving after a delivered event never changes the answer. Then explain exactly which part of your ordering makes a shuffled replay of the same batch produce identical output.

Approach

  1. Separate the two jobs. Deduplication is the UNIQUE on dedupe_key at ingest, because partner feeds resend whole windows; ordering is the fold. Either alone leaves a defect: dedupe without a lattice still lets a stale event win, and a lattice without dedupe double-counts a replayed window in anything that aggregates.
  2. Select the affected legs in one CTE by received_at >= now() - interval '1 hour', then fold over every event for those legs, not only the recent ones. Restricting the fold's input to the recent window is the failure: a lone late picked_up would be the only row considered and the leg would regress.
  3. Fold as a maximum over the lattice: SELECT DISTINCT ON (subject_id) subject_id, observed_status FROM ... ORDER BY subject_id, status_rank DESC, occurred_at DESC, event_id DESC. The equivalent form is ROW_NUMBER() OVER (PARTITION BY subject_id ORDER BY status_rank DESC, occurred_at DESC, event_id DESC) = 1.
  4. The determinism comes from the final event_id tiebreak. Rank and occurred_at alone leave two equal-rank events at the same instant to be resolved by heap order, which changes when rows are re-inserted, so the shuffled replay differs in exactly the cases nobody tests. With a total order over the tuple the fold is a max, which is commutative and associative and therefore order-insensitive.
  5. Do not advance a leg on occurred_at alone. The timestamp is device-reported and can be wrong, which is why the lattice rank leads the ORDER BY and why clock_offset_ms is worth storing: a future-dated event with a low rank then cannot promote itself past a delivered.
  6. Keep superseded rows and mark processing_status = 'superseded' rather than deleting them. The discarded event is usually the one that explains a delivery dispute, and deleting it makes the trace unanswerable.

Follow-up

  • An event arrives a week late for a leg that is already invoiced. Does the fold apply it, reject it as late, or apply it and emit a correction?
  • The stream is partitioned by source_id rather than by subject_id. What ordering guarantee do you actually have, and what breaks?
  • How would you materialise this fold so the customer-facing status read is not running a window function per request?

Explain your process for ensuring code reliability in a system where hardware interaction is critical.

Medium

Approach

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

Follow-up

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

How would you approach designing a software system to monitor high-volume manufacturing data?

Medium

Approach

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

Follow-up

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

Allocate through a regional failover with unknown replication state

Hard · failover · replication lag · consistency · durability

The inventory primary sits in one region with an asynchronous standby in another, typically 2 seconds behind. The region becomes unreachable at peak, with 3,000 allocations and 3,000 ATP reads per second in flight. Decide whether to promote, what happens to allocations that may have committed without replicating, and what ATP answers meanwhile. Then quantify the cost of making the allocation write synchronous instead, given today's 4 ms transaction and about 42 allocations per second on the hottest item-node pair.

Approach

  1. Split the decision by operation class before touching the failover mechanics. ATP reads can move to the standby at once with a stated staleness, because a 2-second-old read only mis-advertises and the allocation write re-checks anyway. Allocations cannot: promoting an asynchronous standby discards up to 2 seconds of committed allocations, several thousand at this rate, and each one is a promise already made to a customer.
  2. Recover those from the requester's durable intent, not from the database that just vanished. If the orchestrator wrote its saga step and an outbox row in one local transaction before calling, every lost allocation can be re-driven after promotion under its original idempotency key, and allocation's partial unique index makes a re-drive that did replicate fail closed instead of double-claiming. Without that intent record the loss is simply unrecoverable, and the honest design refuses allocations until the primary returns rather than pretending otherwise.
  3. Price synchronous replication in the currency that actually binds, which is hot-row throughput rather than average latency. A cross-region synchronous commit adds tens of milliseconds inside the transaction, turning a 4 ms hold into roughly 65 ms and dropping one row from about 250 to about 15 allocations per second, below the 42 demanded - so the promotion tail becomes an outage every promotion. A synchronous standby in a second zone of the same region costs 1-2 ms, leaves the hold near 6 ms and the ceiling near 170 per second, still four times demand.
  4. So the configuration is: synchronous in-region for zero data loss on a node or zone failure, asynchronous cross-region for the regional case, and promotion of the cross-region standby as a deliberate human decision with a written procedure. Automatic promotion of an asynchronous standby is how a regional blip becomes two primaries and a permanently split ledger that reconciliation cannot repair, because both halves are internally consistent.
  5. Define the degraded contracts explicitly for the gap. ATP serves the conservative bucketed answer from the standby with staleness stated; allocate returns a typed unavailability so the orchestrator holds lines in sourcing rather than cancelling or retrying them into a storm. After promotion, replay the outbox, then run the reconciliation re-sum before any node's availability is trusted again, because the re-summed positions are the only evidence of what actually survived.

Worked solution (40 min)

  1. Classify every operation as safe or unsafe on a 2-second-stale replica and write that classification down before choosing a failover posture.
  2. Trace one allocation that committed a second before the outage through promotion, and name the exact durable record that lets you re-drive it.
  3. Recompute the hot-row ceiling under in-region and cross-region synchronous commit and set both against the 42 per second demand.
  4. Write the degraded contracts: what ATP returns, what allocate returns, and what the orchestrator does with each.
  5. Write the post-promotion sequence - outbox replay, reconciliation re-sum, then re-enable - and say what is not trusted until it completes.

Expected result: Reads promoted immediately with stated staleness and writes not; lost allocations recovered by re-driving from the orchestrator's durable intent under their original idempotency keys; synchronous commit placed in a second in-region zone (hold about 6 ms, ceiling near 170 per second against 42 demanded) rather than cross-region (about 65 ms, ceiling near 15); and promotion as a deliberate decision followed by outbox replay and a reconciliation re-sum before availability is trusted.

Verify it yourself

  • Every allocation a customer was told about is either present after promotion or re-drivable from a record held outside the failed database.
  • The cost of synchronous replication is expressed as a change in the hot-row ceiling, not only as added average latency.
  • Nothing in the design permits the old primary and the promoted standby to accept allocations at the same time.
  • The degraded ATP answer errs toward under-promising and says so explicitly.

Follow-up

  • The old primary returns holding writes the standby never saw. What do you do with them, and what does the reconciliation job report about them?
  • How do you rehearse this? Describe a game day that exercises the promotion path without risking an oversell in production.
  • Which of the two regions' clocks do you trust for occurred_at ordering during the switch, and does it matter here?

Allocation p99 hits nine seconds while database CPU stays flat

Medium · lock contention · connection pool · outbox

At 14:10 allocation p99 went from 30 ms to 9 s and an unrelated read service sharing the connection pool began failing. Database CPU and mean statement time are unchanged. pg_stat_activity shows 80 of 100 connections in state 'idle in transaction' with a mean transaction age of 18 s, and lock waits concentrated on about 30 inventory_position rows. In the same minute a carrier's rate-quote p99 went from 180 ms to 22 s. Give the ordered diagnosis, the immediate mitigation, and the structural fix.

Approach

  1. Ask whether the database is working or waiting, because the two look the same from the client. Flat CPU and flat mean statement time with most connections in 'idle in transaction' says the database is idle inside open transactions: the application has begun a transaction and gone off to do something else. That state name is the whole diagnosis and it points at application structure, not at the database.
  2. Tie the holders to the dependency rather than assuming the link. Compare the distribution of now() minus xact_start against the carrier's latency histogram in the same window; when the two track each other, the transaction is wrapping the external call. The 18 s mean transaction age against a 22 s partner p99 is that match.
  3. Locate the contention: join pg_locks to pg_stat_activity, group by the locked tuple, and confirm the waits concentrate on a few dozen hot (item_id, node_id) keys. Demand concentrates on those rows precisely during a promotion, which is also when partner latency degrades, so the two curves multiply rather than add.
  4. Mitigate without a deploy where possible. Set a call timeout on the partner shorter than the lock budget, set idle_in_transaction_session_timeout and lock_timeout so a stall sheds instead of queueing, and give the carrier gateway its own connection pool so one degraded partner cannot drain the pool an unrelated service depends on. A circuit breaker on that partner turns a 22 s wait into an immediate, handled failure.
  5. Fix the structure: take the external call out of the transaction. Write the allocation state change and an outbox row in one short local transaction, let a relay make the rating or tender call asynchronously with a persisted idempotency key, and advance the saga on the result. Lock hold time then equals local work, single-digit milliseconds, and is independent of any partner.
  6. Accept and handle what the outbox costs you. Publication becomes at-least-once, so the consumer and the external effect must both be idempotent, and the caller now sees an asynchronous outcome it must be designed for. That is the trade: bounded lock hold time in exchange for eventual consistency and duplicate delivery you have to absorb.

Follow-up

  • Your mitigation adds a lock_timeout. What does the allocation path do when it fires, and what does the customer see?
  • Even with short transactions, 30 rows absorb most writes on promotion day. How do you keep those keys from serialising everything?
  • How do you test this before the next promotion without a degraded partner to hand?

Suggested practice checklist

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

Day 1 — Coding, one pass at shallow depth

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

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

Day 2 — Design, one pass at shallow depth

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

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

Day 3 — Fundamentals and the practical rounds

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

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

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

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

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

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

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

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

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

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

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

Day 7 — Reassemble the loop

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

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

The estimate and the day you knew

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.

Tell me about a time you had to learn a new technology or domain quickly to solve a problem.

Medium · behavioural and engineering judgement

Approach

  1. Pick a story where you made the decision, not one where you watched it.
  2. Name the disagreement and how you resolved it with evidence.
  3. Close with what you would do differently, concretely.

Follow-up

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

Can you walk me through a complex project you led from conception to completion?

Medium · behavioural and engineering judgement

Approach

  1. Name the disagreement and how you resolved it with evidence.
  2. State the situation in two sentences and spend the rest on the reasoning.
  3. Give the blast radius: what could have broken, and what you measured.

Follow-up

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

How do you handle data integrity when integrating software with physical equipment?

Medium · behavioural and engineering judgement

Approach

  1. Close with what you would do differently, concretely.
  2. State the situation in two sentences and spend the rest on the reasoning.
  3. Name the disagreement and how you resolved it with evidence.

Follow-up

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

What specific aspects of test engineering or systems integration are you most passionate about?

Medium · behavioural and engineering judgement

Approach

  1. State the situation in two sentences and spend the rest on the reasoning.
  2. Pick a story where you made the decision, not one where you watched it.
  3. Close with what you would do differently, concretely.

Follow-up

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