210. Course Schedule II

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:

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.

Hints:

1. 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.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

Solution:

Kahn's algorithm

One of these algorithms, first described by Kahn (1962), works by choosing vertices in the same order as the eventual topological sort. First, find a list of "start nodes" which have no incoming edges and insert them into a set S; at least one such node must exist in a non-empty acyclic graph. Then:

L ← Empty list that will contain the sorted elements
S ← Set of all nodes with no incoming edges
while S is non-empty do
    remove a node n from S
    add n to tail of L
    for each node m with an edge e from n to m do
        remove edge e from the graph
        if m has no other incoming edges then
            insert m into S
if graph has edges then
    return error (graph has at least one cycle)
else 
    return L (a topologically sorted order)

If the graph is a DAG, a solution will be contained in the list L (the solution is not necessarily unique). Otherwise, the graph must have at least one cycle and therefore a topological sorting is impossible.

Reflecting the non-uniqueness of the resulting sort, the structure S can be simply a set or a queue or a stack. Depending on the order that nodes n are removed from set S, a different solution is created. A variation of Kahn's algorithm that breaks ties lexicographically forms a key component of the Coffman–Graham algorithm for parallel scheduling and layered graph drawing.

Code:

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        ArrayList<Integer>[] adj = new ArrayList[numCourses];
        int[] indegree = new int[numCourses];
        int[] order = new int[numCourses];
        
        for (int i = 0; i < numCourses; i++) {
            adj[i] = new ArrayList<Integer>();
        }
        
        for (int[] pre : prerequisites) {
            adj[pre[1]].add(pre[0]);
            indegree[pre[0]]++;
        }
        
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0) {
                queue.offer(i);
            }
        }
        
        int count = 0;
        while (!queue.isEmpty()) {
            int course = queue.poll();
            order[count++] = course;
            for (int neighbor : adj[course]) {
                indegree[neighbor]--;
                if (indegree[neighbor] == 0) {
                    queue.offer(neighbor);
                }
            }
        }
        
        if (count == numCourses) {
            return order;
        }
        
        return new int[0];
    }
}