course-schedule dfs py3
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0207_course-schedule/README.md
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0207_course-schedule/README.md
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There are a total of `numCourses` courses you have to take, labeled from `0` to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`.
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* For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.
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Return `true` if you can finish all courses. Otherwise, return `false`.
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**Example 1:**
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Input: numCourses = 2, prerequisites = [[1,0]]
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Output: true
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Explanation: There are a total of 2 courses to take.
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To take course 1 you should have finished course 0. So it is possible.
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**Example 2:**
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Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
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Output: false
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Explanation: There are a total of 2 courses to take.
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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.
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**Constraints:**
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* `1 <= numCourses <= 2000`
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* `0 <= prerequisites.length <= 5000`
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* `prerequisites[i].length == 2`
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* `0 <= ai, bi < numCourses`
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* All the pairs prerequisites\[i\] are **unique**.
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https://leetcode.com/problems/course-schedule
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0207_course-schedule/python3/dfs.py
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0207_course-schedule/python3/dfs.py
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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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graph = defaultdict(list)
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for course, pre in prerequisites:
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graph[course].append(pre)
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cycles = set()
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visited = set()
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def dfs(course):
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if course in cycles:
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return False
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if course in visited:
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return True
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visited.add(course)
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cycles.add(course)
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for dep in graph[course]:
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if not dfs(dep):
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return False
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cycles.remove(course)
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return True
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for course in range(numCourses):
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cycles = set()
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if not dfs(course):
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return False
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return True
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0207_course-schedule/python3/solution.py
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0207_course-schedule/python3/solution.py
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