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course_schedule.py
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# -*- coding: utf-8 -*-
"""
Created on Fri May 29 23:16:10 2020
@author: johnoyegbite
"""
# SOLVED!
"""
Problem:
There are a total of numCourses courses you have to take, labeled from 0 to
numCourses-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, is it
possible for you to finish all courses?
Example 1:
Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
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.
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:
The input prerequisites is a graph represented by a list of edges, not
adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
1 <= numCourses <= 10^5
"""
class Solution:
def dfs(self, node, visited, graph):
# This node doesn't have a prerequisite
if node not in graph:
return True
# There is a cycle
if node in visited:
return False
visited.add(node)
for child in graph[node]:
if not self.dfs(child, visited, graph):
return False
# We are done with the prerequisite of this node,
# so, remove for other node to use as a prerequisite
visited.remove(node)
# This node's prerequisite has been completed and
# it doesn't have a cycle.
return True
def canFinish(self, numCourses: int, prerequisites) -> bool:
"""
tupe numCourses: int
type prerequisites: List[List[int]]
rype: bool
"""
# Use the idea of Topological Sort.
graph = {}
for course, prereq in prerequisites:
if course in graph:
graph[course].append(prereq)
else:
graph[course] = [prereq]
for course in range(numCourses):
visited = set()
if not self.dfs(course, visited, graph):
return False
return True