[LeetCode] Course Schedule II

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.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

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. Read more about how a graph is represented.

Hints:

1. This problem is equivalent to finding the topological order 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 based on the previous post regarding the problem Course Schedule, which asks for detecting whether a cycle exists in a directed graph.
The modification lies in the following:
We initialize an array of size n (since a valid topological order has size n). In the DFS process, we add one element to the array only after all its dependents have been added, and the element is added to the front of the array. In the following Java code, we initialize idx = n - 1, and do idx-- after adding each element.
Further reading: Topological sorting.

public class Solution {
    private ArrayList<ArrayList<Integer>> list;
    private int[] topOrder;
    private int idx;
    
    public boolean DFS(int i, boolean[] visited, boolean[] recStack) {
        if (visited[i]) {
            return true;
        }
        
        visited[i] = true;
        recStack[i] = true;
        for (int j : list.get(i)) {
            if (recStack[j] || !DFS(j, visited, recStack)) {
                return false;
            }
        }
        recStack[i] = false;
        topOrder[idx] = i;
        idx--;
        return true;
    }
    
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int n = numCourses;
        list = new ArrayList<ArrayList<Integer>>();
        for (int i = 0; i < n; i++) {
            list.add(new ArrayList<Integer>());
        }
        for (int[] pair : prerequisites) {
            list.get(pair[1]).add(pair[0]);
        }
        topOrder = new int[n];
        idx = n - 1;
        boolean[] visited = new boolean[n];
        boolean[] recStack = new boolean[n];
        Arrays.fill(visited, false);
        Arrays.fill(recStack, false);
        for (int i = 0; i < n; i++) {
            if (!DFS(i, visited, recStack)) {
                return new int[0];
            }
        }
        return topOrder;
    }
}