Assessment tips
- Confirm the current Meesho pattern from the drive mail before practising old papers
- Finish one problem or section fully before starting the next
- Revise the fundamentals listed in the round table on this page
How Meesho fresher SDE interviews usually run, campus and off-campus, synthesized from GeeksforGeeks and Medium write-ups. We reorganized long OAs, machine-coding rounds, and final hybrid filters - treat this as a prep map, not a script.
Meesho placement process includes: 1. Application (on-campus via college placement cell, off-campus via Meesho Careers portal, or through referrals), 2. Online Assessment (OA) - 2-3 coding problems (often 90-165 minutes), 3. Technical Interviews - DSA, projects, fundamentals, 4. Machine Coding / LLD - runnable class design, 5. Managerial / HR hybrid - behavioral plus OS/networks/SQL again. Total duration: 2-3 weeks from application to offer letter.
Eligibility (from student reports): 7.0+ CGPA in 10th, 12th, and graduation.
Fresher package ballpark (from student reports): SDE-1: ₹25-35 LPA. Confirm on your offer letter.
| Round | Duration | What they test |
|---|---|---|
| Online Assessment | 90–165 min | 2–3 coding problems + sometimes OS/network MCQs |
| Technical Round 1 | 30–90 min | DSA (graphs, DP, greedy) + project discussion |
| Machine Coding / LLD | 60–90 min | Runnable class design (carpool, inventory, category orders) |
| Technical Round 2 / HM | 45–60 min | SQL schema, CS fundamentals rapid-fire, system design |
| HR / final hybrid | 30–45 min | Behavioural + OS/networks/SQL again + culture fit |
These are rewritten from public student write-ups. Two things recur across almost every Meesho account, so plan around them: the coding round is long and genuinely hard (not a 2-easy-questions screen), and the round most people underestimate is a machine-coding / low-level-design round, not a distributed-systems whiteboard.
Source: Meesho Interview Experience for SDE-1, On-Campus (Virtual) - GeeksforGeeks
Round 1 - coding + MCQs
Round 2 - technical (~1.5 hours)
Round 3 - technical 2 (~1 hour)
Round 4 - HR
Result: Selected.
Takeaways
Source: Meesho SDE-1 Open Campus Interview Experience 2025 - GeeksforGeeks
Result: Not selected, despite clearing every prior round. Worth knowing: reaching HR at Meesho is not the same as an offer.
Takeaways
Source: Meesho interview Experience (SDE-1 Backend) - Medium
Round 1 - OA (120 min, HackerRank)
Round 2 - Machine coding (90 min, live)
Round 3 - Technical + hiring manager (60 min)
Result: Selected (author joined Meesho as SDE).
Takeaway: Backend track may allocate longer machine coding than the 60-minute carpool reports from open campus - prepare both 60- and 90-minute builds.
Source: Meesho SDE-I Backend full experience
OA highlights
Machine / design prompts reported
IngestOrder(date, qty, category) and range queries on order counts.Result: Offer after multi-round funnel.
Takeaway: Meesho backend OAs are not uniform difficulty - if one problem looks like competitive programming, budget time and partial credit strategy.
Source: My Interview Experience with Meesho - SDE-1 Open Campus 2025 - Medium
Takeaway: Prepare fundamentals flashcards for the final day, not only LLD - several 2025 candidates describe losing offers here.
What these accounts have in common
After technical rounds, Meesho usually closes with a short HR / hiring-manager conversation: motivation, location, notice period, and a couple of behavioural stories. Prepare those answers on the dedicated page rather than cramming them into this hub.
Use these as frames, not scripts - interviewers probe follow-ups.
Q. Find the top K frequent elements in an array. What is the optimal approach?
A. First build a frequency map in one pass, which is O(n) time and O(n) space. Then push entries into a min-heap of size K, popping the smallest whenever the heap exceeds K, so the heap always holds the K most frequent seen so far - that is O(n log K) time and O(K) extra space, better than sorting all distinct elements at O(n log n). If the interviewer pushes for strictly linear time, use bucket sort: create an array of n+1 buckets indexed by frequency, place each element in the bucket matching its count, and walk the buckets from the highest index collecting until you have K, giving O(n). Edge cases to state are K larger than the number of distinct elements, and ties in frequency, where you should ask whether any valid answer is acceptable.
Q. Find the shortest path from a source to a destination cell on a grid with blocked cells.
A. Because every move costs the same, use breadth-first search rather than Dijkstra. Push the source into a queue with distance 0 and mark it visited, then repeatedly pop a cell and push its unvisited, in-bounds, unblocked neighbours with distance plus one; the first time you pop the destination, that distance is the shortest path. It is O(rows times columns) time and space since every cell is enqueued at most once. Marking a cell visited at enqueue time rather than dequeue time is what stops duplicates blowing up the queue. If edge weights differed you would need Dijkstra with a priority queue at O(E log V), and if some moves cost 0 and others 1, a deque-based 0-1 BFS is the right tool.
Q. In the machine coding round, how would you structure the classes for a car-pooling system?
A. Model the nouns first, then the service that coordinates them. Entities: User (id, name, contact), Vehicle (registration, model, capacity, owner), Ride (id, driver, vehicle, source, destination, seatsAvailable, status), and Booking (id, ride, rider, seats, status). Behaviour goes into a RideService that offers createRide, searchRides and bookRide, backed by repository interfaces so the in-memory store can be swapped later. Keep matching logic behind a MatchingStrategy interface (nearest-source, cheapest, fewest-detours) so a new rule is a new class rather than an edit to an if-else chain - that is the open-closed principle the grader is looking for. Seat availability must be decremented atomically and validated against capacity, and cancellations must return seats, since those are the edge cases Meesho interviewers probe.
Q. Write a SQL query for the top 3 selling products in each category.
A. Use a window function to rank within each partition, then filter on the rank in an outer query, because window functions cannot be used in a WHERE clause: SELECT category, product, units FROM (SELECT category, product, units, ROW_NUMBER() OVER (PARTITION BY category ORDER BY units DESC) AS rn FROM product_sales) t WHERE rn <= 3. Use RANK() instead of ROW_NUMBER() if tied products should all be returned, and DENSE_RANK() if you do not want gaps in the ranking numbers. On a large sales table an index on (category, units DESC) lets the engine feed rows to the window in order and avoids a full sort.
Q. Is the graph bipartite?
A. BFS/DFS two-coloring: assign color 0 to the start node, flip for neighbors; if a neighbor already has the same color, return false. Check all components. O(V + E). Meesho open-campus 2025 reports a dedicated 30-minute round on this pattern.
Q. Merge overlapping delivery slots.
A. Sort intervals by start, merge when current.start <= last.end, extend last.end = max(last.end, current.end). O(n log n). Relate to seller pickup windows or hub dispatch batches.
Q. What is the difference between preemptive and non-preemptive scheduling, and what happens during a context switch?
A. Under non-preemptive scheduling a running process keeps the CPU until it blocks or exits, as in first-come-first-served or plain shortest-job-first; under preemptive scheduling the kernel can take the CPU away on a timer interrupt or when a higher-priority process becomes ready, as in round-robin and preemptive shortest-remaining-time-first. Preemption gives much better response time for interactive workloads at the cost of more switching overhead. On a context switch the kernel saves the current process’s registers and program counter into its process control block, updates its state, selects the next process, restores that process’s saved context, and switches the page-table base register - which also flushes or invalidates TLB entries. That TLB and cache cost is why an excessively small round-robin time quantum destroys throughput.
Assessment tips
Interview tips
HR tips
Round-wise notes synthesize public Meesho candidate reports, then add prep takeaways and outcome context. Cross-check timings against your drive mail. Citation inputs:
For OA practice, see Meesho placement papers.