Groww Software Engineer Interview Guide

The 60-second brief

As a Software Engineer at Groww, you are at the heart of democratizing financial services in India. Your work directly impacts millions of users, enabling them to invest in stocks, mutual funds, and other financial instruments with simplicity and speed. You aren't just writing code; you are building highly scalable, resilient systems that handle massive transactional volume while maintaining the highest standards of data security and financial integrity. This role is critical because Groww operates in a high-stakes, low-latency environment. You will tackle complex engineering challenges involving distributed systems, real-time data processing, and highly consistent database architectures.

This guide is scoped to a Software Engineer candidate at Groww.

Focus areas

• Data Structures and Algorithms (DSA)
• Problem Solving
• Algorithmic Complexity (Time/Space)

The Groww interview process

Groww candidates report 4 rounds over 3-5 weeks. The stages below are what candidates describe, not a published process.

Technical Screening

Initial assessment to evaluate problem-solving skills and technical knowledge.

What to demonstrate

• Initial assessment to evaluate problem-solving skills and technical knowledge
• Depth in Data Structures and Algorithms (DSA)

How to prepare

• Answer aloud and timed: Given two strings, find the number of distinct subsequences of one string which equals the other.
• Answer aloud and timed: Optimize a naive recursive solution for string matching problems using memoization or bottom-up dynamic programming.

Architectural Discussions

In-depth conversations about system design and architecture.

What to demonstrate

• In-depth conversations about system design and architecture
• Depth in Data Structures and Algorithms (DSA)

How to prepare

• Answer aloud and timed: Discuss the time and space complexity trade-offs of your proposed solution.
• Answer aloud and timed: Explain how you would handle edge cases where the input strings contain duplicate characters.

Cultural Alignment

Assessment of candidate's fit within the company's culture and values.

What to demonstrate

• Assessment of candidate's fit within the company's culture and values
• Depth in Data Structures and Algorithms (DSA)

How to prepare

• Work Data Structures and Algorithms (DSA) until you can explain it without notes
• Work Problem Solving until you can explain it without notes

Final Management Reviews

Final evaluations by management to make a decision on the candidate.

What to demonstrate

• Final evaluations by management to make a decision on the candidate
• Depth in Data Structures and Algorithms (DSA)

How to prepare

• Work Data Structures and Algorithms (DSA) until you can explain it without notes
• Work Problem Solving until you can explain it without notes

Where candidates lose points

Think Aloud

Your interviewer is more interested in your thought process than the final code. Always talk through your logic.

Edge Case Focus

Always test your code against empty inputs, large inputs, and unexpected data formats before declaring it complete.

Value Alignment

Understand the Groww mission of making financial services accessible; showing genuine interest in the product space can set you apart.

Questions & practice

Given two strings, find the number of distinct subsequences of one string which equals the other.

Medium · Data Structures and Algorithms

Approach

1. Restate the input: its shape, its size, and what is guaranteed about it.
2. Name the brute-force solution and its complexity before improving on it.
3. Choose the data structure from the access pattern, not from familiarity.
4. State the target complexity and say which constraint rules the naive version out.

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?

Optimize a naive recursive solution for string matching problems using memoization or bottom-up dynamic progra

Medium · Data Structures and Algorithms

Optimize a naive recursive solution for string matching problems using memoization or bottom-up dynamic programming.

Approach

1. Restate the input: its shape, its size, and what is guaranteed about it.
2. Name the brute-force solution and its complexity before improving on it.
3. Choose the data structure from the access pattern, not from familiarity.
4. State the target complexity and say which constraint rules the naive version out.

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?

Discuss the time and space complexity trade-offs of your proposed solution.

Medium · Data Structures and Algorithms

Approach

1. Restate the input: its shape, its size, and what is guaranteed about it.
2. Name the brute-force solution and its complexity before improving on it.
3. Choose the data structure from the access pattern, not from familiarity.
4. State the target complexity and say which constraint rules the naive version out.

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?

Explain how you would handle edge cases where the input strings contain duplicate characters.

Medium · Data Structures and Algorithms

Approach

1. Restate the input: its shape, its size, and what is guaranteed about it.
2. Name the brute-force solution and its complexity before improving on it.
3. Choose the data structure from the access pattern, not from familiarity.
4. State the target complexity and say which constraint rules the naive version out.

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?

Denormalise tenant onto revisions and backfill it live

Hard · denormalisation · online migration · backfill

resource_revision (revision_id, resource_id, version, actor_user_id, change_kind, patch, request_id, created_at) has 400M rows and no tenant column; tenant_id lives only on resource. Two reads need it: a tenant-scoped audit feed ordered by created_at DESC, and an offboarding purge. Both join back to resource today. Justify adding tenant_id to resource_revision against those two reads, name the anomaly the copy introduces and the constraint that prevents it, then give the ordered migration for a live table taking 1.2k writes/second — the lock each step takes, how the backfill is batched, and where each step stops being reversible. PostgreSQL 16.

Approach

1. Justify from the access path rather than from taste. Without the column, the audit feed either scans resource_revision by created_at and discards other tenants' rows, or resolves the tenant's resource_ids first and probes with them — both proportional to the tenant's whole history rather than to one page. With (tenant_id, created_at DESC, revision_id DESC) it is a seek that stops at 50 rows, and the purge becomes a ranged delete instead of a join.
2. Name the cost exactly: a second copy of a fact can disagree with the first. Make the disagreement unwritable rather than documented — add UNIQUE (resource_id, tenant_id) on resource so it can serve as a foreign-key target, then FOREIGN KEY (resource_id, tenant_id) REFERENCES resource (resource_id, tenant_id) on the revision table. A revision can then only ever carry its parent's tenant.
3. Step one, expand: ALTER TABLE resource_revision ADD COLUMN tenant_id BIGINT NULL, with no default, so it is a catalogue change and no rewrite. It still needs ACCESS EXCLUSIVE for an instant, and that instant queues behind the longest open transaction on the table while every later query queues behind it — set lock_timeout to 2s and retry rather than wait.
4. Step two, dual-write: deploy the writer that populates tenant_id on every new revision while reads still use the join. Reversible by redeploying the previous build, because nothing reads the column yet.

Follow-up

• The backfill is half finished and a rollback is required. What state is the table in, and what does the previous build do with a half-populated column?
• How do you verify the backfill actually finished, given rows are still being inserted while it runs?

Hold a per-tenant active cap against concurrent creates

Hard · write skew · isolation levels · constraints

A tenant on the standard plan may hold at most 50 resources with status='active'. The create handler runs SELECT count(*) FROM resource WHERE tenant_id = $1 AND status = 'active', compares to 50, then inserts. Two creates arrive 3 ms apart on different instances and the tenant lands at 51. Name the anomaly, say whether PostgreSQL 16 READ COMMITTED or REPEATABLE READ prevents it and why, then give an implementation that holds the cap at READ COMMITTED with the exact statements. Finally, say what changes when the cap is 'at most one running export per tenant' on job_run.

Approach

1. Name it: write skew. The two transactions read an overlapping set and write disjoint rows, so there is no row-level conflict for the engine to detect and each commit is individually legal.
2. Rule out the levels precisely. READ COMMITTED takes a fresh snapshot per statement and takes no lock on the counted rows, so both see 49. PostgreSQL's REPEATABLE READ is snapshot isolation: it removes non-repeatable reads and phantoms within the snapshot but still admits write skew, because the anomaly is not a re-read of a changed row, it is a read of a set that a concurrent transaction invalidates. Only SERIALIZABLE closes it, by tracking the read dependency and aborting one transaction with SQLSTATE 40001 — a guarantee that exists only if the application re-runs the whole transaction from the read.
3. Convert the set predicate into a single-row conflict: keep tenant.active_resource_count and run UPDATE tenant SET active_resource_count = active_resource_count + 1 WHERE tenant_id = $1 AND active_resource_count < 50 in the same transaction as the INSERT. Zero affected rows is the cap, returned as 409. The row lock serialises the decision at any isolation level, and contention is bounded to one tenant's row — which is also the fair-scheduling unit, unlike a global counter that would convoy every tenant behind one row.
4. State the cost you just took on: a counter is a second source of truth that can drift, so every path that changes status must adjust it inside the same transaction, and a periodic reconciliation has to exist, with resource_revision as the authority for what the count should have been.

Follow-up

• A resource moves from archived back to active. Which statements change, and what breaks if the counter update and the status change land in different transactions?
• The cap becomes plan-dependent and a plan can change mid-month. Where does the number 50 live, and who reads it?

Cache the tenant listing feed with a bounded staleness window

Easy · caching · cache invalidation · keyset pagination

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

1. 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.
2. 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.
3. 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.
4. 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.

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?

Make the resource write endpoint safe for retrying clients

Easy · idempotency · unique constraints · retries

The edge API accepts POST /v1/resources at 1.2k writes/second peak under a 400 ms p99 budget, and clients retry on timeout. idempotency_key has PRIMARY KEY (tenant_id, idempotency_key) plus request_fingerprint CHAR(64), state in ('in_flight','succeeded','failed'), response_status, response_body, resource_id, locked_until and expires_at. Specify the exact sequence the handler runs: what the second request does when it arrives while the first is still executing, what a retry carrying a different body receives, what happens when a process dies mid-request, and how the table is kept from growing without bound.

Approach

1. Start from the race rather than the happy path. Insert the key row first, in its own short transaction, with state 'in_flight' and locked_until set to now() plus the request deadline. The unique constraint on (tenant_id, idempotency_key) is the arbitration mechanism: exactly one caller commits the insert, and every other caller takes a unique violation (SQLSTATE 23505) and becomes a follower. Reading the table and then inserting is a check-then-act race that both callers pass, and they pass it most often under the load that generates the retries.
2. Give the follower a branch for every state it can observe, with no fall-through to doing the work: fingerprint mismatch means the key is being reused for a different request, so reject with 422 and never replay the stored response; 'succeeded' replays response_status and response_body verbatim; 'failed' allows a fresh attempt; 'in_flight' with locked_until in the future is answered with 409 and a retry hint, or a short bounded wait, because the only alternative is executing the effect twice.
3. Commit the effect and the record of the effect together. The resource insert, its resource_revision row, its outbox_event row and the transition of the key row to 'succeeded' all happen in one transaction, so no crash can leave the work done and the key still 'in_flight'. The earlier insert is deliberately a separate transaction: it has to be visible to a concurrent caller before the work starts, which an uncommitted row is not.
4. Close the wedged-key path. A process that dies after the insert leaves a row that would otherwise block that key forever, which is what locked_until exists for. Reclaim with a single conditional statement, UPDATE ... SET state='in_flight', locked_until=now()+interval WHERE ... AND state='in_flight' AND locked_until < now(), and treat a zero rowcount as losing the reclaim. Two statements that read and then update reintroduce the original race one layer down.

Follow-up

• The stored response body averages 200 KB and this table is now the largest in the database. What do you store instead, and what does a replay return once the body has been pruned?
• A client library generates a fresh idempotency key on every attempt. What breaks, which layer should have caught it, and does the server have any defence?

Read latency spikes on a sixty-second sawtooth

Medium · cache stampede · single-flight · ttl

The cached listing read path serves about 14k reads/second at an 85% hit rate. p99 sits at 35 ms for 57 seconds, jumps to 900 ms for 3, and repeats. During each spike the primary shows several hundred identical listing queries starting within the same millisecond, all carrying one large tenant's id. Cache entries use a 60-second TTL. Give the mechanism, the ordered checks, the fix, and the correctness hazard your fix must not introduce.

Approach

1. Match the period to a configured number before theorising about load. A spike every 60 seconds against a 60-second TTL is an entry expiring, and you confirm it by correlating spike timestamps with the entry's write time rather than with the traffic curve. If the period had matched a cron or a GC interval instead, this is a different investigation.
2. Establish the concurrency of the miss. Several hundred identical queries in one millisecond means the miss path has no coalescing: every request that arrives between expiry and repopulation recomputes. The herd size is that key's arrival rate times its recompute time, so at 1.2k reads/second for the hot key and a 250 ms recompute you expect about 300 concurrent misses, which matches what is observed.
3. Add single-flight on the miss path so one caller per key recomputes under a short-lived lock while the rest wait for its result. Prefer stale-while-revalidate where the read tolerates it: return the expired value immediately and refresh asynchronously, which removes the latency spike rather than serialising it into a queue of waiters.
4. De-synchronise the keys. Write TTLs with jitter, for example 60 seconds plus or minus 10%, so a deploy or a mass invalidation does not align every key on the same second and turn a per-key herd into a fleet-wide one.

Follow-up

• The same sawtooth appears on a key that is invalidated on write rather than expired. Is that the same bug?
• How does your answer change if the recompute takes 4 seconds instead of 250 ms?

Suggested practice checklist

Built from the rounds and topics Groww candidates report.

Day 1 — Map the Groww loop

• Write out the reported sequence: Technical Screening, Architectural Discussions, Cultural Alignment, Final Management Reviews.
• For each round, write one sentence on what it is judging, from the description above, and mark the one you are least ready for.

Deliverable: A one-page map of the 4 reported rounds, with the weakest marked.

Day 2 — Work Data Structures and Algorithms (DSA)

• Spend the session on Data Structures and Algorithms (DSA), which Groww candidates report being tested on.
• Write one worked example in Data Structures and Algorithms (DSA) and time yourself on it.

Deliverable: One timed worked example in Data Structures and Algorithms (DSA).

Day 3 — Work Problem Solving

• Spend the session on Problem Solving, which Groww candidates report being tested on.
• Write one worked example in Problem Solving and time yourself on it.

Deliverable: One timed worked example in Problem Solving.

Day 4 — Work Algorithmic Complexity (Time/Space)

• Spend the session on Algorithmic Complexity (Time/Space), which Groww candidates report being tested on.
• Write one worked example in Algorithmic Complexity (Time/Space) and time yourself on it.

Deliverable: One timed worked example in Algorithmic Complexity (Time/Space).

Day 5 — Answer out loud: Data Structures and Algorithms

• Answer aloud, timed: Given two strings, find the number of distinct subsequences of one string which equals the other.
• Answer aloud, timed: Optimize a naive recursive solution for string matching problems using memoization or bottom-up dynamic programming.

Deliverable: Spoken answers to 2 reported Data Structures and Algorithms question(s), under time.

Day 6 — Rehearse your own examples

• Prepare three examples from your own work where you made the decision, each with the outcome you can quantify.

Deliverable: Three examples written out, each with a number attached.

Day 7 — Dry run for Groww

• Run one full mock under time, then write down the two questions you most want to ask your interviewers.

Deliverable: A completed timed mock and two questions to ask.

Behavioural preparation

Behavioural rounds judge the decision you made and what it cost.

Turn a code review disagreement into a decision

Easy · code review · optimistic concurrency · communication

A colleague's change updates a row with UPDATE resource SET version = version + 1 WHERE resource_id = $1 AND version = $2 and treats an affected-row count of zero as a successful no-op. You read that as a silently lost update; they think returning 200 is friendlier to clients than returning a conflict. Describe how you have handled a review disagreement of this shape: what goes in the comment, when you leave the thread, and who decides. Then write the comment you would leave here, in under 80 words.

Approach

1. Sort the disagreement before writing anything. A silently discarded write is a correctness claim about data; the choice between 409 and 412 is taste. Only the first justifies blocking a merge, and saying which one you are doing is most of the value of the comment.
2. Make the claim reproducible in the comment itself with an interleaving rather than a principle: A reads version 7, B reads version 7, B commits version 8, A's predicate matches zero rows, A is told it succeeded and A's edit is gone.
3. Offer the alternative with its cost attached: return 409 carrying the current version and the revision that won, so the client can re-read and re-apply. Note that automatic retry is not the fix, because a retry re-reads the winner's state and reapplies an intent formed against data that no longer exists.
4. Apply an escalation rule you can state: two round trips on the thread, then a call, and the service's owner decides rather than the reviewer. A reviewer who cannot be overruled is a bottleneck with extra steps.

Follow-up

• Where would you put the test that fails if someone reintroduces the swallowed zero rowcount?
• The author says clients cannot handle a 409. How do you check whether that is true?

Tell callers you do not own that their integration breaks

Medium · deprecation · compatibility · stakeholders

A field in a write endpoint's response must change shape. You own the endpoint; you do not own the four internal callers or the outbound webhook consumers who read it. Describe a deprecation you were responsible for: what you shipped first, how you established who was actually reading the field, the window you gave and what set its length, what you did about the consumer who never moved, and how you decided removal was safe. Name the signal you used, not the announcement you sent.

Approach

1. Establish the reader set empirically rather than from a wiki of owners: per-field usage counters keyed by principal, or access logs attributed to a consumer. State the blind spot of whichever you pick, since a consumer that reads the field only on a monthly job will not appear in a week of logs.
2. Ship additive first. Populate the new field alongside the old one so no reader is forced to move, which is also what keeps a rolling deploy safe, because old and new instances answer the same requests at the same time and a rollback must still find the old shape present.
3. Set the window from the slowest legitimate consumer's release cadence, not from your calendar, and decide separately what to do for a consumer with no release process at all, such as an external webhook endpoint you can only email.
4. Convert silence into evidence before you rely on it: a short, low-traffic removal window that makes a still-dependent consumer fail visibly and loudly while you are watching, rather than at three in the morning after you have moved on.

Follow-up

• How would you detect a consumer that reads the field only during a monthly export?
• One caller refuses to move and has a commercial relationship behind it. What changes in your plan and what does not?

Estimate work you have never done and defend the range

Hard · estimation · backfills · expand-contract

You are asked to estimate a change you have never attempted: add a column to a 100-million-row table, populate it, move reads across, and drop the old shape. Give a range with the assumptions that generate it, including batch size, the signal your backfill throttles on, and wall-clock hours, and name the three unknowns that would move the number most. Then describe a real estimate you gave under comparable ignorance: how you expressed its uncertainty, what you committed to, and how wrong you turned out to be.

Approach

1. Decompose into independently deployable steps before estimating anything: add the column nullable, write both shapes, backfill in batches, verify, move reads, stop writing the old shape, drop it. That is four deploys spread over days, and the calendar estimate is dominated by them rather than by the loop's runtime.
2. Do the arithmetic aloud for the part that has arithmetic in it: batch size times number of batches times per-batch duration, at a write rate the primary can absorb alongside roughly 1.2k writes per second of production traffic. The loop is throttled by replication lag and lock waits, not by how fast it can issue statements.
3. Price the schema step by its lock rather than its statement duration. In PostgreSQL an ALTER TABLE taking ACCESS EXCLUSIVE waits for every open transaction on that table while later queries queue behind it, so a millisecond change issued during a thirty-second analytics query stalls that table for thirty seconds. Adding a nullable column with a non-volatile default avoids a rewrite from version 11; a new index wants CREATE INDEX CONCURRENTLY, which cannot run inside a transaction block and leaves an invalid index behind if it fails.
4. Express the answer as a range whose endpoints each trace to a stated assumption, then name the cheapest experiment that collapses it, which is almost always running one real batch against the real table and multiplying.

Follow-up

• How do you verify the backfill genuinely finished, given rows written by production traffic while it ran?
• Where does the backfill resume from after a worker is killed mid-batch, and what makes that resume point trustworthy?