# 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].

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented

## Solution

If we use the DFS solution of Course Schedule, a valid sequence can easily be recorded.

## Code

``````public int[] findOrder(int numCourses, int[][] prerequisites) {
if(prerequisites == null){
throw new IllegalArgumentException("illegal prerequisites array");
}

int len = prerequisites.length;

//if there is no prerequisites, return a sequence of courses
if(len == 0){
int[] res = new int[numCourses];
for(int m=0; m<numCourses; m++){
res[m]=m;
}
return res;
}

//records the number of prerequisites each course (0,...,numCourses-1) requires
int[] pCounter = new int[numCourses];
for(int i=0; i<len; i++){
pCounter[prerequisites[i][0]]++;
}

//stores courses that have no prerequisites
for(int i=0; i<numCourses; i++){
if(pCounter[i]==0){
}
}

int numNoPre = queue.size();

//initialize result
int[] result = new int[numCourses];
int j=0;

while(!queue.isEmpty()){
int c = queue.remove();
result[j++]=c;

for(int i=0; i<len; i++){
if(prerequisites[i][1]==c){
pCounter[prerequisites[i][0]]--;
if(pCounter[prerequisites[i][0]]==0){
numNoPre++;
}
}

}
}

//return result
if(numNoPre==numCourses){
return result;
}else{
return new int[0];
}
}
``````