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] = [a_i, b_i] indicates that you must take course b_i first if you want to take course a_i.

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 <= a_i, b_i < numCourses
a_i != b_i
All the pairs [a_i, b_i] are distinct.

Solution

/**
 * @param {number} numCourses
 * @param {number[][]} prerequisites
 * @return {number[]}
 */
var findOrder = function(numCourses, prerequisites) {
    // construct adjacency graph
    const graph = Array(numCourses).fill(null).map(() => []);
    const indegree = Array(numCourses).fill(0);
    for (const [vin, vout] of prerequisites) {
        graph[vout].push(vin);
        indegree[vin]++;
    }

    // initialize queue with courses that has no prerequisites
    const queue = [];
    for (let i = 0; i < numCourses; i++) {
        if (!indegree[i]) {
            queue.push(i);
        }
    }

    // test if a topological sort exists using BFS
    const res = [];
    while (queue.length) {
        const v = queue.shift();
        res.push(v);
        for (const u of graph[v]) {
            indegree[u]--;
            if (indegree[u] === 0) {
                queue.push(u);
            }
        }
    }
    return res.length === numCourses ? res : [];
};

We will implement a topological sort solution. Our solution is almost identical to the Course Schedule problem. The only difference is we keep track of the topological order instead of just the count.

Problem​

Solution​

Problem

Solution