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Ids International Software Engineer Interview Guide

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

The 60-second brief

A Software Engineer at Ids International plays a pivotal role in bridging the gap between complex systems engineering and software-defined solutions. This position is essential to the company’s mission of delivering advanced technical capabilities, often within the specialized domains of Unmanned Aircraft Systems (UAS), radio frequency environments, and mission-critical systems. You are not just writing code; you are building the intelligence that drives hardware performance and operational reliability. The work environment at Ids International is highly specialized and technical, requiring engineers who can navigate the intersection of software architecture and physical systems. Success in this role demands a high degree of precision, as your contributions directly influence the performance and safety of high-stakes technical products.

This guide is scoped to a Software Engineer candidate at Ids International.

Focus areas

  • UAS Engineering (Unmanned Aircraft Systems)
  • Radio Frequency (RF) / Cellular Engineering
  • Wireless Communications

The Ids International interview process

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

Initial Screening

The process begins with an initial screening to assess your overall fit for the role.

What to demonstrate

  • The process begins with an initial screening to assess your overall fit for the role
  • Depth in UAS Engineering (Unmanned Aircraft Systems)

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 Discussions

Deeper technical discussions follow, focusing on your problem-solving process and technical capabilities.

What to demonstrate

  • Deeper technical discussions follow
  • Focusing on your problem-solving process and technical capabilities

How to prepare

  • Answer aloud and timed: Describe your process for optimizing software performance within resource-constrained environments.
  • Answer aloud and timed: What is your experience with the lifecycle of UAS software development?

Where candidates lose points

Contextualize your experience

When discussing past projects, always highlight the specific constraints you faced, such as limited hardware resources or strict latency requirements.

Be ready to pivot

If an interviewer challenges your approach, remain calm and explain the trade-offs you made. They are often testing your ability to defend your design choices under pressure.

Stay current

Ensure your knowledge of the specific technical domain (e.g., UAS or RF engineering) is up to date, as the team will expect you to speak the language of their industry.

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?

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?

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

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

Keep soft-deleted accounts from blocking re-registration

Easy · soft delete · unique constraints · engine differences

app_user holds user_id, tenant_id, email CITEXT, password_hash (NULL for SSO principals), email_verified_at, auth_version, status ('invited','active','suspended','deactivated'), created_at, updated_at, deleted_at. Two live accounts for one address inside a tenant must be impossible, but an address freed by a soft delete must be reusable, and the same tenant may delete and re-register it repeatedly. Write the uniqueness DDL for PostgreSQL 16, then the equivalent for MySQL 8 where partial indexes do not exist, and say what each permits once three deleted rows already hold that address.

Approach

  1. Start from what is actually unique: not (tenant_id, email), but (tenant_id, email) among live rows. PostgreSQL says that directly — CREATE UNIQUE INDEX app_user_live_email ON app_user (tenant_id, email) WHERE deleted_at IS NULL. A full constraint over the same two columns burns the address permanently the first time someone deletes an account.
  2. Keep case-insensitivity in the type or the index, never in the application: CITEXT as given, or UNIQUE (tenant_id, lower(email)) as an expression index where the extension is unavailable. A case-sensitive unique column is exactly how two accounts for one human appear.
  3. For MySQL 8 the predicate has to move inside the key: add a discriminator column that is a constant 0 while the row is live and is set to user_id on delete, with UNIQUE (tenant_id, email, deleted_marker). Live rows share the constant and still collide; deleted rows differ from each other and stop colliding.
  4. State the NULL variant and its dependency: leaving the marker NULL for deleted rows also works, because a unique index treats NULLs as distinct — true in MySQL, and true in PostgreSQL only under the default NULLS DISTINCT, which PostgreSQL 15 lets you reverse. Check the polarity against the three existing deleted rows: constant-on-live is what preserves the collision you want, and reversing it silently admits duplicate live accounts.

Follow-up

  • A deleted account re-registers with the same address the next day. Do the old resource rows follow the new user_id, and how does the API keep the two principals apart?
  • How do you honour an erasure request while resource_revision.actor_user_id still references this table?

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?

Describe your process for optimizing software performance within resource-constrained environments.

Medium · Technical and Domain Knowledge

Approach

  1. Clarify what is being asked and what a complete answer contains.
  2. State your assumptions explicitly before working the problem.
  3. Say what you would check first and why it is the highest-information step.
  4. Work from the requirement backwards to the design.

Follow-up

  • What assumption would you test first?
  • How would you know your answer was wrong?

How do you ensure your code remains maintainable when working on complex, multi-layered systems?

Medium · Technical and Domain Knowledge

Approach

  1. Clarify what is being asked and what a complete answer contains.
  2. State your assumptions explicitly before working the problem.
  3. Say what you would check first and why it is the highest-information step.
  4. Work from the requirement backwards to the design.

Follow-up

  • What assumption would you test first?
  • How would you know your answer was wrong?

How do you approach debugging code that interacts directly with hardware or radio frequency modules?

Medium · Technical and Domain Knowledge

Approach

  1. Establish what changed and when, before forming any theory.
  2. Pick a bisection that eliminates candidates whichever way it turns out.
  3. Check the instrumentation before believing the symptom.
  4. Separate the trigger from the cause; the deploy is rarely the bug.

Follow-up

  • What would you look at first, and what would it rule out?
  • How would you tell a cause from a coincidence here?

Suggested practice checklist

Built from the rounds and topics Ids International candidates report.

Day 1 — Map the Ids International loop

  • Write out the reported sequence: Initial Screening, Technical Discussions.
  • 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 2 reported rounds, with the weakest marked.

Day 2 — Work UAS Engineering (Unmanned Aircraft Systems)

  • Spend the session on UAS Engineering (Unmanned Aircraft Systems), which Ids International candidates report being tested on.
  • Write one worked example in UAS Engineering (Unmanned Aircraft Systems) and time yourself on it.

Deliverable: One timed worked example in UAS Engineering (Unmanned Aircraft Systems).

Day 3 — Work Radio Frequency (RF) / Cellular Engineering

  • Spend the session on Radio Frequency (RF) / Cellular Engineering, which Ids International candidates report being tested on.
  • Write one worked example in Radio Frequency (RF) / Cellular Engineering and time yourself on it.

Deliverable: One timed worked example in Radio Frequency (RF) / Cellular Engineering.

Day 4 — Work Wireless Communications

  • Spend the session on Wireless Communications, which Ids International candidates report being tested on.
  • Write one worked example in Wireless Communications and time yourself on it.

Deliverable: One timed worked example in Wireless Communications.

Day 5 — Answer out loud: Technical and Domain Knowledge

  • Answer aloud, timed: Explain your experience with systems integration in a software-defined environment.
  • Answer aloud, timed: How do you approach debugging code that interacts directly with hardware or radio frequency modules?

Deliverable: Spoken answers to 2 reported Technical and Domain Knowledge 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 Ids International

  • 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.

Explain your experience with systems integration in a software-defined environment.

Medium · Technical and Domain Knowledge

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?

What is your experience with the lifecycle of UAS software development?

Medium · Technical and Domain Knowledge

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?

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?