Course Schedule

Problem:

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.

Analysis:

Use topological sort to check whether the graph has cycle.

See the about graph. There is a cycle. Visit 0 first, reduce 1's indegree by 1. Even so, there is no node that has indegree == 0. The DFS can only get 1 node.

Solution:
Use matrix to store graph is easier to implement.

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int[][] graph = new int[numCourses][numCourses]; // j -> i
        int[] indegree = new int[numCourses];
        int count= 0;
        // in degree 
        for (int i = 0; i < prerequisites.length; i++) {
            int pre = prerequisites[i][1];
            int cur = prerequisites[i][0];
            if (graph[cur][pre] == 0) {
                indegree[cur]++;
            }
            graph[cur][pre] = 1;
        }
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0) {
                queue.offer(i);
            }
        }
        
        while(!queue.isEmpty()) {
            int cur = queue.poll();
            count++;
            // find cur's neighbours and reduce degree by 1
            for (int i = 0; i < numCourses; i++) {
                if (graph[i][cur] == 1) {
                    if (--indegree[i] == 0) {
                        queue.offer(i);
                    }
                }
            }
        }

        return count == numCourses;

    }
}