LeetCode - Course Schedule

Question Definition

There are a total of n courses you have to take, labeled from 0 to n - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses? For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

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.

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.

Hints:

1. This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

   Java Solution I (DFS)

   ```java public boolean canFinish(int numCourses, int[][] prerequisites) { ArrayList[] graph = new ArrayList[numCourses]; for(int i=0;i<numCourses;i++) graph[i] = new ArrayList();

   boolean[] visited = new boolean[numCourses]; for(int i=0; i<prerequisites.length;i++){ graph[prerequisites[i][1]].add(prerequisites[i][0]); }

   for(int i=0; i<numCourses; i++){ if(!dfs(graph,visited,i)) return false; } return true; }

private boolean dfs(ArrayList[] graph, boolean[] visited, int course){ if(visited[course]) return false; else visited[course] = true;;

for(int i=0; i<graph[course].size();i++){
    if(!dfs(graph,visited,(int)graph[course].get(i)))
        return false;
}
visited[course] = false;
return true; } ``` ### Java Solution II (BFS) ```java public boolean canFinish(int numCourses, int[][] prerequisites) {
ArrayList[] graph = new ArrayList[numCourses];
int[] degree = new int[numCourses];
Queue queue = new LinkedList();
int count=0;

for(int i=0;i<numCourses;i++)
    graph[i] = new ArrayList();

for(int i=0; i<prerequisites.length;i++){
    degree[prerequisites[i][1]]++;
    graph[prerequisites[i][0]].add(prerequisites[i][1]);
}
for(int i=0; i<degree.length;i++){
    if(degree[i] == 0){
        queue.add(i);
        count++;
    }
}

while(queue.size() != 0){
    int course = (int)queue.poll();
    for(int i=0; i<graph[course].size();i++){
        int pointer = (int)graph[course].get(i);
        degree[pointer]--;
        if(degree[pointer] == 0){
            queue.add(pointer);
            count++;
        }
    }
}
if(count == numCourses)
    return true;
else
    return false; } ```