# 210. Course Schedule II

## Description

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.

## Solutions

Topological Sort of directed graph, in this way, the difficulty increases,

Due to the basis of the previous one https://leetcode.ca/2016-06-24-207-Course-Schedule/, with a slight modification, every time an array is removed from the queue, it is stored in the result.

In the end, if there is a cycle in the directed graph, the number of elements in the result is not equal to the total number of courses, then we clear the result.

• class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
List<Integer>[] g = new List[numCourses];
Arrays.setAll(g, k -> new ArrayList<>());
int[] indeg = new int[numCourses];
for (var p : prerequisites) {
int a = p[0], b = p[1];
++indeg[a];
}
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < numCourses; ++i) {
if (indeg[i] == 0) {
q.offer(i);
}
}
int[] ans = new int[numCourses];
int cnt = 0;
while (!q.isEmpty()) {
int i = q.poll();
ans[cnt++] = i;
for (int j : g[i]) {
if (--indeg[j] == 0) {
q.offer(j);
}
}
}
return cnt == numCourses ? ans : new int[0];
}
}

• class Solution {
public:
vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
vector<vector<int>> g(numCourses);
vector<int> indeg(numCourses);
for (auto& p : prerequisites) {
int a = p[0], b = p[1];
g[b].push_back(a);
++indeg[a];
}
queue<int> q;
for (int i = 0; i < numCourses; ++i) {
if (indeg[i] == 0) {
q.push(i);
}
}
vector<int> ans;
while (!q.empty()) {
int i = q.front();
q.pop();
ans.push_back(i);
for (int j : g[i]) {
if (--indeg[j] == 0) {
q.push(j);
}
}
}
return ans.size() == numCourses ? ans : vector<int>();
}
};

• class Solution:
def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
g = defaultdict(list)
indeg = [0] * numCourses
for a, b in prerequisites:
g[b].append(a)
indeg[a] += 1
q = deque([i for i, v in enumerate(indeg) if v == 0])
ans = [] # added from previous question
while q:
i = q.popleft()
ans.append(i)
# assumption is only one path
# as in question 'You may assume that there are no duplicate edges in the input prerequisites.'
for j in g[i]:
indeg[j] -= 1
if indeg[j] == 0:
q.append(j)
return ans if len(ans) == numCourses else []


• func findOrder(numCourses int, prerequisites [][]int) []int {
g := make([][]int, numCourses)
indeg := make([]int, numCourses)
for _, p := range prerequisites {
a, b := p[0], p[1]
g[b] = append(g[b], a)
indeg[a]++
}
q := []int{}
for i, x := range indeg {
if x == 0 {
q = append(q, i)
}
}
ans := []int{}
for len(q) > 0 {
i := q[0]
q = q[1:]
ans = append(ans, i)
for _, j := range g[i] {
indeg[j]--
if indeg[j] == 0 {
q = append(q, j)
}
}
}
if len(ans) == numCourses {
return ans
}
return []int{}
}

• function findOrder(numCourses: number, prerequisites: number[][]): number[] {
const g: number[][] = Array.from({ length: numCourses }, () => []);
const indeg: number[] = new Array(numCourses).fill(0);
for (const [a, b] of prerequisites) {
g[b].push(a);
indeg[a]++;
}
const q: number[] = [];
for (let i = 0; i < numCourses; ++i) {
if (indeg[i] === 0) {
q.push(i);
}
}
const ans: number[] = [];
while (q.length) {
const i = q.shift()!;
ans.push(i);
for (const j of g[i]) {
if (--indeg[j] === 0) {
q.push(j);
}
}
}
return ans.length === numCourses ? ans : [];
}


• public class Solution {
public int[] FindOrder(int numCourses, int[][] prerequisites) {
var g = new List<int>[numCourses];
for (int i = 0; i < numCourses; ++i) {
g[i] = new List<int>();
}
var indeg = new int[numCourses];
foreach (var p in prerequisites) {
int a = p[0], b = p[1];
++indeg[a];
}
var q = new Queue<int>();
for (int i = 0; i < numCourses; ++i) {
if (indeg[i] == 0) {
q.Enqueue(i);
}
}
var ans = new int[numCourses];
var cnt = 0;
while (q.Count > 0) {
int i = q.Dequeue();
ans[cnt++] = i;
foreach (int j in g[i]) {
if (--indeg[j] == 0) {
q.Enqueue(j);
}
}
}
return cnt == numCourses ? ans : new int[0];
}
}

• impl Solution {
pub fn find_order(num_courses: i32, prerequisites: Vec<Vec<i32>>) -> Vec<i32> {
let n = num_courses as usize;
let mut adjacency = vec![vec![]; n];
let mut entry = vec![0; n];
// init
for iter in prerequisites.iter() {
let (a, b) = (iter[0], iter[1]);
entry[a as usize] += 1;
}
// construct deque & reslut
let mut deque = std::collections::VecDeque::new();
for index in 0..n {
if entry[index] == 0 {
deque.push_back(index);
}
}
let mut result = vec![];
// bfs
while !deque.is_empty() {
// update degree of entry
entry[out_entry as usize] -= 1;
if entry[out_entry as usize] == 0 {
deque.push_back(out_entry as usize);
}
}
}
if result.len() == n {
result
} else {
vec![]
}
}
}