feat(work): add Course Schedule II solution and clean up Course Schedule I implementation

work/1/Medium/Depth-First Search/207.course-schedule.py

@@ -15,38 +15,60 @@ have to first take course 1.
 
 Return true if you can finish all courses. Otherwise, return false.
 
- 
+
 
 Example 1:
 
 Input: numCourses = 2, prerequisites = [[1,0]]
 Output: true
-Explanation: There are a total of 2 courses to take. 
+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. 
+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:
 
-	• 1 <= numCourses <= 2000
+        • 1 <= numCourses <= 2000
 
-	• 0 <= prerequisites.length <= 5000
+        • 0 <= prerequisites.length <= 5000
 
-	• prerequisites[i].length == 2
+        • prerequisites[i].length == 2
 
-	• 0 <= ai, bi < numCourses
+        • 0 <= ai, bi < numCourses
 
-	• All the pairs prerequisites[i] are unique.
+        • All the pairs prerequisites[i] are unique.
 """
 
+
 class Solution:
     def canFinish(self, numCourses: int, prerequisites: list[list[int]]) -> bool:
-        
+        indegree = [0] * numCourses
+        adj = [[] for x in range(numCourses)]
+        for prereq in prerequisites:
+            adj[prereq[1]].append(prereq[0])
+            indegree[prereq[0]] += 1
+
+        queue = []
+
+        for i in range(numCourses):
+            if indegree[i] == 0:
+                queue.append(i)
+
+        visited = 0
+        while queue:
+            node = queue.pop(0)
+            visited += 1
+            for neighbor in adj[node]:
+                indegree[neighbor] -= 1
+                if indegree[neighbor] == 0:
+                    queue.append(neighbor)
+
+        return numCourses == visited