LeetCode 207. Course Schedule

Description

https://leetcode.com/problems/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 <= 105
  • 0 <= prerequisites.length <= 5000
  • prerequisites[i].length == 2
  • 0 <= ai, bi < numCourses
  • All the pairs prerequisites[i] are unique.

Explanation

Use topological sort to keep appending the course with indegree 0 to the queue. In the end, check how many number of courses can be taken.

Python Solution

class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        
        adjacency_list = []
        
        
        indegrees = []
        
        for i in range(numCourses):
            adjacency_list.append([])
            indegrees.append(0)
            
        for prerequisite in prerequisites:
            adjacency_list[prerequisite[0]].append(prerequisite[1])
            indegrees[prerequisite[1]] += 1
            
        
        queue = []

        for i, indegree in enumerate(indegrees):
            if indegree == 0:
                queue.append(i)
                
        count = 0
        while queue:        
            course = queue.pop(0)
            count += 1
            
            for adj_course in adjacency_list[course]:
                indegrees[adj_course] -= 1
                
                if indegrees[adj_course] == 0:
                    queue.append(adj_course)                
            
        return count == numCourses
        
            
  • Time Complexity: ~V + E
  • Space Complexity: ~V + E

where V is the number of courses, and E is the number of dependencies.

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