As a Software Engineer at OCBC Indonesia, you are at the heart of the bank's digital transformation. Your work directly impacts the financial stability and digital experience of millions of customers, ranging from individual mobile banking users to large-scale corporate entities. You will be responsible for building, maintaining, and scaling the robust systems that power modern banking, ensuring that every transaction is secure, efficient, and seamless.
This role is not merely about writing code; it is about solving complex problems within a high-stakes, regulated environment. You will contribute to critical projects, including microservices architecture, payment gateway integrations, and frontend user interfaces. Whether you are working on Java-based backend systems or React-driven frontend experiences, your contributions will directly influence the bank’s ability to remain competitive in the fast-paced fintech landscape.
Expect to work in a collaborative, cross-functional environment where you will engage with product managers, security teams, and business stakeholders. The work is challenging, requiring a balance of technical precision and an understanding of banking-grade security standards. You will have the opportunity to see your work deployed at scale, making this an ideal role for engineers who thrive on high-impact, mission-critical projects.
Initial Screening Call
reportedThe person on this call usually cannot evaluate your code and does not need to. They write a short paragraph, and that paragraph is what a hiring manager skims when deciding who to put on your loop. So the test is not whether your work was hard, it is whether a non-engineer can repeat it correctly. Name systems by what they did rather than by their internal codename, give each project a shape (what was breaking, what you changed, what happened after), and keep the whole walkthrough near ninety seconds. Depth that cannot survive a paraphrase reads as vagueness.
What to demonstrate
- Whether a non-engineer can restate your projects without distorting them, since their paraphrase is what travels to the hiring manager, not your sentences
- Whether each project has a shape rather than a stack list: the failure or constraint, the change you made, the result and how it was measured
- Whether you can say what was yours inside a team project without either inflating it or disappearing into the plural
How to prepare
- Rewrite each headline project as two sentences with no internal system names and no acronyms outside your company, then say them to someone outside engineering and have them repeat them back. Fix whatever came back wrong
- Attach one measured number to each project: the baseline, the change, and the window it was measured over. Where nothing was ever measured, say that plainly rather than reaching for a plausible percentage
- Time the background walkthrough against a clock. If it runs past two minutes, compress the earliest role to a single clause and spend the recovered time on the most recent one
Technical Assessments
reportedInput bounds are the part of the prompt most often skimmed, and they usually contain the answer. They tell you which complexity class is admissible, which narrows the search before you have thought about the problem itself. As a rough planning figure, a compiled language does on the order of 10^8 simple operations per second and an interpreted one roughly an order of magnitude less. So n up to about twenty admits enumerating subsets, a few thousand admits a quadratic pass, and a million admits neither: you need near-linear, or linear with a log factor. If the bounds are missing, ask for them.
What to demonstrate
- Whether the approach is justified by the stated input size rather than by whichever pattern you recognised first
- Whether you ask about the properties that change the algorithm: whether the input arrives sorted, whether duplicates occur, whether values are bounded integers, whether it all fits in memory
- Whether you can name the bottleneck in your own solution and what would remove it, even when you deliberately leave it in place
- Whether a claimed speedup is real, since memoising a recursion only helps when subproblems genuinely overlap and the state can be keyed cheaply
How to prepare
- For each algorithm you rely on, write down the largest n it handles in roughly a second, then check two of those figures by timing them in the language you will actually type in
- For two weeks, write one line naming your target complexity and the bound that justifies it before you write any code, then compare that line with what you ended up submitting
- Practise the conversion backwards: given a required O(n log n), list the mechanisms that get you there (sorting, a heap, an ordered map, divide and conquer) and choose by what the problem needs to query, not by what you used last
Technical Interviews
reportedMost 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.
Engagement with Leadership
reportedAn unlabelled round is first an information problem, and the cheapest information is free. Whoever schedules it can usually tell you how long it runs, who will be in the room and what they work on, whether you will be writing code and in what environment, and whether anything is being sent beforehand. Ask in writing so the answer is on record, then prepare for the two or three formats those answers still leave open instead of betting on one. What separates a strong candidate is not guessing right; it is having an opening that works whichever one it turns out to be.
What to demonstrate
- Whether you can start work from an ambiguous brief, since tolerating a vague scope without stalling is the same thing the job asks for
- Whether the questions you asked beforehand were ones that change your preparation, such as duration, medium and who is joining, rather than ones whose answers you could not have acted on
- Whether you adapt when the round turns out to be something other than what you were told, instead of spending the first ten minutes visibly recalibrating
How to prepare
- Send one short scheduling message asking four things: how long, who is joining and what they work on, whether you will be writing code and where, and whether to prepare anything in advance. Treat a vague reply as real information, since it means the round is loosely structured and you will be shaping it yourself.
- Write one opening that works in any of the formats still open: restate in your own words what you have been asked to do, then ask which of two directions is more useful to them. Say it aloud until it stops sounding recited.
- Set up for the two most likely formats before the call starts, with a blank editor in the language you would choose and a shared document you can type into, so a format surprise costs you nothing in the first minutes
Final Offer
reportedCoding rounds mostly set a floor. They decide whether you clear the bar, not where you land on the ladder. Level tends to come out of the design discussion and the ownership stories, so the question worth auditing beforehand is whether the scope you describe matches the scope of the job. Work that stops at your own service, or a story whose hard part was writing the code rather than getting several people to agree on an interface, reads a level below where you think you are interviewing, and that gap is usually resolved downwards.
What to demonstrate
- Whether the largest thing you describe owning ran end to end — the decision, the migration path, the rollout, and what you did when it went wrong — or stopped at the change you merged
- Whether design answers include what you would not build, what you would defer, and what you would measure before committing, rather than only what the boxes are
- Whether a disagreement in a story was settled with something checkable — a benchmark, a prototype, a written proposal — instead of by seniority or by waiting it out
- Whether you can say which calls you made alone and which you escalated, and why the line sat where it did
How to prepare
- Write your largest piece of owned work as a timeline of decisions — who decided what, when, and what you did when the plan broke — then delete every sentence whose subject is "we" and see how much survives
- Take one system you know well and drill the migration answer: how old and new paths run side by side under live traffic, how you compare their outputs, what the rollback is once writes are going to both, and which step you would not automate
- Map each line of the ladder in the job posting to a specific thing you have done, find the line you cannot support, and prepare the closest evidence you have plus an honest account of the gap
PracHub editorial advice for the preparation topics above.
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.
Paginating with LIMIT/OFFSET over a set that changes while the client is reading it
OFFSET n makes the database produce and discard n rows before returning anything, so the cost of a page grows with its depth rather than with its size and page 500 costs five hundred pages of work. The correctness problem is worse than the cost: if a row is inserted or reordered between two page fetches, rows shift across the offset boundary and are either skipped entirely or returned twice, and neither outcome leaves any trace in the response for the client to detect. Keyset pagination - WHERE (sort_key, id) < ($last_sort_key, $last_id) ORDER BY sort_key DESC, id DESC LIMIT n, backed by an index in exactly that order - reads only the rows it returns and is stable against concurrent inserts. It requires the tie-break column: a timestamp is not unique, and duplicate sort keys straddling a page boundary reintroduce the skip it was adopted to remove.
Designing for a scale nobody asked for
Ask for request rate, data size, read-to-write ratio and expected growth, then size the simplest option first; one relational instance on current hardware covers a large share of real workloads. Reaching for shards, queues and a cache tier before any number has been quoted reads as pattern-matching rather than judgement.
Abandoning working code to chase the optimal solution
Get the straightforward version correct, state its complexity, and only then optimise, keeping the working version until the faster one passes the same cases. A correct quadratic solution with a stated path to linear beats a half-written optimal one that never ran.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Diff a projection against the primary without per-row point reads
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.
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?
Find overlapping job attempts and peak concurrency from lease records
A day of job_run history yields about 50,000,000 attempt records: (job_run_id, job_type, attempt, started_at, finished_at which is NULL when the worker died, lease_expires_at). Leases expire on a clock, so a job that outran its lease ran twice. Produce (a) every job_run_id whose attempts overlapped in wall-clock time and (b) the peak number of simultaneously running attempts per job_type with the minute it occurred. Target O(n log n). State how you treat a NULL finished_at and what clock skew does to your answer.
Approach
- Define the interval before sorting anything: an attempt occupies [started_at, COALESCE(finished_at, lease_expires_at)). finished_at is observed and lease_expires_at is only a promise, so every attempt without a finish contributes an estimate and the whole result is a lower bound on overlap rather than an exact count.
- For peak concurrency, sweep: emit 2n endpoints, sort by (timestamp, kind) with ends ordered before starts at equal timestamps, then walk the sequence maintaining a counter per job_type and record each type's maximum with its timestamp. O(n log n) dominated by the sort, O(n) space, or O(1) extra if the sort is external and the walk streams.
- For overlap detection, do not compare attempts pairwise. A single global sort by (job_run_id, started_at) gives both the grouping and the order; within a group, keep the maximum end seen so far and report an overlap exactly when the next start is less than that running maximum, which is one linear pass after the sort.
- Half-open intervals matter and are easy to get wrong: with closed intervals an attempt ending at the same millisecond another begins reads as concurrency two, and across 50,000,000 records that artefact swamps the real signal.
- State the clock caveat: started_at and finished_at are written by different workers, so under skew of a few hundred milliseconds an apparent overlap shorter than that bound is not evidence. Filter reported overlaps by a minimum duration, or prefer timestamps written by whichever component heartbeats the lease.
- Scale the sort rather than assuming it fits: the sweep emits two endpoints per attempt, so 50,000,000 records become 100,000,000 endpoints, and at roughly 24 bytes each, an 8-byte timestamp plus a 4-byte job_type plus a kind flag padded to alignment, that is about 2.4 GB of sort keys before any scratch space. Either push the ordering into the database behind an index on (job_type, started_at) or run an external merge sort in chunks; the overlap pass sorts n records rather than 2n, so it is the cheaper of the two.
Follow-up
- A handler is not idempotent and you have found 400 overlapping jobs. Which of them actually caused damage, and what would you query to find out?
- Peak concurrency for one job_type is 4 against a configured cap of 4. Is the cap working, or is the data hiding attempts that never started?
- How would you compute both answers incrementally as records arrive rather than in a daily batch?
Track a rolling failure rate per destination for circuit decisions
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.
Worked solution 20 min
- Define the bucket struct and the advance step: take floor(finished_at_ms / 1000), compare with the ring's current second, zero min(delta, 60) buckets forward, then write into the new head.
- Trace a destination that receives 5 attempts, goes silent for 90 seconds, then receives one more, and confirm the rate is computed from one attempt rather than six.
- Compute total memory for 40,000 destinations at 60 buckets of two 4-byte counters, and state what changes if the window widens to 300 seconds.
- Write the open rule as a single predicate combining the minimum-attempt floor with the rate threshold.
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?
Keep soft-deleted accounts from blocking re-registration
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.
Worked solution 20 min
- Create the PostgreSQL partial unique index, insert a live row, soft delete it, and insert the same address again.
- Repeat the delete-and-reinsert cycle three times and confirm three deleted rows coexist with exactly one live row.
- Write the MySQL form with the discriminator, then deliberately reverse the polarity so live rows carry NULL, and show two live duplicates commit.
- Attempt a second live insert on both engines and map the resulting 23505 / ER_DUP_ENTRY to the 409 the handler should return.
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?
Write the update path that detects a concurrent edit
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.
- Refuse to auto-retry the whole PUT. A retry re-reads the winner's state and reapplies an intent formed against data that no longer exists — the silent overwrite the version token was added to detect. Return the conflict; merge field-wise only if the patches are provably disjoint.
- Note that now() is the transaction timestamp in PostgreSQL, so resource.updated_at, the revision's created_at and the outbox row share one instant, which is what later makes reconciliation between the three tables unambiguous.
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?
- Every write now touches a second hot table. How do you keep the outbox insert and its partial index from becoming the write bottleneck at 1.2k writes/second?
Can you explain how you would handle network resolution in a productio…
Can you explain how you would handle network resolution in a production environment?
Approach
- Name the read and write paths separately; they rarely have the same bottleneck.
- Fix the scope first: who calls this, how often, and what they do when it fails.
- State the consistency you need, and where you are willing to be stale.
Follow-up
- What breaks first when traffic grows ten times?
- What would you drop to keep the system up under load?
How do you approach API integration using React?
How do you approach API integration using React?
Approach
- State the consistency you need, and where you are willing to be stale.
- Choose a partition key and say what query it makes expensive.
- Name the failure you are designing for, then the recovery path.
Follow-up
- How does this behave when that dependency is down for an hour?
- What breaks first when traffic grows ten times?
What is Redis and how does it fit into a backend architecture?
What is Redis and how does it fit into a backend architecture?
Approach
- State the consistency you need, and where you are willing to be stale.
- Choose a partition key and say what query it makes expensive.
- Name the failure you are designing for, then the recovery path.
Follow-up
- What breaks first when traffic grows ten times?
- What would you drop to keep the system up under load?
Explain the difference between Singleton and prototype beans in Spring…
Explain the difference between Singleton and prototype beans in Spring Boot.
Approach
- Work from the requirement backwards to the design.
- Clarify what is being asked and what a complete answer contains.
- Say what you would check first and why it is the highest-information step.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
What is a design pattern, and how have you applied it in your past pro…
What is a design pattern, and how have you applied it in your past projects?
Approach
- Say what you would check first and why it is the highest-information step.
- Work from the requirement backwards to the design.
- State your assumptions explicitly before working the problem.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Choose what to break when replication lag reaches forty seconds
Reads are served from two replicas: 14k requests/second, about 85% absorbed by cache, so roughly 2.1k reads/second reach the database. Writes go to the primary at 1.2k/second. A tenant's backfill drives replication lag from under 100 ms to 40 seconds and it is still climbing. Sessions that have just written are pinned to the primary. Decide, endpoint class by endpoint class, whether to serve stale, fail, or route to the primary, and justify each choice with the load it adds to the primary. Then state what you would have built beforehand.
Approach
- Establish blast radius before cause, because mitigation and diagnosis have different deadlines. The decisive arithmetic is what happens if the database reads move to the primary: 2.1k reads/second on top of 1.2k writes/second roughly triples its operation count, on the node already absorbing the backfill that caused this. Reads and writes are not equal in cost, so treat that as an argument against a blanket move rather than as a capacity model - but it is enough to rule out routing everything to the primary.
- Classify endpoints by what staleness costs, not by how important they feel. Reads whose staleness is invisible - listings, search, counters - stay on the replica and return the watermark so the client can tell. Reads that immediately follow that same session's write keep their primary pin, which is a small bounded slice of traffic rather than the whole 2.1k/second. Reads that feed a decision with a side effect - authorisation, quota, the read half of a read-modify-write - must not be stale at all, because a 40-second-old permission row is the stale-permission failure wearing a different costume; those go to the primary or fail.
- Shed instead of queueing. If the must-be-fresh class alone exceeds the primary's headroom, refuse its lowest-value slice with 503 and a retry-after. A request queued behind a saturated primary holds a connection for a client that has already given up, and the retry storm that follows is what turns degradation into an outage. Bound the connection pool per role so the read fallback cannot consume the write path's connections - that bulkhead is the single decision that determines whether writes survive the next ten minutes.
- Attack the cause in parallel, since it is the one thing that can be stopped. The backfill is the load generator. A backfill that reads replication lag as its throttle signal and pauses above a threshold would have made this a non-event, with batch sizes small enough that each batch's write volume is a fraction of what a replica can apply per second. That is most of the answer to what should have existed beforehand.
- Name the mechanism you would prefer over session pinning. Capture the write position at commit and require the read path to be at or past it: compare the primary's pg_current_wal_lsn() at commit time against the replica's pg_last_wal_replay_lsn(), and fall back to the primary only for the specific request that is ahead of the replica. Session pinning is the cheap approximation and it over-pins - every read in the window goes to the primary whether or not it needed to, which is a share of the cost being paid right now.
Worked solution 35 min
- List the endpoints in three buckets - staleness invisible, staleness visible to the writer only, staleness unsafe - and attach the share of the 2.1k reads/second each bucket carries.
- Compute the primary's operation count under each routing option and mark which options are arithmetically available.
- Write the pin rule and its window, then the shed rule: which slice, what status code, what retry-after.
- Write the backfill's throttle predicate against a measured lag value, including its pause threshold and resume condition.
Follow-up
- Lag returns to normal in nine minutes. Which mitigation do you remove first, and which one stays permanently?
- A user reports their change did not save, and the write committed. Trace the path that produces that report and name the signal that would have shown it before the report arrived.
- The replica is 40 seconds behind but otherwise healthy. Do you take it out of rotation? What does that do to the other replica's lag?
Exports duplicate a row range about once a week
Roughly once a week an export writes a file containing a duplicated range of rows. The affected job_run rows show attempt = 1, status = succeeded, one started_at, and a lease_owner naming a different host from the one whose logs show the job starting. Leases last 30 seconds and are heartbeated every 10 from inside the handler; lease_expires_at is computed on the worker and compared against the database's now(). Find the mechanism, and give a fix that holds even if you cannot fix the clocks.
Approach
- Start from the fact that eliminates the obvious answer. attempt = 1 means no retry was recorded, so this is not a re-run after failure; two workers ran the same row concurrently and the takeover path never touched the counter. lease_owner naming a host other than the one that started the job is the same statement from the other side.
- Enumerate the mechanisms that cause a premature takeover, then find the signal that separates them. Either the lease genuinely expired because the heartbeat did not fire, which is what happens when the heartbeat runs on the handler's own thread and the handler makes a long blocking call, or it only appeared expired because two clocks disagree, since lease_expires_at is written from the worker's clock and evaluated against the database's. The discriminator is the distribution: incidents clustered on the longest exports indict the heartbeat, incidents clustered on one host indict skew. Measure both, and measure each host's offset against the database directly.
- Read the reclaim query precisely. In PostgreSQL now() is transaction start time, not statement time, so a reclaimer holding a long transaction compares against an older timestamp than expected; clock_timestamp() is the statement-time function. This is worth ruling in or out before you redesign anything, because it changes which rows look expired.
- Remove the second clock rather than trying to synchronise it. Issue and extend the lease in the database, with lease_expires_at = now() + interval '30 seconds' in both the claim and the heartbeat, so exactly one clock is ever compared and worker skew stops mattering to this predicate.
- Accept that a lease can still expire under a slow worker, because a lease cannot distinguish slow from dead, and fence the work. Carry a monotonically increasing lease generation and make every write the handler performs conditional on still holding it, as UPDATE ... WHERE job_run_id = $1 AND lease_owner = $2 AND lease_generation = $3, so a displaced worker's writes affect zero rows and it aborts instead of duplicating.
- Make the handler's writes idempotent independently of all that: give each exported chunk a natural key of (job_run_id, batch_start) with a unique constraint so a second copy conflicts rather than appends, move the heartbeat off the handler's thread, and increment attempt on takeover so the event is visible in a metric.
Follow-up
- The displaced worker has already streamed half the file to object storage. What makes that side effect safe to repeat?
- You now count takeovers. What alert fires on that counter, and at what threshold?
- What breaks if you simply raise the lease to five minutes?
For someone who has spent the last few years shipping features and reading other people's code, and who has not solved a timed problem from a blank file in a long time. Five days rebuild the primitives and the patterns that sit on them, working from invariants rather than remembered solutions, and the last two attach that back to the rest of the loop.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Rebuild the primitives by implementing them
- Implement a dynamic array with doubling growth and an operation counter, then change the growth rule to add a fixed sixteen slots instead, and time both for n of ten thousand, a hundred thousand and a million. The fixed-increment version resizes n/16 times at O(n) each, so its total work is quadratic; doubling is what makes append amortised constant.
- Implement a hash map with separate chaining and a load-factor resize, then insert ten thousand keys engineered to land in one bucket and record what happens to lookup time, so that average-case O(1) becomes a claim with a stated precondition rather than a reflex.
- For dynamic-array append and hash-map insert, write down which cost is amortised rather than worst-case, which single operation pays the whole bill, and what a system with a hard per-operation deadline would have to do instead.
Deliverable: Two working implementations plus a timing table showing the input at which each structure's advertised complexity stops holding.
Practice prompt ↗Practice prompt ↗Practice prompt ↗Worked solution ↗02Arrays under an invariant: two pointers, sliding window, binary search
- Solve longest-subarray-with-sum-at-most-K using a sliding window, then run it on an input containing negative numbers and watch it return the wrong answer: extending the window only moves the sum monotonically when every element is non-negative, and that precondition is the whole reason the technique works.
- Write the binary search that finds the first index satisfying a predicate rather than an exact value, put the loop invariant above the loop in a comment, and verify termination on the two inputs that break careless versions: the empty range, and a range where every element satisfies the predicate.
- Compute the midpoint as lo + (hi - lo) / 2 and write one line on why the obvious (lo + hi) / 2 is a genuine defect in a fixed-width integer type and a non-issue in a language with arbitrary-precision integers.
Deliverable: Three solved problems, each with its invariant written above the loop, plus one recorded input on which the sliding window is provably wrong.
Practice prompt ↗Practice prompt ↗03Sorting, heaps, and the greedy argument that has to be proved
- Solve one top-k problem three ways, by full sort, by a size-k heap, and by quickselect, then write the values of n and k at which each becomes the right choice, along with quickselect's quadratic worst case and why a randomised pivot makes that unlikely rather than impossible.
- Implement bottom-up heapify and count sift-down steps to confirm it does linear work rather than n log n, because most nodes sit near the bottom of the tree and therefore move only a short distance.
- Take interval scheduling by earliest finishing time and write the exchange argument out in full: given any optimal schedule, swapping in the earliest-finishing interval keeps it feasible and no smaller. Then construct the weighted variant where that same greedy fails and name what has to replace it.
Deliverable: A three-way top-k comparison with measured crossover points, one written exchange argument, and one counterexample to a greedy rule that looks almost identical.
Practice prompt ↗Practice prompt ↗04Recursion, memoisation, and the step to a table
- Take one problem with overlapping subproblems, such as edit distance or coin change, instrument the plain recursion with a call counter to show the blow-up, then add memoisation and re-count.
- Convert the memoised version to a bottom-up table and state the two properties you relied on: each subproblem's result depends only on its arguments, and the dependencies form a DAG you can enumerate in order.
- Rewrite one deep recursion with an explicit stack, then find the input length at which the original hits the interpreter's frame limit, which defaults to about a thousand frames in CPython, so you know when the rewrite is required rather than decorative.
Deliverable: One problem in three forms, naive, memoised and tabulated, with call counts for each and the input length at which recursion depth becomes the binding constraint.
Practice prompt ↗Practice prompt ↗Worked solution ↗05Graphs, where most of the work is choosing the traversal
- Implement BFS and DFS over one adjacency list, then answer for each which finds a shortest path in an unweighted graph and which you would use to detect a cycle in a directed graph, including why the in-progress versus finished distinction matters for the second.
- Implement topological sort by in-degree, feed it a graph containing a cycle, and confirm the failure signature is that fewer than V nodes come out rather than an exception, then note that the order it produces is one of several valid ones.
- Run a shortest-path search on a graph with a single negative edge weight and show the wrong answer, then write the precondition Dijkstra actually needs, non-negative weights, because it finalises a node's distance the first time that node is popped, and name the algorithm you would switch to and its own limit.
Deliverable: A small graph library with BFS, DFS and topological sort, plus two inputs that produce documented wrong answers under the wrong algorithm choice.
Practice prompt ↗Practice prompt ↗06One day for everything that is not an algorithm
- Sketch one system only to the depth a coding-heavy loop tends to reach: the endpoints, what the service stores, and the single query pattern that decides the schema. Stop at twenty-five minutes.
- Prepare the project answer for an interviewer who codes, which means rehearsing the two levels they push to: the specific thing you built, and why you chose that approach over the alternative they will name. Open with a number and be ready to say what it excludes.
- Prepare the answer to what you would do differently, choosing a real technical mistake with a specific fix rather than a complaint about process or staffing.
Deliverable: One design sketch at endpoint-and-schema depth, plus a project answer rehearsed to two levels of follow-up.
Practice prompt ↗Practice prompt ↗07Solve out loud, under time
- Do three timed problems at twenty-five minutes each in a plain editor with no autocomplete and no execution until the end, then tally separately the failures that were syntax and the ones that were approach, because those two numbers call for different fixes.
- Narrate one solution from the first sentence, stating the approach and its complexity before writing any code, and rehearse the sentence you will use when you realise mid-solution that the approach is wrong.
- Re-solve from blank the two problems you were slowest on this week and compare the times against the day they first appeared.
Deliverable: A recording of one fully narrated solution and a tally that separates syntax failures from approach failures.
Practice prompt ↗Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
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.
How can you benefit the team, and what can you bring to the department…
How can you benefit the team, and what can you bring to the department?
Approach
- Name the disagreement and how you resolved it with evidence.
- State the situation in two sentences and spend the rest on the reasoning.
- Close with what you would do differently, concretely.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
Describe a moment you had to go against your values to get something d…
Describe a moment you had to go against your values to get something done.
Approach
- 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.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
Explain your experience with microservices—why use them and how do you…
Explain your experience with microservices—why use them and how do you manage them?
Approach
- Name the disagreement and how you resolved it with evidence.
- State the situation in two sentences and spend the rest on the reasoning.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- What did you decide not to do, and why?
- How did you know your change caused the improvement?
- 01
How can you benefit the team, and what can you bring to the department?
- 02
Describe a moment you had to go against your values to get something done.
- 03
Explain your experience with microservices—why use them and how do you manage them?
Is this an official OCBC Indonesia interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at OCBC Indonesia. Rounds and questions reflect what candidates have reported, not a process OCBC Indonesia has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How long does the interview process typically take?
The process usually spans 3 to 4 weeks from the initial application to the offer stage, though it can vary based on team availability.
PracHub interview research ↗What is the best way to prepare for the technical assessment?
Focus on core data structures, algorithms, and SQL proficiency. Since many assessments are conducted via platforms like Codility, practice solving problems under time constraints.
PracHub interview research ↗Is the culture at OCBC Indonesia collaborative?
Yes, most candidates describe the atmosphere as professional and friendly. You will be expected to work closely with various cross-functional teams.
PracHub interview research ↗How can I stand out in my interview?
Differentiate yourself by focusing on your past projects. Be ready to explain the technical challenges you faced and the specific, measurable results you achieved.
PracHub interview research ↗Sources & methodology 3 sources ↗
Official role evidence, timestamped platform data and clearly labeled preparation advice.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
platform · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
platform · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
platform · Accessed 2026-09-22
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