LeetCode - Course Schedule II

There are a total of n courses you have to take labelled from 0 to n - 1.

Some courses may have prerequisites, for example, if prerequisites[i] = [ai, bi] this means you must take the course bi before the course ai.

Given the total number of courses numCourses and a list of the prerequisite pairs, return the ordering of courses you should take to finish all courses.

If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

 

Example 1:

Input: numCourses = 2, prerequisites = [[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: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [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].
Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]
 

Constraints:

1 <= numCourses <= 2000
0 <= prerequisites.length <= numCourses * (numCourses - 1)
prerequisites[i].length == 2
0 <= ai, bi < numCourses
ai != bi
All the pairs [ai, bi] are distinct.

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        Map<Integer, List<Integer>> adj = new HashMap<>();
        // set up adjacency list
        for (int i = 0; i < numCourses; i++) {
            adj.put(i, new ArrayList<Integer>());
        }
        // put value in the adjacency list
        for (int[] prereq : prerequisites) {
            adj.get(prereq[1]).add(prereq[0]);
        }
        List<Integer> res = new ArrayList<>();
        int[] visited = new int[numCourses];
        for (int i = 0; i < numCourses; i++) {
            if(!dfs(adj, res, visited, i)) {
                return new int[0];
            }
        }
        return listToArray(res);
    }
    // 1 visited
    // 0 unvisited
    // -1 visiting
    public boolean dfs (Map<Integer, List<Integer>> adj, List<Integer> res, int[] visited, int i) {
        if (visited [i] == 1) {
            return true;
        }
        if (visited [i] == -1) {
            return false;
        }
        
        visited [i] = -1;
        
        if (adj.containsKey(i)) {
            for (int j : adj.get(i)) {
                if (!dfs(adj, res, visited, j)) {
                    return false;
                }
            }
        }
        visited [i] = 1;
        res.add(i);
        return true;
        
    }
    
    private int[] listToArray(List<Integer> list) {
        int size = list.size();
        int[] out = new int[list.size()];
        int i = 0;
        for(Integer course : list) {
            // (size - 1) we need reverse order form the first course to the last. DFS returned us from last to first
            out[(size - 1) - i] = course; 
            i++;
        }
        return out;
    }
}