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207. Course Schedule

Description

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

  • For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return true if you can finish all courses. Otherwise, return false.

 

Example 1:

Input: numCourses = 2, prerequisites = [[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: numCourses = 2, prerequisites = [[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.

 

Constraints:

  • 1 <= numCourses <= 2000
  • 0 <= prerequisites.length <= 5000
  • prerequisites[i].length == 2
  • 0 <= ai, bi < numCourses
  • All the pairs prerequisites[i] are unique.

Solutions

Solution 1: Topological Sorting

For this problem, we can consider the courses as nodes in a graph, and prerequisites as edges in the graph. Thus, we can transform this problem into determining whether there is a cycle in the directed graph.

Specifically, we can use the idea of topological sorting. For each node with an in-degree of $0$, we reduce the in-degree of its out-degree nodes by $1$, until all nodes have been traversed.

If all nodes have been traversed, it means there is no cycle in the graph, and we can complete all courses; otherwise, we cannot complete all courses.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of courses and prerequisites respectively.

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class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        g = defaultdict(list)
        indeg = [0] * numCourses
        for a, b in prerequisites:
            g[b].append(a)
            indeg[a] += 1
        cnt = 0
        q = deque(i for i, x in enumerate(indeg) if x == 0)
        while q:
            i = q.popleft()
            cnt += 1
            for j in g[i]:
                indeg[j] -= 1
                if indeg[j] == 0:
                    q.append(j)
        return cnt == numCourses

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class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<Integer>[] g = new List[numCourses];
        Arrays.setAll(g, k -> new ArrayList<>());
        int[] indeg = new int[numCourses];
        for (var p : prerequisites) {
            int a = p[0], b = p[1];
            g[b].add(a);
            ++indeg[a];
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }
        int cnt = 0;
        while (!q.isEmpty()) {
            int i = q.poll();
            ++cnt;
            for (int j : g[i]) {
                if (--indeg[j] == 0) {
                    q.offer(j);
                }
            }
        }
        return cnt == numCourses;
    }
}

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class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> g(numCourses);
        vector<int> indeg(numCourses);
        for (auto& p : prerequisites) {
            int a = p[0], b = p[1];
            g[b].push_back(a);
            ++indeg[a];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.push(i);
            }
        }
        int cnt = 0;
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            ++cnt;
            for (int j : g[i]) {
                if (--indeg[j] == 0) </