Interview guides · Broadsign

Broadsign Software Engineer Interview Guide

Broadsign Software Engineer interview guide: reported questions on HTML, Django, databases, testing and code quality, plus a 7-day prep plan.

The 60-second brief

Broadsign works in digital out-of-home (DOOH) advertising, which means software that decides what content appears on screens in public places. Candidates describe the Software Engineer role as backend work on the systems that schedule that content, play it and measure it, with distributed systems knowledge expected. The work spans the development lifecycle: gathering requirements with product managers, designing, implementing, testing and deploying, and reviewing other engineers' code. Candidates also report that cloud experience (AWS or GCP), containers (Docker or Kubernetes) and earlier ad-tech or high-volume systems work count in your favour but are not required.

This guide is for software engineers preparing for Broadsign, especially backend-leaning candidates. It covers the five questions candidates report (HTML elements, testing and deployment, overriding Django methods, database architecture trade-offs, code quality), the coding assessment and architecture topics candidates describe, and the collaboration stories that the team-wide meet and greets call for.

Candidates report a recruiter screen, then a technical evaluation that mixes online coding assessments with technical deep-dives, in person or remote, and team-wide meet and greets. No numbered sequence is reported, so confirm the order with your recruiter.

Focus areas

Problem solving
Algorithmic thinking
Technical interviewing

Overview

The five reported questions are broad fundamentals that you explain in your own words: the difference between p and div, how you test and deploy in production, a Django concept from your past projects such as overriding methods, trade-offs between database architectures under heavy traffic, and how you keep code quality up. All five sit in the technical and domain knowledge group, so the preparation is spoken answers backed by examples from systems you have actually shipped, not memorised definitions.

Candidates also describe a coding assessment and topics such as algorithmic thinking and problem solving, with memory management, performance, best practices in your main language and testable, modular code as areas to review. Reported example scenarios include refactoring a legacy function and listing the common pitfalls of distributed systems. For the second, be ready to name partial failure, retries that need idempotency, duplicate or out-of-order delivery, clock differences between machines and timeouts, and say what you do about each.

The architecture areas candidates mention are API design and communication patterns, data consistency and state management, and scalability for high-availability systems. Because the product schedules, plays and measures content on screens, it helps to have a basic picture of how a fleet of remote devices could fetch instructions and report back, and to reason about what happens when a device is offline. Treat that as a practice exercise for your own reasoning, not as knowledge of Broadsign's internals.

Candidates report team-wide meet and greets and say collaboration matters alongside technical skill. The stated responsibilities include working with product managers, doing code reviews, taking ownership and spotting technical debt, so prepare stories on each of those and a few questions of your own for the team.

Where to focus your preparation

Candidates describe a recruiter screen followed by a technical evaluation and meet and greets, without a fixed numbered sequence, so this guide treats the loop as one preparation block.

Preparation focus

Candidates report that the process starts with a recruiter screen covering background, salary and interest, then moves to a technical evaluation. That evaluation mixes online coding assessments with technical deep-dives, in person or remote, and team-wide meet and greets with developers, architects and team leads. The order and number of stages are not reported, so ask your recruiter which stages you will face, which language the coding assessment allows and whether any problem will be solved on paper. Prepare in three layers: spoken answers for the five reported fundamentals questions, timed coding and refactoring practice, and stories showing collaboration and ownership.

What to demonstrate

Command of your main language and computer science fundamentals, including memory management and performance
Ability to write testable, modular code and to refactor existing code safely
Architectural thinking: API design, data consistency and scalability
Communication and collaboration, reported as important in the meet and greets

How to prepare

Rehearse spoken answers to the p versus div, testing and deployment, Django method overriding, database architecture and code quality questions, each with an example from your own work
Practise a refactor of a legacy function from your own code: add tests that pin current behaviour first, then restructure in small steps
Write your list of distributed systems pitfalls with a mitigation for each, and prepare two or three stories about code review, ownership and working with product managers

Where candidates lose points

Answering p versus div with only "block versus inline"

Both are block-level by default, so that answer is wrong. The real difference is meaning: p marks a paragraph of text and carries semantics for browsers, screen readers and search engines, while div is a generic container with none. A p cannot contain block-level elements, because the parser closes it early. Reach for div (or a more specific semantic element) when grouping for layout.

Describing a Django override without the pitfalls

Pick a real example, such as overriding save() on a model. Show that you call super().save(*args, **kwargs), and say what the override misses: bulk_create and queryset update() skip save(), and signals still fire separately. Explain why you chose an override over a signal, a manager method or a form clean step.

Answering the database question with "SQL is consistent, NoSQL scales"

Start from the access pattern and the consistency you need. Compare a relational store, a document store and a key-value store on joins, transactions, schema change and horizontal scaling, then add read replicas, caching and sharding, and name what each costs you, such as replica lag or cross-shard queries.

Stopping the testing and deployment answer at "I run unit tests"

Walk through the path from commit to production: unit and integration tests, review, automated build, staged or canary release behind a flag, monitoring after release, and a rollback plan. Mention that schema migrations must stay compatible with the previous version of the code while both run.

Solving the coding or paper problem in silence

Candidates are told to think aloud and to treat a paper problem as a conversation. State your approach and its complexity before writing, name the edge cases, then write. If you do not know something, say how you would find it out.

Questions & practice

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.

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?

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?

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.

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?

Stop tag and share joins from fanning out a page

Medium · join fan-out · semi-joins · many-to-many

resource_tag is (resource_id, tag_id) with PK (resource_id, tag_id); resource_share is (resource_id, shared_with_user_id, permission). The tagged-and-shared listing inner-joins resource to both, filters tenant_id, tag_id = ANY($2) and shared_with_user_id = $3, orders by updated_at DESC and takes 50. Pages come back with fewer than 50 distinct resources and the total in the header is far too high. Explain the row multiplication, rewrite both the page query and the count query so each is correct, and name the index each one needs. PostgreSQL 16.

Approach

Do the arithmetic against the predicates that are actually there. An inner join emits one row per matching child row, and both joins are filtered: tag_id = ANY($2) admits only the requested tags, shared_with_user_id = $3 admits one user's share rows. So a resource holding three of the requested tags and shared with $3 once yields three rows, not one — the multiplier is its count of matching tags times its share rows for that single user, and that second factor is 1 unless the table admits duplicate (resource_id, shared_with_user_id) pairs. LIMIT 50 then limits rows rather than resources, and COUNT(*) counts pairs — the header is the product, not the population.
Reject DISTINCT as the fix. It deduplicates after the product has been built, so the planner must materialise and sort the fanned-out set before the LIMIT can apply, and it leaves any SUM or AVG in the same select list wrong.
Rewrite both filters as semi-joins, keeping resource as the only row source: AND EXISTS (SELECT 1 FROM resource_tag rt WHERE rt.resource_id = r.resource_id AND rt.tag_id = ANY($2)) and the same shape against resource_share. A semi-join stops at the first match per resource and preserves the driving index order, so ORDER BY updated_at DESC, resource_id DESC LIMIT 50 still stops after 50 rows.
Count with the same predicates and no join at all: SELECT count(*) FROM resource r WHERE r.tenant_id = $1 AND r.status = 'active' AND EXISTS (...) AND EXISTS (...). Nothing multiplies a resource, so the number is the population.

Follow-up

The filter changes from 'any of these tags' to 'all of these tags'. Rewrite it and state what it costs relative to the ANY form.
A resource can be shared with the same user twice under different permissions. Does your count change, and should it?

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?