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Commit d1f4047

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Course Schedule
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‎Graphs/207-Course-Schedule.py

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'''
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Leetcode - https://leetcode.com/problems/course-schedule/
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'''
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'''
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207. Course Schedule
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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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'''
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class Solution:
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def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
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def is_cycle(node):
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if visited[node] == 1:
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return True
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if visited[node] == 2:
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return False
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visited[node] = 1
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if node in graph:
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for neigh in graph[node]:
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if is_cycle(neigh):
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return True
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visited[node] = 2
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return False
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graph = {}
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for course, prereq in prerequisites:
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if prereq not in graph:
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graph[prereq] = []
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graph[prereq].append(course)
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visited = [0]*numCourses
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for course in range(numCourses):
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if is_cycle(course):
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return False
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return True
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# T:O(V + E)
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'''
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Explaination -
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for example---
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numCourses = 4
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prerequisites = [[1, 0], [2, 1], [3, 2], [0, 3]]
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Step 1: Initialize Variables and Graph
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-- numCourses is 4, meaning there are 4 courses labeled from 0 to 3.
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-- prerequisites is a list of prerequisite pairs: [1, 0], [2, 1], [3, 2], and [0, 3].
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Step 2: Build the Graph
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We create an adjacency list to represent the directed graph based on the prerequisite pairs:
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Graph:
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0 -> [3]
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1 -> [0]
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2 -> [1]
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3 -> [2]
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Step 3: Perform DFS to Detect Cycles
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We'll perform Depth-First Search (DFS) to traverse the graph and detect cycles. The DFS function will mark nodes as visiting and visited while exploring the graph.
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Starting DFS from each node:
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- Node 0:
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- Marked as visiting (visited[0] = 1)
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- Visit its neighbor 3
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- Node 3:
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- Marked as visiting (visited[3] = 1)
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- Visit its neighbor 2
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- Node 2:
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- Marked as visiting (visited[2] = 1)
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- Visit its neighbor 1
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- Node 1:
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- Marked as visiting (visited[1] = 1)
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- Visit its neighbor 0
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- Node 0:
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- Already visiting (visited[0] = 1)
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- Cycle detected, return True
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Result: A cycle was detected during the DFS traversal, indicating that it's not possible to finish all courses due to the cyclic dependency.
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Final Output: False
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In this example, the code correctly identified that there is a cycle in the graph, which means
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it's not possible to finish all courses with the given prerequisites. The function returns False.
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The cycle [0, 1, 2, 3, 0] was detected during the traversal.
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'''

‎README.md

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- [x] [Graphs](Graphs)
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- [x] [Number of Islands](Graphs/200-Number-of-Islands.py)
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- [x] [Clone Graph](Graphs/133-Clone-Graph.py)
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- [x] [Course Schedule](Graphs/207-Course-Schedule.py)
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