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ARO Tender Consulting Software Engineer Interview Questions & Guide 2026

A prep plan for ARO Tender Consulting software engineer candidates: answer first, read the solution second, repeat the misses later.

The 60-second brief

The Software Engineer role at ARO Tender Consulting bridges the gap between complex infrastructure requirements and the bidding processes that define the firm's business. As a consultant, you are responsible for designing and maintaining the digital frameworks that allow the firm's teams to manage, track, and execute multimillion-dollar bids within the Electrical Infrastructure & Energy Sector. Your work affects the firm's ability to secure major contracts, ensuring that its technical proposals are backed by scalable and efficient software solutions.

Most rounds test how you handle an under-specified problem more than what you recall. A wrong first approach you correct out loud is recoverable; a constraint you assumed silently and were never given is the expensive mistake.

Focus areas

Make integration retries idempotent without target-side idempotency keys
Version an API you cannot force clients to upgrade
Evolve schemas expand-contract across un-upgradable client deployments

Overview

This position is inherently strategic. You will not only be writing code but also collaborating with Bid Managers and Infrastructure Specialists to translate technical requirements into winning bid strategies. The role offers the unique opportunity to work at the intersection of heavy industry and software innovation, requiring a candidate who is as comfortable with complex data architectures as they are with the fast-paced, high-pressure environment of tender management.

The ARO Tender Consulting interview process

ARO Tender Consulting candidates report 2 rounds · ≈ 2-4 weeks. The stages below are what candidates describe, not a published process.

Initial Assessment

Input 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

Deep-Dive Sessions

An unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.

What to demonstrate

Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating

How to prepare

Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes

Where candidates lose points

Forking the product per client instead of building a configuration or extension point

A fork is the cheapest possible answer for one client and the most expensive possible answer for the estate. Maintenance cost grows as O(clients x changes): a single security patch now needs one port per fork, each with its own conflicts, its own test run and its own release window, and the forks drift further apart with every port. The honest rule is that a required fork is a product gap, so it should either become a supported extension point with a stable interface, or be an explicit, time-boxed and separately priced artefact that is never expected to receive upstream fixes.

Long-lived shared credentials for client systems

One static key used by the connector runtime, a debugging script and three engineers cannot be attributed to a person in an audit, cannot be expired at engagement close, and cannot be rotated without breaking an unknown number of consumers at once. The expensive part of the eventual incident is not the leak; it is that rotation requires finding every holder under time pressure, and nothing recorded who they were. Issue per-principal, per-target, short-lived grants from a broker so rotation is a no-op, attribution is automatic, and closure is a TTL rather than a search.

Tests that assert on the implementation rather than the behaviour

Assert on what a caller can observe, not on the number of internal calls or the shape of a private field. A test that breaks on every refactor but still passes when the answer is wrong costs more than it protects.

Saying 'eventually consistent' without naming the anomaly a user would see

Describe the concrete symptom you are choosing to accept: the author reloads and their own comment is missing for two seconds, or two devices show different balances for a minute. The class of consistency model is a technical label; the tolerable anomaly is the actual product decision.

Questions & practice

Collapse connector retry attempts into logical jobs and failure counts

Easy · hash aggregation · idempotency · retries

You are given one UTC day of connector_run rows: (connector_id, idempotency_key, attempt_no, status, failure_class, records_read, records_applied), unique on (connector_id, idempotency_key, attempt_no), up to 5,000,000 rows in arbitrary order. Retries of one logical job share the idempotency_key; the job's outcome is the status of its highest attempt_no. Return, per connector_id: the number of logical jobs, a count of terminally failed jobs by failure_class, and the dedupe hit total, meaning sum(records_read - records_applied) over terminal-succeeded attempts only. One pass over the input; state your memory bound.

Approach

Key a hash map on (connector_id, idempotency_key) and keep only the highest attempt_no seen so far plus that row's status, failure_class and record counts. One pass, O(n) time, O(d) space where d is the number of distinct logical jobs, not O(n).
Treat a repeated (key, attempt_no) as corrupt input and raise, since the table's uniqueness constraint says it cannot happen; silently overwriting hides a double-insert in the runtime.
Fold the d surviving entries into per-connector counters in a second pass over the map, not over the rows. The failure_class histogram is a small fixed-width map per connector because failure_class is an enum.
Compute dedupe hits only from terminal-succeeded attempts. A failed attempt's records_read is work attempted, not work deduplicated, and adding it counts the same source records once per retry.
If d does not fit in memory, partition the input by hash(connector_id, idempotency_key) into p files and aggregate each partition independently. Same O(n) total work, p sequential passes, memory traded for I/O; the partition function must use the full key or a job's attempts split across files.

Worked solution (20 min)

Take six rows: (c=7,K=A,1,failed,client_429,0,0), (c=7,K=A,2,succeeded,null,500,480), (c=7,K=B,1,ambiguous,client_timeout,300,300), (c=7,K=B,2,succeeded,null,300,0), (c=9,K=C,1,failed,schema_mismatch,0,0), (c=9,K=C,2,failed,schema_mismatch,0,0).
Build the map: A -> attempt 2 succeeded, B -> attempt 2 succeeded, C -> attempt 2 failed/schema_mismatch. Three entries from six rows.
Fold per connector: connector 7 has 2 logical jobs, 0 terminal failures; connector 9 has 1 logical job, 1 terminal failure classed schema_mismatch.
Dedupe hits from terminal-succeeded attempts only: (500-480) + (300-0) = 320 for connector 7, 0 for connector 9.
Compute the wrong answer deliberately by counting attempt rows: 6 jobs and 3 failures, which is what a naive GROUP BY on run rows returns.

Expected result: connector 7: 2 logical jobs, no terminal failures, 320 dedupe hits. connector 9: 1 logical job, 1 terminal failure with failure_class='schema_mismatch', 0 dedupe hits. Attempt-level counting would have reported 6 jobs and 3 failures.

Verify it yourself

The sum of terminal statuses across all connectors equals the number of distinct (connector_id, idempotency_key) pairs, here 3.
records_applied <= records_read holds on every row, so the dedupe total is non-negative.
Rerunning the aggregation over a shuffled copy of the input gives identical output, since the highest attempt_no wins regardless of arrival order.
Map size equals 3, strictly less than the 6 input rows, confirming collapse actually happened.

Follow-up

A job's highest attempt is 'cancelled' but an earlier attempt succeeded. What is the job's outcome, and what does that say about who writes the cancel?
Produce this incrementally as runs land instead of as a daily batch: what state do you keep per key, and what happens when an attempt arrives out of order?
records_applied is written by your runtime, not by the client. What would you reconcile it against before anyone trusts the dedupe-hit number?

Top-k longest connector runs from a stream too large to sort

Medium · heap · top-k · streaming

A day of connector_run rows sits in object storage: 200,000,000 rows, unsorted, far larger than memory, and streamable once. Each row has run_id, connector_id, started_at and finished_at, where finished_at is NULL for runs that never reached a terminal state. Return the 200 longest terminated runs by (finished_at - started_at), ties broken in favour of the lower run_id, plus the count of rows with a NULL finished_at. You may not sort the input or hold it in memory. State the time and space complexity you achieve.

Approach

Keep a bounded min-heap of capacity k=200 ordered on (duration, -run_id). The root is then the weakest retained entry: smallest duration, and on a duration tie the largest run_id, which is exactly the one the stated tie-break should evict.
Per row: if finished_at is NULL, increment a counter and skip; otherwise compute the duration and push when the heap holds fewer than k, else compare against the root and replace only if strictly better. O(n log k) worst case, O(k) space, and after the heap warms nearly every row costs a single comparison against the root.
Reject the alternatives explicitly. A full sort is O(n log n) and at 2e8 rows means an external merge sort over hundreds of gigabytes to produce 200 rows. Quickselect is O(n) average but needs the whole array resident, which the constraint forbids.
Handle the NULL as data, not as an edge case: depending on the language, subtracting from NULL either throws or yields a sentinel that lands at the top of the list. Count those rows and report them, because a spike of unterminated runs is its own signal.
To parallelise, give each of p shard readers its own size-k heap and merge the p results. The union of per-shard top-k sets contains the global top-k, because an element beaten by fewer than k rows globally is beaten by fewer than k rows in its own shard, so the merge is exact at O(pk log k).

Follow-up

Now return the top 20 bindings by total run time instead of the top runs. What changes, and how many distinct keys must you hold to make it exact?
The key space is too large to hold an exact aggregate. What does a Misra-Gries summary with m counters actually guarantee about the counts it returns?
Rows now arrive continuously and you want a rolling one-hour top-k. Does the bounded heap still work, and if not, what breaks?

Measure credential exposure past engagement close without double counting

Medium · interval merge · sweep line · credential ttl

For one engagement you have credential_grant rows: (grant_id, principal_id, target_system_id, issued_at, not_after, revoked_at which may be NULL). The cutoff instant T is the engagement's ends_on plus access_grace_days. A grant is live over the half-open interval [issued_at, min(not_after, revoked_at)). There are up to 200,000 grants across target systems. For each target_system_id, return the total wall-clock time after T during which at least one grant was live, plus the disjoint segments that make it up. Grants that overlap must be counted once.

Approach

Clip each grant to [max(issued_at, T), min(not_after, coalesce(revoked_at, +inf))) and discard any interval whose start is not strictly before its end. O(n), and it removes every grant that already expired inside the engagement window.
Bucket the survivors by target_system_id and sort each bucket by start. O(n log n) overall, and sorting dominates the whole algorithm.
Sweep each bucket once, holding one open segment [s,e): if the next start is greater than e, emit [s,e) and open a new segment, otherwise set e = max(e, next_end). O(n) after the sort, O(1) working state, O(number of emitted segments) output.
Total exposure is the sum of emitted segment lengths, which is the measure of the union. Summing per-grant durations instead answers a different question and is unbounded above by the elapsed wall clock.
Produce a second, conservative figure that ignores revoked_at and uses not_after alone. revoked_at records that you asked for revocation; whether the client's identity provider honoured it is not something this table knows, and a self-contained token validated offline stays valid to its own expiry regardless.
If a bucket does not fit in memory, sort externally and sweep the stream, or push end timestamps into a min-heap keyed by end and pop those below the current start. Same O(n log n), memory proportional to the maximum number of concurrently live grants.

Follow-up

Three grants overlap and one of them belongs to an offboarded contractor. What must the sweep emit so exposure can be attributed per principal?
Which of your two numbers is the real bound on access if the client system validates tokens offline, and what does that imply about issuing TTLs in the first place?
Run this across 20,000 engagements as a nightly job with a fixed memory budget. What changes in the shape of the computation?

Return the latest attempt per connector idempotency key

Easy · distinct-on · indexing · idempotency

connector_run stores one row per attempt: run_id, connector_id, idempotency_key, attempt_no SMALLINT, status (queued, running, succeeded, failed, ambiguous, cancelled, abandoned), failure_class, watermark_from, watermark_to, started_at, finished_at, unique on (connector_id, idempotency_key, attempt_no). For one connector_id, write the query returning the highest-numbered attempt per idempotency_key, keeping only those keys whose latest attempt is not succeeded, ordered by watermark_from. Give the index that serves it, and say what the plan does without that index.

Approach

Use DISTINCT ON (idempotency_key) with ORDER BY idempotency_key, attempt_no DESC. PostgreSQL requires the ORDER BY to lead with exactly the DISTINCT ON expressions, and that requirement is what makes the row picked deterministic rather than whichever the scan happened to reach first.
Apply the not-succeeded filter outside, in a wrapping SELECT. Putting it in the inner WHERE changes the question from 'keys whose latest attempt failed' to 'keys that have any non-succeeded attempt', which reports a key that failed twice and then succeeded, and is the single most common wrong answer here.
Know the portable alternative: row_number() OVER (PARTITION BY idempotency_key ORDER BY attempt_no DESC) in a subquery, filtered = 1 outside. A window function cannot be filtered in WHERE because WHERE is evaluated before the window, and PostgreSQL has no QUALIFY clause, so the subquery is mandatory rather than stylistic.
Index (connector_id, idempotency_key, attempt_no DESC). With connector_id fixed by equality, the index supplies the exact order DISTINCT ON needs and the Sort node disappears. It does not reduce rows read: through PostgreSQL 17 there is no loose index scan, so the executor still walks every attempt in range rather than skipping to each next distinct key.
Give the complexity honestly: without the index the plan is a scan plus an O(n log n) sort over the connector's attempts; with it, O(n) ordered reads and constant working memory, where n is attempts in range, not distinct keys. If the distinct keys are a tiny fraction of attempts, a recursive skip-scan is the next step.

Worked solution (15 min)

Create the table with a handful of rows across three idempotency keys, including one key with attempts 1 failed, 2 ambiguous, 3 succeeded.
Write the inner DISTINCT ON query alone and confirm it returns one row per key.
Wrap it and add WHERE status <> 'succeeded' plus ORDER BY watermark_from in the outer query.
Create the index and compare EXPLAIN output before and after.

Expected result: SELECT * FROM (SELECT DISTINCT ON (idempotency_key) idempotency_key, attempt_no, status, failure_class, watermark_from, watermark_to FROM connector_run WHERE connector_id = $1 ORDER BY idempotency_key, attempt_no DESC) latest WHERE status <> 'succeeded' ORDER BY watermark_from;

Verify it yourself

The key whose attempts run failed, ambiguous, succeeded is absent from the result, proving the filter runs after the pick.
Move the filter into the inner WHERE and watch that key reappear as attempt 2 — the wrong answer, visible on a four-row fixture.
EXPLAIN shows an Index Scan with no Sort node once (connector_id, idempotency_key, attempt_no DESC) exists.

Follow-up

attempt_no is not monotonic in time because a retry was queued out of order. Does your query still pick the right row, and which column should it actually order by?
How would you return the latest attempt per key across all connectors in one environment without the query degrading to a full sort?
An ambiguous latest attempt means the target may or may not have been written. What does this query feed, and should ambiguous be treated as failed here?

Explain why the failed-run backlog query ignores three indexes

Medium · indexing · query planning · partial indexes

connector_run holds roughly six million rows over three years. Its indexes are (connector_id, started_at DESC); (started_at) WHERE status = 'failed'; and ((finished_at AT TIME ZONE 'UTC')::date). The operator backlog query is SELECT run_id, connector_id, failure_class, started_at FROM connector_run WHERE environment_id = $1 AND status = ANY($2) AND started_at >= now() - interval '7 days' ORDER BY started_at DESC LIMIT 50, with $2 bound to {failed, ambiguous}. It sequentially scans. Explain index by index why each is unusable, then give the one index you would add.

Approach

Index one leads on connector_id, which this query never constrains. A B-tree can only be probed from a prefix of its key columns, so with no equality on connector_id it degenerates into a full index scan plus heap fetches, which the planner will rightly cost above a sequential scan. PostgreSQL 18 added B-tree skip scan for an unconstrained prefix, so state your server version before assuming this is fatal; it helps only when that prefix has few distinct values, and hundreds of connectors is already too many to rely on it.
Index two is partial on status = 'failed', and the rows the query wants are partly not in it at all. Predicate implication runs one way: the query's predicate must imply the index's. status IN ('failed','ambiguous') is weaker than status = 'failed' and implies nothing, so no amount of inlining the parameter helps. It would serve only the failed branch of a UNION ALL.
Index three is an expression index on a function of finished_at, while the query filters started_at as a plain range. Even on the right column, a range predicate on the bare column cannot use an index on a function of it. Note also why the AT TIME ZONE 'UTC' wrapper is present: finished_at::date on a timestamptz is STABLE rather than IMMUTABLE because it depends on the session TimeZone, so PostgreSQL refuses to index it without a fixed zone.
Add (environment_id, started_at DESC) WHERE status IN ('failed','ambiguous'): equality on the scoping column first so the scan starts in the right place, the ordering column second so ORDER BY started_at DESC LIMIT 50 stops after fifty index entries instead of sorting the week, and partial on the status set so the index covers the small failing minority rather than six million rows.
Confirm the partial predicate is provable for the query's parameterised form. Under a custom plan the array parameter is substituted as a constant and the prover handles the ScalarArrayOpExpr; under a generic plan it stays a Param and the proof fails, so check plan_cache_mode or the sixth execution of a prepared statement before declaring victory.
Say what you deliberately did not do. Adding INCLUDE (run_id, connector_id, failure_class) to chase an index-only scan pays only while the visibility map is current, which ties the query's latency to whether autovacuum is keeping up with the write rate — a dependency worth accepting consciously, not by accident.

Follow-up

The same query is also run without the environment filter, for an estate-wide backlog. Does your index still help, and what would you add instead?
How does row-level security on this table interact with your index, given that policy quals are applied before non-leakproof user quals?
Six months from now the partial index covers 30 percent of the table. What signal tells you to change it?

How do you balance the need for rapid feature deployment with the necessity of long-term system stability?

Medium

Approach

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

Follow-up

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

How do you approach designing a system that must handle high-concurrency data inputs during a critical bid deadline?

Medium

Approach

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

Follow-up

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

Design the error contract for engagement-scoped credential minting

Easy · error-taxonomy · credentials · fail-closed · http-semantics

The Access Broker mints short-lived credentials for client-owned systems. Its callers are the connector runtime, which is unattended and retries automatically, and an operator CLI with a person waiting. A request names an engagement, a target system, a set of scopes and a requested TTL. The broker refuses past the engagement's end date plus its recorded grace, caps any requested TTL at the remaining window, and fails closed when its own dependencies are unavailable. Design the response contract: the success shape, the failure classes each caller must tell apart, the status codes that carry them, and the required action per class.

Approach

Fix the success shape first, because the interesting case hides there. Return grant_id, the granted not_after, the scopes actually granted and the correlation id. Capping is not an error: a request for twelve hours two hours before the engagement ends is a 201 whose not_after is two hours out. Make it contractual that the caller schedules its next mint from the returned not_after and never from the value it asked for, or it will keep using a credential the broker already considers dead and discover it as a mid-run auth failure.
Classify refusals by what the caller does next, not by where they came from: terminal for good (engagement closed, terminated, or past end plus grace), terminal until something changes (scope not entitled, unknown target system), caller-fix (malformed scopes or an unparseable TTL), and retry-later (rate limited, or the broker failing closed on its own dependency). Map them to 403, 404, 422, 429 with Retry-After and 503 with Retry-After, each carrying a distinct stable code the connector switches on.
Keep fail-closed on the 5xx side deliberately. If a dependency outage returns 403, an outage is indistinguishable from an offboarding, and the unattended connector concludes a live engagement is over and stops scheduling it. The availability requirement stated for this service is exactly why the distinction has to be visible in the status code rather than only in a message.
Separate 'engagement closed' from 'engagement suspended' with different codes even though both are 403. The first means stop scheduling this binding permanently and raise it to a human; the second may succeed after reinstatement, so the correct behaviour is a long backoff. One code covering both forces the connector to guess, and it will guess the expensive way.
Serve both callers from one body: a stable machine-readable code plus a human message. The CLI renders the message; the connector switches on the code and records failure_class ('auth_expired' versus 'our_bug'), so failures stay attributable to an owner. Write an issuance or refusal record on every path, including refusals, and echo issued_by_request_id in the response, so a later question about who held access at a given minute has one identifier crossing both systems.

Worked solution (20 min)

Write the success body's fields and, beside each, which caller reads it and what it does with it; delete any field neither caller acts on.
List every reason the broker can refuse, then write the required action for the unattended connector and for the operator separately. Wherever the two differ, that is a separate stable code.
Assign status codes, and check that no reason lands on a code whose generic client behaviour contradicts the action you require.
Write the connector's handling as a switch on the stable code and confirm only the rate-limited and transient classes retry.

Expected result: A 201 carrying grant_id, the granted not_after, granted scopes and the correlation id; a refusal table in which engagement-closed, engagement-suspended, scope-not-entitled, unknown-target, malformed-request, rate-limited and broker-unavailable each have a distinct stable code; retries only on 429 and 503; and an audit record written on every path including refusals.

Verify it yourself

A TTL longer than the remaining engagement window succeeds with a shortened not_after rather than failing, and the caller's next mint is scheduled from the returned value.
Taking the broker's dependency down yields a retryable 5xx, and nothing in the connector's handling reads that as the engagement being over.
Every refusal leaves a record an auditor can find by correlation id, and the same id appears in the response the caller logged.

Follow-up

The broker returns 503 for ninety seconds during a dependency failure. What does the unattended connector do, and how do you keep that from polluting the per-binding failure rate you alert on?
A caller requests a twelve-hour TTL two hours before the engagement ends. What do you return, and what specifically breaks if you return an error instead?

Reports intermittently return another client's rows under load

Hard · tenant isolation · row-level security · connection pooling · concurrency

A shared-pool deployment scopes every query with row-level security: the application runs SET app.client_id at the start of each request, and policies compare client_id against current_setting('app.client_id'). Twice this week a client's report contained rows belonging to a different client. Staging cannot reproduce it, single-request replays are clean, and the SQL text is identical in the good and bad cases. Connections come from a pool shared by all requests. Give the ordered diagnosis, the reproduction, and the fix.

Approach

Order the hypotheses by what the evidence already excludes. Identical SQL text and clean single-request replays rule out a missing predicate in that endpoint, so the variable is not the query, it is the value the policy reads when the query runs. Check, in this order: is RLS enabled and forced on the table; is the serving role actually subject to it; and only then, what does current_setting return at execution time.
Confirm the policy is in force before blaming it. Policies do not apply to the table owner unless FORCE ROW LEVEL SECURITY is set, and never apply to a role with BYPASSRLS or to a superuser. If traffic is served by the owning role, the policy has been decorative all along and the leak is not intermittent at all, only intermittently noticed.
Instrument the value, not the code. As the first statement inside every transaction, read current_setting('app.client_id', true) and compare it with the tenant on the request, emitting a counter on mismatch. A defect invisible to replays shows up here within minutes of real concurrent traffic.
Explain why pooling makes it intermittent. A plain SET is session-scoped and survives COMMIT, so the backend returns to the pool carrying the previous request's value. This needs no transaction-pooling proxy; an ordinary application connection pool is sufficient. The next checkout that sets the value late, or skips it on some path, inherits the previous tenant, and row-level security then filters flawlessly for the wrong client.
Reproduce deterministically rather than by load-testing and hoping. Cap the pool at one connection, issue two requests for different tenants back to back, and have the second issue its first query before its SET. Zero rows would be the fail-closed outcome; the other tenant's rows is the leak, and the difference between those two results is exactly whether the previous value was left behind.
Fix in layers. Use set_config('app.client_id', $1, true) or SET LOCAL inside the transaction so the value dies at COMMIT; assert it matches the request immediately before the first query rather than trusting an upstream middleware; set FORCE ROW LEVEL SECURITY; and serve traffic with a role that neither owns the tables nor holds BYPASSRLS, reserving the owning role for migrations.

Follow-up

You owe a breach notification within contractual time. What query bounds which client pairs and which rows were exposed, and what in the schema makes that answerable at all?
A background job has no inbound request and therefore no tenant. Where does its tenant come from, and what stops it from being where this happens next?
How does the same failure mode look on the dedicated and client-managed tiers, and what does that tell you about where isolation should live?

Suggested practice checklist

For someone fluent in a dynamic language who has shipped real work but has never had to say what the runtime is doing underneath. The week is built on measuring and deliberately breaking things, because the questions that expose this background are the ones where the interviewer asks why a second time.

Day 1 — Measure before reasoning

Take a slow piece of your own code, write down in advance where you believe the time goes, then profile it and record how wrong the guess was. The cost is usually an allocation you did not notice or an accidental quadratic membership test.
Replace one list membership test inside a loop with a set and measure at a thousand, ten thousand and a hundred thousand elements, confirming the shape of the curve rather than only that it got faster.
Write down the three quantities you can now measure instead of assert: wall time, peak memory, and call count for the function you suspected.

Deliverable: A before-and-after profile of real code plus a written note on the size of the gap between the guess and the measurement.

Day 2 — References, copies, and the bugs they produce

Write the function with a mutable default argument, call it three times, and explain the accumulating result: the default is evaluated once when the function is defined, so every call shares one object.
Build a nested structure, take a shallow copy, mutate an inner element, and show that both views changed, because a shallow copy duplicates the container and not the elements. Then fix it with a deep copy and state the cost you just accepted.
Write two functions, one mutating its argument in place and one rebinding the local name, and predict the caller's view of each before running it. That single distinction produces most of the bugs that pass their tests.

Deliverable: Three small programs whose output you predicted correctly before running, each with a one-line statement of the rule underneath.

Day 3 — Types, once, in a language that checks them

Port one module you have already written, roughly a hundred lines, into a statically typed language, and record every place the compiler demanded an answer your original had left implicit: a value that can be absent, a numeric width, a case never handled.
Write the same signature in both languages and state what the static one guarantees before the program runs and what it does not, since it will not save you from a wrong algorithm or an index out of range.
Write the difference between an interface satisfied by declaration and one satisfied structurally, with one case each where the other approach would miss the mistake.

Deliverable: One module in two languages plus a list of the questions the type checker forced you to answer.

Day 4 — Concurrency, starting with what actually runs at the same time

Run the same CPU-bound function across four threads and four processes and measure both. Under the default CPython build the threaded version will not speed up, because only one thread executes bytecode at a time; the process version will. Check which build you are on first, since free-threaded builds remove that lock and change the result.
Then run a blocking I/O workload across four threads and measure it speeding up, because the interpreter releases that lock around blocking calls, which is why treating threads as useless is wrong as a general claim.
Build the lost update: two threads each incrementing a shared counter a hundred thousand times, and show a final value below the expected sum, because an increment is a load, an add and a store and the thread can be suspended between them. Fix it with a lock and then measure what the lock costs.

Deliverable: Three measurements, threads against processes on CPU work, threads on I/O work, and a demonstrated lost update, each with the mechanism written underneath.

Day 5 — Debugging as a procedure rather than an instinct

Work one real failure as a bisection: find a revision or an input size where it is good and one where it is bad, halve repeatedly, and state the two assumptions bisection needs, that the property changes exactly once across the range and that the test is reliable.
Minimise one failing input to the smallest version that still fails, and record how many rounds it took.
Keep a hypothesis log for one bug in three columns, what I believe, what would disprove it, what I observed, and stop yourself the first time you are about to change two things at once.

Deliverable: One bug worked to root cause with a written hypothesis log and a minimised reproducing input.

Day 6 — Tests that catch the bug you are about to write

Implement an LRU cache with a capacity bound, then write the three test cases that would catch an off-by-one in eviction: insert exactly capacity items and assert nothing was evicted, insert one more and assert the least recently used key is the one gone, and read an old key just before that insert so the eviction victim changes.
Add a property test comparing your implementation against a deliberately slow reference, an ordered list scanned linearly, over a few thousand random operation sequences, because a slow reference finds the cases you would not have thought to write.
Write one numeric test that fails under exact equality and passes with a tolerance, and state why the tolerance has to be relative rather than absolute once the magnitudes grow.

Deliverable: An LRU implementation with three boundary tests, one property test against a slow reference, and one tolerance-based numeric test.

Day 7 — Debug something broken, out loud

Have someone plant three defects in a two-hundred-line program, an off-by-one, a shared mutable state bug, and a wrong error-handling path, then find them while narrating, under a fixed rule: state the hypothesis before touching anything.
Time each one and record which tool found it, reading, a printed value, a debugger, or a test, because the question asked in interviews is how you would find it rather than what it was.
Write the sentence you will use when you do not yet know the cause, one that names the next measurement instead of offering a guess.

Deliverable: A recorded debugging session with time-to-find per defect and the method that found each.

Know where your numbers came from

Bring the two or three numbers the story rests on and know how they were collected. A p99 whose timer starts inside your handler excludes the time a request spent queued, so it can sit flat while users wait longer. Give the window, the percentile and what the measurement left out, or drop the number.

Describe a situation where you disagreed with a technical decision made by a team member; how did you resolve the conflict?

Medium · behavioural and engineering judgement

Approach

State the situation in two sentences and spend the rest on the reasoning.
Name the disagreement and how you resolved it with evidence.
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?

Tell me about a time you had to explain a complex technical limitation to a non-technical stakeholder, such as a Bid Manager.

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

Describe a time you had to refactor a significant portion of a codebase; what was your approach to minimizing downtime?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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

What steps do you take to ensure your work aligns with the broader business objectives of ARO Tender Consulting?

Medium · behavioural and engineering judgement

Approach

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

Follow-up

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