Solve LeetCode 207: Course Schedule in Python with a Kahn topological sort approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.
This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.
| Difficulty | Medium |
|---|---|
| Topic | Graph General |
| Reusable pattern | Kahn topological sort |
| Complexity | O(V+E) time and O(V+E) space |
What the problem is testing
Count prerequisites as indegrees, repeatedly remove zero-indegree courses, and reduce the indegrees of dependent courses.
Algorithm
- Count prerequisites as indegrees, repeatedly remove zero-indegree courses, and reduce the indegrees of dependent courses.
- Maintain this invariant: The queue contains exactly the currently schedulable courses with no remaining prerequisites.
- Continue until every input item or reachable state has been resolved, then return the accumulated result.
Python solution
from collections import Counter, defaultdict, deque, OrderedDict
import random
class Solution:
def canFinish(self, numCourses, prerequisites):
graph = [[] for _ in range(numCourses)]; indegree = [0] * numCourses
for course, prerequisite in prerequisites:
graph[prerequisite].append(course); indegree[course] += 1
queue = deque(i for i, degree in enumerate(indegree) if degree == 0)
completed = 0
while queue:
course = queue.popleft(); completed += 1
for following in graph[course]:
indegree[following] -= 1
if indegree[following] == 0: queue.append(following)
return completed == numCoursesWhy this is correct
The proof follows the maintained state: The queue contains exactly the currently schedulable courses with no remaining prerequisites. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.
Complexity
O(V+E) time and O(V+E) space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.
Edge cases
Processing fewer than all courses proves a directed cycle.
Tested reference code
This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.
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