Course Schedule II

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, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].


Naive Thinking: Everything the same with Course Schedule I . Return the order of topological sort instead of whether can it be passed. Use an arraylist to store the order first. And then check if the size meets the number of courses. Then copy the arraylist to output array.

 public class Solution {  
   public int[] findOrder(int numCourses, int[][] prerequisites) {  
       
     // initilization  
     int[] rslt = new int[0];  
     List<Integer> order = new ArrayList<Integer>();  
     Queue<Integer> zero_pre_courses = new LinkedList<Integer>();  
     List<Course> courses = new ArrayList<Course>();  
     for(int i = 0;i < numCourses;i++) courses.add(new Course(i));  
       
     // read in prerequisite infomation  
     for(int i = 0;i < prerequisites.length;i++){  
       courses.get(prerequisites[i][0]).prerequisite.add(prerequisites[i][1]);  
       courses.get(prerequisites[i][1]).advanced.add(prerequisites[i][0]);  
     }  
       
     // group zero_prerequisite courses  
     for(int i = 0;i < numCourses;i++)  
       if(courses.get(i).prerequisite.isEmpty()) zero_pre_courses.add(i);  
       
     // topological sort  
     while(!zero_pre_courses.isEmpty()){  
       int curNum = zero_pre_courses.poll();  
       order.add(curNum);  
       Iterator<Integer> iter = courses.get(curNum).advanced.iterator();  
       while(iter.hasNext()){  
         int advNum = iter.next();  
         courses.get(advNum).prerequisite.remove(curNum);  
         if(courses.get(advNum).prerequisite.isEmpty()) zero_pre_courses.add(advNum);  
       }  
     }  
       
     // generate result  
     if(order.size() == numCourses){  
       rslt = new int[numCourses];  
       for(int i = 0;i < numCourses;i++)  
         rslt[i] = order.get(i);  
     }  
       
     return rslt;  
   }  
     
   class Course{  
     Set<Integer> prerequisite;  
     Set<Integer> advanced;  
     int label;  
     Course(int l){  
       this.label = l;  
       this.prerequisite = new HashSet<Integer>();  
       this.advanced = new HashSet<Integer>();  
     }  
   }  
 }  

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.

Naive Thinking: The description leads to a topological sort solution. But to execute a topological sort is not easy. If I directly use the 2D array on each iteration, I need to go though all 2D vectors. That's definitely not a good run time. The input 2D array needs to be preprocessed before implementing topological sort on it. After several trial, I find out that use a extra Course object with both prerequisite and advanced course attributes sounds.

 public class Solution {  
   public boolean canFinish(int numCourses, int[][] prerequisites) {  
     boolean exit = true;  
     List<Course> courses = new ArrayList<Course>();  
     Queue<Course> zero_pre_courses = new LinkedList<Course>();  
       
     // initialize courses list  
     for(int i = 0;i < numCourses;i++) courses.add(new Course(i));  
       
     // read in course prerequisite relationship  
     for(int i = 0;i < prerequisites.length;i++){  
       courses.get(prerequisites[i][0]).advanced.add(prerequisites[i][1]);  
       courses.get(prerequisites[i][1]).prerequisite.add(prerequisites[i][0]);  
     }  
       
     // group all zero prerequisite courses  
     for(int i = 0;i < numCourses;i++)  
       if(courses.get(i).prerequisite.isEmpty()) zero_pre_courses.add(courses.get(i));  
       
     // topological sort  
     while(!zero_pre_courses.isEmpty()){  
       Course course = zero_pre_courses.poll();  
       Iterator<Integer> iter = course.advanced.iterator();  
       while(iter.hasNext()){  
         int advancedCourseLabel = (int)iter.next();  
         Course advancedCourse = courses.get(advancedCourseLabel);  
         advancedCourse.prerequisite.remove(course.label);  
         if(advancedCourse.prerequisite.isEmpty()) zero_pre_courses.add(advancedCourse);  
       }  
       course.label = -1; // set label to -1 as the course is took  
     }  
       
     // check if all courses are took  
     for(int i = 0;i < numCourses;i++)  
       if(courses.get(i).label!=-1) exit = false;  
       
     return exit;  
       
   }  
   class Course{  
     Set<Integer> prerequisite;  
     Set<Integer> advanced;  
     int label;  
     Course(int l){  
       label = l;  
       prerequisite = new HashSet<Integer>();  
       advanced = new HashSet<Integer>();  
     }  
   }  
 }