Leetcode #1462: Course Schedule IV

In this guide, we solve Leetcode #1462 Course Schedule IV in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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 ai first if you want to take course bi.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Depth-First Search, Breadth-First Search, Graph, Topological Sort

Intuition

The data forms a graph, so we should explore nodes and edges systematically.

A traversal ensures we visit each node once while maintaining the needed state.

Approach

Build an adjacency list and traverse with BFS or DFS.

Aggregate results as you visit nodes.

Steps:

Build the graph.
Traverse with BFS/DFS.
Accumulate the required output.

Example

Input: numCourses = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: The pair [1, 0] indicates that you have to take course 1 before you can take course 0.
Course 0 is not a prerequisite of course 1, but the opposite is true.

Python Solution

class Solution:
    def checkIfPrerequisite(
        self, n: int, prerequisites: List[List[int]], queries: List[List[int]]
    ) -> List[bool]:
        f = [[False] * n for _ in range(n)]
        for a, b in prerequisites:
            f[a][b] = True
        for k in range(n):
            for i in range(n):
                for j in range(n):
                    if f[i][k] and f[k][j]:
                        f[i][j] = True
        return [f[a][b] for a, b in queries]

Complexity

The time complexity is $O(n^3)$ , and the space complexity is $O(n^2)$ , where $n$ is the number of nodes. The space complexity is $O(n^2)$ , where $n$ is the number of nodes.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Python Solution

class Solution:
    def checkIfPrerequisite(
        self, n: int, prerequisites: List[List[int]], queries: List[List[int]]
    ) -> List[bool]:
        f = [[False] * n for _ in range(n)]
        for a, b in prerequisites:
            f[a][b] = True
        for k in range(n):
            for i in range(n):
                for j in range(n):
                    if f[i][k] and f[k][j]:
                        f[i][j] = True
        return [f[a][b] for a, b in queries]

Leetcode #1462: Course Schedule IV

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #1462: Course Schedule IV

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview