course-schedule-ii¶

Description¶

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] = [a_i, b_i] indicates that you must take course b_i first if you want to take course a_i.

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 <= a_i, b_i < numCourses
a_i != b_i
All the pairs [a_i, b_i] are distinct.

Solution(Python)¶

from collections import defaultdict, deque


class Solution:
    def findOrder(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: List[int]
        """

        # Prepare the graph
        adj_list = defaultdict(list)
        indegree = {}
        for dest, src in prerequisites:
            adj_list[src].append(dest)

            # Record each node's in-degree
            indegree[dest] = indegree.get(dest, 0) + 1

        # Queue for maintainig list of nodes that have 0 in-degree
        zero_indegree_queue = deque(
            [k for k in range(numCourses) if k not in indegree])

        topological_sorted_order = []

        # Until there are nodes in the Q
        while zero_indegree_queue:

            # Pop one node with 0 in-degree
            vertex = zero_indegree_queue.popleft()
            topological_sorted_order.append(vertex)

            # Reduce in-degree for all the neighbors
            if vertex in adj_list:
                for neighbor in adj_list[vertex]:
                    indegree[neighbor] -= 1

                    # Add neighbor to Q if in-degree becomes 0
                    if indegree[neighbor] == 0:
                        zero_indegree_queue.append(neighbor)

        return (
            topological_sorted_order
            if len(topological_sorted_order) == numCourses
            else []
        )