Leetcode 207: Course Schedule

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.

 

public class Solution {
    public bool CanFinish(int numCourses, int[,] prerequisites) {
        
        var dependencies = new List<int>[numCourses];
        
        for (int i = 0; i < prerequisites.GetLength(0); i++)
        {
            int c = prerequisites[i, 0], d = prerequisites[i, 1];
            
            // a course is dependent on itself
            if (c == d) return false;
            
            if (dependencies[c] == null) dependencies[c] = new List<int>();
            
            dependencies[c].Add(d);
        }
        
        var visited = new bool[numCourses];
        var done = new bool[numCourses];
        
        for (int i = 0; i < numCourses; i++)
        {
            if (!DFS(new bool[numCourses], done, dependencies, i))
            {
                return false;
            }
        }
        
        return true;
    }
    
    private bool DFS(bool[] visited, bool[] done, List<int>[] dependencies, int c)
    {
        if (done[c]) return true;
        if (visited[c]) return false;
        
        visited[c] = true;
        
        if (dependencies[c] == null)
        {
            done[c] = true;
            return true;
        }
        else
        {
            foreach (var d in dependencies[c])
            {
                if (!DFS(visited, done, dependencies, d)) return false;
            }
            
            done[c] = true;
            return true;
        }
    }
}