210. Course Schedule II

Problem

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 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.

Solution (Java)

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int i = 0; i < numCourses; i++) {
            graph.put(i, new ArrayList<>());
        }
        for (int[] classes : prerequisites) {
            graph.get(classes[0]).add(classes[1]);
        }
        List<Integer> output = new ArrayList<>();
        Map<Integer, Boolean> visited = new HashMap<>();
        for (int c : graph.keySet()) {
            if (dfs(c, graph, visited, output)) {
                return new int[0];
            }
        }
        int[] res = new int[output.size()];
        for (int i = 0; i < output.size(); i++) {
            res[i] = output.get(i);
        }
        return res;
    }

    private boolean dfs(
            int course,
            Map<Integer, List<Integer>> graph,
            Map<Integer, Boolean> visited,
            List<Integer> output) {
        if (visited.containsKey(course)) {
            return visited.get(course);
        }
        visited.put(course, true);
        for (int c : graph.get(course)) {
            if (dfs(c, graph, visited, output)) {
                return true;
            }
        }
        visited.put(course, false);
        output.add(course);
        return false;
    }
}

Solution (Javascript)

/**
 * @param {number} numCourses
 * @param {number[][]} prerequisites
 * @return {number[]}
 */
var findOrder = function(numCourses, prerequisites) {
    // Initialize result array
    const result = new Array(numCourses).fill(0)
    const inDegree = new Array(numCourses).fill(0);

    for(const pre of prerequisites) {
        inDegree[pre[0]]++
    }

    const zeroDegree = [];

    for(let i = 0; i < numCourses; i++) {
        if(inDegree[i]===0) {
            zeroDegree.push(i);
        }
    }

    // Topological sort not possible
    if(zeroDegree.length === 0) return []

    let i = 0
    while(zeroDegree.length) {
        const course = zeroDegree.pop()
        // Add course to the result array 
        result[i++] = course
        for(const pre of prerequisites) {
            if(course === pre[1]) {
                inDegree[pre[0]]--
                if(inDegree[pre[0]]===0) {
                    zeroDegree.push(pre[0])
                }
            }
        }
    }

    // Topological sort not possible
    for(const num of inDegree) {
        if(num!== 0) return []
    }

    return result;
};

Explain:

nope.

Complexity:

• Time complexity :
• Space complexity :