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?

Example 1:

Input: 2, [[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: 2, [[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.

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.

// BFS
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) { // time: O(V + E); space: O(V + E)
    vector<vector<int> > graph(numCourses);
    vector<int> indegree(numCourses, 0);
    for (auto& pre : prerequisites) {
        graph[pre[1]].push_back(pre[0]);
        ++indegree[pre[0]];
    }
    queue<int> q;
    for (int i = 0; i < numCourses; ++i) {
        if (indegree[i] == 0) q.push(i);
    }
    while (!q.empty()) {
        int t = q.front(); q.pop();
        for (int a : graph[t]) {
            if (--indegree[a] == 0) q.push(a);
        }
    }
    for (int in : indegree) {
        if (in) return false;
    }
    return true;
}

// DFS
bool dfs(vector<vector<int> >& graph, vector<int>& visited, int i) {
    if (visited[i] == -1) return false;
    if (visited[i] == 1) return true;
    visited[i] = -1;
    for (int a : graph[i]) {
        if (!dfs(graph, visited, a)) return false;
    }
    visited[i] = 1;
    return true;
}
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) { // time: O(V + E); space: O(V + E)
    vector<vector<int> > graph(numCourses); // adjacency lists
    vector<int> visited(numCourses, 0); // 0: univisted, 1: visited, -1: conflict
    for (auto& pre : prerequisites) {
        graph[pre[1]].push_back(pre[0]);
    }
    for (int i = 0; i < numCourses; ++i) {
        if (!dfs(graph, visited, i)) return false;
    }
    return true;
}