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Joseph Luce
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Create 207_course_schedule.md
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# 207. Course Schedule
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## DFS Recursive Solution
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- Runtime: O(N)
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- Space: O(N)
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- N = Number of courses
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When the problem has relationships, such as a course needing more than one prerequisite, you can think of a graph.
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In this case, its a directed graph.
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The core of the problem is to find a cycle in the graph, as example 2 of the problem shows that.
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We will need to create a graph, as it is not provided to us, it can be an adjacent list or a matrix, doesn't matter.
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For any dfs, you will need a global visited and a local visited.
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The global visited will tell us if we need to dfs starting at this node, this is to reduce run-time, else it will be O(N^N).
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The local visited is for when we are traversing the graph via. dfs and looking for cycles.
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I decided to use a dictionary to simplify the code, -1 will be used during the dfs, then after the dfs, changed into a 1, showing that its already visited and has no cycles.
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```
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from collections import defaultdict
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class Solution:
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def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
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adj_list = self.create_adj_list(prerequisites)
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visited = defaultdict(int)
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for node in adj_list:
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if not self.dfs(node, adj_list, visited):
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return False
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return True
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def dfs(self, node, adj_list, visited):
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if visited[node] == -1: # currently visiting, cycle
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return False
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if visited[node] == 1: # already visited, no cycle
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return True
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visited[node] = -1
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for neighbor in adj_list[node]:
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if not self.dfs(neighbor, adj_list, visited):
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return False
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visited[node] = 1
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return True
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def create_adj_list(self, prereqs):
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adj_list = defaultdict(list)
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for course, prereq in prereqs:
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adj_list[course].append(prereq)
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adj_list[prereq]
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return adj_list
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```

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