Went straight for DFS and fell apart on the visited state logic.
Model the courses and prerequisites as a directed graph where edges represent dependencies. Then, detect whether the graph contains a cycle using either Kahn's algorithm (BFS-based topological sort) or DFS with recursion stack. If a cycle exists, it's impossible to complete all courses; otherwise, it's possible.
Pro tip: Discuss the trade-offs between Kahn's algorithm and DFS: Kahn's is iterative and avoids recursion depth issues, while DFS can be simpler to implement but may hit stack limits for large graphs. Also, mention that this problem is equivalent to checking if the course dependency graph is a DAG (Directed Acyclic Graph).
Confirm that the input is a number of courses (n) and a list of prerequisite pairs [a, b] meaning b must be taken before a. Ensure understanding that we need to return true if all courses can be finished, false otherwise.
Construct an adjacency list where each course points to its dependents (or prerequisites). Also compute the in-degree for each node if using Kahn's algorithm.
Use either Kahn's algorithm (repeatedly remove nodes with in-degree 0 and count processed nodes) or DFS with a recursion stack to detect back edges. If all nodes are processed (Kahn's) or no back edge is found (DFS), there is no cycle.
If a cycle is detected, return false (impossible to complete all courses). Otherwise, return true.
State that both approaches run in O(V + E) time and O(V + E) space, where V is the number of courses and E is the number of prerequisites.
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