Interview guides · Aptech

Aptech Software Engineer Interview Guide

Aptech Software Engineer interview guide: reported rounds, questions and how candidates describe the process.

The 60-second brief

A Software Engineer at Aptech plays a foundational role in delivering high-quality educational technology solutions and software services. You will be responsible for translating complex requirements into robust, scalable code, ensuring that the systems supporting Aptech remain performant and reliable. Your work directly impacts how users interact with our digital infrastructure, making your technical proficiency and problem-solving abilities vital to our operational success. This role requires a blend of deep technical knowledge and a practical mindset. Whether you are working on database architecture, refining API integrations, or building out frontend components, you are expected to maintain clean, maintainable code standards.

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

Focus areas

Object-Oriented Programming (OOP)
SQL (MySQL)
Database Joins

The Aptech interview process

Aptech candidates report 5 rounds over 4-6 weeks. The stages below are what candidates describe, not a published process.

Initial Screening

The process begins with an initial screening to evaluate basic qualifications.

What to demonstrate

The process begins with an initial screening to evaluate basic qualifications
Depth in Object-Oriented Programming (OOP)

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.

Technical Screening

Candidates undergo technical screenings to assess their coding and problem-solving skills.

What to demonstrate

Candidates undergo technical screenings to assess their coding and problem-solving skills
Depth in Object-Oriented Programming (OOP)

How to prepare

Answer aloud and timed: What are the core differences between languages like C, Java, and Python?
Answer aloud and timed: How do you approach implementing RESTful API structures?

Machine Tests

Candidates complete machine tests where they perform specific tasks under time constraints.

What to demonstrate

Candidates complete machine tests where they perform specific tasks under time constraints
Depth in Object-Oriented Programming (OOP)

How to prepare

Answer aloud and timed: Can you explain the folder structure and MVC patterns in frameworks like Laravel or CodeIgniter?
Answer aloud and timed: Explain your process for designing a database structure for a new application.

Management Interviews

Interviews with management to evaluate fit within the team and organizational culture.

What to demonstrate

Interviews with management to evaluate fit within the team and organizational culture
Depth in Object-Oriented Programming (OOP)

How to prepare

Answer aloud and timed: How do you approach complex MySQL join queries and stored procedures?
Answer aloud and timed: What are your strategies for implementing caching mechanisms and using tools like Memcached?

Final Assessment

The final assessment consolidates all evaluations to determine candidate suitability.

What to demonstrate

The final assessment consolidates all evaluations to determine candidate suitability
Depth in Object-Oriented Programming (OOP)

How to prepare

Answer aloud and timed: How would you handle SSL certificate installation and configuration?
Answer aloud and timed: How do you handle tight deadlines and high-pressure project scenarios?

Where candidates lose points

Show your work

When solving problems, talk through your thought process out loud. This helps interviewers understand your logic, even if you run into a snag.

Understand the "Why

Don't just memorize definitions. Be prepared to explain why you would use one technology over another.

Focus on basics

Many of our technical rounds start with fundamental questions about C, Java, or Python. Ensure your foundational knowledge is sharp.

Professionalism matters

Always maintain a professional demeanor throughout the process, from initial contact to the final interview round.

Questions & practice

Identify the heaviest tenants in a five-minute window under memory pressure

Medium · top-k · heavy hitters · streaming

The edge service handles about 3,000 requests per second across roughly 50,000 tenants, peaking near 9,000. Expose the 50 heaviest tenants by request count over the trailing five minutes so limits can be tightened before one tenant's backfill starves the fleet. You may not retain five minutes of raw records. Give the exact solution and its memory, then the bounded-memory approximation with its error stated as a formula, and say which you would ship and at what tenant cardinality that choice changes.

Approach

Do the exact version first, because it is affordable at this cardinality: a ring of 300 one-second counters per tenant, advanced lazily, is 1,200 bytes of counters per tenant and roughly 60 to 90 MB for 50,000 tenants with overhead. Carry a running total and subtract the bucket you overwrite so a window read is O(1) rather than 300 adds.
Extract the top 50 with a size-k min-heap over the tenant sums: O(d log k) for d tenants, against O(d log d) to sort them all. Maintaining the heap continuously instead of on query requires a tenant-to-heap-index map, because incrementing a count already inside the heap means sifting from a known position, and without that map you rebuild the heap on every request.
State the approximation precisely rather than gesturing at sketches. Misra-Gries with m counters retains every item whose true count exceeds N/(m+1), and each retained count underestimates the truth by at most N/(m+1). With m = 1,000 and N = 900,000 requests in the window the error is roughly 900 requests, which is fine for spotting a tenant sending 50,000 and useless for ranking two tenants 200 apart.
Say what breaks when the window slides: Misra-Gries and Space-Saving are insert-only and cannot be decremented as records age out. The workable construction is one summary per sub-window, say ten seconds, with 30 summaries merged at query time, and the merged error is the sum of the per-summary errors, so the bound degrades linearly in the number of sub-windows.

Follow-up

The heaviest tenant is heavy because of one export job rather than user traffic. Should the limiter treat those as the same tenant?
Two tenants sit tied at the boundary of the top 50. Does your answer flap, and does the flapping matter?

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

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.
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.
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.
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?

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

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.
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.
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.
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?

How do you approach complex MySQL join queries and stored procedures?

Medium · Database & System Design

Approach

Name the grain you start from and join outward from it.
Check whether any join is one-to-many before aggregating, or the sums inflate.
Say which index the query would use, and what makes it unusable.
Handle the rows that do not match: that is usually the actual question.

Follow-up

How does the query change if that join becomes one-to-many?
What happens to this when the table is ten times larger?

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

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.
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.
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.
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?