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.

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.

Solution:

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<Set<Integer>> graph = new ArrayList<>();
        for(int i=0; i<numCourses; i++) 
            graph.add(new HashSet<Integer>());
        int[] in = new int[numCourses];
        for(int[] e : prerequisites) 
            if(graph.get(e[1]).add(e[0])) 
                in[e[0]]++;
        Queue<Integer> visit = new LinkedList<>();
        for(int i=0; i<numCourses; i++) 
            if(in[i]==0) visit.add(i);
        int n = 0;
        while(!visit.isEmpty()){
            Integer u = visit.poll(); /* u->v */
            n++;
            for(Integer v : graph.get(u)){
                in[v]--;
                if(in[v]==0) visit.add(v);
            }
        }
        return n==numCourses;
    }
}