Course Scheduling from Leetcode - graph

Question

This is my solution to Course Scheduling Problem from leetcode. I am looking for any suggestions to improve my code, even slightest ones.
Here is the question:
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.
Example 1:
Input: 2, [[1,0]]
Output: [0,1]
Explanation: 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] .
Example 2:
Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: 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].
Here is my solution:
class Solution:
def findOrder(self, numCourses, prerequisites):
"""
:type numCourses: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
#Convert prerequisites into an adjacency list
adj = []
for i in range(numCourses):
adj.append(set())
for pair in prerequisites:
adj[pair[0]].add(pair[1])
def DFSHelper(s):
visited.add(s)
stack.add(s)
for neighbor in adj[s]:
# if neighbor vertex has never been visted before, there is no way it could be a backedge.
# visit this unvisited vertex
if(neighbor not in visited):
if(not DFSHelper(neighbor)):
return False
Sorted.append(neighbor)
else:
if(neighbor in stack):
return False
stack.remove(s)
return True
visited = set()
stack = set()
Sorted = []
for j in range(len(adj)):
if(j not in visited):
if(not DFSHelper(j)):
print(j)
return []
Sorted.append(j)
return Sorted
I first converted given prerequisites list into an adjacency list representation of graph, then did topological sorting of the graph. I used DFS recursively to topologically sort the graph. The list Sorted stores the result of sorting. While doing DFS I also checked if the graph contains any cycle, if it does just return []. For purpose of checking cycle I maintained a set called stack that stores all the vertices that are currently in call stack.

Answer 1

This is a simple question first create a graph and then find topological sorting on nodes.
If topological order contains all nodes then we have our answer else not possible to finish all the courses.
class Solution {
public int[] findOrder(int n, int[][] prerequisites) {
List<Integer>[] g = new ArrayList[n];
for(int i = 0; i < n; i++)g[i] = new ArrayList();
int[] deg = new int[n];
for(int[] e: prerequisites) {
g[e[1]].add(e[0]);
deg[e[0]]++;
}
Queue<Integer> q = new LinkedList();
for(int i = 0; i < n; i++) {
if(deg[i] == 0)q.add(i);
}
int[] res = new int[n];
int idx = 0;
while(!q.isEmpty()) {
int u = q.poll();
res[idx++] = u;
for(int v: g[u]) {
deg[v]--;
if(deg[v] == 0) q.add(v);
}
}
return idx == n ? res: new int[0];
}}