There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.
Return true if you can finish all courses. Otherwise, return false.

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take. 
To take course 1 you should have finished course 0. So it is possible.
Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: 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.
 

Constraints:

1 <= numCourses <= 105
0 <= prerequisites.length <= 5000
prerequisites[i].length == 2
0 <= ai, bi < numCourses
All the pairs prerequisites[i] are unique.

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        // check corner cases
        if (prerequisites.length == 0) {
            return true;
        }
        // set up adjacency list
        Map<Integer, List<Integer>> adj = new HashMap<>();
        for (int i = 0; i < numCourses; i++) {
            adj.put(i, new ArrayList<Integer>());
        }
        // populate the adjacency list with all nodes' neighbors
        for (int i = 0; i < prerequisites.length; i++) {
            adj.get(prerequisites[i][1]).add(prerequisites[i][0]); // directed graph, so only one way
        }
        // create a visted array where 
        // 0 = unvisisted
        // -1 = visisting
        // 1 = visisted
        int[] visited = new int[numCourses];
        for (int i = 0; i < numCourses; i++) {
            if(!dfs(adj, visited, i)) {
                return false;
            }
        }
        return true;
    }
    
    private boolean dfs (Map<Integer, List<Integer>> adj, int[] visited, int i) {
        // cycle dected, return false
        if (visited[i] == -1) {
            return false;
        }
        // previous visted, no need to visist again, return true
        if (visited[i] == 1) {
            return true;
        }
        // currently visiting
        visited[i] = -1;
        if (adj.containsKey(i)) {
            for (int neighbor : adj.get(i)) {
                // dfs into it's neighbors and retrun false if cycle was found
                if(!dfs(adj, visited, neighbor)) {
                    return false;
                }
            }
        }
        // if all the neighbors are visited and no cycle was found, 
        // we mark this node as visited and return true
        visited[i] = 1;
        return true;
    }
}