As a Software Engineer at Uber Freight, you are not just writing code; you are building the digital infrastructure that powers the global supply chain. You will work on complex, high-scale systems that connect shippers and carriers, optimizing the flow of goods across the country. Your work directly impacts the efficiency of logistics, helping to solve real-world problems like capacity matching, pricing algorithms, and route optimization.
This role requires a blend of technical rigor and a product-focused mindset. You will often find yourself navigating high-stakes environments where reliability and performance are paramount. Whether you are scaling microservices or designing fault-tolerant systems, you are contributing to a platform that is transforming the transportation industry through data-driven innovation and engineering excellence.
Recruiter Screening
reportedHalf of this call is the part candidates treat as small talk: start date, notice period, work authorisation and its timing, location and time zone, on-call, and the number. Those are what kill offers late, after several engineers have each spent a day. Surfacing a hard constraint now costs you nothing and occasionally buys you something, since a loop compressed to fit a competing deadline can usually only be arranged if it is asked for early. The common failure is deflecting the compensation question twice, then discovering at offer stage that the band never reached your number.
What to demonstrate
- Whether your hard constraints are compatible with the role before a loop gets booked: earliest start, notice period, what authorisation you hold and when it needs action, days on site, willingness to carry a pager
- Whether you give a compensation range with something behind it, such as current total compensation or a competing timeline, rather than leaving the band untested
- Whether your stated timeline is real, since a competing deadline raised now is something scheduling can sometimes work around and the same deadline raised at offer stage usually is not
How to prepare
- Write each constraint down in one line before the call and state them as facts rather than negotiating them live under a question you were not expecting
- Set your range from two or three current data points for that level and location, and name the structure you are quoting in, so the number is comparable to the one they are holding
- If another process is running, say where it stands and by when, and ask directly whether this loop can be scheduled inside that window
Technical Assessment
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
Full-Day Loop
reportedA day like this is several different games in a row, and the expensive mistake is carrying the previous one into the next room. Coding rewards narrow precision and finishing inside a timer. Design rewards breadth, stated assumptions and naming what you are deliberately not building. Behavioural rewards specificity about people and decisions. Candidates who over-engineer a coding problem they were supposed to finish, or who start sketching class hierarchies before anyone has agreed what the system has to do, are usually still playing the last round. Between rooms, name out loud which game the next one is.
What to demonstrate
- Whether the coding round ends with something that runs and has been traced against a degenerate input, rather than an extensible design that was never finished
- Whether a design discussion opens by agreeing on traffic shape, read-to-write ratio and what is allowed to be stale, instead of proceeding from an architecture you arrived with
- Whether a behavioural answer names a person, a disagreement and what you did about it, rather than describing the system the story happened inside
- Whether the opening habits still appear late in the day: restating the problem, asking for constraints, saying the plan before typing
How to prepare
- Book three mocks of different types back to back on one afternoon and ask each interviewer afterwards which round you answered in the wrong mode
- Write a three-line opening script per round type — coding: restate, name the approach and its cost, then type; design: ask for scale, read-write mix and what must not break; behavioural: name the person, the stakes and the decision — and run it off a card so the switch is mechanical rather than remembered
- Practise coding with a timer you do not extend, stopping when it stops, so the trained reflex is to finish a correct solution rather than to keep improving one
2 candidate reports. Individual accounts describe a particular role and hiring cycle.
Uber Freight Software Engineer Interview Experience — DSA, a File System LLD, and a URL Shortener Design
Round 1 — DSA There is a car and N riders. Each rider has two constraints: L[i] — the minimum number of other riders this rider is willing to share the ride with. H[i] — the maximum number of other riders this rider is willing to share the ride with. If k riders are selected, every selected rider must satisfy L[i] <= k - 1 <= H[i]. The goal is to maximize the number of riders who can ride togethe…
Read full experienceUber Freight Software Engineer Interview Experience — Real-Time Restaurant Analytics Dashboard System Design Round
System Design Design a real-time Restaurant Analytics Dashboard System requirements: show order volume, top dishes, and revenue for the past 1h, 1d, and 7d. Data must update in real time, and throughput needs to support 100K orders/sec. Primary store schema design: determine the primary storage's tables, partition key, and columns. Cache key design: determine the Redis key pattern and the corresp…
Read full experiencePracHub editorial advice for the preparation topics above.
Floating-point quantities and implicit unit-of-measure conversions
Binary floating point cannot represent most decimal fractions exactly, so a chain of pallet-to-case-to-each conversions accumulates a residue that a later rounding turns into a unit that was created or destroyed, breaking conservation with no concurrency involved at all. The bug is slow and non-local: it surfaces as a cycle-count variance weeks later, at a node nobody changed, and it is untraceable because the arithmetic looked correct at every individual step. The fix is to hold quantities as integers in the smallest transacting unit, carry the UoM explicitly on every row that carries a number, store conversion factors as integers that are versioned in time, and reject a conversion that does not divide exactly instead of rounding it. The same discipline applies to money on the freight side, where an accessorial in floating-point dollars produces invoice reconciliation differences of a cent that cost more to investigate than the freight.
Judging delivery windows and cut-offs in UTC
A delivery appointment, a receiving cut-off and a same-day promise are all statements about local wall-clock time at a specific stop, so evaluating them in UTC misclassifies on-time performance for any leg that crosses a zone, and misses cut-offs by an hour on the two days a year a region changes offset. The subtlety that catches experienced engineers is that converting a future local time to a UTC instant at write time bakes in the offset rules as they were known then: when a jurisdiction changes its rules, an appointment stored that way silently moves, which is why a future window is better stored as local wall time plus an IANA zone identifier and resolved at read time, while a past event is correctly stored as an instant. Duration arithmetic needs the same care, because a transit that spans a DST boundary is genuinely 23 or 25 hours long in local terms even though the elapsed instant count is unchanged. Storing only a date with no zone - common on partner feeds - is the version of this that cannot be fixed downstream, so the zone has to be captured at ingestion.
Writing code before the input contract is pinned down
Before the first line, state the types, the size bounds, whether duplicates, negatives or an empty input are possible, whether the input is sorted, whether you may mutate it, and what the function returns when nothing matches. Every one of those answers changes the code, and discovering one at minute twenty costs a rewrite you no longer have time for.
Reading the constraints as preamble rather than as part of the problem
The bounds are usually there to eliminate the obvious approach: n up to 10^5 makes an O(n^2) scan roughly 10^10 operations, far outside any per-test time budget, and an input larger than memory rules out loading it at all. When a bound is not given, ask for it, then say out loud which approach it kills.
Choose a category, try a prompt, then open its approach, worked solution or follow-up when you need it.
Implement a function to handle load balancing across multiple server n…
Implement a function to handle load balancing across multiple server nodes.
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.
- Walk one small example through your approach before writing the whole thing.
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?
Write a solution for a concurrency problem involving shared resources.
Write a solution for a concurrency problem involving shared resources.
Approach
- Name the brute-force solution and its complexity before improving on it.
- 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.
Follow-up
- Which test case would catch an off-by-one here?
- How does this change if the input no longer fits in memory?
Describe the time complexity of your approach and suggest potential op…
Describe the time complexity of your approach and suggest potential optimizations.
Approach
- Name the brute-force solution and its complexity before improving on it.
- Choose the data structure from the access pattern, not from familiarity.
- 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?
- Which test case would catch an off-by-one here?
Given a list of shipments, how would you group them to minimize transi…
Given a list of shipments, how would you group them to minimize transit time?
Approach
- Name the brute-force solution and its complexity before improving on it.
- 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.
Follow-up
- What is the worst case, and how likely is it on real data?
- Which test case would catch an off-by-one here?
Fold late out-of-order status events into a monotonic leg state
Leg statuses form the chain planned < tendered < accepted < picked_up < in_transit < delivered, with exception and cancelled outside it. You ingest 300 million observation_event rows a day carrying subject_id (leg), observed_status, occurred_at, received_at, clock_offset_ms and dedupe_key; five percent arrive over an hour late and a tail arrives a day late. Write the merge that produces each leg's current status, prove the result depends only on the set of events and not their order, and state the per-event cost.
Approach
- Define the per-leg state as a tuple ordered totally:
(status_rank, occurred_at, source_priority, event_id), and definemerge(a, b)as the maximum under that order. Rank leads;occurred_atand the rest only break ties. The proof is then one line: maximum over a total order is idempotent, commutative and associative, so the state space withmergeis a join-semilattice and the fold over any permutation of the event set yields the same value. That is also why no global sort is needed — the O(n log n) disappears and the fold is O(1) per event, O(distinct legs) state. - Rank before time, deliberately.
occurred_atis device-reported and sometimes wrong; if it led the comparison, the leg's status would be handed to whichever device has the most badly skewed clock. Ranking first means a bad clock can only reorder events within one rank. Correctoccurred_atbyclock_offset_mswhere it was measured, and where the offset exceeds a threshold keep the observation but bar it from winning a tie — the event is still evidence, it just stops being authority. - Make the lattice total, because
exceptionandcancelledare not points on the chain andmaxover a partial order is undefined for incomparable pairs. Modelcancelledas an absorbing top element, and carryexceptionas a boolean flag merged by OR alongside the rank rather than as a rank of its own — otherwise an exception at rank 4 either blocksdeliveredat rank 6 or is erased by it, and both are wrong. A product of semilattices is a semilattice, so the order-insensitivity argument still holds over the tuple. - Deduplicate on
dedupe_keyeven though the merge is idempotent. Idempotence covers a byte-identical redelivery; a partner replaying a window with a freshevent_idis not identical, and the tie-break would let the replay displace the original for no reason. Dedupe on the stable key, then fold — both halves are needed, and conflating them is the usual defect. - Partition the stream by
subject_idso all events for one leg land on one partition: broker ordering guarantees are per-partition only, and while a commutative merge makes interleaving harmless, a concurrent read-modify-write on the same leg row is still a lost update. Where that cannot be guaranteed, make the write a compare-and-set on a version column, or a conditional update that only raises the state. Because the state is mergeable, shard maps combine cleanly and replaying the whole day converges to the same answer — which is what makes reprocessing safe. - Say where genuine regressions live: a delivered shipment that is refused and returns is a new leg, not a rank decrease. Monotonicity is only sound if real reversals are modelled as new objects, and keeping every superseded observation is what lets you answer the dispute later, so the fold discards nothing — it marks losers
superseded.
Worked solution 40 min
- Write the rank table and the tuple comparison explicitly, including where
cancelledand theexceptionflag sit, before any pipeline code. - Implement
merge(a, b)as a pure function and test the three algebraic laws directly:merge(a, a) == a,merge(a, b) == merge(b, a), andmerge(merge(a, b), c) == merge(a, merge(b, c)). - Build a 200-event batch across five legs including an out-of-order
picked_upafterdelivered, a duplicateddedupe_key, onecancelled, oneexception, and one event with a three-hour clock offset. - Fold the batch, then fold 50 random permutations of it and assert every run produces identical per-leg state.
- Re-feed the entire batch a second time and assert the state is unchanged, which is idempotence end to end rather than only in the merge function.
Follow-up
- An event arrives a week late for a leg already closed and invoiced. Does it change the status, the on-time metric, both, or neither?
- Two events for one leg have the same rank and the same
occurred_atfrom different sources. What decides, and is that decision stable across a replay? - How would you test order-insensitivity rather than assert it?
Two scan-event queries that stopped using their indexes
observation_event is range-partitioned monthly on occurred_at, with a btree on (subject_type, subject_id, occurred_at), a BRIN on occurred_at and a GiST on geo. Rows are inserted in received_at order and events are routinely hours late. Two queries regressed: (A) WHERE occurred_at::date = current_date AND observed_status = 'delivered'; (B) a backfill scanning WHERE received_at >= now() - interval '1 day'. For each, say why no index is used and no partition is pruned, and give the fix.
Approach
- Query A wraps the column in a cast, so the predicate is over occurred_at::date and not over occurred_at. A btree or BRIN on the bare column cannot be probed, and the planner has no range to prune partitions with, so it plans every monthly partition. Fix it by rewriting to a half-open range: occurred_at >= :start AND occurred_at < :start + interval '1 day', with :start computed by the caller from the relevant IANA zone.
- Do not reach for an expression index as the first fix here. The timestamptz-to-date cast is STABLE rather than IMMUTABLE because it depends on the session TimeZone, so PostgreSQL refuses to index it; indexing ((occurred_at AT TIME ZONE 'UTC')::date) is accepted but silently pins the day boundary to UTC, which is the wrong day for any stop outside it.
- Query B filters a column that is neither indexed nor the partition key, so it is a full scan of every partition by construction. The subtler half is why the BRIN cannot rescue it: BRIN stores a min and max per block range and only pays when physical order correlates with the column. Rows arrive in received_at order while the indexed column is occurred_at, and late events interleave hours-old timestamps with current ones, so each range's min and max span days and almost every range qualifies.
- Measure the correlation rather than assuming it: SELECT attname, correlation FROM pg_stats WHERE tablename = 'observation_event'. Near 1 means BRIN earns its keep, near 0 means it is a full scan with extra steps. Expect received_at near 1 and occurred_at well below it on this insert pattern.
- Fix B with a btree on received_at, which stays compact because it is correlated with insert order, or better, drive the backfill from a monotonic cursor it already has such as event_id, which needs no new index on a table taking tens of thousands of inserts per second.
- Confirm with EXPLAIN (ANALYZE, BUFFERS) and read the plan, not the clock: check how many partitions appear, whether rows removed by filter dwarfs rows returned, and whether shared buffers read matches the size of one partition or of the table.
Worked solution 25 min
- Load a few million rows across three monthly partitions with occurred_at deliberately shuffled against insert order.
- EXPLAIN (ANALYZE, BUFFERS) query A as written and record the partitions scanned and the buffers read.
- Rewrite A as a half-open range and re-run, comparing partitions scanned and buffers.
- Query pg_stats for the correlation of occurred_at and received_at, then add a btree on received_at and re-plan query B.
Follow-up
- Where would you put the partition boundary given events arrive up to a week late, and what happens to a row whose occurred_at falls in a partition that has already been detached?
- The delivered-today query is asked per node in local time. Does that change your partitioning, your index, or only the caller?
- Adding a btree to this table costs write amplification. How would you decide whether the backfill is worth it?
Add a required unit column and uniqueness to a live ledger
inventory_movement holds 4.2 billion rows on PostgreSQL 15 and takes about 2,000 inserts per second around the clock. Two changes are required with no downtime: add uom_code TEXT NOT NULL, whose truthful value must come from the conversion factors in force at each row's occurred_at rather than from today's item master; and make idempotency_key UNIQUE, although duplicates predating the retry fix already exist. Give the ordered steps, the lock each one takes, its expected duration and its rollback.
Approach
- ADD COLUMN with a non-volatile DEFAULT has been metadata-only since PostgreSQL 11, so no rewrite, but it still takes ACCESS EXCLUSIVE for an instant. That lock queues behind any long-running reader while blocking every statement behind it, so one 10-minute analytics scan turns a millisecond change into a 10-minute outage. Set lock_timeout to 2s and retry in a loop rather than waiting.
- Reach NOT NULL without a scan under an exclusive lock: ADD CONSTRAINT ... CHECK (uom_code IS NOT NULL) NOT VALID, which is instant, then VALIDATE CONSTRAINT, which takes only SHARE UPDATE EXCLUSIVE and runs concurrently with inserts, then SET NOT NULL, which PostgreSQL 12 and later proves from the validated check and so skips its own scan. Drop the redundant check afterwards.
- The truthful backfill is the long pole and must read the versioned conversion factor as of each row's occurred_at, not the current item master, or a supplier moving a case from 12 units to 10 retroactively reinterprets last quarter. Batch by primary-key range, a few thousand rows per transaction, with a sleep between batches and a resumable cursor in a control table. At 5,000 rows a second, 4.2 billion rows is roughly ten days of throttled work: that is a schedule, not a failure, and it must not block the deploy.
- The UNIQUE index cannot be built until the duplicates are gone. CREATE UNIQUE INDEX CONCURRENTLY aborts on the first violation and leaves an INVALID index behind that is ignored by queries but still carries update overhead on every insert, and it can only be removed with DROP INDEX CONCURRENTLY. So: find the duplicates, resolve each one as a compensating reversal rather than a DELETE, since every duplicate key is a double-applied movement and deleting it fixes the constraint while leaving the position wrong.
- Then CREATE UNIQUE INDEX CONCURRENTLY, which makes two passes over the table and waits out concurrent transactions, so budget hours here, and note it cannot run inside a transaction block: a migration framework that wraps every step in BEGIN will fail on this one. Promote it with ALTER TABLE ... ADD CONSTRAINT ... UNIQUE USING INDEX, which is a brief ACCESS EXCLUSIVE.
- Question whether the constraint should cover all history. At roughly 40 bytes per key plus overhead, a btree over 4.2 billion rows is in the hundreds of gigabytes, while a retry never arrives a year later. A partial unique index bounded by a fixed occurred_at literal, rotated forward on a schedule, gives the same protection for a fraction of the size, and the predicate must be a literal because now() is not immutable.
Follow-up
- Sequence the application deploy against these steps. When exactly may the code start writing uom_code, and when may it start requiring it?
- The backfill is on day six and the conversion-factor table is itself corrected. What happens to the rows already written?
- How do you roll back after SET NOT NULL is in place and half the traffic depends on the column?
What strategies would you use to ensure high availability in the event…
What strategies would you use to ensure high availability in the event of a regional service outage?
Approach
- Fix the scope first: who calls this, how often, and what they do when it fails.
- Name the read and write paths separately; they rarely have the same bottleneck.
- Choose a partition key and say what query it makes expensive.
Follow-up
- How does this behave when that dependency is down for an hour?
- What breaks first when traffic grows ten times?
Design a real-time tracking system for thousands of active trucks.
Design a real-time tracking system for thousands of active trucks.
Approach
- Name the read and write paths separately; they rarely have the same bottleneck.
- Name the failure you are designing for, then the recovery path.
- Choose a partition key and say what query it makes expensive.
Follow-up
- What breaks first when traffic grows ten times?
- How does this behave when that dependency is down for an hour?
Paginate an append-only movement ledger under continuous writes
A partner's nightly reconciliation job pulls every inventory_movement row for its nodes since its last run, while the table is being appended to throughout. The columns available are movement_id BIGINT (sequence-assigned), node_id, occurred_at (device-reported) and recorded_at (insert time). Specify the cursor format, the ordering key, the page-size policy and the delivery guarantee you promise across job restarts. State explicitly what happens to a row inserted today whose occurred_at is eighteen hours in the past, and how the client resumes after a crash mid-page.
Approach
- Use a keyset cursor, not OFFSET: WHERE node_id = ANY($1) AND movement_id > $2 ORDER BY movement_id LIMIT n, served by an index on (node_id, movement_id). Each page costs O(log n + k); OFFSET re-walks the skipped prefix at O(offset + k), so page 900 of a nightly pull is the page that times out.
- Order by insert order, never by occurred_at. A handheld that reconnects after a partition inserts movements stamped hours earlier; a cursor advanced past that instant will never return them. occurred_at is a filter and a display field, recorded_at and movement_id are the feed order - conflating the two is how a reconciliation silently misses a day of picks.
- Name the sequence-gap hazard and close it. The sequence hands out movement_id at INSERT, but transactions do not commit in id order, so a reader that takes max(id) from a page can step over a row that commits afterwards with a lower id. Either hold the cursor behind a watermark older than the longest writer transaction (bounded by statement_timeout plus idle_in_transaction_session_timeout) or read a commit-ordered source such as logical decoding or a transactional outbox, and say which you chose.
- Make the cursor opaque - encode the ordering key plus a schema tag - so you can change the key later without a coordinated client release. Return next_cursor and has_more; do not return a total count, which is a second full scan and is stale before it is serialised.
- State the guarantee honestly: at-least-once across restarts, because a client that crashes after processing a page and before persisting the cursor will re-request it. Delivery is idempotent on the partner side through movement_id, and exactly-once over a network is not on offer.
Worked solution 20 min
- Write the page query and the index it requires, then state the cost per page in terms of index height and page size.
- Trace a late insert: a device buffers at 09:00, reconnects at 15:00, the row lands with occurred_at 09:00 and a current movement_id. Confirm it is delivered under your ordering key.
- Trace two interleaved writers to show the sequence-gap skip concretely: txn A takes id 100 and commits late, txn B takes 101 and commits first, a reader pages past 101.
- Pick the watermark lag or the commit-ordered feed, and write the number or the mechanism down rather than leaving it as an assumption.
- Write the resume test: page the table while a writer inserts continuously, kill the client mid-page, resume from the persisted cursor, and assert no row is missing.
Follow-up
- The partner asks for 10,000-row pages and still times out. What do you change before raising the limit?
- They ask for a changes-since-timestamp parameter instead of a cursor. What breaks, and what do you offer instead?
- They need to re-read a three-day window to settle a dispute. How does that coexist with the cursor contract?
Duplicate bookings and second trailers after a partner slowdown
A partner's booking API ran at 45 s p99 for ninety minutes. Afterwards 214 shipment_leg rows carry two booking references, 41 stops received two trailers, and finance holds duplicate invoices. booking_idempotency_key is generated and persisted on the leg before the call and is sent to the partner, whose documented dedupe window is 60 seconds. The HTTP client retries timeouts three times, and a sweeper re-tenders any leg still in status 'planned' after ten minutes. Explain how duplicates happened despite the key, and give the corrected flow.
Approach
- Establish that the duplicates are timeouts rather than double-clicks. Join the gateway request log to shipment_leg for the 214 legs and check whether each duplicate pair carries the same idempotency key and how far apart the calls landed. Same key, calls minutes apart, first attempt ending in a client timeout is the ambiguous-outcome signature; different keys would mean the key is being regenerated and is a different bug.
- Name why the key did not help. The key was correct and persisted, but deduplication was delegated to the partner, whose window is 60 seconds. A retry at 45 s may land inside it; the third retry does not, and the ten-minute sweeper re-tender certainly does not, so the partner sees a new request and books a second truck. A short partner window plus a long retry horizon is an at-least-once system wearing an at-most-once label.
- Identify the missing state. The leg goes planned to tendered only on a 2xx, so a timeout leaves it exactly as if nothing was sent, which is false: the partner may have committed before the connection dropped. The state machine needs an explicit in-doubt status between requested and confirmed, and nothing may issue a new tender for a leg in that state.
- Replace re-issuing with asking. On any ambiguous outcome, move the leg to in-doubt and let a sweeper resolve it by querying the partner for that key or for bookings against our shipment reference, and only tender again when the query definitively returns nothing. Where the partner offers no query, the manifest or the status feed is the authority and resolution is slower, so the in-doubt state must be allowed to persist rather than being timed out into a retry.
- Move retry ownership out of the process. A retry decorator around an HTTP client disappears on deploy or crash, so the persisted state is the only thing that can drive recovery: a sweeper reading legs in requested or in-doubt, with bounded attempts, backoff and an alert on in-doubt depth. The call site should not retry an irreversible effect at all.
- Deal with the 214 that already exist, since code cannot undo them. Reconcile against the partner's manifest, cancel the duplicate where the cancellation window still allows it, and for legs already executed record the second cost as a compensating entry and dispute the invoice. State plainly that this is a business process with a cost, which is exactly why the guard belongs before the call.
Follow-up
- The partner offers no lookup by your key and no search by reference. What is your resolution path and how long may a leg stay in doubt?
- Which other effects in this system have the same shape and the same missing state?
- How do you test this without a degraded partner, and what does the fault injection have to simulate that a plain timeout does not?
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 ↗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.
Keep one story where the bad call was yours rather than a dependency's or a manager's. Name the check that would have caught it, whether you added that check afterwards, and whether it has fired since. Answers that route blame outward end the conversation early; answers that end in a guardrail someone still relies on tend to open it up.
Tell me about a time you had to resolve a significant technical disagr…
Tell me about a time you had to resolve a significant technical disagreement with a stakeholder.
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.
- Close with what you would do differently, concretely.
Follow-up
- What would you do differently if you ran that again?
- What did you decide not to do, and why?
How do you balance the need for rapid feature delivery with the necess…
How do you balance the need for rapid feature delivery with the necessity of reducing technical debt?
Approach
- Give the blast radius: what could have broken, and what you measured.
- 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?
Reverse an allocation design after peak contention
You chose optimistic concurrency for allocation: read the position, then UPDATE inventory_position SET qty = qty - :n, version = version + 1 WHERE version = :v, retrying on an affected-row count of zero. It was correct and fast in load tests with spread keys. At peak, a few hundred hot item-node pairs absorbed most write traffic, retries amplified, and allocation p99 went past the checkout budget. Describe a decision you reversed under production evidence: what you originally reasoned, the measurement that forced the change, what you replaced it with, and what you would have measured before committing.
Approach
- Say why the original choice was reasonable, because a reversal story is only useful if the first decision was defensible. Compare-and-set avoids holding a lock across the read, has no deadlock surface, and is uncontended on the long tail of keys, which is most keys most of the time.
- Name the mechanism of the failure rather than calling it contention. On PostgreSQL under READ COMMITTED, a conflicting UPDATE does not fail fast: it blocks on the row lock until the other transaction commits, then re-evaluates its predicate against the new row version and reports zero rows affected. Each loser therefore pays a full lock wait before learning it must retry, so with k writers queued on one pair the work is quadratic in k across the burst, and the retry loop adds round trips rather than avoiding waits.
- Bring the measurement that settled it, not the anecdote: attempts per successful allocation on the hottest pairs, the distribution of write traffic across item-node keys, and the p99 contribution of lock wait time separated from query time. Load tests with spread keys cannot show any of this, which is the real lesson and the thing you would run differently.
- State the replacement and its cost. Serialising each hot key behind a single writer with a bounded queue converts an unbounded retry storm into a bounded wait plus explicit shedding, at the cost of a new component, a routing decision and a failure mode when the writer for a key is unavailable. SELECT ... FOR UPDATE is the smaller change and trades the retry loop for an in-database queue that still consumes a connection per waiter.
- Describe the migration, since reversing a write path in production is where these stories become concrete: route only the measured hot keys first, keep both paths live behind a per-key decision, and verify with the same attempts-per-success metric before widening.
- Close on what you would have measured before committing, and be specific: key skew from production traffic, not from a synthetic generator, is the input the original decision was missing.
Follow-up
- Under REPEATABLE READ on PostgreSQL that same conflict raises a serialization failure instead. What changes in your retry code and your error budget?
- One key becomes so hot that even the single writer saturates. How do you shed load without overselling?
- An allocation must span two positions atomically. What breaks in your replacement design if those rows live on different shards?
- 01
Tell me about a time you had to resolve a significant technical disagreement with a stakeholder.
- 02
How do you balance the need for rapid feature delivery with the necessity of reducing technical debt?
- 03
You chose optimistic concurrency for allocation: read the position, then UPDATE inventory_position SET qty = qty - :n, version = version + 1 WHERE version = :v, retrying on an affected-row count of zero. It was correct and fast in load tests with spread keys. At peak, a few hundred hot item-node pairs absorbed most write traffic, retries amplified, and allocation p99 went past the checkout budget. Describe a decision you reversed under production evidence: what you originally reasoned, the measurement that forced the change, what you replaced it with, and what you would have measured before committing.
Is this an official Uber Freight interview guide?
No. It is PracHub's own research and practice material for the Software Engineer role at Uber Freight. Rounds and questions reflect what candidates have reported, not a process Uber Freight has published, and they change over time. Confirm the current format and scope with your recruiter.
PracHub interview research ↗How difficult are the interviews?
The interviews are considered challenging and are designed to test your depth in both coding and system design. Expect a high bar for both technical accuracy and architectural clarity.
PracHub interview research ↗What is the most effective way to prepare for the coding rounds?
Focus on solving problems that require you to think about edge cases and efficiency. Don't just practice "easy" problems; focus on medium-to-hard problems that mimic real-world engineering constraints.
PracHub interview research ↗How much time should I spend on system design?
Dedicate significant time to understanding distributed system trade-offs. Being able to explain *why* you chose a specific database or caching strategy is more important than the specific tool itself.
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 2026-09-24 - 02PracHub Software Engineer practice ↗
Cross-company practice questions for this role.
PracHub page · Accessed 2026-09-24 - 03PracHub interview preparation framework ↗
The framework the preparation plan follows.
PracHub page · Accessed 2026-09-24