As a Software Engineer at AngloGold Ashanti, you are not just writing code; you are building the digital backbone of a global leader in the mining industry. Your work directly impacts how complex, large-scale mining operations—ranging from deep-level underground sites to expansive open-pit projects—function safely and efficiently. You will contribute to mission-critical systems that optimize production, ensure worker safety, and manage the complex data streams inherent in a modern, sustainable mining environment.
This role requires a unique blend of technical expertise and a pragmatic, field-aware mindset. You will often find yourself collaborating with mining superintendents, geotechnical engineers, and operational staff to bridge the gap between abstract software requirements and the physical realities of the mine site. Whether you are automating reporting, integrating sensor data, or developing tools to improve operational throughput, your contribution is vital to maintaining AngloGold Ashanti’s competitive edge and operational excellence.
Initial Screening
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
Assessments
reportedThe same problem is scored by two different mechanisms depending on the format, and preparing for one does not cover the other. With a person watching, partial progress is visible and a hint is a correction you can absorb; silence is the expensive failure, because nobody can read a half-written function. With an automated grader there is no partial credit for what you were about to do, nobody to ask, and the worked examples in the prompt are the entire specification. Read them as a contract, down to whether an empty result should be an empty list or no output at all.
What to demonstrate
- In a live session, whether your commentary tracks what your hands are doing, and whether a hint redirects you or gets defended against
- In an automated one, whether you cover the cases the examples do not show, since the hidden cases are where the score moves
- Whether you manage the clock on purpose: abandoning an approach that is not converging while there is still time to write something simpler that finishes
How to prepare
- Have someone hand you a problem and feed you one deliberately wrong hint. Practise testing it against a concrete case instead of accepting or rejecting it on authority.
- Do one timed run a week in a plain browser editor with autocomplete, linting and your own snippets switched off, which is closer to what these environments give you
- For the automated format, write the harness before the solution: a main that feeds the worked examples plus an empty and a single-element case and prints expected against actual, so a wrong submission is caught by you first
Assessment Center
reportedWhat this round decides is narrow: whether you can produce code that runs and is correct on inputs nobody showed you. An elegant solution that does not compile scores below a plain one that does, so write a correct brute force first, say out loud that you know its cost, and improve it with the working version still on screen. What separates strong answers is who finds the broken case. Trace your own code against an empty input, a single element, and duplicate keys before you say you are finished, because being told is far more expensive than noticing.
What to demonstrate
- Whether degenerate inputs get checked without being asked for: an empty collection, one element, every element equal, and the extreme value the input type allows
- Whether the complexity you state matches the code you actually wrote, including a sort or a copy sitting inside a loop
- Whether the finished answer is verified against the worked examples before you call it done, rather than assumed correct because the code reads correctly
How to prepare
- Take five problems you have already solved and, without running anything, write down what each returns for empty input, a single element, and all-duplicates. Then run them and count how many you predicted wrong.
- Drill the brute force as its own skill: on ten problems, write only the obviously-correct slow version and time how long it takes to get it passing. If that is more than a few minutes, that is what to practise, not the optimal version.
- Add a fixed last step before you submit anything, reading only the loop bounds and the initial value of each accumulator, which is where most off-by-one errors live
PracHub editorial advice for the preparation topics above.
Assuming an isolation level prevents the anomaly you actually have
Isolation levels are named by the SQL standard but implemented differently, so any claim about one is only true of a named engine. PostgreSQL defaults to READ COMMITTED, where every statement takes a fresh snapshot, so two statements inside one transaction can legitimately disagree about the same row. Its REPEATABLE READ is snapshot isolation: it removes non-repeatable and phantom reads but permits write skew, where two transactions each read a set, each conclude their own write is safe, both commit, and the combined result violates a constraint that no single row expresses. Only SERIALIZABLE closes that, and it closes it by aborting a transaction with a serialization failure (SQLSTATE 40001), which means the guarantee is theoretical unless the application has a retry loop. InnoDB's REPEATABLE READ is a different mechanism again - plain SELECTs read a consistent snapshot while locking reads and writes see the latest committed row - so a read-modify-write inside one transaction can act on a value that the transaction's own earlier SELECT never returned.
Choosing an index from the columns a query mentions rather than from how it filters and orders
A composite B-tree index on (a, b, c) can be seeked only as a left prefix: equality on a, then equality on b, then a range or an ordering on c. A query that filters on b alone cannot seek into it at all and at best gets a full scan of the index; a query that filters a and ranges on b gets no benefit from c, because the index is only sorted by c within a fixed (a, b) pair. The practical consequence is that one index per column is close to useless for multi-predicate queries while a single correctly ordered composite index turns a scan into a lookup. The ordering half is what gets missed: if the index cannot satisfy the ORDER BY, the database must read every matching row and sort before the limit can apply, so a LIMIT 20 over a million matching rows still reads a million rows.
Assuming the bug is in the framework
Suspect your own code first: read the stack trace top to bottom, check which versions are actually installed rather than which ones you believe are, and reproduce in isolation before blaming a library that thousands of people run daily. When the fault really is upstream, you need that minimal reproduction to say so credibly anyway.
Retrying a write that is not safe to repeat
A timeout tells you nothing about whether the server applied the write, so a blind retry of a create or a charge can duplicate it. Either make the operation idempotent, with a caller-supplied key the server deduplicates on or a conditional update, or do not retry it; and use exponential backoff with jitter so the retries of many clients do not synchronise into a second outage.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Archive a resource graph without breaking live references or recursing
Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.
Approach
- Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
- Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
- Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
- Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.
- Store the graph as CSR rather than a map of lists: an offsets array of V+1 8-byte entries plus E 8-byte targets is about 80 MB at this size, where boxed adjacency lists cost several times that and lose cache locality on every hop.
- Run Kahn over the condensation for the order in O(V+E). If the emitted count is short of the component count the condensation step itself is wrong, since a condensation cannot contain a cycle, which makes the check free.
Worked solution 30 min
- Write the edge-loading query with the tenant predicate on both endpoints and state what it does with a cross-tenant edge.
- Implement iterative Tarjan with an explicit stack and confirm on a three-node cycle that it emits one component of size three.
- Build the transpose restricted to the visited set and mark every node with an in-edge from outside it as refused, carrying the referrer id.
- Run Kahn over the condensation and verify the emitted order against the referrer-before-referenced rule.
- Size the CSR arrays for 2,000,000 nodes and 8,000,000 edges and compare against a boxed adjacency map.
Follow-up
- The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
- The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?
- An edge points at a resource in another tenant. Is that a refusal, an error, or an alert?
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?
Identify the heaviest tenants in a five-minute window under memory pressure
The edge service handles about 3,000 requests per second across roughly 50,000 tenants, peaking near 9,000. Expose the 50 heaviest tenants by request count over the trailing five minutes so limits can be tightened before one tenant's backfill starves the fleet. You may not retain five minutes of raw records. Give the exact solution and its memory, then the bounded-memory approximation with its error stated as a formula, and say which you would ship and at what tenant cardinality that choice changes.
Approach
- Do the exact version first, because it is affordable at this cardinality: a ring of 300 one-second counters per tenant, advanced lazily, is 1,200 bytes of counters per tenant and roughly 60 to 90 MB for 50,000 tenants with overhead. Carry a running total and subtract the bucket you overwrite so a window read is O(1) rather than 300 adds.
- Extract the top 50 with a size-k min-heap over the tenant sums: O(d log k) for d tenants, against O(d log d) to sort them all. Maintaining the heap continuously instead of on query requires a tenant-to-heap-index map, because incrementing a count already inside the heap means sifting from a known position, and without that map you rebuild the heap on every request.
- State the approximation precisely rather than gesturing at sketches. Misra-Gries with m counters retains every item whose true count exceeds N/(m+1), and each retained count underestimates the truth by at most N/(m+1). With m = 1,000 and N = 900,000 requests in the window the error is roughly 900 requests, which is fine for spotting a tenant sending 50,000 and useless for ranking two tenants 200 apart.
- Say what breaks when the window slides: Misra-Gries and Space-Saving are insert-only and cannot be decremented as records age out. The workable construction is one summary per sub-window, say ten seconds, with 30 summaries merged at query time, and the merged error is the sum of the per-summary errors, so the bound degrades linearly in the number of sub-windows.
- Choose and defend it: at 50,000 tenants the exact rings cost under 100 MB in a process that already holds more, so ship exact. Keep the sketch for the case that actually motivates it, a per-principal or per-IP key where cardinality runs to millions and is not bounded by anything you control.
- Raise the fleet problem before it is asked: each of 20 to 40 instances sees only its share, and the top 50 of one shard is not the top 50 of the fleet. Either aggregate counts centrally or accept that a per-instance threshold multiplied by instance count is the limit you are really enforcing.
Follow-up
- The heaviest tenant is heavy because of one export job rather than user traffic. Should the limiter treat those as the same tenant?
- Two tenants sit tied at the boundary of the top 50. Does your answer flap, and does the flapping matter?
- You switch to per-principal keys and cardinality goes to 10 million. Walk through what changes.
Replace offset paging on the resource feed with keyset
resource holds resource_id, tenant_id, owner_user_id, title, body_ref, version, status ('draft','active','archived','deleted'), created_at, updated_at, deleted_at, with an index on (tenant_id, status, updated_at DESC, resource_id DESC). The listing endpoint returns active resources for one tenant, newest update first, 50 per page, today with LIMIT 50 OFFSET n. Tenants reach page 400 and rows are created while they read. Write the keyset query, define what the cursor carries and how it is encoded, and say which part of the index each predicate uses. Assume PostgreSQL 16.
Approach
- Name the two failures separately. OFFSET 20000 makes the server produce and discard 20,000 rows, so page cost grows with depth rather than with page size. Independently, any write that changes how many rows sort above the offset moves the window between two fetches, and the direction decides which anomaly you get: an insert lands at the head of updated_at DESC and pushes already-returned rows down past the boundary, so they are returned a second time; a delete above the offset, or a row whose updated_at is bumped above the cursor, pulls rows up and one is never returned at all. Nothing in the response reveals either.
- Write the seek: WHERE tenant_id = $1 AND status = 'active' AND (updated_at, resource_id) < ($2, $3) ORDER BY updated_at DESC, resource_id DESC LIMIT 50. The row-value comparison is one index range rather than a disjunction, and both columns are NOT NULL, which is what makes that comparison well defined.
- Map each predicate onto the index: tenant_id and status are equality on the leading columns, (updated_at, resource_id) is the range, and the ORDER BY matches the index order so no Sort node appears and the scan stops after 50 rows. The DESC in the definition only matters for mixed directions — a plain ascending btree on the same columns is read backwards for this query.
- Put both sort columns in the cursor and nothing the client can tamper with into another tenant: base64 of (updated_at, resource_id), validated server-side, with tenant_id taken from the principal.
- State the residual honestly. Keyset is stable against concurrent inserts and deletes, but not against a row whose updated_at changes mid-scroll — that row moves in the ordering and can be seen twice. If the feed must be a snapshot, order by an immutable key or bound the page set with updated_at <= the cursor's start value.
- Keep a total out of the page path. A tenant-wide COUNT(*) is the scan keyset just removed; fetch LIMIT 51 and return has_more instead.
Worked solution 25 min
- Seed one tenant with 500k active resources, 2% of them sharing an identical updated_at.
- Time LIMIT 50 OFFSET 0 against OFFSET 20000 and record rows-read from EXPLAIN (ANALYZE, BUFFERS) for each.
- Page the whole set with the keyset query while an insert-only writer adds 100 rows/second, collecting resource_ids, and repeat the run with OFFSET.
- Repeat both runs under a second writer profile that also deletes 20 rows/second from pages already returned and bumps updated_at on 20 more, and diff each collected id set against the rows that existed for the whole run.
- Remove resource_id from the cursor so the seek degrades to updated_at < $2, and re-run the tie-heavy section of the feed.
Follow-up
- The client asks for 'jump to page 400'. What do you offer instead, and what does the honest version cost?
- Sort order becomes user-selectable across four columns. How many indexes is that, and which would you refuse to add?
- What does the cursor do when the row it points at has since been deleted?
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.
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?
What is your understanding of common mining methods, such as sublevel …
What is your understanding of common mining methods, such as sublevel stopping?
Approach
- Clarify what is being asked and what a complete answer contains.
- State your assumptions explicitly before working the problem.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Can you describe a challenging project you have worked on and how you …
Can you describe a challenging project you have worked on and how you managed the obstacles?
Approach
- Work from the requirement backwards to the design.
- State your assumptions explicitly before working the problem.
- 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?
Rebuild the search projection while it serves nine thousand queries
The read-model service answers about 9k queries/second at a 120 ms p99 from a projection built off the event log, normally under 2 seconds behind. A mapping change forces a full rebuild from resource_revision, during which apply lag rises to minutes. Writes continue at 1.2k/second and events at 4k/second. Design the rebuild: how the new index is populated and cut over, how position is tracked per log partition, what the API returns alongside results so a client can tell a stale answer from a current one, and the criterion for cutting over.
Approach
- Build into a second index and swap an alias rather than mutating the live one. The rebuild is then reversible by pointing the alias back, so a bad mapping costs a wasted rebuild instead of an outage. The price is peak storage for two full copies and double apply load during catch-up, and both numbers should be stated up front rather than discovered when disk fills.
- Track position per log partition, not globally. Order is guaranteed only within an aggregate's partition, so progress is a vector, and the only number safe to publish is taken from the least advanced partition - the slowest one is what bounds completeness. Publishing the most advanced partition's position declares the projection current while another partition sits twenty minutes behind.
- Make apply idempotent so the backfill and the live tail can overlap without a freeze. Each document records the aggregate_version it reflects, and any event at or below that version is discarded. This is why the event carries the full fact and not a delta: a delta cannot be discarded safely, and a consumer that calls back to read current state applies a state newer than the event it is processing, which is how a projection ends up with changes applied out of order.
- Turn lag into a contract instead of a surprise. Return the watermark with every result set, and return the version produced by a write so the client can compare the two. A client that wrote version 7 and receives results at a watermark older than its own commit can show that its change is still landing, rather than rendering the previous value as current. Blocking the read until the projection catches up would convert a staleness problem into an availability problem at 9k queries/second, and choosing not to do that is the trade.
- Define the cutover numerically. Writes are untouched by the rebuild - they commit to the primary and land in the outbox - so the only coupling is apply throughput. If catch-up applies slower than the 4k events/second arriving, it never converges. The cutover criterion is that lag is measurably decreasing and below a stated threshold, not that the backfill loop reached the end of its range.
Worked solution 30 min
- Write the rebuild sequence: snapshot source, populate the second index, attach the live tail, verify, swap alias, retire the old index - naming what is reversible at each step.
- Define the per-partition position record and the single watermark derived from it, and show what each would report when one partition stalls.
- Write the idempotent apply rule using the document's recorded aggregate_version, and replay one event twice against it.
- State the cutover threshold in lag terms and the measurement that shows convergence rather than completion.
Follow-up
- The rebuilt index disagrees with the primary tables for 300 documents. Which is authoritative, and how do you decide without freezing writes?
- The rebuild doubles load on the log and pushes the projection p99 from 120 ms to 400 ms. What do you throttle, and which signal sets how much?
- Clients start polling until the watermark passes their write. What does that do at 9k queries/second, and what do you offer instead?
Listing latency scales with page size, not with filters
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.
- Keep the tenant predicate in the batched query. The per-row version was implicitly scoped because its ids came from tenant-scoped rows; a batched user_id = ANY(...) with no tenant_id is an unscoped read that behaves correctly only as long as the id list is trustworthy.
- Re-measure at 10, 50 and 200 rows and confirm the statement count is constant. Latency should now track bytes returned.
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?
- The latest-revision actor is only used to render an avatar. Make the case for denormalising it onto resource, and name the write anomaly that introduces.
Day one measures instead of guessing, under a fixed rubric, and the remaining hours are allocated in proportion to the gaps before any studying begins. The allocation is deliberately not renegotiated midweek, because the area that feels worst on day three is usually the one that is moving.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Diagnostic, scored before you study anything
- Sit a 110-minute diagnostic in four blocks: forty-five minutes on two coding problems, twenty-five on one design prompt taken to interface and data model, twenty of short-answer fundamentals, and twenty delivering two behavioural answers aloud.
- Score each block from 0 to 3 on a fixed rubric where 3 is correct and fluent, 2 is correct but slow or prompted, 1 is partially correct and 0 is stuck, grading the artifact rather than how the attempt felt.
- Allocate days two to five in proportion to 3 minus each block's score, write the allocation down, and commit to leaving it alone.
Deliverable: A scored rubric and a fixed hour allocation for the rest of the week.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Largest gap: find the boundary rather than the subject
- Split the weakest area into named sub-skills and rate each separately. For coding those are restating the problem, choosing the structure, stating the invariant, turning the invariant into loop bounds, handling empty and single-element input, and accounting for complexity out loud.
- Attempt three items positioned just above where the rating drops off, and for each write the first move you failed to make.
- Re-attempt one of them from blank four hours later with nothing open.
Deliverable: A sub-skill map with the two blocking sub-skills circled.
Practice prompt ↗Practice prompt ↗03Drill the blocking sub-skill by repeating the shape
- Do eight short repetitions of the same shape rather than eight different problems, so what gets practised is the pattern and not the puzzle.
- State the rule you now hold in one sentence, then test it against a case built to break it, a sliding window over an array containing negative values, or a cache-aside read path whose invalidation message is dropped.
- Have someone else read your one-sentence rule and find the precondition you left out.
Deliverable: One rule statement with its preconditions attached and one counterexample that would have caught the incomplete version.
Practice prompt ↗Practice prompt ↗04Second gap, plus maintenance on the strongest area
- Run the same sub-skill decomposition on the second-largest gap in half the time.
- Spend twenty-five timed minutes on the block you scored highest, choosing the hardest item you can still finish rather than a warm-up.
- Write whether each area fails you on recall, on setup, or on execution, and set the fix accordingly: repetition for recall, a written checklist for setup, timed work for execution.
Deliverable: A second sub-skill map plus a one-line failure diagnosis for each area.
Practice prompt ↗Practice prompt ↗Worked solution ↗05The gap that is not a skill
- Record one technical and one behavioural answer, then count two things in the playback: seconds before your first clarifying question, and sentences you began without knowing where they would end.
- Practise saying that you do not know, followed by how you would find out, without letting it soften into a guess, and practise stating a complexity or an estimate before being asked for it.
- Redeliver one answer under a hard ninety-second cap, which forces structure ahead of detail.
Deliverable: Two recordings with a counted reduction in time-to-first-question.
Practice prompt ↗Practice prompt ↗06Retest under day-one conditions
- Sit the same 110-minute structure with new prompts of comparable difficulty and score it on the identical rubric.
- For any block that did not move, change the method rather than adding hours: a block stuck at 1 usually means the practice was too varied, not too short.
- Write down which single block you would still lose the offer on.
Deliverable: A second scored rubric placed beside the first, with one named remaining risk.
Practice prompt ↗Practice prompt ↗07Full loop under interview conditions
- Run a sixty-minute mock over the two blocks that moved least, with an interviewer briefed to interrupt and change direction mid-answer.
- Write the recovery script for going blank: restate the question, state your assumption, name the first thing you would check.
- Say every rule from the week aloud without reading it, and cut any you cannot state in a single sentence, since a rule you have to reconstruct mid-answer will not survive an interruption.
Deliverable: A one-page card holding the recovery script and only the rules you could state from memory.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
Conflict answers where you were right and everyone came round are the weakest ones. Stronger: the evidence you went and collected, what would have changed your mind, and what you did in the weeks after the call went against you. Implementing a design you argued against, properly, is a specific and checkable behaviour.
Describe a time when you had to deal with a difficult colleague.
Describe a time when you had to deal with a difficult colleague.
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.
Follow-up
- How did you know your change caused the improvement?
- What did you decide not to do, and why?
What do you know about our company values and operations?
What do you know about our company values and operations?
Approach
- Pick a story where you made the decision, not one where you watched it.
- Name the disagreement and how you resolved it with evidence.
- Close with what you would do differently, concretely.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
What are your expectations regarding the work environment (e.g., FIFO …
What are your expectations regarding the work environment (e.g., FIFO or site-based roles)?
Approach
- Close with what you would do differently, concretely.
- 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?
- What would you do differently if you ran that again?
Why do you want to work for AngloGold Ashanti?
Why do you want to work for AngloGold Ashanti?
Approach
- Name the disagreement and how you resolved it with evidence.
- 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.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
- 01
Describe a time when you had to deal with a difficult colleague.
- 02
What do you know about our company values and operations?
- 03
What are your expectations regarding the work environment (e.g., FIFO or site-based roles)?
- 04
Why do you want to work for AngloGold Ashanti?
Is this an official AngloGold Ashanti interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at AngloGold Ashanti. Rounds and questions reflect what candidates have reported, not a process AngloGold Ashanti 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 can vary, but it is often described as thorough. From the initial screening to the final interview, expect several weeks of engagement, including potential assessment center days that can last several hours.
PracHub interview research ↗What differentiates successful candidates?
Successful candidates show genuine interest in AngloGold Ashanti’s mission and demonstrate the ability to collaborate effectively in team-based exercises. Being able to connect your technical skills to the practical needs of a mine site is a major advantage.
PracHub interview research ↗Is the technical interview very difficult?
Feedback suggests that the technical portion is manageable for those with a solid foundation. The focus is often on your thought process and experience rather than just obscure algorithms.
PracHub interview research ↗What is the work culture like?
The culture is professional, helpful, and highly collaborative. You will find that interviewers and staff are generally supportive, even during the more rigorous group assessment stages.
PracHub interview research ↗Sources & methodology 3 sources ↗
No official company page is cited. Rounds and questions come from candidate reports and PracHub editorial material; each source shows the date it was read.
- 01PracHub interview research ↗
PracHub editorial research into this company and role, maintained with this guide. Candidate-reported, not an employer publication.
PracHub page · Accessed 2026-09-22 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-22 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-22
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