Interview guides · Dixit Infotech Services

Dixit Infotech Services Software Engineer Interview Guide

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

The 60-second brief

A Software Engineer at Dixit Infotech Services plays a pivotal role in maintaining the technical backbone of our clients' operations. Whether you are working as a Network Engineer, Voice Engineer, or Desktop Support Engineer, your primary objective is to ensure seamless connectivity, robust infrastructure performance, and high-availability systems. You are the frontline expert tasked with diagnosing complex technical issues and implementing scalable solutions that keep business processes running without interruption. This role is critical because Dixit Infotech Services prides itself on delivering reliable IT infrastructure and support services. You will be directly involved in managing network environments, troubleshooting voice systems, or overseeing desktop configurations, requiring a blend of deep technical knowledge and a service-oriented mindset.

This guide is scoped to a Software Engineer candidate at Dixit Infotech Services.

Focus areas

Windows
Microsoft Office
Communication (Interview Process)

The Dixit Infotech Services interview process

Dixit Infotech Services candidates report 2 rounds over 2-4 weeks. The stages below are what candidates describe, not a published process.

Technical Screening

Initial assessment to verify baseline technical knowledge.

What to demonstrate

Initial assessment to verify baseline technical knowledge
Depth in Windows

How to prepare

Answer aloud and timed: Explain the difference between TCP and UDP and when to use each.
Answer aloud and timed: How do you troubleshoot a situation where a user cannot reach a specific website?

Deep-Dive Discussions

In-depth conversations with senior engineering staff or team leads.

What to demonstrate

In-depth conversations with senior engineering staff or team leads
Depth in Windows

How to prepare

Answer aloud and timed: Describe the process of configuring a VLAN on a network switch.
Answer aloud and timed: What are the common steps to resolve a "No Dial Tone" issue in a VOIP environment?

Where candidates lose points

Be honest about your gaps

If you don't know the answer to a technical question, explain how you would go about finding it rather than guessing.

Prepare your stories

Use the STAR method (Situation, Task, Action, Result) to frame your responses to behavioral questions.

Focus on the "Why

When discussing past projects, emphasize why you chose a specific solution over others.

Understand the role

Ensure you know exactly which track you are interviewing for, as the expectations for a Voice Engineer differ significantly from those for a Desktop Support Engineer.

Questions & practice

Merge partitioned event streams into one ordered feed with bounded lateness

Hard · k-way merge · watermarks · out-of-order streams

The read-model service consumes 64 log partitions carrying about 4,000 events per second in total. Each partition is ordered within itself, but partitions drift by up to 30 seconds, and the activity feed must present a tenant's events in occurred_at order. Produce the merge. State its complexity, the buffer it requires in events and in bytes, what happens when one partition is idle, and what you do with an event that arrives after you have already emitted its position. Payloads average 1 KB.

Approach

Merge with a min-heap over the 64 partition heads keyed on (occurred_at, event_id): O(log P) per event and O(n log P) overall. The tie-break on event_id is what makes the output deterministic when two partitions carry the same millisecond, which matters because the feed is paginated and a non-deterministic order reorders pages under the reader.
Emitting the heap head is only correct once every partition has produced everything up to that timestamp, so the emit condition is a watermark: the minimum across partitions of the highest occurred_at seen, less the allowed lateness. Events are held until the watermark passes them, which is what turns individually ordered streams into a jointly ordered one.
Size the buffer from the lateness rather than guessing: 4,000 events per second times 30 seconds is 120,000 buffered events, and at 1 KB each about 120 MB of heap. That number is the real price of the ordering guarantee and belongs in front of whoever asked for it.
Handle the idle partition explicitly, because it fails the feed rather than corrupting it: a partition with no traffic never advances its own maximum, so the watermark freezes and output stops entirely. Either every partition emits a periodic idle marker carrying the broker's current time, or the watermark falls back to wall clock for a partition silent beyond a threshold.

Follow-up

The lateness budget is raised to five minutes. What is the new buffer, and what besides memory changes?
The consumer restarts. Where does it resume from, and what does the feed look like for the first 30 seconds?

Diff a projection against the primary without per-row point reads

Hard · reconciliation · range hashing · throttling

The listing projection has drifted and some rows show a stale version. The primary holds 40,000,000 resource rows across 12,000 tenants while serving 1,200 writes and 14,000 reads per second. The obvious repair, reading each resource row and comparing its version against the projection, is correct and would eventually finish. Explain precisely why it is unacceptable here, then give a diff that finds the differing rows, state its complexity, and make it safe to run against a live primary. Replication lag is usually under 100 ms and is not bounded.

Approach

Quantify the naive cost rather than calling it slow: 40,000,000 point reads at even 0.5 ms each is over five hours serialised, and the only lever is concurrency, which is exactly what you cannot spend. The primary's pool is sized for the write path, and 40,000,000 random reads evict the buffer cache that sustains the 85 percent cache hit rate, so the audit degrades the system it is auditing.
Replace random access with one ordered pass per side. Both sides can be read in (tenant_id, resource_id) order, which is a sequential scan on each and a merge join in O(n) time and O(1) memory. For a dense diff that is the whole answer, and it reads the primary once instead of 40,000,000 times.
For the expected sparse case, compare range hashes instead of rows: partition the key space, compute per range an order-independent aggregate over hash(resource_id, version), compare aggregates, and descend only into ranges that differ. With d differing rows and branching factor B, at most d ranges mismatch per level, so the drill-down examines O(d log_B(n/d)) ranges and reads full rows only in mismatching leaves.
Aggregate with a sum modulo 2^64 or a multiset hash, never XOR. XOR is order-independent but self-cancelling, so two rows wrong in the same way, or a row duplicated on one side, leave the range aggregate matching and the range is declared clean.

Follow-up

The diff reports 900 stale rows. How do you decide between patching those rows and rebuilding the projection from resource_revision?
Same job, but the projection lives in a search index that cannot be scanned in key order. What changes?

Track a rolling failure rate per destination for circuit decisions

Easy · sliding window · ring buffer · circuit breaker

The egress service delivers about 1,500 webhooks per second across roughly 40,000 destinations, each call bounded by a 10 second timeout. Maintain, per destination, the failure rate over the trailing 60 seconds so a caller can ask before dispatch whether the circuit should open. Attempts arrive as (destination_id, finished_at_ms, outcome). Requirement: amortised O(1) per attempt, with total memory bounded by the destination count rather than by traffic. Give the structure, its exact memory, and the rule that stops a destination with three attempts from opening a circuit.

Approach

Name the exact-deque version and then reject it as the default. Holding timestamps and advancing a tail pointer past anything older than now minus 60 seconds is a correct two-pointer window at amortised O(1) per attempt, but its memory tracks in-window traffic, so one destination in a retry storm holds hundreds of thousands of entries while thousands of quiet destinations hold none.
Use a ring of 60 one-second buckets per destination, each bucket a pair of counters for attempts and failures. On an attempt, advance the ring by the elapsed whole seconds, zeroing at most min(elapsed, 60) buckets, then increment the head. That is amortised O(1) with a fixed footprint per destination.
State the footprint: 60 buckets times two 4-byte counters is 480 bytes of payload per destination, so 40,000 destinations is roughly 20 to 25 MB with per-entry overhead, bounded by the catalogue rather than by the rate. The cost is granularity, since the oldest bucket ages out in whole seconds, which is far tighter than the decision needs.
Require a minimum sample before the circuit may open. A destination with three attempts and three failures reads as 100 percent and is not evidence; a floor of roughly 20 attempts in the window makes the ratio meaningful, and below that floor use a run of consecutive failures as the trigger instead.

Follow-up

The fleet is 30 instances and each sees roughly a thirtieth of a destination's traffic. Where does the rate actually live, and what does a per-instance answer get wrong?
A destination answers in 9.5 seconds and succeeds. It is not failing but it is consuming your per-destination concurrency. What signal should open the circuit here?

Explain why the owner filter ignores the listing index

Medium · composite index · query plans · sort elimination

The only index on resource is (tenant_id, status, updated_at DESC, resource_id DESC). A new endpoint returns one user's resources across all statuses, newest created first: WHERE tenant_id = $1 AND owner_user_id = $2 ORDER BY created_at DESC LIMIT 20. On a tenant with 2M rows it takes 900 ms and EXPLAIN shows a sort above a large scan. Explain precisely why the existing index cannot serve it, give the index that can, and state which of these the new index still will not help: owner_user_id alone across tenants; the same query ordered by updated_at. PostgreSQL 16.

Approach

Separate the two jobs an index does. For filtering, a composite btree is seekable only on a left prefix, so with no predicate on status the scan can at best range over tenant_id and test owner_user_id per row; PostgreSQL 16 has no btree skip scan to jump the unconstrained column.
For ordering, the index is sorted by (status, updated_at) within a tenant and not by created_at, so the LIMIT cannot stop early: every matching row is read and then sorted. That is the 'Sort Method: top-N heapsort' line, and it is why the plan reads 2M rows to answer with 20.
Derive the replacement from the access path — equality, equality, then the ordering column: CREATE INDEX CONCURRENTLY ON resource (tenant_id, owner_user_id, created_at DESC). The scan seeks to the (tenant, owner) range and walks 20 entries in order, so the Sort node disappears along with the row-read.
Treat INCLUDE (title, status) as conditional, not free. An index-only scan still visits the heap for any row whose page is not marked all-visible, so on a table taking 1.2k writes/second the win depends on autovacuum keeping the visibility map current, and the wider index costs more on every insert.

Follow-up

90% of rows are status='active'. Would a partial index WHERE status = 'active' change your answer, and for which of the three queries?
A dashboard runs this for 40 owners in one page load. What changes about the design?

Write the update path that detects a concurrent edit

Medium · optimistic concurrency · transactional outbox · audit history

resource carries version INT NOT NULL DEFAULT 1. resource_revision holds revision_id, resource_id, version, actor_user_id, change_kind, patch JSONB, request_id, created_at with UNIQUE (resource_id, version). outbox_event holds aggregate_type, aggregate_id, aggregate_version, event_type, payload, status. A PUT carries the version the client read. Write the exact statements for the single transaction that applies the edit, records the revision and enqueues 'resource.updated', and give the handler's branch on zero affected rows. Then say what PostgreSQL 16 does under READ COMMITTED when two of these updates hit one row at once.

Approach

One transaction, three writes, no network call inside it: UPDATE resource SET title = $3, version = version + 1, updated_at = now() WHERE resource_id = $1 AND tenant_id = $4 AND version = $2; then INSERT the resource_revision row at version $2 + 1; then INSERT the outbox_event row at the same aggregate_version. The event goes to a table rather than a broker because no transaction spans both.
Branch on the affected-row count before doing anything else. Zero has three causes — stale version, wrong tenant, row gone — so re-read once and map to 409 carrying the current version, or 404 for an id outside the caller's tenant, which also stops the endpoint confirming that another tenant's id exists.
State the engine behaviour instead of assuming it. Under READ COMMITTED the second UPDATE blocks on the row lock, and when the first commits PostgreSQL re-evaluates the WHERE clause against the newly committed row, so the version predicate now fails and the statement reports zero rows. Under REPEATABLE READ the identical collision raises SQLSTATE 40001 instead, so the handler must fold both shapes into one conflict response.
Keep UNIQUE (resource_id, version) even though the predicate already serialises writers. It is what makes a lost update unwritable if any other path ever reaches the revision table, and it converts a logic bug into 23505 rather than into a silently missing history row.

Follow-up

A client sends the version it read ten minutes ago and the resource has moved three versions. What is in your 409 so it can resolve the conflict without a full re-fetch?
Two editors, two disjoint fields, no overlap. Does your answer still refuse the second write, and should it?