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I wrote a solution to Course Schedule II - LeetCode

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

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

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

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: 2, [[1,0]] 
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished 

Example 2:

Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.


2. You may assume that there are no duplicate edges in the input prerequisites.

my solution

class Solution:
 def findOrder(self,numCourses, prerequirements):
 """
 :type numCourse: int
 :type prerequirements: List[List[int]]
 :rtype:bool
 """
 #if not prerequirements: return True 
 # if numCourses == None and len(prerequirements) == 0: return True

 L = []

 in_degrees = [ 0 for _ in range(numCourses)] #index as node
 #graph = [[]] * numCourses
 graph = [[] for _ in range(numCourses)]

 #Construct the graph 
 for u, v in prerequirements:
 graph[v].append(u) #index as node
 in_degrees[u] += 1 
 logging.debug(f"graph: graph")
 logging.debug(f"in_degrees in_degrees")
 #
 Q = [i for i in range(len(in_degrees)) if in_degrees[i]==0] #collect nodes without pre-edges
 logging.debug(f"Q: Q")

 while Q: #while Q is not empty 
 start = Q.pop()#remove a node from Q
 L.append(start) #add n to tail of L
 logging.debug(f"L: L")

 for v in graph[start]:#for each node v with a edge e 
 in_degrees[v] -= 1 #remove edge 
 if in_degrees[v] == 0:
 Q.append(v)
 logging.debug(f"indegree: in_degrees")
 #check there exist a cycle
 for i in range(len(in_degrees)): #if graph has edge 
 if in_degrees[i] > 0: 
 return [] 
 logging.debug(f"L: L")
 return L

TestCase:

class MyCase(unittest.TestCase):
 def setUp(self):
 self.solution1 = Solution()
 self.solution2 = Solution2()

 def test_bfs1(self):
 numCourse = 2
 prerequirements = [[1,0]] 
 check = self.solution1.findOrder(numCourse, prerequirements)
 logging.debug(f"check")
 answer = [0, 1] 
 self.assertEqual(check, answer)


 def test_bfs2(self):
 numCourse = 4
 prerequirements = [[1,0],[2,0],[3,1],[3,2]]
 check = self.solution1.findOrder(numCourse, prerequirements)
 logging.debug(f"check")
 answer = [[0,1,2,3], [0,2,1,3]] 
 self.assertIn(check, answer)


 def test_bfs3(self):
 numCourse = 2
 prerequirements = []
 check = self.solution1.findOrder(numCourse, prerequirements)
 logging.debug(f"check")
 answer = [1,0]
 self.assertEqual(check, answer)


 def test_bfs4(self):
 numCourse = 2
 prerequirements = [[0,1],[1,0]]
 check = self.solution1.findOrder(numCourse, prerequirements)
 logging.debug(f"check")
 answer = []
 self.assertEqual(check, answer)

Get a low score

Runtime: 56 ms, faster than 57.28% of Python3 online submissions for Course Schedule II.
Memory Usage: 14 MB, less than 51.41% of Python3 online submissions for Course Schedule II.