Course Schedule

There are numCourses courses labeled from 0 through numCourses - 1. Each pair [course, prerequisite] means that prerequisite must be completed before course.
Return true if all courses can be completed. Return false if the prerequisite relationships contain a directed cycle.

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

Examples
Example 1
numCourses = 2
prerequisites = [[1,0]]
return = true
Course 0 can be completed before course 1.
Example 2
numCourses = 2
prerequisites = [[1,0],[0,1]]
return = false
Each course requires the other first, so the graph contains a cycle.
Example 3
numCourses = 4
prerequisites = [[1,0],[2,1],[3,2]]
return = true
The courses can be completed in the order 0, 1, 2, 3.

Constraints
1 <= numCourses <= 2000
0 <= prerequisites.length <= 5000
Every pair contains two valid course labels from 0 through numCourses - 1.

Build an adjacency list from the pairs, with an edge from prerequisite to course. Then pick a cycle detection method. The cleanest is Kahn's algorithm, a BFS topological sort. Count in-degrees, push every course with in-degree 0 into a queue, pop one, decrement its neighbors, and push any that hit 0. If you process all numCourses, return true. Otherwise a cycle blocked some courses, so return false. The DFS alternative uses three states per node: unvisited, visiting, done. Hitting a visiting node means a cycle. The common pitfall is using a plain visited set, which flags diamond shapes as cycles. Another is forgetting that courses with no prerequisites still count. Constraints are small, up to 2000 courses and 5000 edges, so O(V+E) is easily fast enough. If your mind goes blank mid-assessment, StealthCoder can surface the full solution so you don't lose the round over syntax.

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