Answer: (iii) $O(V+E)$

Explanation:

• Topological sort via DFS: run DFS on DAG, push vertex onto stack when finished (post-order), then reverse. Each vertex visited once, each edge examined once.
• Time = $O(V)$ for initialization + $O(V+E)$ for DFS = $O(V+E)$.
• Kahn's algorithm (BFS with indegree) is also $O(V+E)$.
• $O(V^2)$ / $O(E^2)$ would be for adjacency matrix without adjacency lists; with adjacency list it's linear.

Correct choice: (iii) $O(V+E)$