Course Schedule

There are numCourses courses labeled from 0 to numCourses - 1. Each pair [course, prerequisite] means the prerequisite must be completed before the course.
Return true if all courses can be completed, or false if the prerequisite graph contains a cycle.
Interview Follow-up
The interviewer asked follow-up questions about alternative solutions, edge cases, and complexity analysis.

Function
canFinish(numCourses: int, prerequisites: int[][]) → boolean

Examples
Example 1
numCourses = 2
prerequisites = [[1,0]]
return = true
Example 2
numCourses = 2
prerequisites = [[1,0],[0,1]]
return = false
Each course requires the other first, creating a cycle.

Build an adjacency list from the prerequisites, then detect a cycle. Two clean ways. First, Kahn's algorithm: compute in-degrees, push every course with in-degree 0 into a queue, pop and decrement neighbors, and count how many you process. If the count equals numCourses, return true. Second, DFS with three states: unvisited, visiting, visited. Hitting a visiting node means a cycle. The common pitfall is using a plain visited set, which flags diamond shapes as cycles when they aren't. Another is flipping the edge direction, though for a pure cycle check it doesn't matter. Watch for empty prerequisites and disconnected components, since you must start from every node. Complexity is O(V + E) time and space. If you freeze mid-assessment, StealthCoder is the hedge that reads the problem and hands you the code.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.