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MAY-29 2020 Challenge

Course Schedule


 There are a total of numCourses courses you have to take, labeled from 0 to numCourses-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: 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:


The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
1 <= numCourses <= 10^5

Solution in jAvA:



class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if (numCourses <= 0) {
            return true;
        }
        
        if (prerequisites == null || prerequisites.length == 0) {
            return true;
        }
        
        // First transform the edge list to adj. list
        Map<Integer, List<Integer>> adjList = new HashMap<>();
        for (int[] edge : prerequisites) {
            if (adjList.containsKey(edge[0])) {
                List<Integer> neighbors = adjList.get(edge[0]);
                neighbors.add(edge[1]);
                adjList.put(edge[0], neighbors);
            } else {
                List<Integer> neighbors = new ArrayList<Integer>();
                neighbors.add(edge[1]);
                adjList.put(edge[0], neighbors);
            }
        }
        
        int[] visited = new int[numCourses];
        // Check if the graph contains a circle, if yes, return false.
        for (int i = 0; i < numCourses; i++) {
            if (hasCircles(i, visited, adjList)) {
                return false;
            }
        }
        
        return true;
    }
    
    private boolean hasCircles(int vertexId, int[] visited, Map<Integer, List<Integer>> adjList) {
        if (visited[vertexId] == -1) {
            return true;
        }
        
        if (visited[vertexId] == 1) {
            return false;
        }
        
        visited[vertexId] = -1;
        
        List<Integer> neighbors = adjList.get(vertexId);
        if (neighbors != null) {
            for (int neighbor : neighbors) {
                if (hasCircles(neighbor, visited, adjList)) {
                    return true;
                }
            }
        }
        
        visited[vertexId] = 1;
        
        return false;
    }
}
Course Schedule

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