207.md

207. Course Schedule

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, is it possible for you to finish all courses?

For example:

2, [[1,0]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
click to show more hints.

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

my thoughts:

1. as in hints, check if there is a topological cycle by counting visited nodes.

my solution:

**********
class Solution:
    def canFinish(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        if len(prerequisites) > numCourses * (numCourses - 1) // 2:
            return False
        d = {}
        for p in prerequisites:
            d[p[0]] = d.get(p[0], []) + [p[1]]
            
        if len(d) >= numCourses:
            return False
        
        for key in d:
            pre = d[key]
            count = 1
            while pre and count <= numCourses:
                temp = []
                for p in pre:
                    if p not in d:
                        continue
                    else:
                        temp.extend(d[p])
                    count += 1
                pre = temp
            if pre:
                return False
        return True

my comments:

from other ppl's solution:

1. N/A