Course Schedule II

Problem

https://leetcode.com/problems/course-schedule-ii/

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:sub:`i`, b:sub:`i`] indicates that you must take course b:sub:`i` first if you want to take course a:sub:`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:sub:`i`, b:sub:`i`< numCourses

• a:sub:`i`!= b:sub:`i`

• All the pairs [a:sub:`i`, b:sub:`i`] are distinct.

Pattern

Depth-First Search, Breadth-First Search, Graph Theory, Topological Sort

Approaches

Topological Sort

Explanation

We can use topological sort to produce an ordering from a graph that respects the order of prerequisites. First, convert the list of prerequisites from an edge list into an adjacency list while also tracking the in degree of each node in a dictionary in_degree (the number of remaining prereqs). We can only start a course when all its prereqs are finished, i.e. in-degree = 0. We add these to a queue. As we take each course by popping from the queue, we reduce the in-degree of the downstream courses by 1. If those courses reach an in-degree of 0, then we add them to the queue. The set of taken courses is a valid course path.

Code

from collections import deque


def findOrder(numCourses: int, prerequisites: list[list[int]]) -> list[int]:
    """Return an ordering of courses that satisfies ``prerequisites``, or an
    empty list if it is impossible to finish all courses.
    """
    graph = {course: [] for course in range(numCourses)}
    in_degrees = {course: 0 for course in range(numCourses)}

    for [a, b] in prerequisites:
        graph[b].append(a)
        in_degrees[a] += 1

    queue = deque()
    for node, in_degree in in_degrees.items():
        if in_degree == 0:
            queue.append(node)

    courses = []
    while queue:
        node = queue.popleft()
        courses.append(node)

        for neighbor in graph[node]:
            in_degrees[neighbor] -= 1
            if in_degrees[neighbor] == 0:
                queue.append(neighbor)

    if len(courses) == numCourses:
        return courses
    else:
        return []

Test

>>> from course_schedule_ii__topological_sort import findOrder
>>> findOrder(2, [[1, 0]])
[0, 1]
>>> findOrder(4, [[1, 0], [2, 0], [3, 1], [3, 2]]) in ([0, 1, 2, 3], [0, 2, 1, 3])
True
>>> findOrder(1, [])
[0]

Complexity

\(V\) is the number of courses and \(E\) is the length of prerequisites.

Measure	Complexity	Notes
Time	\(O(V + E)\)	every edge and vertex is visited once
Auxiliary Space	\(O(V + E)\)	we create a graph dictionary which stores all the edges and a taken courses array which stores all the courses

course_schedule_ii__topological_sort.findOrder(numCourses: int, prerequisites: list[list[int]]) → list[int]

Return an ordering of courses that satisfies prerequisites, or an empty list if it is impossible to finish all courses.