How do you solve Course Schedule II on LeetCode?

Course Schedule II (LeetCode #210) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + p) time complexity and O(n) space complexity. Key insight: Understanding graph representation is crucial for solving problems involving dependencies.

What is the brute force solution for Course Schedule II?

The brute force approach for Course Schedule II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of course orders and checking each one to see if it satisfies the prerequisites. This is simple but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n + p).

What algorithm or data structure is used to solve Course Schedule II?

Course Schedule II uses the following concepts: Depth-First Search, Breadth-First Search, Graph Theory, Topological Sort. The recommended patterns to study are: Graph Theory, Topological Sort, BFS/DFS.

What are the interview tips for Course Schedule II?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and approach to the interviewer as you work through the solution. Practice implementing both brute force and optimal solutions to strengthen your understanding.

What are common mistakes when solving Course Schedule II?

Failing to account for cycles in the graph, leading to incorrect assumptions about course completion. Not properly managing in-degrees, which can result in missing courses that should be processed.

#210

Course Schedule II

Medium

Depth-First SearchBreadth-First SearchGraph TheoryTopological SortGraph TheoryTopological SortBFS/DFS

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + p)
Space	O(1)	O(n)

💡

Intuition

Time O(n + p)Space O(n)

The optimal solution uses topological sorting to efficiently determine the order of courses based on their prerequisites. This method leverages graph theory to ensure we respect the dependencies.

⚙️

Algorithm

4 steps

1Step 1: Build a graph using an adjacency list and track the in-degrees of each course.
2Step 2: Use a queue to process courses with zero in-degrees (no prerequisites).
3Step 3: While the queue is not empty, remove a course, add it to the result, and decrease the in-degrees of its neighbors. If any neighbor's in-degree reaches zero, add it to the queue.
4Step 4: If the result contains all courses, return it; otherwise, return an empty array (indicating a cycle).

solution.py23 lines

1# Full working Python code
2from collections import defaultdict, deque
3
4def findOrder(numCourses, prerequisites):
5    graph = defaultdict(list)
6    in_degree = [0] * numCourses
7    
8    for a, b in prerequisites:
9        graph[b].append(a)
10        in_degree[a] += 1
11    
12    queue = deque([i for i in range(numCourses) if in_degree[i] == 0])
13    order = []
14    
15    while queue:
16        course = queue.popleft()
17        order.append(course)
18        for neighbor in graph[course]:
19            in_degree[neighbor] -= 1
20            if in_degree[neighbor] == 0:
21                queue.append(neighbor)
22    
23    return order if len(order) == numCourses else []

ℹ

Complexity note: The time complexity is O(n + p) where n is the number of courses and p is the number of prerequisites. We traverse each course and each prerequisite once.

1Understanding graph representation is crucial for solving problems involving dependencies.
2Topological sorting is a powerful technique for ordering tasks based on prerequisites.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.