LeetCode 210. Course Schedule II

Description

https://leetcode.com/problems/course-schedule-ii/

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

  • For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

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.

Explanation

Similar to the problem Course Schedule I, we can use topological sort to solve this problem. Base on the indegree values for the courses to find out the result.

Python Solution

class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        
        results = []
        
        adjacent_list = []
        indegrees = []
        
        for i in range(numCourses):
            adjacent_list.append([])
            indegrees.append(0)
            
            
        for prerequisite in prerequisites:            
            adjacent_list[prerequisite[1]].append(prerequisite[0])
            indegrees[prerequisite[0]] += 1
            
        queue = []
        
        for i, indegree in enumerate(indegrees):
            if indegree == 0:
                queue.append(i)
                
        while queue:
            course = queue.pop(0)
            results.append(course)
            
            for adj_course in adjacent_list[course]:
                indegrees[adj_course] -= 1
                
                if indegrees[adj_course] == 0:
                    queue.append(adj_course)
                                    
        if len(results) < numCourses:
            return []
        
        return results
        
  • Time Complexity: ~V + E
  • Space Complexity: ~V + E

where V is the number of courses, and E is the number of prerequisites.

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