Interview guides · Aramex

Aramex Software Engineer Interview Questions & Guide 2026

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

The 60-second brief

As a Software Engineer at Aramex, you are at the heart of the digital transformation driving one of the world’s most recognized logistics and transportation providers. Your work directly impacts the efficiency of global supply chains, affecting how millions of packages are tracked, sorted, and delivered daily. You will be responsible for building robust, scalable software solutions that enable Aramex to maintain its competitive edge in a fast-paced, high-volume industry.

Ask how many rounds there are and what each one is before you plan your weeks, because the composition is what your hours should follow. A loop with three coding rounds and one short design conversation deserves a different split from the reverse, and whoever schedules it will usually say plainly which it is.

Focus areas

Trace a symptom to a mechanism under load
Choose indexes from the query's access path
Bound every outbound call with a timeout

Overview

This role requires a blend of technical precision and architectural foresight. Whether you are optimizing backend services, refining data structures, or integrating complex logistics APIs, you are tasked with solving real-world problems that have tangible, physical-world consequences. The engineering culture at Aramex values resilience and performance, making this an ideal environment for developers who thrive when tackling high-stakes, large-scale technical challenges.

Aramex operates in a high-pressure, mission-critical environment. Your technical solutions must not only be elegant but also highly reliable to ensure seamless global operations.

The Aramex interview process

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

HR Screening

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

Technical Evaluations

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.

Where candidates lose points

Letting a slow dependency consume unbounded concurrency

The failure that takes a service down is usually not an error but a delay. A dependency answering in thirty seconds instead of fifty milliseconds holds each request's worker or connection six hundred times longer, and since required concurrency is arrival rate times latency, a fleet sized for sixty in-flight requests now needs thirty-six thousand to sustain the same rate - so it queues, and requests whose clients have already abandoned them still occupy resources. Retries make it precisely worse: a policy of three attempts triples the load on a dependency at the exact moment it is least able to serve, which is how one slow dependency becomes an outage of everything sharing that pool. Containment is four specific things - a timeout on every outbound call shorter than the caller's remaining budget, a bounded pool per dependency so one cannot starve the others, backoff with full jitter rather than a fixed delay so retries do not resynchronise, and a circuit that stops sending once the failure rate makes an attempt pointless.

Assuming an isolation level prevents the anomaly you actually have

Isolation levels are named by the SQL standard but implemented differently, so any claim about one is only true of a named engine. PostgreSQL defaults to READ COMMITTED, where every statement takes a fresh snapshot, so two statements inside one transaction can legitimately disagree about the same row. Its REPEATABLE READ is snapshot isolation: it removes non-repeatable and phantom reads but permits write skew, where two transactions each read a set, each conclude their own write is safe, both commit, and the combined result violates a constraint that no single row expresses. Only SERIALIZABLE closes that, and it closes it by aborting a transaction with a serialization failure (SQLSTATE 40001), which means the guarantee is theoretical unless the application has a retry loop. InnoDB's REPEATABLE READ is a different mechanism again - plain SELECTs read a consistent snapshot while locking reads and writes see the latest committed row - so a read-modify-write inside one transaction can act on a value that the transaction's own earlier SELECT never returned.

Issuing one query per row of a result set

Fetch related rows in a single batched query keyed by the ids you already hold, or join them into the original query. A per-row round trip multiplies network latency by the row count, and it looks perfectly fine against the ten rows in your development database.

Going silent while thinking

Narrate the candidates and why you are discarding them, even in fragments: sorting first would make this a two-pointer scan, but it destroys the original indices, which the output needs. From the other side of the table, a candidate thinking hard and a candidate stuck are indistinguishable until one of them speaks.

Questions & practice

Merge partitioned event streams into one ordered feed with bounded lateness

Hard · k-way merge · watermarks · out-of-order streams

The read-model service consumes 64 log partitions carrying about 4,000 events per second in total. Each partition is ordered within itself, but partitions drift by up to 30 seconds, and the activity feed must present a tenant's events in occurred_at order. Produce the merge. State its complexity, the buffer it requires in events and in bytes, what happens when one partition is idle, and what you do with an event that arrives after you have already emitted its position. Payloads average 1 KB.

Approach

Merge with a min-heap over the 64 partition heads keyed on (occurred_at, event_id): O(log P) per event and O(n log P) overall. The tie-break on event_id is what makes the output deterministic when two partitions carry the same millisecond, which matters because the feed is paginated and a non-deterministic order reorders pages under the reader.
Emitting the heap head is only correct once every partition has produced everything up to that timestamp, so the emit condition is a watermark: the minimum across partitions of the highest occurred_at seen, less the allowed lateness. Events are held until the watermark passes them, which is what turns individually ordered streams into a jointly ordered one.
Size the buffer from the lateness rather than guessing: 4,000 events per second times 30 seconds is 120,000 buffered events, and at 1 KB each about 120 MB of heap. That number is the real price of the ordering guarantee and belongs in front of whoever asked for it.
Handle the idle partition explicitly, because it fails the feed rather than corrupting it: a partition with no traffic never advances its own maximum, so the watermark freezes and output stops entirely. Either every partition emits a periodic idle marker carrying the broker's current time, or the watermark falls back to wall clock for a partition silent beyond a threshold.
Choose the late-event policy from what the projection is keyed on. The projection upserts on (aggregate_id, aggregate_version) and discards a version it has already applied, so a late event is safe to apply out of order and correctness never depended on the merge at all. Apply it, recompute the affected feed page, and count lateness so the 30-second budget can be re-derived from data rather than folklore.
Say what the merge does not buy: ordering is guaranteed within one aggregate by the log's partitioning, and no watermark makes the cross-aggregate order authoritative. Two events from different aggregates in the same millisecond have no true order, so the feed's order is a presentation choice that must be stable rather than correct.

Follow-up

The lateness budget is raised to five minutes. What is the new buffer, and what besides memory changes?
The consumer restarts. Where does it resume from, and what does the feed look like for the first 30 seconds?
One partition is ten minutes behind because its producer is slow. Do you stall the feed or emit without it?

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.
Expire idle destinations, or memory grows with every destination ever seen rather than with the live set. Hold the rings in a bounded LRU keyed on destination_id and treat a miss as no history, which is the correct default for an endpoint that has been silent for a minute.
Keep the half-open probe out of the window arithmetic. After the circuit opens, one probe per interval decides whether to close it, and folding that single success into a window that still holds a 100 percent failure history would reopen the destination on one data point.

Follow-up

The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?
How would you make the window survive a process restart, and is it worth the cost?

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

Hard · reconciliation · range hashing · throttling

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

Approach

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

Worked solution (35 min)

Compute the naive cost explicitly at 40,000,000 reads and 0.5 ms each, then at 100 concurrent, and state what those connections do to a pool already carrying 1,200 writes per second.
Write the merge-join version over (tenant_id, resource_id) and state its memory.
Define the range aggregate: the range key, the per-row hash input, and the combining function, with one sentence excluding XOR.
Work an example with 40,000,000 rows, branching factor 256 and 5 differing rows, and count the ranges examined.
Add the watermark filter and the resume point, and name the throttle signal the loop watches.

Expected result: A rejection of per-row point reads backed by the time cost and the cache-eviction argument, a single ordered merge join as the dense-case answer at O(n) time and O(1) memory, a range-hash drill-down examining O(d log_B(n/d)) ranges using a sum or multiset hash rather than XOR, a watermark excluding recently written rows, and a resumable throttled run loop.

Verify it yourself

Injecting one known stale row makes exactly one leaf range mismatch and the drill-down reaches it.
Duplicating a row on one side is detected, which is precisely the case XOR would miss.
Running the diff twice against an idle system reports zero mismatches both times.

Follow-up

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

Keep soft-deleted accounts from blocking re-registration

Easy · soft delete · unique constraints · engine differences

app_user holds user_id, tenant_id, email CITEXT, password_hash (NULL for SSO principals), email_verified_at, auth_version, status ('invited','active','suspended','deactivated'), created_at, updated_at, deleted_at. Two live accounts for one address inside a tenant must be impossible, but an address freed by a soft delete must be reusable, and the same tenant may delete and re-register it repeatedly. Write the uniqueness DDL for PostgreSQL 16, then the equivalent for MySQL 8 where partial indexes do not exist, and say what each permits once three deleted rows already hold that address.

Approach

Start from what is actually unique: not (tenant_id, email), but (tenant_id, email) among live rows. PostgreSQL says that directly — CREATE UNIQUE INDEX app_user_live_email ON app_user (tenant_id, email) WHERE deleted_at IS NULL. A full constraint over the same two columns burns the address permanently the first time someone deletes an account.
Keep case-insensitivity in the type or the index, never in the application: CITEXT as given, or UNIQUE (tenant_id, lower(email)) as an expression index where the extension is unavailable. A case-sensitive unique column is exactly how two accounts for one human appear.
For MySQL 8 the predicate has to move inside the key: add a discriminator column that is a constant 0 while the row is live and is set to user_id on delete, with UNIQUE (tenant_id, email, deleted_marker). Live rows share the constant and still collide; deleted rows differ from each other and stop colliding.
State the NULL variant and its dependency: leaving the marker NULL for deleted rows also works, because a unique index treats NULLs as distinct — true in MySQL, and true in PostgreSQL only under the default NULLS DISTINCT, which PostgreSQL 15 lets you reverse. Check the polarity against the three existing deleted rows: constant-on-live is what preserves the collision you want, and reversing it silently admits duplicate live accounts.
Say what a soft delete must do besides setting deleted_at: increment auth_version so existing tokens stop validating, leave resource.owner_user_id and resource_revision.actor_user_id intact, and accept that the address is retained — erasure is a different requirement answered by scrubbing the column, not by a DELETE that would break those references.

Follow-up

A deleted account re-registers with the same address the next day. Do the old resource rows follow the new user_id, and how does the API keep the two principals apart?
How do you honour an erasure request while resource_revision.actor_user_id still references this table?
What changes if a user may hold membership in two tenants?

Find version gaps and relay lag with window functions

Medium · window functions · event tables · frame semantics

outbox_event holds event_id, aggregate_type, aggregate_id, aggregate_version, event_type, payload, status ('pending','published','dead'), attempts, created_at, published_at. A projection is missing rows and you must decide whether the relay skipped events or the consumer dropped them. Write three queries over the last seven days: one listing every aggregate_id whose published aggregate_version sequence has a hole, one giving per-day counts with a running total, and one returning the newest published event per aggregate. For each, say where the window function is evaluated relative to WHERE and LIMIT. PostgreSQL 16.

Approach

Gaps: compute lead(aggregate_version) OVER (PARTITION BY aggregate_id ORDER BY aggregate_version) in a subquery, then filter next_version <> aggregate_version + 1 in the outer query. Window functions are evaluated after WHERE, GROUP BY and HAVING and before the outer ORDER BY and LIMIT, so the predicate cannot sit in the same WHERE clause and PostgreSQL 16 has no QUALIFY.
Say what the seven-day filter does to the answer: it truncates every partition, so the first row per aggregate has no predecessor inside the window and a hole spanning the boundary is invisible. Widen the window, or join to resource.version as the authority for the true maximum.
Running total: SELECT date_trunc('day', created_at) AS d, count() AS n, sum(count()) OVER (ORDER BY date_trunc('day', created_at) ROWS UNBOUNDED PRECEDING). An aggregate inside a window call is legal because grouping runs before windowing. The grouping key is unique per row here so ROWS and RANGE agree, but write the frame anyway — over ungrouped rows with tied timestamps the default RANGE frame pulls in every peer row and the total jumps.
Newest per aggregate: DISTINCT ON (aggregate_id) ... ORDER BY aggregate_id, aggregate_version DESC is the cheap PostgreSQL-only form when an index matches that order; row_number() OVER (PARTITION BY aggregate_id ORDER BY aggregate_version DESC) = 1 is the portable form and needs a subquery for the same evaluation-order reason as the gap query.
Interpret rather than report: no gaps plus a normal p95 of published_at - created_at points at the consumer; gaps or a fat lag tail point at the relay; rows still 'pending' with attempts > 0 point at neither, because they never left the database.
Be explicit that the partial index on (created_at, event_id) WHERE status = 'pending' does not serve any of these — they read published rows. Name the index a recurring monitor would need, and say why a query run twice a year may not deserve one.

Worked solution (30 min)

Write the three queries against seven days of data and confirm each returns without error.
In a scratch copy, delete one middle event for a single aggregate and confirm the gap query names that aggregate and the versions either side.
Run a running total over ungrouped rows ordered by date_trunc('second', created_at), once with the default frame and once with ROWS, and record where the two series diverge.
Compare the DISTINCT ON and row_number() plans on the same data and record rows-read for each.

Expected result: The gap query returns one row per hole with the version on each side; the daily query returns seven rows whose running total ends at the seven-day count; the newest-per-aggregate query returns exactly one row per aggregate_id present in the window. The RANGE running total is higher than the ROWS one at every tied second.

Verify it yourself

The per-day counts sum to COUNT(*) over the same predicate.
Re-running the gap query after reinserting the deleted event returns zero rows.
count(*) of the newest-per-aggregate result equals count(DISTINCT aggregate_id) over the same filter.

Follow-up

Relay failover redelivers events. Does a duplicate break the gap query, and how would you detect one from this table alone?
Turn the gap check into a continuous monitor rather than a query someone runs after an incident. What does it watch?
The consumer claims it never received event 4,812,006. What do you look at, in what order?

What are the different operational identifiers in OOPS?

Medium

Approach

Clarify what is being asked and what a complete answer contains.
Work from the requirement backwards to the design.
Say what you would check first and why it is the highest-information step.

Follow-up

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

Describe the difference between method overloading and overriding.

Medium

Approach

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

Follow-up

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

Make the resource write endpoint safe for retrying clients

Easy · idempotency · unique constraints · retries

The edge API accepts POST /v1/resources at 1.2k writes/second peak under a 400 ms p99 budget, and clients retry on timeout. idempotency_key has PRIMARY KEY (tenant_id, idempotency_key) plus request_fingerprint CHAR(64), state in ('in_flight','succeeded','failed'), response_status, response_body, resource_id, locked_until and expires_at. Specify the exact sequence the handler runs: what the second request does when it arrives while the first is still executing, what a retry carrying a different body receives, what happens when a process dies mid-request, and how the table is kept from growing without bound.

Approach

Start from the race rather than the happy path. Insert the key row first, in its own short transaction, with state 'in_flight' and locked_until set to now() plus the request deadline. The unique constraint on (tenant_id, idempotency_key) is the arbitration mechanism: exactly one caller commits the insert, and every other caller takes a unique violation (SQLSTATE 23505) and becomes a follower. Reading the table and then inserting is a check-then-act race that both callers pass, and they pass it most often under the load that generates the retries.
Give the follower a branch for every state it can observe, with no fall-through to doing the work: fingerprint mismatch means the key is being reused for a different request, so reject with 422 and never replay the stored response; 'succeeded' replays response_status and response_body verbatim; 'failed' allows a fresh attempt; 'in_flight' with locked_until in the future is answered with 409 and a retry hint, or a short bounded wait, because the only alternative is executing the effect twice.
Commit the effect and the record of the effect together. The resource insert, its resource_revision row, its outbox_event row and the transition of the key row to 'succeeded' all happen in one transaction, so no crash can leave the work done and the key still 'in_flight'. The earlier insert is deliberately a separate transaction: it has to be visible to a concurrent caller before the work starts, which an uncommitted row is not.
Close the wedged-key path. A process that dies after the insert leaves a row that would otherwise block that key forever, which is what locked_until exists for. Reclaim with a single conditional statement, UPDATE ... SET state='in_flight', locked_until=now()+interval WHERE ... AND state='in_flight' AND locked_until < now(), and treat a zero rowcount as losing the reclaim. Two statements that read and then update reintroduce the original race one layer down.
Size the retention before it becomes an incident. At the 1.2k/second peak a 24-hour window is on the order of 100 million rows, and the table grows with traffic rather than with data, so the sweep is load-bearing. Batched deletes driven by an index on expires_at keep it bounded. Range-partitioning by day and dropping whole partitions is cheaper, but PostgreSQL requires every partition-key column to appear in the primary key, so the key becomes (tenant_id, idempotency_key, created_at) and a retry that straddles the boundary no longer collides with its original. State that cost rather than meeting it later.

Worked solution (20 min)

Draw a timeline with the retry arriving 150 ms into a 900 ms first attempt, and mark the instant each row becomes visible to the other transaction.
Write the insert-first statement, then a five-row branch table for the follower: fingerprint mismatch, 'succeeded', 'failed', live 'in_flight', expired 'in_flight' - one action each.
Write the single conditional UPDATE that reclaims an expired in_flight row, and argue why two concurrent reclaimers cannot both see a non-zero rowcount.
Compute rows per day at peak, name the index the sweep uses, and check the sweep rate against the insert rate.

Expected result: The key row is inserted at 'in_flight' in its own transaction before any work; the unique violation identifies the follower. The follower replays the stored response on 'succeeded', returns 422 on a fingerprint mismatch, and waits or returns 409 on a live 'in_flight', never performing the work itself. The effect, its revision row, its outbox row and the transition to 'succeeded' commit in one transaction. Expired in_flight rows are reclaimed by one conditional UPDATE, and expired keys are swept in batches.

Verify it yourself

The retry creates no second resource row, whether it arrives during the first attempt or after it.
A retry with the same key and a changed body is rejected rather than served the first attempt's response.
Killing the process between the key insert and the commit leaves a key that is reclaimable within locked_until rather than one that is stuck.
For create operations, every key row in state 'succeeded' carries a resource_id that exists.

Follow-up

The stored response body averages 200 KB and this table is now the largest in the database. What do you store instead, and what does a replay return once the body has been pruned?
A client library generates a fresh idempotency key on every attempt. What breaks, which layer should have caught it, and does the server have any defence?
The first attempt succeeded but its response was lost, and the client retries 30 hours later, after the key expired. What does the second attempt do, and is that acceptable?

Edge instances grow 400 MB per hour until the nightly restart

Hard · memory leak · heap analysis · metric cardinality · gc

Edge API instances start at 700 MB resident and grow about 400 MB/hour; a nightly rolling restart has hidden it for weeks. Growth continues unchanged when request rate halves overnight, p99 degrades in the last hours before an instance is recycled, and heap used immediately after a forced full GC rises monotonically. The service holds no product state. Name the discriminating measurement that separates the plausible causes, give the most likely cause, and give the fix and how you would verify it.

Approach

Separate resident memory from live heap first, because they fail differently. Resident size can grow from fragmentation, native buffers or thread stacks while the heap is flat; heap used after a full GC rising monotonically is the measurement that says objects are reachable and not being released. You already have it, so this is retention, not fragmentation, and that closes off half the candidate list.
Use the rate's independence from traffic as the discriminator. Growth that continues at half the request rate rules out per-request objects that are merely slow to collect and points at a structure that grows with distinct values observed rather than with call volume. Write the candidates that have that property: a metrics registry keyed on a high-cardinality label, an unevicted cache, an interner, a per-key lock map.
Take two heap snapshots an hour apart and diff by retained size, reading the dominator tree, not by allocation count or instance count. Expect one root holding a map with millions of entries, then follow the reference chain to the code that inserts and never removes. Allocation profilers point at churn, which is the wrong signal here.
The candidate that fits this service is an observability label carrying an identifier, such as a request path recorded before templating so that /v1/resources/48213 becomes its own metric series. That grows with distinct ids seen, is independent of rate, and explains the late p99 degradation, since GC cost rises with the size of the live set.
Fix by bounding cardinality at the source: template the path to /v1/resources/{id} before it becomes a label, move tenant id from a label to a log field or an exemplar, and cap the registry with a bounded map that evicts. Add a cardinality ceiling that fails loudly in a lower environment rather than growing quietly in production.
Verify with a soak rather than a restart. Hold one instance out of the nightly recycle for 48 hours with the fix and compare post-GC heap and series count against an unfixed control taking the same traffic.

Follow-up

Post-GC heap is now flat but resident size still creeps. What are you looking at, and does it matter?
How would you have detected this before an OOM, given the nightly restart masked the trend?
That label is what makes one dashboard useful. How do you keep the dashboard and lose the leak?

Suggested practice checklist

Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.

Day 1 — Diagnostic, scored before you study anything

Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.

Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.

Day 2 — Largest gap: find the boundary rather than the subject

Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
Re-attempt one of them from blank four hours later with nothing open.

Deliverable: A sub-skill map with the two blocking sub-skills circled.

Day 3 — Drill the blocking sub-skill by repeating the shape

Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
Have someone else read your one-sentence rule and find the precondition you left out.

Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.

Day 4 — Second gap, plus maintenance on the strongest area

Run the same sub-skill decomposition on the second-largest gap in half the time.
Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.

Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.

Day 5 — The gap that is not a skill

Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.

Deliverable: Two recordings with a counted reduction in time-to-first-question.

Day 6 — Retest under day-one conditions

Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
Write down which single block you would still lose the offer on.

Deliverable: A second scored rubric placed beside the first, with one named remaining risk.

Day 7 — Full loop under interview conditions

Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.

Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.

The bug you decided not to fix

Engineers over-index on what they repaired. A stronger answer covers something you knowingly left broken: the alert you tuned down, the data inconsistency you documented instead of chasing, the cleanup you deferred past two quarters. Give the reasoning and the condition that would have reopened it, so it reads as a decision and not as neglect.

What is the most challenging project you’ve worked on and how did you handle it?

Medium · behavioural and engineering judgement

Approach

Pick a story where you made the decision, not one where you watched it.
Name the disagreement and how you resolved it with evidence.
State the situation in two sentences and spend the rest on the reasoning.

Follow-up

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

Reverse your own decision and price the reversal

Medium · reversibility · measurement · migrations

Describe a technical decision you made and later reversed. Pick one that cost something: a service you split and merged back, a cache you added and removed, an index you created that pushed the planner onto a worse plan, a projection you rebuilt from scratch. State what you believed when you decided, the measurement that changed your mind, how long the wrong version ran in production, and what the reversal cost in migrations, dual writes, and a deprecation window for callers you did not own.

Approach

State the original rationale without irony, in the version you would still defend given what was known then. If it is not defensible, the story is about carelessness rather than judgement, and a different example serves you better.
Give the measurement that moved with a before and after: the p99 that did not improve, the cache hit rate that sat at 40%, the plan that flipped to a sequential scan once the table passed a size you can name.
Cost the reversal in steps, not adjectives: expand-and-contract deploys, the dual-write window, the callers who had to be notified, the rows already written in the wrong shape that had to be backfilled or abandoned.
Distinguish reversal from rewrite by naming what you kept. Most good reversals preserve the schema or the interface and undo one decision inside it, which is also why they were affordable.
Finish on the process change: the smallest experiment that would have produced the same measurement in a day, and why you did not run it the first time.

Follow-up

What in that decision was irreversible, and did you know it was irreversible when you made it?
How did you tell the people who had already built on top of the original decision?
What do you now measure before committing to a change of this size?

Turn a code review disagreement into a decision

Easy · code review · optimistic concurrency · communication

A colleague's change updates a row with UPDATE resource SET version = version + 1 WHERE resource_id = $1 AND version = $2 and treats an affected-row count of zero as a successful no-op. You read that as a silently lost update; they think returning 200 is friendlier to clients than returning a conflict. Describe how you have handled a review disagreement of this shape: what goes in the comment, when you leave the thread, and who decides. Then write the comment you would leave here, in under 80 words.

Approach

Sort the disagreement before writing anything. A silently discarded write is a correctness claim about data; the choice between 409 and 412 is taste. Only the first justifies blocking a merge, and saying which one you are doing is most of the value of the comment.
Make the claim reproducible in the comment itself with an interleaving rather than a principle: A reads version 7, B reads version 7, B commits version 8, A's predicate matches zero rows, A is told it succeeded and A's edit is gone.
Offer the alternative with its cost attached: return 409 carrying the current version and the revision that won, so the client can re-read and re-apply. Note that automatic retry is not the fix, because a retry re-reads the winner's state and reapplies an intent formed against data that no longer exists.
Apply an escalation rule you can state: two round trips on the thread, then a call, and the service's owner decides rather than the reviewer. A reviewer who cannot be overruled is a bottleneck with extra steps.
Close in writing wherever the decision lands, so the next reader finds the reasoning in the code or the ticket instead of in a collapsed review thread.

Follow-up

Where would you put the test that fails if someone reintroduces the swallowed zero rowcount?
The author says clients cannot handle a 409. How do you check whether that is true?
How do you handle the same review comment when the author is more senior than you and in a hurry?