How do you solve Course Schedule IV on LeetCode?

Course Schedule IV (LeetCode #1462) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n²) space complexity. Key insight: Understanding graph representation is crucial.

What is the brute force solution for Course Schedule IV?

The brute force approach for Course Schedule IV is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each query by performing a depth-first search (DFS) for each course to see if it can reach the target course. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n³).

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

Course Schedule IV uses the following concepts: Depth-First Search, Breadth-First Search, Graph Theory, Topological Sort. The recommended patterns to study are: Graph Traversal, Dynamic Programming.

What are the interview tips for Course Schedule IV?

Clarify the problem requirements before jumping to code. Think about edge cases, such as no prerequisites or circular dependencies. Explain your thought process clearly while coding.

What are common mistakes when solving Course Schedule IV?

Not considering indirect prerequisites. Confusing the direction of prerequisites (i.e., assuming course A is a prerequisite of course B when it's the opposite).

#1462

Course Schedule IV

Medium

Depth-First SearchBreadth-First SearchGraph TheoryTopological SortGraph TraversalDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³)
Space	O(1)	O(n²)

💡

Intuition

Time O(n³)Space O(n²)

The optimal solution uses a graph representation and performs a transitive closure using the Floyd-Warshall algorithm to determine reachability between all pairs of courses in O(n³) time, which is efficient given the constraints.

⚙️

Algorithm

4 steps

1Step 1: Create a graph representation of the courses and their prerequisites.
2Step 2: Initialize a reachability matrix where reach[i][j] is true if course i is a prerequisite of course j.
3Step 3: Use the Floyd-Warshall algorithm to update the reachability matrix based on the prerequisites.
4Step 4: Answer each query in O(1) time using the reachability matrix.

solution.py15 lines

1# Full working Python code
2def canFinish(numCourses, prerequisites, queries):
3    reach = [[False] * numCourses for _ in range(numCourses)]
4    for a, b in prerequisites:
5        reach[a][b] = True
6
7    for k in range(numCourses):
8        for i in range(numCourses):
9            for j in range(numCourses):
10                reach[i][j] = reach[i][j] or (reach[i][k] and reach[k][j])
11
12    result = []
13    for u, v in queries:
14        result.append(reach[u][v])
15    return result

ℹ

Complexity note: The time complexity is O(n³) due to the three nested loops in the Floyd-Warshall algorithm. The space complexity is O(n²) because we maintain a 2D reachability matrix.

1Understanding graph representation is crucial.
2Transitive relationships can be efficiently computed using algorithms like Floyd-Warshall.

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