207. 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:

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 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.
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:

This is a topological order question.

It asks whether the courses can be formed a topological order.

Therefore we use a count variable to store the number of courses that can form such order.

If it equals to the total number of courses, we return true.



Code:

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<Integer>[] adj = new List[numCourses];
        for (int i = 0; i < numCourses; i++) {
            adj[i] = new ArrayList<Integer>();
        }
        int[] degree = new int[numCourses];
        for (int[] pair : prerequisites) {
            int courseFrom = pair[1];
            int courseTo = pair[0];
            adj[courseFrom].add(courseTo);
            degree[courseTo]++;
        }
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < degree.length; i++) {
            if (degree[i] == 0) {
                queue.offer(i);
            }
        }
        int count = 0;
        while (!queue.isEmpty()) {
            int course = queue.poll();
            count++;
            for (int nei : adj[course]) {
                degree[nei]--;
                if (degree[nei] == 0) {
                    queue.offer(nei);
                }
            }
        }
        return count == numCourses;
    }
}