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Published: Mar 23, 2021 License: MIT Imports: 0 Imported by: 0

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210. Course Schedule II (Medium)

There are a total of n courses you have to take labelled from 0 to n - 1.

Some courses may have prerequisites, for example, if prerequisites[i] = [ai, bi] this means you must take the course bi before the course ai.

Given the total number of courses numCourses and a list of the prerequisite pairs, return the ordering of courses you should take to finish all courses.

If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].

Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]

 

Constraints:

  • 1 <= numCourses <= 2000
  • 0 <= prerequisites.length <= numCourses * (numCourses - 1)
  • prerequisites[i].length == 2
  • 0 <= ai, bi < numCourses
  • ai != bi
  • All the pairs [ai, bi] are distinct.

[Depth-first Search] [Breadth-first Search] [Graph] [Topological Sort]

Similar Questions

  1. Course Schedule (Medium)
  2. Alien Dictionary (Hard)
  3. Minimum Height Trees (Medium)
  4. Sequence Reconstruction (Medium)
  5. Course Schedule III (Hard)

Hints

Hint 1 This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
Hint 2 Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Hint 3 Topological sort could also be done via BFS.

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