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210 Course Schedule II

Leetcode

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]

Note:

  1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
  2. You may assume that there are no duplicate edges in the input prerequisites.

:::tip Idea Topological sorting :::

Solution 1: Topological Sorting

Time complexity: \(O(V+E)\) Space complexity: \(O(V+E)\)

// Runtime: 83 ms
class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        ArrayList<ArrayList<Integer>> graph = new ArrayList<>();

        for (int i = 0; i < numCourses; ++i)
            graph.add(new ArrayList<Integer>());

        for (int i = 0; i < prerequisites.length; ++i) {
            int course = prerequisites[i][0];
            int prerequisite = prerequisites[i][1];
            graph.get(course).add(prerequisite);
        }

        int[] visited = new int[numCourses];
        List<Integer> ans = new ArrayList<Integer>();
        Integer index = numCourses;
        for (int i = 0; i < numCourses; ++i)
            if (dfs(i, graph, visited, ans)) return new int[0];

        return ans.stream().mapToInt(i->i).toArray();
    }

    private boolean dfs(int curr, ArrayList<ArrayList<Integer>> graph, int[] visited, List<Integer> ans) {
        if (visited[curr] == 1) return true;
        if (visited[curr] == 2) return false;

        visited[curr] = 1;
        for (int next : graph.get(curr))
            if (dfs(next, graph, visited, ans)) return true;

        visited[curr] = 2;
        ans.add(curr);

        return false;
    }
}