Interview guides · Ares Management

Ares Management Software Engineer Interview Questions & Guide 2026

Caching practice for Ares Management software engineer prep: settle the staleness budget before choosing an invalidation strategy.

The 60-second brief

As a Software Engineer at Ares Management, you play a vital role in building and maintaining the advanced technical infrastructure that powers a global alternative investment manager. Your work directly supports critical business units—ranging from Credit Technology and Legal & Compliance to Infrastructure Debt and HR Tech. By developing robust applications, optimizing data pipelines, and ensuring seamless systems integration, you enable the firm to manage complex financial portfolios with speed and precision.

The behavioural round is a technical round in narrative form. Prepare it by collecting specifics you actually owned, such as a design you argued against, an incident you diagnosed, or a decision you later reversed, rather than by rehearsing phrasing.

Focus areas

Build at-least-once pipelines with explicit deduplication horizons
Bound blast radius with per-tenant concurrency limits
Make every write idempotent under client retries

Overview

The problem spaces you will encounter involve a unique blend of enterprise software development and financial domain complexity. Whether you are improving database response times for high-volume transactions, engineering resilient network architectures, or building scalable internal platforms, your contributions directly impact business efficiency and data integrity. The scale and global reach of Ares Management mean that your solutions must be secure, highly available, and capable of handling complex financial workflows.

This role requires a balance of strong technical execution and domain curiosity. You will collaborate closely with quantitative analysts, product managers, and business stakeholders who expect reliable, high-performance technology. Expect a rigorous, fast-paced environment where problem-solving and clean code are paramount, offering an exceptional platform to grow your career at the intersection of technology and alternative asset management.

The Ares Management interview process

Ares Management candidates report 6 rounds · ≈ 4-6 weeks. The stages below are what candidates describe, not a published process.

Recruiter Screen

The title covers product work, platform work, infrastructure, mobile and frontend, and those are different jobs with different loops behind them. A screening call is the cheapest place to find out which one the seat is, and asking reads as experienced rather than fussy. The questions that separate them: what the team is on call for, what the last three projects were, and whether any round happens inside an existing repository instead of a blank file. Then say which of that you have done and which you have not. Claiming the whole posting is the fastest way to be found out one round later.

What to demonstrate

Whether you can locate your experience inside one flavour of the role honestly instead of claiming the entire requirements list
Whether you name what you have not done, which an experienced screener reads as a level signal and can plan the loop around
Whether what you want next matches what the seat is: someone who wants greenfield work landing on a team that mostly operates an existing system is a hire that leaves within the year

How to prepare

Mark every line of the posting as done, adjacent or new, and write one sentence for each adjacent line naming the closest thing you actually built
Split your last two years into rough percentages across feature work, operating and debugging live systems, and design or review, so a question about scope gets numbers rather than adjectives
Bring three questions that discriminate between seats: what the team is paged for, how much of the work is changing existing code versus standing up something new, and what shipped in the last quarter

Aptitude Assessment

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.

Technical Evaluations

What this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.

What to demonstrate

Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly

How to prepare

Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live

Coding Abilities Test

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

System Design Knowledge

Because the format is not fixed, the first job in the room is classification. Listen to the opening question and decide what it is: a probe into work you have already described, a fresh problem to solve now, or a conversation about how you operate. Each wants a different register, and the common failure is forcing a rehearsed structure onto a question that did not ask for it. Running a full design ritual on a ten-minute debugging question reads as not listening. When you cannot tell which it is, ask how long they want to spend and answer at that depth.

What to demonstrate

Whether the shape of your answer matches the question, so a yes-or-no gets answered before it is justified and an open prompt gets a direction before a detour
Whether you check how much depth is wanted instead of deciding for them, and whether you stop when the answer is complete rather than continuing until someone interrupts
Whether you can be redirected in the middle of an answer without restarting it from the beginning
Whether a question outside your experience gets an honest boundary followed by reasoning from what you do know, instead of a confident answer with nothing behind it

How to prepare

Rehearse one project at three lengths, roughly thirty seconds, three minutes, and a full walkthrough at the depth of a design review, and practise switching between them when someone interrupts mid-telling
Have someone ask you five questions of deliberately mixed type in one sitting without telling you the types, and score only whether you identified each one correctly before you started answering
Draft the sentence you will use to check depth, along the lines of asking whether the short version is useful here or they want the detail, and use it in a real conversation this week so the day of the round is not its first outing

Leadership and HR Discussions

When a round has no standard shape, it is often there because something is still open: an area no earlier conversation reached, a round where the signal came out mixed, or a decision someone is not ready to make alone. Work out which by going back over what each earlier round actually covered rather than how it felt, and arrive able to give evidence on that point without being asked twice. Weak answers replay the loop's earlier material at the same depth. Strong ones go a level deeper and stay consistent with what you already said.

What to demonstrate

Whether your account of a project matches the one you gave earlier in the loop, since what you said before may be available to whoever runs this round
Whether you can go a level deeper on something already covered, reaching the decision and its alternatives rather than repeating the summary
Whether you state your own uncertainty accurately, including parts of a system you did not build and decisions you inherited, instead of claiming even ownership across all of it
Whether you can answer a question you handled poorly earlier by naming what you missed, rather than delivering a polished second version as if the first had not happened

How to prepare

Reconstruct the loop on one page: for each round, the questions you were asked and the answer you actually gave, not the better one you thought of afterwards. The gaps on that page are your best available guess at why this round exists.
Take the two claims you made earlier that carry the most weight and assemble the backing for each: the measurement, the date, what broke, the decision you would make differently now.
Write down the three facts about your work that must not drift between tellings, such as team size, timeline and your own role, and check your stories against that list rather than trusting recall under pressure

Where candidates lose points

Serialising a tenant's writes through select ... for update on a single counter row

It is the first change that makes a counter correct, and it caps that tenant's write throughput at roughly one divided by the lock hold time. A transaction that takes the lock, makes a network call and then commits holds it for the entire round trip: at 2 ms that is about 500 writes per second for the whole tenant, and the largest tenants are exactly the ones that exceed it. The damage then spreads, because every waiter holds a database connection while it queues, so one hot tenant drains the shared pool and the symptom presents as a site-wide latency incident rather than as a lock problem. The repairs are to shrink the critical section to a single statement, to shard the counter into per-(tenant, hour) or per-(tenant, bucket) rows and sum on read, or to batch in memory and flush periodically while accepting the bounded loss that batching implies.

Holding money in a floating-point type, or rounding it more than once

Binary floating point cannot represent 0.01 or 0.1 exactly, so sums drift and two code paths that should agree disagree by cents nobody can trace back. The fix is integer minor units or an exact decimal type end to end, with sub-cent rates expressed as scaled integers such as micro-units, because a per-request price genuinely is smaller than a cent. The second half of the trap is rounding position: rounding each line and then summing gives a different total from summing and rounding once, and half-up and half-even diverge systematically across many lines, so rounding must happen at one named place and every downstream reader must carry the rounded value rather than recompute it from quantity and rate.

Check-then-act on shared state

Read, decide, write is not safe under concurrency unless the decision and the write are one atomic step: a unique constraint with conflict handling, a compare-and-set, or a row lock held for the whole transaction. Two requests can both pass the existence check before either inserts, which shows up as duplicate rows under load and never in a single-threaded test.

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

Seal an hour under late data with bounded memory

Hard · watermarks · late-data · quantile-sketch · conditional-write

Metering ingest reads 256 partitions at 10,000 to 40,000 events/second. Events carry occurred_at and ingested_at, and during a producer replay the gap between them is hours. Seal each UTC hour once no more than 50 parts per million of that hour's eventual quantity can still arrive, using memory that does not grow with the size of the replay. Define the watermark, the lateness parameter and how you measure it, the structure holding open hours, and the write that performs the seal. State what an idle partition does to your watermark.

Approach

Two clocks, two jobs. Bucket by occurred_at, because that is the hour the customer is billed for, and advance the watermark on ingested_at, because that is what the fold has consumed and what source_max_ingested_at records. Conflating them is what makes late data invisible.
The global watermark is the min over partitions of each partition's committed ingested_at, not the max: the fold is trustworthy only as far as the slowest partition. The consequence is that one idle partition pins the watermark forever and nothing seals, so an idle partition must promote its watermark to wall clock after a stated idle timeout, and that timeout becomes a correctness parameter, because a partition that is slow rather than idle gets sealed past.
Choose the lateness L from the measured distribution of ingested_at - occurred_at, weighted by quantity rather than by event count. The target is 50 ppm of the hour's quantity, and a replay is rare in events while carrying disproportionate mass, so an event-weighted quantile picks an L that is comfortably wrong at exactly the moment it matters.
Measure that quantile in bounded memory. A Greenwald-Khanna summary gives epsilon-approximate quantiles in O((1/epsilon) log(epsilon n)) space; a t-digest costs more per merge but has relative error that tightens at the tails, which is the half of the distribution you are reading at p99.99. Keep a separate summary per tenant class, because one tenant's batch importer is not the population.
Hold open hours in a min-heap keyed by hour_start. When the watermark advances, pop every hour with hour_end + L < W and seal it: O(log H_open) per advance and O(1) amortised per event to touch its bucket. Memory is open hours multiplied by distinct (tenant, workspace, sku) keys, so cap the number of simultaneously open hours and spill the oldest into usage_rollup_hourly as status='open' with a revision bump. While an hour is open the row is upsertable, so the store is your overflow.
The seal itself is a conditional write: update ... set status='sealed', sealed_at=now() where status='open' returning .... Two sealers race on every restart, and the loser must see zero rows and stop rather than write a second value. After the seal, an event for that hour is not an upsert but an adjustment, and source_max_ingested_at is what proves it arrived afterwards.

Worked solution (40 min)

Replay a day of events with a synthetic lateness distribution: 99.9% under two minutes, plus a 0.05% tail at four to six hours that carries 3% of total quantity.
Compute the p99.99 lateness two ways, event-weighted and quantity-weighted, and put the two numbers side by side.
Implement the min-heap of open hours with the watermark as the min over 256 partitions, then stall one partition for 20 minutes and observe what seals.
Set the idle-partition timeout to 60 seconds, repeat the stall, and measure how much quantity arrives after the seal.
Attempt the seal from two workers at once and confirm the conditional update lets exactly one through.

Expected result: The quantity-weighted p99.99 is hours larger than the event-weighted one. Choosing L from the event-weighted number lets roughly the tail's 3% of quantity land after the seal, 600 times the 50 ppm target. With the min watermark and no idle timeout, the stalled partition blocks all sealing; with a 60-second timeout, the stall is sealed past and its events arrive late.

Verify it yourself

Post-seal quantity as a fraction of the hour's total is at or below 50 ppm at the chosen L, measured rather than assumed.
The second sealer's update reports zero rows affected and writes nothing.
Heap size equals the number of open hours and does not grow with the replay's event count.

Follow-up

A replay starts during the sealing window for a period you are about to close. What do you do, and what is the customer-visible consequence of each option?
Your measured quantity-weighted p99.99 lateness is six hours and the invoice must be issued at 02:00 UTC on the first. How do you reconcile those two numbers?
How would you detect that L has drifted before it costs you an hour's quantity?

Schedule ordered webhook retries with a heap of subscription queues

Medium · heap · scheduling · backoff · head-of-line-blocking

Design the in-memory scheduler for webhook delivery. Up to 20 million rows sit in status pending or failed_retryable across 200,000 subscriptions, each row carrying next_attempt_at and attempt_count, and each endpoint having a circuit breaker. Deliveries for one subscription must be attempted in order, so at most one attempt per subscription may be in flight. Support due(now), complete(delivery, outcome) and insert(delivery) in O(log S), where S is the subscription count rather than the delivery count. Give the backoff formula you schedule retries with.

Approach

Key the global heap by subscription, not by delivery. Each subscription owns a FIFO of its due deliveries in event order; the heap holds one entry per eligible subscription, keyed by its head's next_attempt_at. That is 200,000 heap entries instead of 20 million, and it makes the one-in-flight rule structural rather than a check somebody can forget.
due(now): peek the minimum. If its key is in the future, sleep until then instead of spinning. Otherwise pop it, move the subscription into an in-flight set, and do not re-push it. A subscription absent from the heap cannot be dispatched twice, which is precisely how ordering is preserved.
complete: on success, drop the head and re-push the subscription keyed by its new head, or leave it out when the queue empties. On a retryable failure, increment attempt_count and set next_attempt_at = now + uniform(0, min(cap, base * 2^attempt)), sampled uniformly across the whole interval. That is full jitter; deterministic backoff re-synchronises the herd you just created.
Circuit breaker: park the subscription in a second heap keyed by its half-open time, so an endpoint dead for six hours costs one heap entry and zero attempts rather than consuming worker slots. Admit exactly one probe at half-open and close the breaker only on its success.
Say the price of the ordering guarantee out loud. One in-flight attempt per subscription means an endpoint answering in 10 seconds drains at 0.1 deliveries/second however many workers you run, and its backlog grows until it recovers. If the customer does not need order, allow k in flight and document delivery as unordered; that is the trade, and it is a product decision.
All three operations are O(log S) with O(S) resident heap memory and the queues themselves backed by the store. The database-backed equivalent is a partial index on (subscription_id, next_attempt_at) where status in ('pending','failed_retryable') claimed with FOR UPDATE SKIP LOCKED, and the write-back must be fenced on lease_token so a worker that stalled and resumed cannot overwrite a newer attempt.

Follow-up

One subscription has 4 million queued deliveries. What stops it from starving the other 199,999, and what does your heap look like under that load?
A customer requests redelivery of last Tuesday's events. Where do those rows enter your structure, and what keeps them from reordering live traffic?
The process restarts. How much state do you rebuild, and what stops every subscription from being attempted in the same second?

Hold a tenant to a trailing sixty-second request limit

Medium · sliding-window · two-pointer · rate-limiting · tenant-skew

The gateway must hold each tenant to R requests in any trailing 60 seconds, in aggregate across three regions and every pod, within a budget of under 10 ms added p99. Peak is 30,000 requests/second across 200,000 active tenants, and traffic is heavily skewed toward a handful of them. Give an exact single-process algorithm with its amortised per-request cost and its memory per tenant, then a bounded-memory approximation and the worst-case overshoot it actually admits. Say what the distributed version does when the counter store is unreachable.

Approach

Exact, single process: a per-tenant deque of request timestamps. On arrival, pop from the front while front <= now - 60s, then admit if the remaining length is below R and push. Each timestamp is pushed once and popped once, so the cost is O(1) amortised. The O(R) version is the one that re-filters the whole deque on every request.
Quote the memory. R = 1,000 across 200,000 active tenants is up to 2 x 10^8 timestamps at 8 bytes, about 1.6 GB, and that is the worst case rather than the mean, because the long tail of small tenants holds almost nothing. Skew helps you here and hurts you in the sharding decision.
Bounded alternative, with its real bound stated: a fixed 60-second counter is O(1) memory but admits close to 2R across a 60-second span straddling a boundary. The weighted two-bucket estimate, prev * (60 - elapsed)/60 + cur, is better on smooth traffic but assumes the previous window's arrivals were uniform; an adversary packing them at the end of that window is undercounted and can still approach 2R. Say that rather than calling it exact.
Token bucket is the usual gateway answer and a different contract: O(1) state per tenant (tokens, last_refill), a sustained rate, and a deliberate burst allowance equal to the bucket size. Choose it when a burst is acceptable and the log when the limit is contractual.
Distributed: the limit is per tenant in aggregate, so a local bucket of R/N per pod is wrong in both directions under skew. A tenant landing on one pod is throttled at R/N, and a tenant spread evenly across pods exceeds R. The shared check must be a single atomic round trip, one script or one increment-and-compare, never read-then-write, and it must fit inside the 10 ms p99 budget.
Decide the unavailable case in advance and write it down. Failing open keeps the product up and lets a tenant exceed its limit for the duration; failing closed converts a counter-store outage into a full outage. Most gateways fail open on rate limits and closed on authorisation, and those are two separate decisions made separately.

Follow-up

One tenant sends 40% of all traffic. What does that do to a single counter key, and what do you shard on instead?
Quotas rather than rate limits: the check is select used; if used < limit then insert. Name the isolation level that still permits the overshoot, and the two fixes.
How do you return an accurate Retry-After from the exact algorithm without a second scan?

Paginate a tenant's delivery export without skipping rows

Easy · keyset pagination · indexing · multi-tenancy

A customer exports webhook_delivery: delivery_id (bigint identity), subscription_id, tenant_id, event_id, status, attempt_count, next_attempt_at, created_at, delivered_at, updated_at. The endpoint runs select ... where tenant_id = $1 order by created_at desc limit 100 offset $2, and customers report rows missing from exports taken while new deliveries are being inserted. Write the replacement query and the index that supports it, paging a tenant's deliveries newest first at constant cost per page. State why updated_at cannot be the cursor column.

Approach

Name the defect precisely. OFFSET is a position in a result set that is recomputed on every request, so a row inserted ahead of the window shifts everything back by one and the next page starts after a row the client never received. Nothing errors and no identifier gap appears, so the loss is silent.
Replace the position with a value predicate over a stable, unique, indexed ordering: where tenant_id = $1 and (created_at, delivery_id) < ($2, $3) order by created_at desc, delivery_id desc limit 100. The row comparison is load-bearing: created_at alone is not unique, so ties straddling a page boundary are dropped or repeated, which is the same bug in a smaller window.
Index (tenant_id, created_at, delivery_id). PostgreSQL scans a btree in either direction, so an all-DESC ORDER BY is served by an ASC index read backwards and no DESC modifiers are needed; they only matter when the ORDER BY mixes directions. Confirm the plan has no Sort node above the index scan, or the LIMIT stops being an early exit.
Price both forms: keyset is one index descent plus 100 adjacent leaf entries per page, constant regardless of depth, while OFFSET still produces and discards every skipped row, so page N costs time proportional to N times the page size and a deep page on a large table goes from milliseconds to seconds.
Rule out updated_at as the cursor from the precondition, not from taste: a cursor column must never change value for a row already paged past. updated_at moves on every delivery attempt, so a row the client already emitted re-enters a later page and is exported twice. created_at and delivery_id are immutable, which is the whole qualification.

Follow-up

The client wants a snapshot as of one instant rather than a live tail. Compare a repeatable-read transaction held open, an added created_at <= $snapshot bound, and a materialised export table.
A retention job deletes deliveries older than 90 days. What does a client mid-walk see, and does keyset pagination help at all?
The customer wants to resume an export from yesterday's last cursor. What must be true of the cursor for that to be safe?

Decide which facts an invoice line copies instead of joining

Medium · denormalisation · immutable records · foreign keys · rounding

invoice_line_item already denormalises tenant_id, which is reachable through invoice_id, and stores amount_minor even though quantity times unit_price_micros would recompute it. A reviewer asks you to normalise both away, and separately asks whether the tenant's legal name and billing address should be copied onto the invoice header. Decide each case. For every field you keep denormalised, name the read pattern or the invariant that justifies it, the anomaly the copy can develop, and the mechanism that prevents that anomaly here.

Approach

Split the question into two kinds of copy, because they fail differently. A copy of a currently mutable fact is a cache: it drifts and needs invalidation. A copy of a fact frozen at write time is not a cache at all, it is the record of what happened, and normalising it away destroys information the source no longer holds.
Keep tenant_id on the line. It costs 8 bytes, it leads every index on the table so no read is ever accidentally cross-tenant, and it turns a wrong join into an empty result rather than another tenant's money. Prevent the drift structurally: a unique constraint on invoice (invoice_id, tenant_id) plus a composite foreign key from the line on (invoice_id, tenant_id) makes a mismatched pair impossible, so the database enforces agreement instead of a code review.
Keep amount_minor. Rounding must happen exactly once, at a named site, with a stated mode (half-even here). If readers recompute from quantity and unit_price_micros, every reader owns a rounding decision, and half-up and half-even diverge systematically across thousands of lines rather than cancelling out. A check constraint can bound the stored value but deliberately cannot re-derive it.
Copy the legal name and billing address onto the invoice header, written once and never updated. The statement must show what was true when it was sealed, and the tenant record will change afterwards. This is a snapshot for the same reason source_rollup_watermark is stored per line: without it, nobody can reconstruct what the customer was told.
Name the read pattern that pays for all of it. Rendering, dispute response and export are per-tenant, per-period reads over thousands of lines that would otherwise join back to slowly changing dimensions that no longer hold the historical value. The write side is a once-per-period batch, so the extra columns cost nothing that matters.
Concede the case where the reviewer is right: a mutable operational attribute such as the tenant's current plan name has no business on a line. If a report wants it, join. If a statement needs the plan as of the period, that is another snapshot and it belongs on the header with the rest.

Worked solution (25 min)

Write the DDL: unique (invoice_id, tenant_id) on invoice, the composite FK from the line, and a comment on each denormalised column saying whether it is a snapshot or a cache.
Attempt to insert a line whose tenant_id differs from its invoice's and confirm the foreign key rejects it.
Rename a tenant, re-render a sealed invoice, and confirm the rendered name is the one stored on the header.
Recompute amount_minor from quantity times unit_price_micros for a thousand synthetic lines rounding half-up, sum both ways, and record the divergence from the stored half-even values.

Expected result: tenant_id kept and enforced by a composite foreign key to (invoice_id, tenant_id); amount_minor kept because rounding happens once with a stated mode; legal name and billing address snapshotted onto the invoice header and never updated; mutable operational attributes left normalised. Each decision paired with its read pattern, its anomaly, and the mechanism that prevents the anomaly.

Verify it yourself

A line cannot be attached to an invoice belonging to another tenant, and the rejection comes from the database rather than the application.
Re-rendering a sealed invoice after editing the tenant record produces identical output.
Rounding per line and rounding once at the end differ measurably over a thousand lines, which is the argument for storing the value.
No column on the line is recomputed by any reader.

Follow-up

Write the composite foreign key and the unique constraint it requires on the parent. What does it cost on every line insert, and what does it do to a bulk load?
A tenant is renamed after being invoiced. Which rows change, and what does the customer see on last quarter's PDF?
Where does currency live, and what breaks if a tenant's billing currency changes between two periods?

How would you change a port number of an HTTP service without disrupting active users on the network?

Medium · system design and architecture

Approach

Choose a partition key and say what query it makes expensive.
Name the failure you are designing for, then the recovery path.
State the consistency you need, and where you are willing to be stale.

Follow-up

What breaks first when traffic grows ten times?
How does this behave when that dependency is down for an hour?

If you had an unlimited budget, what enterprise network would you build and using what specific equipment?

Medium · system design and architecture

Approach

Name the read and write paths separately; they rarely have the same bottleneck.
Choose a partition key and say what query it makes expensive.
Fix the scope first: who calls this, how often, and what they do when it fails.

Follow-up

What would you drop to keep the system up under load?
How does this behave when that dependency is down for an hour?

What is your approach to handling disaster recovery, SAN administration, and multi-vertical system redundancy?

Medium · system design and architecture

Approach

State the consistency you need, and where you are willing to be stale.
Choose a partition key and say what query it makes expensive.
Fix the scope first: who calls this, how often, and what they do when it fails.

Follow-up

What would you drop to keep the system up under load?
How does this behave when that dependency is down for an hour?

Can you walk through your project code from a recent development cycle and explain your architectural decisions?

Medium

Approach

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

Follow-up

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

How would you approach Selenium validation and implement object-oriented programming concepts in your current stack?

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.
Work from the requirement backwards to the design.

Follow-up

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

Webhook fan-out with per-endpoint isolation and backoff

Medium · fan-out · backpressure · circuit breaker · ordering

One domain event fans out to every matching subscription, producing a webhook_delivery row per (subscription_id, event_id, redelivery_seq). Peak unique event rate is 20k/second; attempts run five to ten times that once fan-out and retries are counted. One customer endpoint has returned 503 for six hours and its backlog holds days of events; every other customer must be unaffected. Design the delivery system: how a worker claims work, the backoff schedule, the per-endpoint circuit breaker, the queue partitioning, and whether you offer ordering per subscription. State the delivery guarantee in one sentence.

Approach

State the guarantee first, because it determines the rest: at-least-once with a stable event_id, and the consumer documented as responsible for idempotency. Exactly-once over HTTP is not deliverable - the 200 can be lost after the customer has already committed - so any design that promises it is either lying or is really offering at-most-once.
Partition work per subscription rather than into one global pool, with a concurrency cap per subscription. With a shared pool, the endpoint that has been dead for six hours consumes workers on retries that will fail, and every other customer's delivery latency rises: head-of-line blocking across tenants is the exact failure being designed against here.
Claim by compare-and-set with a fencing token: UPDATE webhook_delivery SET status = 'in_flight', lease_token = $new, leased_until = now() + interval '60 seconds' WHERE delivery_id = $1 AND status IN ('pending','failed_retryable') AND (leased_until IS NULL OR leased_until < now()), and make the terminal write carry AND lease_token = $new so a paused worker's late write is rejected rather than overwriting a newer attempt. Find due work through the partial index on next_attempt_at WHERE status IN ('pending','failed_retryable'), so the scan is proportional to live rows rather than to the terminal rows that outnumber them by orders of magnitude.
Use full jitter: sleep uniformly in [0, min(cap, base x 2^(attempt-1))]. Plain exponential backoff hands a recovering endpoint its entire backlog as one synchronised herd and knocks it over again; full jitter de-correlates it. Then check the schedule actually spans the retention you promise - with base 1 s and a 3,600 s cap, twenty attempts have an expected total elapsed time of only about 4.6 hours, so a twenty-four-hour promise needs roughly fifty-nine attempts or a larger cap.
Trip a circuit per endpoint on consecutive failures or a failure ratio over a rolling window: stop dispatching, push next_attempt_at out or mark new deliveries dropped_circuit_open, and half-open with exactly one probe rather than a batch. Bound the backlog explicitly with a per-subscription cap or retention, and decide in advance whether a recovered endpoint receives six hours of events at full rate or a pointer telling it to fetch what it missed.
Offer ordering only as an opt-in mode of one in-flight attempt per subscription, and price it honestly: with parallel attempts a retried event overtakes a newer one, so ordering requires serialisation, and serialisation means one slow endpoint blocks its own queue entirely. That converts a shared problem into that customer's own problem, which is the right place for it, but it is still a real cost.

Worked solution (35 min)

Compute the attempt rate: 20k unique events/second x mean fan-out x retry multiplier, and size worker pools and the per-subscription concurrency cap from it.
Write the claim statement and the terminal write, and point at the clause that rejects a resumed worker's stale write.
Tabulate the backoff for attempts 1 to 20 with base 1 s and cap 3,600 s, take the expected value of each full-jitter sleep as half its ceiling, and sum to get total expected elapsed coverage.
Decide the policy for a subscription down six hours: events buffered, bytes held, and what the customer actually receives when it returns.

Expected result: At-least-once with a stable event_id; per-subscription lanes with a concurrency cap; lease-and-fence claiming; full jitter whose schedule matches the retention promised - with base 1 s and a one-hour cap, twenty attempts covers only about 4.6 hours of expected elapsed time, so twenty-four hours needs roughly fifty-nine attempts or a larger cap; a per-endpoint breaker with a single half-open probe; ordering offered only as opt-in single-in-flight with head-of-line blocking named as its price.

Verify it yourself

Sum the expected sleeps in your schedule and compare against the retention window you promised the customer. If the two numbers disagree, one of them is wrong and it is usually the promise.
Execute the terminal write with a stale lease_token and confirm it affects zero rows. If it succeeds, the lease was doing the work alone and the token is decorative.
The dead endpoint's backlog must not be visible in any queue another subscription's workers poll. If one index scan serves all subscriptions, the isolation is nominal.

Follow-up

The endpoint recovers. Does it receive six hours of events at full rate, and what does that do to it?
Trace the exact code path by which an event belonging to one tenant could be signed and sent to another tenant's endpoint.
A customer insists they never received an event your row marks delivered. What evidence do you have, and what does payload_digest let you prove?

One tenant's counter writes stall the whole connection pool

Medium · lock contention · connection pool · multi-tenancy

A change that made a per-tenant usage counter correct now produces site-wide latency whenever one large tenant writes: unrelated endpoints time out waiting for a connection while database CPU stays low and no statement is slow. The change wraps the counter update in a transaction that takes SELECT ... FOR UPDATE on one row, calls an external pricing service, then updates and commits. Give an ordered checklist, the arithmetic that bounds that tenant's write rate, and three repairs with the cost each one accepts.

Approach

Separate waiting from working. Low database CPU alongside high application latency points at a queue, so instrument connection-acquisition wait separately from query execution time; that queue forms in the application and is invisible in database metrics, which is why the database looks healthy throughout.
Confirm the lock rather than assuming it: sample waiting sessions and group by wait event, relation and tuple. Contention concentrated on one tuple belonging to one tenant is the signature; a deadlock would instead show the database aborting transactions after its detection timeout, which is not happening here.
Do the arithmetic out loud. Throughput on a serialised row is one divided by the lock hold time, and the hold spans the external call, so a 20 ms pricing call caps that tenant near 50 writes per second no matter how many pods run. Every waiter also holds a pooled connection while it queues, so the shared pool drains and unrelated tenants fail at acquisition.
Repair one: shrink the critical section to a single statement with the price resolved before the transaction opens. Cost is a stale price for the duration of one request and a second round trip; benefit is a hold time measured in the database's own execution time.
Repairs two and three change where the contention lives rather than how long it is held. Sharding the counter into per-(tenant, bucket) rows and summing on read multiplies write throughput by the shard count, at the cost of an aggregate on every read and a shard count you must size against the largest tenant rather than the median. Accumulating in memory and flushing periodically removes the per-write round trip entirely, paid for with a bounded loss window on crash, which is acceptable for a rate limiter and not for a billing counter.
Contain independently of which repair wins: a separate pool or per-tenant concurrency cap for this write class, a statement timeout low enough that a pathological query dies before it accumulates waiters, and an idle-in-transaction timeout so a stuck client cannot pin a connection and its locks.

Follow-up

What would a genuine deadlock look like here, which two code paths would produce one, and how does the database's response differ from what you observed?
If a transaction-pooling proxy sits in front of the database, which of your three repairs changes behaviour, and what stops working that would have worked on a direct connection?
The counter also enforces a quota. Why is SELECT the count and then INSERT still wrong after you have fixed the contention?

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.

What the behavioural round is actually measuring

Every story you tell gets read for blast radius and judgement: what could have broken, who else it touched, what you knew at the moment you decided. Nobody can audit your code in an hour, so they audit your reasoning instead. Pick work where the call was genuinely yours and the consequences were real enough to remember.

Can you share an example of how you handle ambiguous technical requirements and cross-functional friction?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Pick a story where you made the decision, not one where you watched it.
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?

Why are you interested in technology roles supporting alternative asset classes like infrastructure debt or credit?

Medium · behavioural and engineering judgement

Approach

Name the disagreement and how you resolved it with evidence.
Pick a story where you made the decision, not one where you watched it.
Give the blast radius: what could have broken, and what you measured.

Follow-up

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

What are three words your coworkers would use to describe you and your working style?

Medium · behavioural and engineering judgement

Approach

Give the blast radius: what could have broken, and what you measured.
Close with what you would do differently, concretely.
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?