Question

Formatted question description: https://leetcode.ca/all/207.html

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] = [a_i, b_i] indicates that you must take course b_i first if you want to take course a_i.

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 <= a_i, b_i < numCourses
All the pairs prerequisites[i] are unique.

Algorithm

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).

Code

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;
    }
}

// OJ: https://leetcode.com/problems/course-schedule/
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
    bool canFinish(int n, vector<vector<int>>& E) {
        vector<vector<int>> G(n);
        vector<int> indegree(n);
        for (auto &e : E) {
            G[e[1]].push_back(e[0]);
            ++indegree[e[0]];
        }
        queue<int> q;
        for (int i = 0; i < n; ++i) {
            if (indegree[i] == 0) q.push(i);
        }
        while (q.size()) {
            int u = q.front();
            q.pop();
            --n;
            for (int v : G[u]) {
                if (--indegree[v] == 0) q.push(v);
            }
        }
        return n == 0;
    }
};

'''
>>> 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

############

class Solution(object):
  def canFinish(self, numCourses, prerequisites):
    """
    :type numCourses: int
    :type prerequisites: List[List[int]]
    :rtype: bool
    """

    def dfs(start, parent, visited, graph):
      visited[start] = 1
      for nbr in graph[start]:
        if visited[nbr] == 1:
          return False
        if dfs(nbr, start, visited, graph) == False:
          return False
      visited[start] = 2
      return True

    graph = [[] for _ in range(0, numCourses)]
    for pre in prerequisites:
      start, end = pre
      graph[start].append(end)

    visited = [0 for _ in range(0, numCourses)]

    for pre in prerequisites:
      start, end = pre
      if visited[start] == 0:
        if dfs(start, None, visited, graph) == False:
          return False
    return True

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, v := range indeg {
		if v == 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 {
    let g = Array.from({ length: numCourses }, () => []);
    let indeg = new Array(numCourses).fill(0);
    for (let [a, b] of prerequisites) {
        g[b].push(a);
        ++indeg[a];
    }
    let q = [];
    for (let i = 0; i < numCourses; ++i) {
        if (!indeg[i]) {
            q.push(i);
        }
    }
    let cnt = 0;
    while (q.length) {
        const i = q.shift();
        ++cnt;
        for (let 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;
    }
}

207 - Course Schedule

Question

Algorithm

Kahn’s Algorithms：

Tarjan’s Algorithms：

Code

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Algorithm

Kahn’s Algorithms：

Tarjan’s Algorithms：

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