LeetCode - Course Schedule

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.

• For example, the pair [0, 1], indicates that to take course 0 you 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. 
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:

• 1 <= numCourses <= 2000
• 0 <= prerequisites.length <= 5000
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• All the pairs prerequisites[i] are unique.

 

ChatGPT's Solution:

from collections import deque, defaultdict


class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        # Step 1: Build the graph and calculate in-degrees
        graph = defaultdict(list)
        in_degree = [0] * numCourses

        for course, prereq in prerequisites:
            graph[prereq].append(course)
            in_degree[course] += 1
        
        # Step 2: Initialize the queue with courses that have no prerequisites (in-degree 0)
        queue = deque()
        for i in range(numCourses):
            if in_degree[i] == 0:
                queue.append(i)
        
        # Step 3: Perform BFS and reduce in-degree of dependent courses
        visited_courses = 0
        while queue:
            course = queue.popleft()
            visited_courses += 1

            # Visit all courses that depend on the current course
            for dependent_course in graph[course]:
                in_degree[dependent_course] -= 1
                if in_degree[dependent_course] == 0:
                    queue.append(dependent_course)
        
        # If we have visited all the courses, it means we can finish them
        return visited_courses == numCourses