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This question is commonly asked for Software Engineer candidates at Verkada during technical interviews.

Find user who can access every camera | Verkada Coding Question

Quick Overview

This question evaluates reasoning about access-control and transitive relationships, testing skills in graph modeling, reachability, and set operations within permission hierarchies.

Find user who can access every camera

Company: Verkada

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Onsite

You are given permissions data describing which **users** and **groups** have ownership access to which **cameras**, and which entities are members of which **groups**. Each permission record is a triple: - `(entity_id, relationship, object_id)` Where: - `relationship` is one of: - `"camera_owner"`: `entity_id` (a user or a group) directly has access to `object_id` (a camera). - `"group_member"`: `entity_id` (a user or a group) is a member of `object_id` (a group). ### Access rules - A user can access a camera if they: - own it directly (`user -> camera_owner -> camera`), or - are a member of a group that owns it, including through **nested group membership**. - Groups can be nested to multiple levels. - There are **no cycles** in group membership. - Users can be members of multiple groups. - Some users may not belong to any group. ### Task Return the ID of the **single Admin user**—defined as the *only* user who can access **every camera referenced anywhere in the data** (i.e., every `camera_*` appearing as the `object_id` of a `"camera_owner"` record). It is guaranteed that **exactly one** such admin user exists. ### Input - `permissionsData`: an array/list of triples `(entity_id, relationship, object_id)`. - `entity_id` is a string like `"user_1"` or `"group_2"`. - `object_id` is a string like `"camera_3"` (for `camera_owner`) or `"group_7"` (for `group_member`). ### Output - A string user ID (e.g., `"user_1"`) for the admin user. ### Example Given: - `("user_1", "camera_owner", "camera_1")` - `("user_1", "group_member", "group_1")` - `("group_1", "camera_owner", "camera_2")` - `("group_1", "group_member", "group_2")` - `("group_2", "camera_owner", "camera_3")` Then `user_1` can access `camera_1` directly, `camera_2` via `group_1`, and `camera_3` via `group_1 -> group_2`, so `user_1` would be the admin in this dataset. ### Constraints - Must run within typical interview-scale limits (e.g., a few seconds).

Quick Answer: This question evaluates reasoning about access-control and transitive relationships, testing skills in graph modeling, reachability, and set operations within permission hierarchies.

You are given a list of permission triples (entity_id, relationship, object_id). A record with relationship 'camera_owner' means the entity directly has access to a camera. A record with relationship 'group_member' means the entity belongs to a group. Entities can be users or groups, and groups can be nested to multiple levels without cycles. A user can access a camera if they own it directly or inherit access through any chain of group memberships. Return the ID of the single admin user: the only user who can access every camera that appears as the object_id of any 'camera_owner' record.

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Examples

Input: [('user_1', 'camera_owner', 'camera_1'), ('user_1', 'group_member', 'group_1'), ('group_1', 'camera_owner', 'camera_2'), ('group_1', 'group_member', 'group_2'), ('group_2', 'camera_owner', 'camera_3'), ('user_2', 'group_member', 'group_1')]

Expected Output: 'user_1'

Explanation: user_1 has camera_1 directly, camera_2 through group_1, and camera_3 through group_1 -> group_2. user_2 only gets camera_2 and camera_3.

Input: [('user_9', 'camera_owner', 'camera_42')]

Expected Output: 'user_9'

Explanation: This is the single-record edge case. The only referenced camera is camera_42, and user_9 owns it directly.

Input: [('group_1', 'camera_owner', 'camera_1'), ('group_2', 'camera_owner', 'camera_2'), ('user_1', 'group_member', 'group_1'), ('user_1', 'group_member', 'group_2'), ('user_2', 'group_member', 'group_1'), ('user_3', 'group_member', 'group_2')]

Expected Output: 'user_1'

Explanation: user_1 combines access from two groups and reaches both cameras. user_2 and user_3 each reach only one.

Input: [('group_3', 'camera_owner', 'camera_4'), ('group_2', 'group_member', 'group_3'), ('group_1', 'group_member', 'group_2'), ('user_7', 'group_member', 'group_1'), ('user_7', 'camera_owner', 'camera_1'), ('group_1', 'camera_owner', 'camera_2'), ('group_2', 'camera_owner', 'camera_3'), ('user_8', 'group_member', 'group_2')]

Expected Output: 'user_7'

Explanation: user_7 gets camera_1 directly and inherits camera_2, camera_3, and camera_4 through group_1 -> group_2 -> group_3. user_8 only gets camera_3 and camera_4.

Input: [('group_1', 'camera_owner', 'camera_1'), ('group_1', 'camera_owner', 'camera_1'), ('group_2', 'camera_owner', 'camera_2'), ('group_1', 'group_member', 'group_2'), ('user_1', 'group_member', 'group_1'), ('user_1', 'group_member', 'group_1'), ('user_2', 'camera_owner', 'camera_1')]

Expected Output: 'user_1'

Explanation: Duplicate records should not change the result. user_1 gets camera_1 from group_1 and camera_2 through group_1 -> group_2, while user_2 only has camera_1.

Quick Overview

Find user who can access every camera

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