As a Software Engineer at Apollo Global Management, you are stepping into a role that sits at the intersection of high-stakes finance and cutting-edge technology. Apollo is an alternative asset manager, and its engineering teams build the scalable systems behind billion-dollar investment decisions, complex portfolio management, and global operations. Your work affects Apollo's ability to generate alpha, manage risk, and scale its business lines.
In this position, you will tackle complex architectural challenges, optimize heavy data pipelines, and build platforms that Apollo's quantitative analysts, portfolio managers, and operations teams rely on daily. The scale of Apollo's data and the critical nature of its trading and investment workflows make performance, reliability, and precision essential. You will not just be writing code; you will be solving intricate business problems through technical innovation.
Expect a fast-paced, intellectually demanding environment. You will collaborate with brilliant minds across engineering and finance, navigating ambiguous requirements to deliver highly optimized solutions. If you thrive on deep technical problem-solving and want to see your code drive tangible business outcomes in the global financial markets, this role will offer you unparalleled opportunities for growth and impact.
Recruiter Phone Screen
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 Assessment
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.
Technical Deep-Dive Interviews
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
Behavioral Probing
reportedWhat you say here is written down by each interviewer and compared afterwards, so the unit of evaluation is a claim someone else could check, not a well-told narrative. Two things make a story checkable: detail only a participant would hold, and a clean line around which part was yours. Vague ownership is the usual failure and it is usually accidental, because engineers say we about the team's work and we about their own, so the thing they personally built disappears into the plural. Name the part you wrote, and name who did the rest.
What to demonstrate
- Whether your details are ones a participant would hold and an observer would not: the constraint that ruled out the obvious approach, the first attempt that failed, the person who objected and on what grounds
- Whether ownership survives a direct question, since a follow-up to we decided is routinely who decided, and an answer that stays plural at that point is read as the work belonging to someone else
- Whether the numbers you quote are ones you would say identically to a former colleague with the dashboard open
How to prepare
- Go through each story replacing every we with either I or a named role (the on-call engineer, the reviewer, the other team) and check the story still holds together. Wherever it stops making sense you have found a part you cannot actually speak to
- Open the artefacts for two of your stories, the pull request, the design doc, the incident notes, and read them for dates and figures you have been rounding in the retelling. Correct your version to match
- For each story write the single sentence you would least want repeated to a former teammate, then either make it accurate or take it out
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.
Running a schema change as though the lock lasts as long as the statement
In PostgreSQL an ALTER TABLE that needs an ACCESS EXCLUSIVE lock must first wait for every open transaction touching that table, and while it waits, later queries needing a conflicting lock queue behind it rather than overtaking it. A DDL statement that would execute in milliseconds, issued while a thirty-second analytics query is open, therefore stalls all traffic on that table for thirty seconds: the outage length is set by the longest open transaction, not by the change. The defences are specific and worth knowing by name - set lock_timeout low and retry rather than queue, add columns without a volatile default so no table rewrite occurs (from version 11 a non-volatile default is a metadata-only change), build indexes with CREATE INDEX CONCURRENTLY while accepting that it cannot run inside a transaction block and leaves an invalid index behind if it fails, and add constraints as NOT VALID followed by a separate VALIDATE CONSTRAINT, which takes a weaker lock.
A cache with no invalidation story
Say how an entry goes stale, how long you can serve it stale, and what happens when many requests miss the same key at the same instant. One popular key expiring under load sends every concurrent request to the origin together; single-flight coalescing, jittered expiry, or serving stale while revalidating are the standard answers.
Answering a debugging question with a guess instead of a bisection
Give a procedure that halves the search space at each step: confirm the symptom reproduces, establish the last known-good version, input or timestamp, then bisect over commits, over the data, or over the layers of the request path. A plausible cause with no way to confirm it is the same move whether it happens to be right or wrong, which is why it scores nothing.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Write a recursive function to process a nested data structure, and exp…
Write a recursive function to process a nested data structure, and explain its space complexity.
Approach
- State the target complexity and say which constraint rules the naive version out.
- Walk one small example through your approach before writing the whole thing.
- Restate the input: its shape, its size, and what is guaranteed about it.
Follow-up
- What is the worst case, and how likely is it on real data?
- How does this change if the input no longer fits in memory?
Implement a solution to traverse a complex graph structure efficiently…
Implement a solution to traverse a complex graph structure efficiently.
Approach
- State the target complexity and say which constraint rules the naive version out.
- Restate the input: its shape, its size, and what is guaranteed about it.
- Name the brute-force solution and its complexity before improving on it.
Follow-up
- How does this change if the input no longer fits in memory?
- What is the worst case, and how likely is it on real data?
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.
Worked solution 25 min
- Size the exact structure: 300 one-second counters per tenant across 50,000 tenants, plus the running-total trick that makes a window read O(1).
- Write the top-k extraction with a size-50 min-heap and compare its complexity against sorting all 50,000 sums.
- Substitute N = 900,000 and m = 1,000 into N/(m+1) and state in requests what the sketch can and cannot distinguish.
- Write the sub-window merge for the sliding case and state the resulting bound for 30 merged summaries.
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?
Find version gaps and relay lag with window functions
outbox_event holds event_id, aggregate_type, aggregate_id, aggregate_version, event_type, payload, status ('pending','published','dead'), attempts, created_at, published_at. A projection is missing rows and you must decide whether the relay skipped events or the consumer dropped them. Write three queries over the last seven days: one listing every aggregate_id whose published aggregate_version sequence has a hole, one giving per-day counts with a running total, and one returning the newest published event per aggregate. For each, say where the window function is evaluated relative to WHERE and LIMIT. PostgreSQL 16.
Approach
- Gaps: compute lead(aggregate_version) OVER (PARTITION BY aggregate_id ORDER BY aggregate_version) in a subquery, then filter next_version <> aggregate_version + 1 in the outer query. Window functions are evaluated after WHERE, GROUP BY and HAVING and before the outer ORDER BY and LIMIT, so the predicate cannot sit in the same WHERE clause and PostgreSQL 16 has no QUALIFY.
- Say what the seven-day filter does to the answer: it truncates every partition, so the first row per aggregate has no predecessor inside the window and a hole spanning the boundary is invisible. Widen the window, or join to resource.version as the authority for the true maximum.
- Running total: SELECT date_trunc('day', created_at) AS d, count() AS n, sum(count()) OVER (ORDER BY date_trunc('day', created_at) ROWS UNBOUNDED PRECEDING). An aggregate inside a window call is legal because grouping runs before windowing. The grouping key is unique per row here so ROWS and RANGE agree, but write the frame anyway — over ungrouped rows with tied timestamps the default RANGE frame pulls in every peer row and the total jumps.
- Newest per aggregate: DISTINCT ON (aggregate_id) ... ORDER BY aggregate_id, aggregate_version DESC is the cheap PostgreSQL-only form when an index matches that order; row_number() OVER (PARTITION BY aggregate_id ORDER BY aggregate_version DESC) = 1 is the portable form and needs a subquery for the same evaluation-order reason as the gap query.
- Interpret rather than report: no gaps plus a normal p95 of published_at - created_at points at the consumer; gaps or a fat lag tail point at the relay; rows still 'pending' with attempts > 0 point at neither, because they never left the database.
- Be explicit that the partial index on (created_at, event_id) WHERE status = 'pending' does not serve any of these — they read published rows. Name the index a recurring monitor would need, and say why a query run twice a year may not deserve one.
Follow-up
- Relay failover redelivers events. Does a duplicate break the gap query, and how would you detect one from this table alone?
- Turn the gap check into a continuous monitor rather than a query someone runs after an incident. What does it watch?
- The consumer claims it never received event 4,812,006. What do you look at, in what order?
Write Python OOP code to design a scalable parking lot or financial le…
Write Python OOP code to design a scalable parking lot or financial ledger system.
Approach
- Work from the requirement backwards to the design.
- Say what you would check first and why it is the highest-information step.
- State your assumptions explicitly before working the problem.
Follow-up
- What assumption would you test first?
- How would you know your answer was wrong?
Provide pseudocode demonstrating how you would handle multithreading a…
Provide pseudocode demonstrating how you would handle multithreading and prevent race conditions in a high-throughput application.
Approach
- 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.
- Work from the requirement backwards to the design.
Follow-up
- How would you know your answer was wrong?
- What assumption would you test first?
Cache the tenant listing feed with a bounded staleness window
GET /v1/resources returns one tenant's resources ordered by updated_at DESC, 20 per page, at 14k requests/second peak against a 120 ms p99. The table carries the index (tenant_id, status, updated_at DESC, resource_id DESC) and writes land on the primary at 1.2k/second. Design the read path: the pagination contract, the cache key and value, what a write invalidates, and the staleness a user can observe. State the request rate that actually reaches the database, and the one repopulation race that deleting on write does not close.
Approach
- Settle the pagination contract first, because it decides what is cacheable. OFFSET makes the database produce and discard the skipped rows, so page 500 costs five hundred pages of work, and rows inserted between two fetches shift across the boundary and are skipped or repeated with nothing in the response to reveal it. The cursor is the row value of the last row returned: WHERE tenant_id = $1 AND status = $2 AND (updated_at, resource_id) < ($3, $4) ORDER BY updated_at DESC, resource_id DESC LIMIT 21. That is a row-value comparison, not updated_at < $3 AND resource_id < $4, which is a different and wrong predicate.
- Confirm the index actually serves it: equality on the two leading columns, then a range on the pair that follows in exactly the index's sort order, so the plan is an index scan that touches 21 entries with no sort node. Requesting 21 to return 20 is how has_more is answered without a count. resource_id is not decoration - updated_at is not unique, and without the tie-break two rows sharing a timestamp at a page boundary are the skip that keyset pagination was adopted to remove.
- Key the cache on every value the predicate reads: tenant_id, status, cursor and page size. A key that omits tenant_id is a cross-tenant disclosure, and no test running against a single tenant's data will show it.
- Be honest that a write does not invalidate one key. An update moves its row to the head of the ordering, so it invalidates the first page and every cursor page whose range spans the row's old and new position, which is not enumerable. Cache the first page per (tenant_id, status) - that is where the traffic is - with a short TTL, invalidate it on write, and serve deep cursor pages uncached from a replica, since each is already a 21-row index scan and they are rare.
- Name the residual race and the real bound. A reader that loaded rows before the write can populate the cache after the invalidation deleted the key, so the delete is not a staleness bound; the TTL is. Choose the TTL as the staleness a listing can tolerate, and jitter expiries so a busy tenant's keys do not all expire together and stampede the replica. Then do the arithmetic: at an 85% hit rate, 14k requests/second is about 2.1k database reads/second across two replicas, and that is the number capacity planning uses.
Worked solution 20 min
- Write the keyset predicate and match it column by column against the index, marking which columns are equality, which is the range, and which satisfies the ORDER BY.
- Compare rows examined for page 1 and page 500 under OFFSET and under keyset, and state both numbers.
- Write the cache key template and the first-page invalidation the write path performs.
- Write the interleaving in which a stale value is written into the cache after the invalidation, and identify what bounds it.
Follow-up
- The tenant writes and immediately lists. What does it see, and what is the smallest change that makes its own write visible without sending all 14k requests/second to the primary?
- A tenant has 4 million resources and a client walks every page nightly. What does that do to the cache hit rate, and should that traffic share this path at all?
- Sort order becomes configurable - by title, by created_at. What happens to the index set and to the cache key space?
p99 jumped on one listing filter while p50 stayed flat
After a release that added an owner_user_id filter to the resource listing, p99 rose from 90 ms to 1.9 s while p50 stayed at 40 ms. Traffic and row counts are unchanged. resource carries the index (tenant_id, status, updated_at DESC, resource_id DESC). The new query filters tenant_id and owner_user_id, orders by updated_at DESC, resource_id DESC, and takes 20 rows. On PostgreSQL, explain the shape of the regression, prove it from a query plan, and give the index you would add.
Approach
- Start from the shape. A flat p50 with a moved p99 means a subset of requests changed cost, not all of them, so the first job is naming the subset. Bucket the endpoint's latency by the tenant's row count; the natural hypothesis is that large tenants are a small share of requests and all of the tail.
- Get the plan for the new query on a large tenant with EXPLAIN (ANALYZE, BUFFERS). Expect an index scan over the tenant's range, a filter discarding most of it, then a Sort feeding the Limit, possibly reporting Sort Method: external merge Disk. Read actual rows on the scan node, not estimated.
- Explain why the existing index cannot serve it. A composite B-tree is seekable only as a left prefix, and with no equality predicate on status the scan cannot treat updated_at as an ordering, because rows in the tenant's range are ordered by status first. Everything matching must be read and sorted before LIMIT 20 can apply, so a tenant with 400,000 rows pays 400,000 rows to return 20.
- Add (tenant_id, owner_user_id, updated_at DESC, resource_id DESC). Equality on the first two columns leaves the index ordered by updated_at within that pair, so the plan becomes an index scan that stops after 20 rows with no Sort node. PostgreSQL can scan a B-tree backwards, so the DESC markers matter only if the two sort columns ever disagree in direction; keeping them explicit documents the order the keyset cursor depends on.
- Price the fix. This is a fourth index on a table taking about 1.2k writes/second, and every insert and version bump maintains it. Justify it against the query it serves, and check whether it makes an existing index redundant, which here it does not, since the original still serves the status-filtered default listing.
- Re-measure per tenant-size bucket rather than in aggregate. A fleet-wide p99 can improve while the largest tenant is still on the old plan.
Follow-up
- The endpoint paginates with OFFSET. What does page 500 cost with your index, and what does the keyset version cost?
- How would you have caught this before release, given that a 10,000-row seed database produces the same plan shape at an unnoticeable cost?
- If a fourth index were unacceptable on write grounds, what else could serve this query?
Four days sample coding, design, fundamentals and the practical rounds at deliberately shallow depth, which is enough to surface the topics you did not know were in scope. That map, rather than a guess made on day one, decides where the last three days go.
Prepare, practise & reflect
One practical outcome each day. Spend longer where you need it.
0 / 7 done01Coding, one pass at shallow depth
- Solve one problem from each of six families, an array with two pointers, hash counting, binary search, a tree traversal, a graph traversal and one dynamic program, under a hard twenty-minute cap with no extensions, marking each finished, late, or stalled.
- For every stall, write the exact move you could not make rather than the subject, so the note reads could not turn the recurrence into a loop rather than bad at dynamic programming.
- Fix nothing today. The value of the pass is the unfixed record.
Deliverable: Six timed attempts marked finished, late or stalled, each stall carrying a named blocking move.
Practice prompt ↗Practice prompt ↗Worked solution ↗02Design, one pass at shallow depth
- Spend twenty minutes each on three different shapes, a read-heavy feed, a write-heavy ingest path, and something needing a transaction across two entities, stopping each at requirements, interface and data model.
- After each, write the first question you could not answer, which is usually a number you could not estimate or a failure mode you had no vocabulary for.
- Mark which of the three you would be most relieved not to be asked, and treat that as data rather than as a preference.
Deliverable: Three shallow designs, each with the first unanswerable question written at the bottom.
Practice prompt ↗Practice prompt ↗03Fundamentals and the practical rounds
- Answer eight short questions in writing at four minutes each, covering the material that fills the gaps between the big rounds: what happens between a URL and a rendered page, what an index costs on write, when a process is preferable to a thread, and what conditions a deadlock requires.
- Do one thirty-minute practical task of the kind a take-home compresses: read an unfamiliar two-hundred-line file and write what it does, what you would change, and the one thing you remain unsure of.
- Score every answer fluent, correct but slow, or absent, and keep the absent ones visible.
Deliverable: Eight scored short answers and one written reading of unfamiliar code.
Practice prompt ↗Practice prompt ↗04The rounds that are about you, and the map
- Deliver three behavioural answers aloud against a timer, a conflict, a failure you owned, and a decision made without enough information, marking any that ran past three minutes or contained no number.
- Assemble the map: every marked item from days one to three on a single page, sorted by how likely it is to appear in your loop rather than by how uncomfortable it felt.
- Choose exactly two areas for the remaining three days and write down what you are deliberately abandoning.
Deliverable: A one-page scored map of the whole surface area with two areas chosen and the rest explicitly abandoned.
Practice prompt ↗Practice prompt ↗Worked solution ↗05First chosen area, to the depth you skipped
- Work the higher-ranked area in four focused blocks, choosing items one level above where you stalled rather than repeating what already works.
- After each block write the rule you extracted in one sentence with its precondition attached, since a rule carrying no precondition is exactly what fails under a variation.
- Re-attempt the day-one or day-two item that exposed this area and compare against the original timing.
Deliverable: Four worked blocks, a timed re-attempt against the original, and three one-sentence rules with preconditions.
Practice prompt ↗Practice prompt ↗06Second chosen area, where the gap is coverage rather than speed
- Treat the second area differently from the first. Day five drilled something you could already half-do; this one is usually a topic you had simply never met, so build one worked reference example end to end and keep it, rather than attempting six problems badly.
- Write down the vocabulary you were missing on day two or three, five terms at most, each with the one sentence that makes it usable in an answer rather than the textbook definition.
- Redo the shallow attempt that exposed this area and note whether you now fail later in the problem, because moving the failure point is the realistic gain from a single day and is worth more than a score that did not change.
Deliverable: One worked reference example for the newly covered area, a five-term vocabulary list, and a note on where the failure point moved.
Practice prompt ↗07Reassemble the loop
- Sit two rounds back to back with no gap, ordering them so the area you chose second comes last, because the map was built from rested, isolated attempts and the loop will reach your weaker area when you are already spent.
- Write where the second round suffered from the first, which is normally the point at which structure collapses into narration.
- Reduce the week to one page holding only the rules you can state without reading them.
Deliverable: Mock notes on cross-round carryover plus a one-page card of rules you can recite from memory.
Practice prompt ↗Worked solution ↗Expand any day for tasks and deliverables. Your progress is saved on this device.
For anything that touched live traffic, be ready to say how you would have undone it: a flag, a staged rollout, dual writes with the old path still authoritative. Once the old column is dropped or the source rows are overwritten there is no reverse, so name what you kept a copy of and for how long.
Tell me about a time you spent significant hours on a project, only to…
Tell me about a time you spent significant hours on a project, only to have the requirements change drastically. How did you handle it?
Approach
- 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.
- Name the disagreement and how you resolved it with evidence.
Follow-up
- How did you know your change caused the improvement?
- What would you do differently if you ran that again?
How do you handle highly ambiguous technical questions when you don't …
How do you handle highly ambiguous technical questions when you don't have all the necessary information?
Approach
- State the situation in two sentences and spend the rest on the reasoning.
- Name the disagreement and how you resolved it with evidence.
- Pick a story where you made the decision, not one where you watched it.
Follow-up
- What would you do differently if you ran that again?
- How did you know your change caused the improvement?
Describe a situation where you had to push back against a senior stake…
Describe a situation where you had to push back against a senior stakeholder or manager regarding a technical decision.
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.
- Close with what you would do differently, concretely.
Follow-up
- What did you decide not to do, and why?
- What would you do differently if you ran that again?
- 01
Tell me about a time you spent significant hours on a project, only to have the requirements change drastically. How did you handle it?
- 02
How do you handle highly ambiguous technical questions when you don't have all the necessary information?
- 03
Describe a situation where you had to push back against a senior stakeholder or manager regarding a technical decision.
Is this an official Apollo Global Management interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Apollo Global Management. Rounds and questions reflect what candidates have reported, not a process Apollo Global Management has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How difficult are the initial technical assessments?
The initial HackerRank or take-home assessments are highly rigorous. You should expect LeetCode Medium to Hard problems, complex SQL queries requiring specific syntax (like unions), and OOP challenges. Time management is critical, as brute-force solutions will often fail the hidden test cases.
PracHub interview research ↗What is the typical timeline for the interview process?
The timeline can vary significantly. Some candidates complete the process in a few weeks, while others experience longer wait times between rounds. Because hiring decisions involve senior leadership, periods of silence can occur while teams deliberate. Patience and polite follow-ups are recommended.
PracHub interview research ↗Are take-home projects common, and how much time should I spend on them?
Yes, depending on the specific team, you may be asked to complete a take-home project. These can be intensive, sometimes requiring several hours of focused work over a weekend. Treat these projects as a demonstration of your best production-quality code.
PracHub interview research ↗Who will I be interviewing with during the onsite or Superday?
You will typically meet with a mix of peers, engineering managers, and senior leadership. It is not uncommon to have a technical round with a CTO or VP. Prepare to adjust your communication style depending on whether you are speaking with a fellow developer or a high-level executive.
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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