Equall (LTCV Credit) Software Engineer Interview Guide

The 60-second brief

As a Software Engineer at Equall (LTCV Credit), you are at the intersection of high-scale engineering and complex financial systems. Your work directly impacts our Loan Management System, a critical component of our infrastructure that manages the lifecycle of credit products. You will be responsible for building robust, scalable, and secure services that handle financial transactions with high precision and reliability. This role is not just about writing code; it is about architectural stewardship. You will contribute to the design of systems that must remain highly available and consistent under significant load. Because we operate in the fintech space, your ability to balance performance with strict data integrity and regulatory compliance will define your success and your impact on our mission to modernize credit accessibility.

This guide is scoped to a Software Engineer candidate at Equall (LTCV Credit).

Focus areas

• Low-Level Design (LLD)
• System Design
• Entity Modeling

The Equall (LTCV Credit) interview process

Equall (LTCV Credit) candidates report 3 rounds over 3-5 weeks. The stages below are what candidates describe, not a published process.

Initial Screens

Begin with a review of your past experiences and technical background.

What to demonstrate

• Begin with a review of your past experiences and technical background
• Depth in Low-Level Design (LLD)

How to prepare

• Be able to walk your CV end to end in two minutes, and say why this company specifically.
• Have your salary expectations, notice period and location constraints ready, and ask for the rest of the loop in writing.

Problem-Solving Assessment

Engage in practical assessments that reflect real challenges faced by the team.

What to demonstrate

• Engage in practical assessments that reflect real challenges faced by the team
• Depth in Low-Level Design (LLD)

How to prepare

• Answer aloud and timed: Explain the trade-offs between a monolithic and microservices architecture for a payment processing service.
• Answer aloud and timed: How do you handle database sharding when dealing with transactional financial data?

Technical Deep-Dive

Defend your technical decisions in real-time while addressing ambiguity in requirements.

What to demonstrate

• Defend your technical decisions in real-time while addressing ambiguity in requirements
• Depth in Low-Level Design (LLD)

How to prepare

• Answer aloud and timed: Design a service to track real-time order status and update downstream stakeholders.
• Work Low-Level Design (LLD) 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 answer. Narrate your assumptions and the trade-offs you are considering.

Ask Clarifying Questions

Never start designing before you have fully understood the constraints (e.g., traffic volume, read/write ratio, latency requirements).

Be Opinionated but Flexible

Have a strong rationale for your architectural choices, but be open to pivoting if the interviewer points out a flaw in your logic.

Focus on Reliability

In the context of LTCV Credit, data integrity and system reliability are paramount. Always mention how you handle failures and data consistency.

Questions & practice

Archive a resource graph without breaking live references or recursing

Medium · graph traversal · topological order · tenant isolation

Resources reference other resources within a tenant; for the largest tenant the reference table holds up to 2,000,000 nodes and 8,000,000 edges. Archiving a resource must archive everything reachable from it that nothing outside the set still references, refuse when a live external referrer exists, and terminate when references form cycles, which they legitimately do. Produce the archive order and the refusal list, targeting O(V+E). Say what stops the traversal crossing a tenant boundary, and why recursion is the wrong control structure at this size.

Approach

1. Load the subgraph with the tenant predicate on both endpoints of the edge, not only on the side you started from. Scoping the left table alone is the classic cross-tenant leak: one mis-entered edge then pulls another tenant's resources into the traversal and, worse, into the archive.
2. Traverse iteratively with an explicit stack. A 2,000,000-node graph can hold a chain deep enough to exhaust a native stack in the low tens of thousands of frames, and that failure is a process crash rather than an error you can return.
3. Treat cycles as data rather than corruption: compute strongly connected components with Tarjan in O(V+E) using its own explicit stack, then condense. The condensation is a DAG, so a topological order over it gives the archive order, and every member of a component archives in one transaction because no order within a cycle is valid.
4. Decide refusals with reverse edges. A candidate is archivable only if every in-edge originates inside the candidate set, so build the transpose or count in-degrees restricted to the visited set, and emit each blocked resource with the id of the external referrer, which is the only part of the answer an operator can act on.

Follow-up

• The graph is read in one query and the archive writes a minute later. What can change in between, and how do you make the write safe?
• The candidate set is 400,000 resources. Is that one transaction, and if not, what does a half-finished archive look like to a reader?

Canonicalise a request body into a stable idempotency fingerprint

Medium · parsing · canonicalisation · hashing

idempotency_key.request_fingerprint is a SHA-256 over the method, path and canonicalised body, and a retry whose fingerprint differs must be rejected with 422 rather than served the stored response. Write the canonicaliser. Bodies are JSON up to 256 KB nested at most 32 levels; clients vary key order, whitespace and unicode escaping, and some send 64-bit ids as JSON numbers. Produce a deterministic byte string such that semantically identical bodies match and any semantic difference does not. State your complexity and name two normalisations you refuse to perform.

Approach

1. Parse once into a tree, then re-serialise under fixed rules: object keys sorted, array order preserved, one escaping convention, no insignificant whitespace. Parsing is O(n) and sorting keys is O(k log k) per object, so O(n log n) overall with O(depth) stack, and the 32-level cap is enforced during parsing because hostile nesting is how a canonicaliser becomes a stack overflow.
2. Sort keys by their UTF-8 bytes and say why the obvious implementation is wrong in some runtimes: a default string comparison that orders by UTF-16 code units places surrogate pairs, meaning code points from U+10000 up, below U+E000 to U+FFFF, which is not UTF-8 byte order, so two services written in different languages disagree on the same document.
3. Do not re-encode numbers through a double. IEEE-754 binary64 represents integers exactly only up to 2^53, so normalising a 19-digit id through a float changes it, and 1 against 1.0 cannot be reconciled without deciding whether they are the same value. Preserve the literal token, and require ids as strings at the API boundary if you want them comparable.
4. Reject duplicate keys rather than picking one. JSON permits them and parsers disagree, most keeping the last, so any choice you make ties the fingerprint to a parser detail that the code handling the request does not necessarily share.

Follow-up

• A client sends the same logical request with an extra field your API ignores. Same key, different fingerprint, so you return 422. Is that the right answer?
• Where does the fingerprint get computed relative to request decompression and the body-size limit?

Collapse a redelivered event batch into per-aggregate high-water marks

Easy · hashing · at-least-once · aggregation

You drain a batch of up to 5,000,000 events, each (aggregate_id BIGINT, aggregate_version INT, event_type, payload). The log guarantees order within one aggregate only; the batch merges 64 partitions, and a relay failover has redelivered a range, so an older version for an aggregate can appear after a newer one. Given a map of last_applied_version per aggregate, produce the events worth applying, at most one per (aggregate_id, version), plus the count discarded. Target O(n) time. State the memory for 2,000,000 distinct aggregates and what you do when it does not fit.

Approach

1. One pass, one hash map from aggregate_id to the highest version kept, and a discard counter. An event whose version is at or below last_applied_version for its aggregate is dropped without further work, which is the whole reason the event carries its version rather than a delta. O(n) expected time, O(d) space in distinct aggregates.
2. Keep the maximum, never the last occurrence. The redelivered range means the final appearance of an aggregate in the batch can be an older version than one seen earlier in the same batch, so last-wins applies stale state over newer state and the projection regresses with no error anywhere.
3. Cost the memory instead of calling it large: an 8-byte key plus a 4-byte version is 12 bytes of payload, and an open-addressed table held at a 0.7 load factor costs roughly 17 bytes per entry before per-slot metadata, so 2,000,000 aggregates is tens of megabytes in a native layout and several times that in a runtime that boxes both key and value.
4. If the distinct set exceeds memory, partition on hash(aggregate_id) mod P and reduce each partition independently. Every event for one aggregate hashes to the same partition, so the per-partition result is exact and the merge is concatenation rather than a second reduction.

Follow-up

• The payload is a patch rather than a snapshot, so applying only the highest version loses the intermediate changes. What changes in your reduction?
• How do you detect that version 7 arrived while version 6 was never delivered, and what should the consumer do about the gap?

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?

Stop tag and share joins from fanning out a page

Medium · join fan-out · semi-joins · many-to-many

resource_tag is (resource_id, tag_id) with PK (resource_id, tag_id); resource_share is (resource_id, shared_with_user_id, permission). The tagged-and-shared listing inner-joins resource to both, filters tenant_id, tag_id = ANY($2) and shared_with_user_id = $3, orders by updated_at DESC and takes 50. Pages come back with fewer than 50 distinct resources and the total in the header is far too high. Explain the row multiplication, rewrite both the page query and the count query so each is correct, and name the index each one needs. PostgreSQL 16.

Approach

1. Do the arithmetic against the predicates that are actually there. An inner join emits one row per matching child row, and both joins are filtered: tag_id = ANY($2) admits only the requested tags, shared_with_user_id = $3 admits one user's share rows. So a resource holding three of the requested tags and shared with $3 once yields three rows, not one — the multiplier is its count of matching tags times its share rows for that single user, and that second factor is 1 unless the table admits duplicate (resource_id, shared_with_user_id) pairs. LIMIT 50 then limits rows rather than resources, and COUNT(*) counts pairs — the header is the product, not the population.
2. Reject DISTINCT as the fix. It deduplicates after the product has been built, so the planner must materialise and sort the fanned-out set before the LIMIT can apply, and it leaves any SUM or AVG in the same select list wrong.
3. Rewrite both filters as semi-joins, keeping resource as the only row source: AND EXISTS (SELECT 1 FROM resource_tag rt WHERE rt.resource_id = r.resource_id AND rt.tag_id = ANY($2)) and the same shape against resource_share. A semi-join stops at the first match per resource and preserves the driving index order, so ORDER BY updated_at DESC, resource_id DESC LIMIT 50 still stops after 50 rows.
4. Count with the same predicates and no join at all: SELECT count(*) FROM resource r WHERE r.tenant_id = $1 AND r.status = 'active' AND EXISTS (...) AND EXISTS (...). Nothing multiplies a resource, so the number is the population.

Follow-up

• The filter changes from 'any of these tags' to 'all of these tags'. Rewrite it and state what it costs relative to the ANY form.
• A resource can be shared with the same user twice under different permissions. Does your count change, and should it?

Design a Food Delivery Platform (focusing on core entities and relationships).

Medium · System Design & LLD

Approach

1. Fix the scope first: who calls this, how often, and what they do when it fails.
2. Name the read and write paths separately; they rarely have the same bottleneck.
3. Choose a partition key and say what query it makes expensive.
4. State the consistency you need, and where you are willing to be stale.

Follow-up

• What breaks first when traffic grows ten times?
• How does this behave when that dependency is down for an hour?

How would you design a schema for a Loan Management System to handle high-concurrency repayments?

Medium · System Design & LLD

Approach

1. Fix the scope first: who calls this, how often, and what they do when it fails.
2. Name the read and write paths separately; they rarely have the same bottleneck.
3. Choose a partition key and say what query it makes expensive.
4. State the consistency you need, and where you are willing to be stale.

Follow-up

• What breaks first when traffic grows ten times?
• How does this behave when that dependency is down for an hour?

Explain the trade-offs between a monolithic and microservices architecture for a payment processing service.

Medium · System Design & LLD

Approach

1. Fix the scope first: who calls this, how often, and what they do when it fails.
2. Name the read and write paths separately; they rarely have the same bottleneck.
3. Choose a partition key and say what query it makes expensive.
4. State the consistency you need, and where you are willing to be stale.

Follow-up

• What breaks first when traffic grows ten times?
• How does this behave when that dependency is down for an hour?

Design a service to track real-time order status and update downstream stakeholders.

Medium · System Design & LLD

Approach

1. Fix the scope first: who calls this, how often, and what they do when it fails.
2. Name the read and write paths separately; they rarely have the same bottleneck.
3. Choose a partition key and say what query it makes expensive.
4. State the consistency you need, and where you are willing to be stale.

Follow-up

• What breaks first when traffic grows ten times?
• How does this behave when that dependency is down for an hour?

One customer endpoint stalls deliveries to every other destination

Medium · bulkhead · timeouts · saturation

The egress service delivers about 1.5k webhooks/second across 40,000 destinations, with a per-destination concurrency cap of 4 and a 10-second connect-plus-read timeout. Throughput falls to 300/second, queue depth climbs, and p99 delivery latency for unaffected destinations goes from 200 ms to minutes, while the error rate barely moves. One tenant holds 900 destination rows whose URLs share a hostname that now answers in 9.5 seconds. Explain the mechanism with the arithmetic, then give the containment in the order you would apply it.

Approach

1. Look at saturation before errors. A flat error rate with collapsing throughput says nothing is failing, things are waiting, so the first signal to pull is in-flight request count or pool wait time rather than the error counter. This is the distinction that decides the whole investigation.
2. Group in-flight work by resolved host, not by destination id. The cap is keyed per destination row, so 900 rows sharing one hostname buy 3,600 concurrent slots against a single host, each held for 9.5 seconds. The bulkhead was never a bulkhead for that host, and grouping by the wrong dimension is why the dashboard looked healthy.
3. Do the arithmetic in both directions. Required concurrency is arrival rate times latency, so 1.5k/second at 200 ms needs about 300 in flight, which is entirely consumed by 3,600 slow slots; conversely whatever concurrency is left sustains rate equals concurrency divided by 9.5 seconds, which is the 300/second you are seeing. Matching both numbers is what promotes this from a plausible story to the mechanism.
4. Explain why the circuit breaker never helped. It opens on consecutive failures, and a 9.5-second response inside a 10-second timeout is a success. Slow is not failing, so an error-rate breaker cannot see this; you need a slow-call ratio, a deadline propagated from the caller's remaining budget, or a concurrency limiter.

Follow-up

• The host recovers to 80 ms. How long does the queue take to drain, and what does the drain do to the recovered host?
• Where should the 10-second timeout number actually come from?

Suggested practice checklist

Built from the rounds and topics Equall (LTCV Credit) candidates report.

Day 1 — Map the Equall (LTCV Credit) loop

• Write out the reported sequence: Initial Screens, Problem-Solving Assessment, Technical Deep-Dive.
• 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 3 reported rounds, with the weakest marked.

Day 2 — Work Low-Level Design (LLD)

• Spend the session on Low-Level Design (LLD), which Equall (LTCV Credit) candidates report being tested on.
• Write one worked example in Low-Level Design (LLD) and time yourself on it.

Deliverable: One timed worked example in Low-Level Design (LLD).

Day 3 — Work System Design

• Spend the session on System Design, which Equall (LTCV Credit) candidates report being tested on.
• Write one worked example in System Design and time yourself on it.

Deliverable: One timed worked example in System Design.

Day 4 — Work Entity Modeling

• Spend the session on Entity Modeling, which Equall (LTCV Credit) candidates report being tested on.
• Write one worked example in Entity Modeling and time yourself on it.

Deliverable: One timed worked example in Entity Modeling.

Day 5 — Answer out loud: System Design & LLD

• Answer aloud, timed: Design a Food Delivery Platform (focusing on core entities and relationships).
• Answer aloud, timed: How would you design a schema for a Loan Management System to handle high-concurrency repayments?

Deliverable: Spoken answers to 2 reported System Design & LLD 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 Equall (LTCV Credit)

• 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.

How do you handle database sharding when dealing with transactional financial data?

Medium · System Design & LLD

Approach

1. Pick a story where you made the decision, not one where you watched it.
2. State the situation in two sentences and spend the rest on the reasoning.
3. Give the blast radius: what could have broken, and what you measured.
4. Name the disagreement and how you resolved it with evidence.

Follow-up

• What would you do differently if you ran that again?
• How did you know your change caused the improvement?

Ship under a deadline and bound the debt you chose

Medium · pagination · technical debt · tradeoffs

You have four days to ship a tenant-facing listing endpoint. The version you would defend uses keyset pagination over (tenant_id, status, updated_at DESC, resource_id DESC); the version you can finish uses LIMIT/OFFSET with no matching index. Describe a deadline call you actually made of this shape: what you shipped, what you knowingly deferred, how you bounded the damage with a mechanism rather than an intention, and the specific numeric condition that would force the follow-up. Name who you told and where you wrote it down.

Approach

1. Name the deferred failure precisely instead of calling it slow. OFFSET n makes the database produce and discard n rows, so cost grows with page depth; without an index matching the sort, every matching row is read and sorted before the limit applies; and rows inserted between two page fetches shift across the boundary so items are skipped or repeated with nothing in the response to signal it.
2. Bound the blast radius with something mechanical rather than a promise: cap maximum page depth, cap page size, restrict the endpoint to one internal caller, or keep it behind a flag. State which failure each cap removes and which it leaves standing.
3. Attach a number to the trigger and wire it to an alarm: the first tenant crossing N resources, or the endpoint's p99 crossing its share of the 400 ms budget, so the debt announces itself instead of waiting to be remembered.
4. Write it where the next engineer looks, which is the code and the ticket, not a chat message: what was deferred, why, the cap, and the trigger.

Follow-up

• At what page depth does the offset version breach your latency budget, given your page size and row counts?
• What breaks first when you switch to keyset pagination later, and what does a client holding an old page token see?

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?