Course Schedule II

LeetCode Q 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.

Example 1:
Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]] ; Output: [0,1,2,3] or [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] .

Note:

• The input prerequisites is a graph represented by a list of edges, not adjacency matrices.
• You may assume that there are no duplicate edges in the input prerequisites.

Solution

Solution 1 : BFS

Code:

int[] res;
int count;
public int[] findOrder(int numCourses, int[][] reqs) {
	res = new int[numCourses];
	count = 0;

	// build graph
	List<Integer>[] graph = new ArrayList[numCourses];
	int[] order = new int[numCourses];

	for (int i = 0; i < numCourses; i++)
		graph[i] = new ArrayList<Integer>();
	for (int[] req: reqs) { 
		graph[req[1]].add(req[0]);
		order[req[0]]++;    
	}
	
	Queue<Integer> que = new LinkedList<>();
	for (int i = 0; i < numCourses; i++) {
		if (order[i] == 0) que.offer(i);
	}

	while (!que.isEmpty()) {
		int curr = que.poll();
		res[count++] = curr;

		for (int next: graph.get(curr)) {
			order[next]--;
			if (order[next] == 0) que.offer(next);
		}
	}

	return count == numCourses ? res : new int[0];
	
}

Solution 2 : DFS

Code:

int[] res;
int count;
public int[] findOrder(int numCourses, int[][] reqs) {
	res = new int[numCourses];
	count = numCourses - 1;

	// build graph
	List<Integer>[] graph = new ArrayList[numCourses];
	int[] order = new int[numCourses];

	for (int i = 0; i < numCourses; i++)
		graph[i] = new ArrayList<Integer>();
	for (int[] req: reqs)
		graph[req[1]].add(req[0]);
	
	int[] status = new int[numCourses];

	for (int i = 0; i < numCourses; i++) {
		if (status[i] == 2) continue;
		if (hasCycle(graph, i, status))
			return new int[0];
	}

	return res;
}

private boolean hasCycle (List<Integer>[] graph, int course, int[] status) {
	if (status[course] == 1) return false;

	status[course] = 1;
	for (int next: graph[course]) {
		if (status[i] == 2) continue;
		if (hasCycle(graph, next, status))
			return true;
	}

	status[course] = 2;
	res[count--] = course;

	return false;
}