# Kahn's algorithms to arrange CouseSchedule

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}")

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}")

def test_bfs3(self):
numCourse = 2
prerequirements =  []
check = self.solution1.findOrder(numCourse, prerequirements)
logging.debug(f"{check}")

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


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.

• Your code uses less time and less memory than a half of all submissions in the same language. It's certainly not a low score. – CiaPan Apr 3 '19 at 6:59

• The docstring is wrong. findOrder returns a list, not a boolean.

• Few inefficiencies:

• The cycle checking implementation is way suboptimal. The algorithm terminates successfully if every node lands in L. It is enough to test that len(L) = len(in_degrees). No need to loop.

• You don't need Q. You may work directly with L. Less memory, and less copying. Admittedly, the resulting code may look a bit scary (list is modified while being iterated on), but append appears to be safe. You may find this discussion interesting.

• [i for i in range(len(in_degrees)) if in_degrees[i]==0] doesn't look Pythonic. Consider

    [i for i, n in enumerate(in_degrees) if n == 0]

• To my taste, the code is commented too much.

- Nitpick. Strictly speaking, the problem statement doesn't guarantee that the courses IDs are dense. A schedule may look like 3, [[100, 200], [200, 300]]. In such case case graph[v].append(u) will raise an IndexError. Consider a dictionary instead of a list as a graph representation.

• Strictly speaking, the problem statement does guarantee it: There are a total of n courses you have to take, labeled from 0 to n-1. However, it does not specify the possible range for n; specifically, it doesn't guarantee n will fit in the standard int type. – CiaPan Apr 3 '19 at 7:06
• @CiaPan Thanks. Misread the PS. Edited. – vnp Apr 3 '19 at 7:16