# 207. Course Schedule

## 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 true if you can finish all courses. Otherwise, return false.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.


Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.


Constraints:

• 1 <= numCourses <= 2000
• 0 <= prerequisites.length <= 5000
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• All the pairs prerequisites[i] are unique.

## Solutions

Solution 1: Topological Sorting

For this problem, we can consider the courses as nodes in a graph, and prerequisites as edges in the graph. Thus, we can transform this problem into determining whether there is a cycle in the directed graph.

Specifically, we can use the idea of topological sorting. For each node with an in-degree of $0$, we reduce the in-degree of its out-degree nodes by $1$, until all nodes have been traversed.

If all nodes have been traversed, it means there is no cycle in the graph, and we can complete all courses; otherwise, we cannot complete all courses.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of courses and prerequisites respectively.

### Kahn’s Algorithms：

https://en.wikipedia.org/wiki/Topological_sorting#Kahn’s_algorithm

BFS based，

• start from with vertices with 0 incoming edge，insert them into list S，at the same time we remove all their outgoing edges，
• after that find new vertices with 0 incoming edges and go on.

### Tarjan’s Algorithms：

• loop through each node of the graph in an arbitrary order，
• initiating a depth-first search that terminates when it hits any node that has already been visited
• since the beginning of the topological sort or the node has no outgoing edges (i.e. a leaf node).
• public class Course_Schedule {

public class Solution_bfs {
public boolean canFinish(int numCourses, int[][] prerequisites) { // Kahn's Algorithm
if(numCourses <= 0) {
return false;
}

if(prerequisites == null || prerequisites.length == 0) {
return true;
}

int[] inDegree = new int[numCourses];

// 1. setup indgree count
for(int[] edge : prerequisites) {
inDegree[edge[0]]++;
}

Queue<Integer> queue = new LinkedList<>();

// 2. start from node with no indgree, i.e. no prerquisites for this course
for(int i = 0; i < inDegree.length; i++) {
if(inDegree[i] == 0) {
queue.offer(i);
}
}

List<Integer> result = new ArrayList<>();

while(!queue.isEmpty()) {
int currentCourse = queue.poll();

for(int[] edge : prerequisites) {
if(edge[1] == currentCourse) { // if a course requires current course

if(--inDegree[edge[0]] == 0) // -1, since current course is taken, and fulfill course edge[0]
queue.offer(edge[0]);
}
}
}

return result.size() == numCourses;
}
}

public class Solution_dfs {
public boolean canFinish(int numCourses, int[][] prerequisites) {

if(prerequisites == null || prerequisites.length < 2) {
return true;
}

// 0 for not visited，1 for globally visited，-1 for visisted AND on current path
int[] isVisited = new int[numCourses];
List<List<Integer>> graph = new ArrayList<>();

// 1. build graph
for (int i = 0; i < numCourses; i++) {
}
for (int[] each: prerequisites) {
}

// 2. dfs
for (int i = 0; i < numCourses; i++) {
if (!dfs(i, isVisited, graph)) {
return false;
}
}

return true;
}

private boolean dfs(int courseIndex, int[] isVisited, List<List<Integer>> graph) {
if (isVisited[courseIndex] == 1) {
return true;
}
if (isVisited[courseIndex] == -1) {
return false; // cycle found
}

isVisited[courseIndex] = -1;
for (Integer next: graph.get(courseIndex)) {
if(!dfs(next, isVisited, graph)) {
return false;
}
}

isVisited[courseIndex] = 1;
return true;
}
}

}

//////

class Solution {
public boolean canFinish(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 cnt = 0;
while (!q.isEmpty()) {
int i = q.poll();
++cnt;
for (int j : g[i]) {
if (--indeg[j] == 0) {
q.offer(j);
}
}
}
return cnt == numCourses;
}
}

• class Solution {
public:
bool canFinish(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);
}
}
int cnt = 0;
while (!q.empty()) {
int i = q.front();
q.pop();
++cnt;
for (int j : g[i]) {
if (--indeg[j] == 0) {
q.push(j);
}
}
}
return cnt == numCourses;
}
};

• '''
>>> from collections import defaultdict

>>> a = defaultdict(int)
>>> a
defaultdict(<type 'int'>, {})
>>>
>>> a['hehehe']
0
>>> a
defaultdict(<type 'int'>, {'hehehe': 0})

>>> a = defaultdict(list)
>>> a
defaultdict(<type 'list'>, {})
>>> a['hehehe']
[]
>>> a
defaultdict(<type 'list'>, {'hehehe': []})
'''
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
g = defaultdict(list)
indeg = [0] * numCourses
for a, b in prerequisites:
g[b].append(a)
indeg[a] += 1
cnt = 0
q = deque([i for i, v in enumerate(indeg) if v == 0])
while q:
i = q.popleft()
cnt += 1
for j in g[i]:
indeg[j] -= 1
if indeg[j] == 0:
q.append(j)
return cnt == numCourses


• func canFinish(numCourses int, prerequisites [][]int) bool {
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)
}
}
cnt := 0
for len(q) > 0 {
i := q[0]
q = q[1:]
cnt++
for _, j := range g[i] {
indeg[j]--
if indeg[j] == 0 {
q = append(q, j)
}
}
}
return cnt == numCourses
}

• function canFinish(numCourses: number, prerequisites: number[][]): boolean {
const g: number[][] = new Array(numCourses).fill(0).map(() => []);
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);
}
}
let cnt = 0;
while (q.length) {
const i = q.shift()!;
cnt++;
for (const j of g[i]) {
if (--indeg[j] == 0) {
q.push(j);
}
}
}
return cnt == numCourses;
}


• public class Solution {
public bool CanFinish(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 cnt = 0;
while (q.Count > 0) {
int i = q.Dequeue();
++cnt;
foreach (int j in g[i]) {
if (--indeg[j] == 0) {
q.Enqueue(j);
}
}
}
return cnt == numCourses;
}
}

• use std::collections::VecDeque;

impl Solution {
pub fn can_finish(num_course: i32, prerequisites: Vec<Vec<i32>>) -> bool {
let num_course = num_course as usize;
// The graph representation
let mut graph: Vec<Vec<i32>> = vec![vec![]; num_course];
// Record the in degree for each node
let mut in_degree_vec: Vec<i32> = vec![0; num_course];
let mut q: VecDeque<usize> = VecDeque::new();
let mut count = 0;

// Initialize the graph & in degree vector
for p in &prerequisites {
let (from, to) = (p[0], p[1]);
graph[from as usize].push(to);
in_degree_vec[to as usize] += 1;
}

// Enqueue the first batch of nodes with in degree 0
for i in 0..num_course {
if in_degree_vec[i] == 0 {
q.push_back(i);
}
}

// Begin the traverse & update through the graph
while !q.is_empty() {
// Get the current node index
let index = q.front().unwrap().clone();
// This course can be finished
count += 1;
q.pop_front();
for i in &graph[index] {
// Update the in degree for the current node
in_degree_vec[*i as usize] -= 1;
// See if can be enqueued
if in_degree_vec[*i as usize] == 0 {
q.push_back(*i as usize);
}
}
}

count == num_course
}
}