[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.
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.
click to show more hints.
Hints:
  1. This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
  2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
  3. Topological sort could also be done via BFS.

Thought process:
Topological sort:
  1. Build graph:
    1. Use a hash map of int -> set of int to represent graph. The key of the map is a course, the value is a set of courses which have the key course as a prerequisite.
    2. Use another hash map of int -> int to keep track of the in-degrees of each course.
    3. Iterate through the edge list. Populate the graph and the in-degree map.
  2. Sort:
    1. Iterate through the in-degrees. Offer courses with no prerequisites to a queue.
    2. Topological sort. Keep track of courses polled from queue.
    3. If the number of sorted courses equals total number of courses, there is no circular dependency. So it's possible to finish all courses and vice versa.

Solution 1 (BFS):
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public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        Map<Integer, Set<Integer>> graph = new HashMap<>();
        Map<Integer, Integer> indegrees = new HashMap<>();
        
        for (int[] edge : prerequisites) {
            Set<Integer> uppers = graph.getOrDefault(edge[1], new HashSet<>());
            uppers.add(edge[0]);
            graph.put(edge[1], uppers);
            
            int indegree = indegrees.getOrDefault(edge[0], 0);
            indegree++;
            indegrees.put(edge[0], indegree);
            indegrees.put(edge[1], indegrees.getOrDefault(edge[1], 0));
        }
        
        Queue<Integer> queue = new LinkedList<>();
        for (int course : indegrees.keySet()) {
            if (indegrees.get(course) == 0) {
                queue.offer(course);
            }
        }
        
        int sorted = 0;
        while (!queue.isEmpty()) {
            int course = queue.poll();
            sorted++;
            
            if (graph.containsKey(course)) {
                for (int upper : graph.get(course)) {
                    int indegree = indegrees.get(upper) - 1;
                    indegrees.put(upper, indegree);
                    if (indegree == 0) {
                        queue.offer(upper);
                    }
                }
            }
        }
        return sorted == indegrees.size();
    }
}

Time complexity:
Building graph and in-degrees takes O(E). Topological sort takes O(V + E). The overall time complexity is O(V + E).

 Solution 2 (DFS):

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class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        Map<Integer, Set<Integer>> graph = new HashMap<>();
        for (int i = 0; i < numCourses; i++) {
            graph.put(i, new HashSet<>());
        }
        for (int[] edge : prerequisites) {
            graph.get(edge[1]).add(edge[0]);
        }
        
        boolean[] visited = new boolean[numCourses];
        for (int i = 0; i < numCourses; i++) {
            if (hasCycle(i, graph, new boolean[numCourses], visited)) {
                return false;
            }
        }
        return true;
    }
    
    private boolean hasCycle(int course, Map<Integer, Set<Integer>> graph, boolean[] onPath, boolean[] visited) {
        if (onPath[course]) {
            return true;
        }
        
        onPath[course] = true;
        for (int neighbor : graph.get(course)) {
            if (!visited[neighbor] && hasCycle(neighbor, graph, onPath, visited)) {
                return true;
            }
        }
        visited[course] = true;
        
        return false;
    }
}

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