feat(work): implement Course Schedule II solution

work/1/Medium/Depth-First Search/210.course-schedule-ii.py

@@ -17,7 +17,7 @@ Return the ordering of courses you should take to finish all courses.
 If there are many valid answers, return any of them. If it is
 impossible to finish all courses, return an empty array.
 
- 
+
 
 Example 1:
 
@@ -42,23 +42,55 @@ Example 3:
 Input: numCourses = 1, prerequisites = []
 Output: [0]
 
- 
+
 
 Constraints:
 
-	• 1 <= numCourses <= 2000
+        • 1 <= numCourses <= 2000
 
-	• 0 <= prerequisites.length <= numCourses * (numCourses - 1)
+        • 0 <= prerequisites.length <= numCourses * (numCourses - 1)
 
-	• prerequisites[i].length == 2
+        • prerequisites[i].length == 2
 
-	• 0 <= ai, bi < numCourses
+        • 0 <= ai, bi < numCourses
 
-	• ai != bi
+        • ai != bi
 
-	• All the pairs [ai, bi] are distinct.
+        • All the pairs [ai, bi] are distinct.
 """
 
+
 class Solution:
     def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]:
-        
+        prereq = {c: [] for c in range(numCourses)}
+        for crs, pre in prerequisites:
+            prereq[crs].append(pre)
+
+        # course has 3 possible states:
+        # visited -> crs has been added to output
+        # visiting -> crs not added to output, but added to cycle
+        # unvisited -> crs not added to output nor cycle
+        output = []
+        visit, cycle = set(), set()
+
+        def dfs(crs):
+            if crs in cycle:
+                return False
+            if crs in visit:
+                return True
+
+            cycle.add(crs)
+            for pre in prereq[crs]:
+                if dfs(pre) == False:
+                    return False
+
+            cycle.remove(crs)
+            visit.add(crs)
+            output.append(crs)
+            return True
+
+        for c in range(numCourses):
+            if dfs(c) == False:
+                return []
+
+        return output