0207: Course Schedule

Problem Statement

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.

Code Solution

from collections import deque

class Solution:
    def canFinish(self, numCourses, prerequisites):
        G = {ix: [] for ix in range(numCourses)}
        P = {ix: 0 for ix in range(numCourses)}
        for pr in prerequisites:
            v, u = pr[0], pr[1]
            G[u].append(v)
            P[v] = P[v] + 1
        iset = []
        for ix in P:
            if P[ix] == 0:
                iset.append(ix)
        dq = deque(iset)
        vset = set(iset)
        count = 0
        while dq:
            curr = dq.popleft()
            count = count + 1
            for adj in G[curr]:
                P[adj] = P[adj] - 1
                if P[adj] == 0:
                    if adj not in vset:
                        vset.add(adj)
                        dq.append(adj)
        return count == numCourses