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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.

Kahn’s Algorithms：

https://en.wikipedia.org/wiki/Topological_sorting#Kahn’s_algorithm

BFS based，

  • start from with vertices with 0 incoming edge，insert them into list S，at the same time we remove all their outgoing edges，
  • after that find new vertices with 0 incoming edges and go on.

Tarjan’s Algorithms：

https://en.wikipedia.org/wiki/Tarjan%27s_strongly_connected_components_algorithm DFS based，

  • loop through each node of the graph in an arbitrary order，
  • initiating a depth-first search that terminates when it hits any node that has already been visited
  • since the beginning of the topological sort or the node has no outgoing edges (i.e. a leaf node).
  • public class Course_Schedule {

    public class Solution_bfs {
        public boolean canFinish(int numCourses, int[][] prerequisites) { // Kahn's Algorithm
            if(numCourses <= 0) {
                return false;
            }

            if(prerequisites == null || prerequisites.length == 0) {
                return true;
            }

            int[] inDegree = new int[numCourses];

            // 1. setup indgree count
            for(int[] edge : prerequisites) {
                inDegree[edge[0]]++;
            }

            Queue<Integer> queue = new LinkedList<>();

            // 2. start from node with no indgree, i.e. no prerquisites for this course
            for(int i = 0; i < inDegree.length; i++) {
                if(inDegree[i] == 0) {
                    queue.offer(i);
                }
            }

            List<Integer> result = new ArrayList<>();

            while(!queue.isEmpty()) {
                int currentCourse = queue.poll();
                result.add(currentCourse);

                for(int[] edge : prerequisites) {
                    if(edge[1] == currentCourse) { // if a course requires current course

                        if(--inDegree[edge[0]] == 0) // -1, since current course is taken, and fulfill course edge[0]
                            queue.offer(edge[0]);
                    }
                }
            }

            return result.size() == numCourses;
        }
    }

    public class Solution_dfs {
        public boolean canFinish(int numCourses, int[][] prerequisites) {

            if(prerequisites == null || prerequisites.length < 2) {
                return true;
            }

            // 0 for not visited，1 for globally visited，-1 for visisted AND on current path
            int[] isVisited = new int[numCourses];
            List<List<Integer>> graph = new ArrayList<>();

            // 1. build graph
            for (int i = 0; i < numCourses; i++) {
                graph.add(new ArrayList<>());
            }
            for (int[] each: prerequisites) {
                graph.get(each[1]).add(each[0]);
            }

            // 2. dfs
            for (int i = 0; i < numCourses; i++) {
                if (!dfs(i, isVisited, graph)) {
                    return false;
                }
            }

            return true;
        }

        private boolean dfs(int courseIndex, int[] isVisited, List<List<Integer>> graph) {
            if (isVisited[courseIndex] == 1) {
                return true;
            }
            if (isVisited[courseIndex] == -1) {
                return false; // cycle found
            }

            isVisited[courseIndex] = -1;
            for (Integer next: graph.get(courseIndex)) {
                if(!dfs(next, isVisited, graph)) {
                    return false;
                }
            }

            isVisited[courseIndex] = 1;
            return true;
        }
    }

}

//////

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;
    }
}

  • 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) {
                    q.push(j);
                }
            }
        }
        return cnt == numCourses;
    }
};

  • '''
>>> from collections import defaultdict

>>> a = defaultdict(int)
>>> a
defaultdict(<type 'int'>, {})
>>>
>>> a['hehehe']
0
>>> a
defaultdict(<type 'int'>, {'hehehe': 0})


>>> a = defaultdict(list)
>>> a
defaultdict(<type 'list'>, {})
>>> a['hehehe']
[]
>>> a
defaultdict(<type 'list'>, {'hehehe': []})
'''
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, v in enumerate(indeg) if v == 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


  • func canFinish(numCourses int, prerequisites [][]int) bool {
	g := make([][]int, numCourses)
	indeg := make([]int, numCourses)
	for _, p := range prerequisites {
		a, b := p[0], p[1]
		g[b] = append(g[b], a)
		indeg[a]++
	}
	q := []int{}
	for i, x := range indeg {
		if x == 0 {
			q = append(q, i)
		}
	}
	cnt := 0
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		cnt++
		for _, j := range g[i] {
			indeg[j]--
			if indeg[j] == 0 {
				q = append(q, j)
			}
		}
	}
	return cnt == numCourses
}

  • function canFinish(numCourses: number, prerequisites: number[][]): boolean {
    const g: number[][] = new Array(numCourses).fill(0).map(() => []);
    const indeg: number[] = new Array(numCourses).fill(0);
    for (const [a, b] of prerequisites) {
        g[b].push(a);
        indeg[a]++;
    }
    const q: number[] = [];
    for (let i = 0; i < numCourses; ++i) {
        if (indeg[i] == 0) {
            q.push(i);
        }
    }
    let cnt = 0;
    while (q.length) {
        const i = q.shift()!;
        cnt++;
        for (const j of g[i]) {
            if (--indeg[j] == 0) {
                q.push(j);
            }
        }
    }
    return cnt == numCourses;
}


  • public class Solution {
    public bool CanFinish(int numCourses, int[][] prerequisites) {
        var g = new List<int>[numCourses];
        for (int i = 0; i < numCourses; ++i) {
            g[i] = new List<int>();
        }
        var indeg = new int[numCourses];
        foreach (var p in prerequisites) {
            int a = p[0], b = p[1];
            g[b].Add(a);
            ++indeg[a];
        }
        var q = new Queue<int>();
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.Enqueue(i);
            }
        }
        var cnt = 0;
        while (q.Count > 0) {
            int i = q.Dequeue();
            ++cnt;
            foreach (int j in g[i]) {
                if (--indeg[j] == 0) {
                    q.Enqueue(j);
                }
            }
        }
        return cnt == numCourses;
    }
}

  • use std::collections::VecDeque;

impl Solution {
    #[allow(dead_code)]
    pub fn can_finish(num_course: i32, prerequisites: Vec<Vec<i32>>) -> bool {
        let num_course = num_course as usize;
        // The graph representation
        let mut graph: Vec<Vec<i32>> = vec![vec![]; num_course];
        // Record the in degree for each node
        let mut in_degree_vec: Vec<i32> = vec![0; num_course];
        let mut q: VecDeque<usize> = VecDeque::new();
        let mut count = 0;

        // Initialize the graph & in degree vector
        for p in &prerequisites {
            let (from, to) = (p[0], p[1]);
            graph[from as usize].push(to);
            in_degree_vec[to as usize] += 1;
        }

        // Enqueue the first batch of nodes with in degree 0
        for i in 0..num_course {
            if in_degree_vec[i] == 0 {
                q.push_back(i);
            }
        }

        // Begin the traverse & update through the graph
        while !q.is_empty() {
            // Get the current node index
            let index = q.front().unwrap().clone();
            // This course can be finished
            count += 1;
            q.pop_front();
            for i in &graph[index] {
                // Update the in degree for the current node
                in_degree_vec[*i as usize] -= 1;
                // See if can be enqueued
                if in_degree_vec[*i as usize] == 0 {
                    q.push_back(*i as usize);
                }
            }
        }

        count == num_course
    }
}

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