course-schedule-ii.py

# Time:  O(|V| + |E|)
# Space: O(|E|)

# 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.
#
# 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 the correct course order is [0,1]
#
# 4, [[1,0],[2,0],[3,1],[3,2]]
# 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:
# The input prerequisites is a graph represented by a list of edges, not adjacency matrices.
# Read more about how a graph is represented.
#
# Hints:
# 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.
# 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.
#

import collections


class Solution(object):
    def findOrder(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: List[int]
        """
        res, zero_in_degree_queue, in_degree, out_degree = [], collections.deque(), {}, {}

        for i, j in prerequisites:
            if i not in in_degree:
                in_degree[i] = set()
            if j not in out_degree:
                out_degree[j] = set()
            in_degree[i].add(j)
            out_degree[j].add(i)

        for i in xrange(numCourses):
            if i not in in_degree:
                zero_in_degree_queue.append(i)

        while zero_in_degree_queue:
            prerequisite = zero_in_degree_queue.popleft()
            res.append(prerequisite)

            if prerequisite in out_degree:
                for course in out_degree[prerequisite]:
                    in_degree[course].discard(prerequisite)
                    if not in_degree[course]:
                        zero_in_degree_queue.append(course)

                del out_degree[prerequisite]

        if out_degree:
            return []

        return res