What is the optimal time complexity of solving Course Schedule?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Course Schedule?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Course Schedule?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Course Schedule?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Course Schedule

Learn how to solve the 'Course Schedule' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

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.

Write a function canFinish(numCourses: int, prerequisites: List[List[int]]) -> bool.

Constraints

•1 <= numCourses <= 2000
•0 <= len(prerequisites) <= 5000
•prerequisites[i].length == 2
•0 <= ai, bi < numCourses

Examples

Example 1

Input

numCourses = 2, prerequisites = [[1,0]]

Output

True

Explanation

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

To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. This is a cycle and therefore impossible.

Need a Hint?

Represent graph node connections as an adjacency list/matrix, then use standard BFS or DFS graph traversal.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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