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Endurance Warranty Services Software Engineer Interview Guide

Endurance Warranty Services Software Engineer interview guide: reported rounds, questions and how candidates describe the process.

The 60-second brief

A Software Engineer at Endurance Warranty Services plays a vital role in maintaining and evolving the enterprise software systems that power our auto repair coverage products. As we continue to scale our operations, your work directly impacts how we serve our customers, protect them from costly breakdowns, and maintain our position as a leader in the vehicle service contract industry. You will be at the heart of our technical operations, building features that streamline our services and ensure high availability for our internal and external users. This role offers a unique opportunity to work in a fast-paced environment where you will contribute to production-level code from your first days on the team. Whether you are a new graduate or an early-career professional, you will be expected to leverage a modern stack—primarily C#/.NET and React—to solve complex business problems.

This guide is scoped to a Software Engineer candidate at Endurance Warranty Services.

Focus areas

C#
.NET
React

Where to focus your preparation

No round sequence has been reported for Endurance Warranty Services. Confirm the format with your recruiter.

Preparation focus

No round sequence has been reported for this company, so confirm the format with your recruiter and work the reported questions below.

What to demonstrate

Breadth across the topics this company reports testing
Whether you confirm the format before preparing for it

How to prepare

Ask the recruiter for the sequence, the duration of each stage and whether you will be writing code
Work the reported questions below and time yourself

Where candidates lose points

Communicate your thought process

During technical rounds, explain why you are making a specific choice. We are as interested in your reasoning as we are in the final code.

Prepare for remote interaction

Since this is a remote role, ensure your workspace is set up for clear audio and video communication before your interviews.

Research our product

Understanding the vehicle service contract industry will give you a significant advantage when discussing how your code contributes to the business.

Be honest about your skills

If you haven't used a specific tool, express your willingness and strategy for learning it quickly.

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.

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?

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

Medium · top-k · heavy hitters · streaming

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

Approach

Do the exact version first, because it is affordable at this cardinality: a ring of 300 one-second counters per tenant, advanced lazily, is 1,200 bytes of counters per tenant and roughly 60 to 90 MB for 50,000 tenants with overhead. Carry a running total and subtract the bucket you overwrite so a window read is O(1) rather than 300 adds.
Extract the top 50 with a size-k min-heap over the tenant sums: O(d log k) for d tenants, against O(d log d) to sort them all. Maintaining the heap continuously instead of on query requires a tenant-to-heap-index map, because incrementing a count already inside the heap means sifting from a known position, and without that map you rebuild the heap on every request.
State the approximation precisely rather than gesturing at sketches. Misra-Gries with m counters retains every item whose true count exceeds N/(m+1), and each retained count underestimates the truth by at most N/(m+1). With m = 1,000 and N = 900,000 requests in the window the error is roughly 900 requests, which is fine for spotting a tenant sending 50,000 and useless for ranking two tenants 200 apart.
Say what breaks when the window slides: Misra-Gries and Space-Saving are insert-only and cannot be decremented as records age out. The workable construction is one summary per sub-window, say ten seconds, with 30 summaries merged at query time, and the merged error is the sum of the per-summary errors, so the bound degrades linearly in the number of sub-windows.

Follow-up

The heaviest tenant is heavy because of one export job rather than user traffic. Should the limiter treat those as the same tenant?
Two tenants sit tied at the boundary of the top 50. Does your answer flap, and does the flapping matter?

Collapse a redelivered event batch into per-aggregate high-water marks

Easy · hashing · at-least-once · aggregation

You drain a batch of up to 5,000,000 events, each (aggregate_id BIGINT, aggregate_version INT, event_type, payload). The log guarantees order within one aggregate only; the batch merges 64 partitions, and a relay failover has redelivered a range, so an older version for an aggregate can appear after a newer one. Given a map of last_applied_version per aggregate, produce the events worth applying, at most one per (aggregate_id, version), plus the count discarded. Target O(n) time. State the memory for 2,000,000 distinct aggregates and what you do when it does not fit.

Approach

One pass, one hash map from aggregate_id to the highest version kept, and a discard counter. An event whose version is at or below last_applied_version for its aggregate is dropped without further work, which is the whole reason the event carries its version rather than a delta. O(n) expected time, O(d) space in distinct aggregates.
Keep the maximum, never the last occurrence. The redelivered range means the final appearance of an aggregate in the batch can be an older version than one seen earlier in the same batch, so last-wins applies stale state over newer state and the projection regresses with no error anywhere.
Cost the memory instead of calling it large: an 8-byte key plus a 4-byte version is 12 bytes of payload, and an open-addressed table held at a 0.7 load factor costs roughly 17 bytes per entry before per-slot metadata, so 2,000,000 aggregates is tens of megabytes in a native layout and several times that in a runtime that boxes both key and value.
If the distinct set exceeds memory, partition on hash(aggregate_id) mod P and reduce each partition independently. Every event for one aggregate hashes to the same partition, so the per-partition result is exact and the merge is concatenation rather than a second reduction.

Follow-up

The payload is a patch rather than a snapshot, so applying only the highest version loses the intermediate changes. What changes in your reduction?
How do you detect that version 7 arrived while version 6 was never delivered, and what should the consumer do about the gap?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

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

Approach

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

Follow-up

A client sends the version it read ten minutes ago and the resource has moved three versions. What is in your 409 so it can resolve the conflict without a full re-fetch?
Two editors, two disjoint fields, no overlap. Does your answer still refuse the second write, and should it?

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.

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?