Course Schedule

There are a total of

ncourses 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:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.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.
- 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.

Topological sort BFS approach

- for all node p in graph:
- if p has no incoming edges:
- push p in queue

- if p has no incoming edges:
- while queue is not empty:
- p = queue.front()
- queue.pop()
- for all v, v is p’s neighbor:
- remove edge between p and v
- if v has no incoming edges:
- push v into queue

- if graph has edges:
- return error (exists circle)

- else:
- has no circle

Topological sort DFS approach

sort(graph g)

- set all nodes as unmarked
- for all un-permanent marked node v in graph:
- visit(v)

visit(node v)

- if v is temporally marked:
- return false (circle exists)

- if v is not permanently marked:
- temporally mark v
- for all neighbors q of v:
- visit(q)

- un temporally mark v
- permanent mark v

class Solution { public: bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) { //[0, numCourses - 1] //topological sort. vector<int> incomingNum(numCourses, 0); vector<vector<int> > graph(numCourses, vector<int>()); queue<int> q; //construct directed graph using edges for(auto edge : prerequisites){ incomingNum[edge.first]++; graph[edge.second].push_back(edge.first); } //find all nodes whoes incomingNum is 0 for(int i = 0; i < numCourses; i++){ if(incomingNum[i] == 0){ q.push(i); } } while(!q.empty()){ int idx = q.front(); q.pop(); //find all neighbours, remove edges //if incomingNum of the neighbour is 0, push it into queue vector<int> neighbours = graph[idx]; for(auto neighbour: neighbours){ incomingNum[neighbour]--; if(incomingNum[neighbour] == 0){ q.push(neighbour); } } graph[idx].clear(); } //if there are still some edges remain, cicle exists for(auto node: graph){ if(!node.empty()) return false; } return true; } };