Find user who can access every camera

Q: Find user who can access every camera

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

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Q: What difficulty level is this interview question?

This is a medium difficulty Coding & Algorithms question, commonly asked during Onsite rounds at Verkada.

Q: What role is this question designed for?

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

Question

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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).