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.
Hints:
- 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.
- There are several ways to represent a graph. For example, the input prerequisites is a graph represented by a list of edges. Is this graph representation appropriate?
- Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
- Topological sort could also be done via BFS.
Solution 1:
DFS
class DiGraph { int V; List<Integer>[] adj; public DiGraph(int v) { V = v; adj = new List[v]; for(int i=0; i<v; i++) { adj[i] = new ArrayList<>(); } } public void addEdge(int u, int v) { adj[u].add(v); } } public boolean canFinish(int numCourses, int[][] prerequisites) { DiGraph g = new DiGraph(numCourses); for(int[] courses:prerequisites) { g.addEdge(courses[1], courses[0]); } boolean[] visited = new boolean[numCourses]; for(int i=0; i<numCourses; i++) { if(!canFinish(g, visited, i)) return false; } return true; } private boolean canFinish(DiGraph g, boolean[] visited, int u) { visited[u] = true; for(int v:g.adj[u]) { if(visited[v] || !canFinish(g, visited, v)) return false; } visited[u] = false; return true; }