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IBC Software Engineer Interview Guide
IBC Software Engineer interview guide: reported rounds, questions and how candidates describe the process.
The 60-second brief
As a Software Engineer at IBC, you are at the intersection of robust financial systems and modern development practices. This role is critical to the organization’s ability to deliver secure, scalable, and efficient banking solutions that serve a diverse client base. You will be responsible for building and maintaining the digital infrastructure that powers IBC’s operations, ensuring that the technology stack remains reliable while adapting to new market demands. This position is particularly interesting for engineers who thrive in environments where they can gain exposure to a wide variety of frameworks and architectural patterns. Because IBC operates within the banking sector, you will face unique challenges related to data integrity, security, and performance.
This guide is scoped to a Software Engineer candidate at IBC.
Focus areas
- Java
- Fullstack Development
- Banking Domain Knowledge
Where to focus your preparation
No round sequence has been reported for IBC. 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
Be an ambassador
Treat every interaction as an opportunity to show your professionalism; the company views the interview as a reflection of how you will treat colleagues.
Communicate honestly
If you do not know an answer, focus on how you would go about finding the solution rather than guessing.
Understand the business
Researching the banking industry and understanding why IBC is an interesting place to work will help you stand out.
Questions & practice
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?
Canonicalise a request body into a stable idempotency fingerprint
Medium · parsing · canonicalisation · hashing
idempotency_key.request_fingerprint is a SHA-256 over the method, path and canonicalised body, and a retry whose fingerprint differs must be rejected with 422 rather than served the stored response. Write the canonicaliser. Bodies are JSON up to 256 KB nested at most 32 levels; clients vary key order, whitespace and unicode escaping, and some send 64-bit ids as JSON numbers. Produce a deterministic byte string such that semantically identical bodies match and any semantic difference does not. State your complexity and name two normalisations you refuse to perform.
Approach
- Parse once into a tree, then re-serialise under fixed rules: object keys sorted, array order preserved, one escaping convention, no insignificant whitespace. Parsing is O(n) and sorting keys is O(k log k) per object, so O(n log n) overall with O(depth) stack, and the 32-level cap is enforced during parsing because hostile nesting is how a canonicaliser becomes a stack overflow.
- Sort keys by their UTF-8 bytes and say why the obvious implementation is wrong in some runtimes: a default string comparison that orders by UTF-16 code units places surrogate pairs, meaning code points from U+10000 up, below U+E000 to U+FFFF, which is not UTF-8 byte order, so two services written in different languages disagree on the same document.
- Do not re-encode numbers through a double. IEEE-754 binary64 represents integers exactly only up to 2^53, so normalising a 19-digit id through a float changes it, and 1 against 1.0 cannot be reconciled without deciding whether they are the same value. Preserve the literal token, and require ids as strings at the API boundary if you want them comparable.
- Reject duplicate keys rather than picking one. JSON permits them and parsers disagree, most keeping the last, so any choice you make ties the fingerprint to a parser detail that the code handling the request does not necessarily share.
Follow-up
- A client sends the same logical request with an extra field your API ignores. Same key, different fingerprint, so you return 422. Is that the right answer?
- Where does the fingerprint get computed relative to request decompression and the body-size limit?
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?
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?
Denormalise tenant onto revisions and backfill it live
Hard · denormalisation · online migration · backfill
resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.
Approach
- Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
- Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
- Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
- Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.
Follow-up
- The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
- How do you verify the backfill actually finished, given rows are still being inserted while it runs?
Relay committed events to the log without gaps or reordering
Medium · transactional outbox · at-least-once · ordering
outbox_event rows are written in the same transaction as the state change and carry aggregate_type, aggregate_id, aggregate_version, payload and status, with a partial index on (created_at, event_id) WHERE status = 'pending'. The relay publishes about 4k events/second to a partitioned append-only log keyed by aggregate_id, with one leader per partition range holding a lease. Consumers must never miss an event; they may see one twice. Design the claim-publish-mark loop, and state exactly what consumers observe when a leader's lease expires while it is mid-batch.
Approach
- Claim with SELECT ... WHERE status='pending' ORDER BY created_at, event_id LIMIT $batch FOR UPDATE SKIP LOCKED inside a transaction. SKIP LOCKED lets several relay workers share a range without serialising on each other's rows, and the partial index keeps the claim proportional to the backlog rather than to a table that is overwhelmingly published rows. At 4k/second a batch of 500 is eight claims per second, each an index scan of 500 entries.
- Publish before marking, never the reverse, and say why it is a choice. Marking first loses the event outright if the process dies in the gap, and the loss is silent - nothing remains to retry, and it surfaces later as a projection missing a row. Publishing first can repeat the event, and repetition is what every consumer is already built to survive. That single ordering is the whole at-least-once guarantee.
- Preserve the only ordering on offer. Partition by aggregate_id and never publish two events for one aggregate concurrently: claim in (created_at, event_id) order and publish sequentially within an aggregate, or hash aggregate_id to a worker slot. Order across aggregates is not available at any price here, which is why the event carries aggregate_version and the full fact rather than a delta - a consumer can then discard what it has already applied without coordinating with anyone.
- State the failover behaviour precisely, because it is the consistency-versus-availability decision in this design. A lease expires because the holder is slow, and no mechanism distinguishes that from dead, so for the length of the lease window two leaders can publish the same claimed batch. The system accepts duplicates to avoid stalling publication for every aggregate in the range whenever one worker pauses. Consumers deduplicate on (aggregate_id, aggregate_version) and drop anything at or below what they have applied.
Follow-up
- The relay is down 40 minutes and 9.6 million rows are pending. What does catch-up do to the primary, and what changes in the claim loop to survive it?
- A consumer insists it never received an event. Which single query settles whether the relay lost it, and what does each answer look like?
Design the async export contract a client can resume safely
Hard · async operations · idempotency · leases
A tenant asks for a CSV of every resource. The work runs for minutes on the worker fleet through a job_run row carrying a lease, an attempt count and a dedupe_key, far past the edge's 400 ms budget. Callers are a browser that polls and a script that walks away and checks later. Specify what the submit call returns, the operation resource and its states, how a duplicate submit is handled, how a client learns about completion, what cancellation means given that a lease can expire mid-run, and how the result is fetched and when it expires.
Approach
- Split the API in two. Submit returns 202 with an operation id and a location to poll, and never blocks on the work. The operation is a real resource with its own lifecycle - queued, running, succeeded, failed, cancelled - plus attempt, a monotonic progress figure, and a terminal error drawn from the same code taxonomy the synchronous endpoints use, so a client needs one error vocabulary rather than two.
- Deduplicate at submit using job_run.dedupe_key, unique over (job_type, dedupe_key) while status is 'queued' or 'running': a repeat submit of the same logical export returns 200 with the existing operation instead of 202 with a new one, and the partial index deliberately permits a legitimate re-run once the first has finished. Pair it with the request's idempotency key so an HTTP-level retry of the submit is exact rather than merely similar.
- Tell the poller how to poll: Retry-After on the polling response, a minimum interval enforced at the edge, and a documented maximum lifetime after which an operation is reaped. Polling is the contract of record; the webhook is the fast path, and both must lead to the same terminal state, so a client that receives the completion event and then polls anyway sees no contradiction.
- Be exact about cancellation. A cancel request records intent; it cannot stop work already executing. The handler reads the flag at checkpoints, and because a lease expires on a clock that cannot distinguish a dead worker from a slow one, a second copy may start after the cancel was recorded - so the handler re-reads the flag immediately after claiming the lease. 'cancelled' becomes terminal only when no lease is outstanding; reporting it earlier shows a client a stopped job while a worker is still writing output.
Follow-up
- An operation has said 'running' for 40 minutes and the worker is gone. What does the client see, and which columns in job_run decide that?
- Two tenants each submit 50 exports at once. What in this contract stops one of them delaying the other?
Listing latency scales with page size, not with filters
Easy · n+1 · latency · query batching
The tenant listing endpoint reads resource filtered by tenant_id and status, ordered by updated_at DESC, and returns each row plus the owner's display name from app_user and the actor of that resource's latest resource_revision. p99 is 55 ms at 10 rows per page and 1.4 s at 200. Database telemetry shows 401 statements per request, each under 1 ms, and nothing in the slow-query log. Diagnose the cause and give the fix, stating the statement count per request and the p99 you expect afterwards.
Approach
- Read the counters before forming a theory. 401 statements for 200 rows is one driver query plus two per row, and sub-millisecond execution with an empty slow-query log rules out a bad plan. The time is round trips, which is why it is invisible in every per-query metric and scales with rows returned rather than with filter selectivity.
- Name the two per-row statements from their normalised text: a single-row app_user lookup by user_id, and a resource_revision lookup by resource_id ordered by version DESC LIMIT 1. Confirm by dropping those two response fields and watching the statement count fall to one. That locates the calls in the serialisation layer, not the repository.
- Check that the arithmetic accounts for the whole gap. Measure one round trip to the replica in isolation; 400 trips at roughly 3 ms of network plus 0.2 ms of execution is about 1.3 s on top of a 55 ms baseline, which matches. If the multiplication had fallen short, the N+1 would only be part of the story and you would keep looking.
- Batch both lookups. Collect owner_user_ids and resource_ids from the driver query, then issue WHERE tenant_id = $1 AND user_id = ANY($2) for the users, and PostgreSQL's SELECT DISTINCT ON (resource_id) ... WHERE resource_id = ANY($2) ORDER BY resource_id, version DESC for the latest revision, which the UNIQUE (resource_id, version) index serves directly. On an engine without DISTINCT ON, use a lateral join or a row_number window. Three statements per request at any page size.
Follow-up
- The page size is capped at 200 today. What breaks first if it is raised to 2,000, and is it still this bug?
- How do you stop the next N+1 from reaching production, given that no individual query is slow and the endpoint's tests pass?
Suggested practice checklist
Built from the topics and questions IBC candidates report; no round sequence has been reported.
Day 1 — Establish the IBC format
- No round sequence has been reported, so ask your recruiter for the sequence, the duration of each stage and whether you will write code.
Deliverable: A written reply from your recruiter confirming the format.
Day 2 — Work Java
- Spend the session on Java, which IBC candidates report being tested on.
- Write one worked example in Java and time yourself on it.
Deliverable: One timed worked example in Java.
Day 3 — Work Fullstack Development
- Spend the session on Fullstack Development, which IBC candidates report being tested on.
- Write one worked example in Fullstack Development and time yourself on it.
Deliverable: One timed worked example in Fullstack Development.
Day 4 — Work Banking Domain Knowledge
- Spend the session on Banking Domain Knowledge, which IBC candidates report being tested on.
- Write one worked example in Banking Domain Knowledge and time yourself on it.
Deliverable: One timed worked example in Banking Domain Knowledge.
Day 5 — Answer out loud: Behavioral and Professional Alignment
- Answer aloud, timed: Talk about the experience that you have had.
- Answer aloud, timed: What are your strengths?
Deliverable: Spoken answers to 2 reported Behavioral and Professional Alignment question(s), under time.
Day 6 — Rehearse your own examples
- Prepare three examples from your own work where you made the decision, each with the outcome you can quantify.
Deliverable: Three examples written out, each with a number attached.
Day 7 — Dry run for IBC
- Run one full mock under time, then write down the two questions you most want to ask your interviewers.
Deliverable: A completed timed mock and two questions to ask.
Behavioural preparation
Behavioural rounds judge the decision you made and what it cost.
Talk about the experience that you have had.
Medium · Behavioral and Professional Alignment
Approach
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
What are your strengths?
Medium · Behavioral and Professional Alignment
Approach
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
How do you handle collaborative projects within a team?
Medium · Behavioral and Professional Alignment
Approach
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
Why are you interested in working in the banking sector specifically?
Medium · Behavioral and Professional Alignment
Approach
- Pick a story where you made the decision, not one where you watched it.
- State the situation in two sentences and spend the rest on the reasoning.
- Give the blast radius: what could have broken, and what you measured.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?