LeetCode #1462 — MEDIUM

Course Schedule IV

Move from brute-force thinking to an efficient approach using topological sort strategy.

The Problem

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] = [a_i, b_i] indicates that you must take course a_i first if you want to take course b_i.

For example, the pair [0, 1] indicates that you have to take course 0 before you can take course 1.

Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c.

You are also given an array queries where queries[j] = [u_j, v_j]. For the j^th query, you should answer whether course u_j is a prerequisite of course v_j or not.

Return a boolean array answer, where answer[j] is the answer to the j^th query.

Example 1:

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.

Example 2:

Input: numCourses = 2, prerequisites = [], queries = [[1,0],[0,1]]
Output: [false,false]
Explanation: There are no prerequisites, and each course is independent.

Example 3:

Input: numCourses = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
Output: [true,true]

Constraints:

2 <= numCourses <= 100
0 <= prerequisites.length <= (numCourses * (numCourses - 1) / 2)
prerequisites[i].length == 2
0 <= a_i, b_i <= numCourses - 1
a_i != b_i
All the pairs [a_i, b_i] are unique.
The prerequisites graph has no cycles.
1 <= queries.length <= 10⁴
0 <= u_i, v_i <= numCourses - 1
u_i != v_i

Step 01

Brute Force Baseline

Problem summary: 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. For example, the pair [0, 1] indicates that you have to take course 0 before you can take course 1. Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c. You are also given an array queries where queries[j] = [uj, vj]. For the jth query, you should answer whether course uj is a prerequisite of course vj or not. Return a boolean array answer, where answer[j] is the answer to the jth query.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Topological Sort

Example 1

2
[[1,0]]
[[0,1],[1,0]]

Example 2

2
[]
[[1,0],[0,1]]

Example 3

3
[[1,2],[1,0],[2,0]]
[[1,0],[1,2]]

Step 02

Core Insight

What unlocks the optimal approach

Imagine if the courses are nodes of a graph. We need to build an array isReachable[i][j].
Start a bfs from each course i and assign for each course j you visit isReachable[i][j] = True.
Answer the queries from the isReachable array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1462: Course Schedule IV
class Solution {
    public List<Boolean> checkIfPrerequisite(int n, int[][] prerequisites, int[][] queries) {
        boolean[][] f = new boolean[n][n];
        for (var p : prerequisites) {
            f[p[0]][p[1]] = true;
        }
        for (int k = 0; k < n; ++k) {
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    f[i][j] |= f[i][k] && f[k][j];
                }
            }
        }
        List<Boolean> ans = new ArrayList<>();
        for (var q : queries) {
            ans.add(f[q[0]][q[1]]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1462: Course Schedule IV
func checkIfPrerequisite(n int, prerequisites [][]int, queries [][]int) (ans []bool) {
	f := make([][]bool, n)
	for i := range f {
		f[i] = make([]bool, n)
	}
	for _, p := range prerequisites {
		f[p[0]][p[1]] = true
	}
	for k := 0; k < n; k++ {
		for i := 0; i < n; i++ {
			for j := 0; j < n; j++ {
				f[i][j] = f[i][j] || (f[i][k] && f[k][j])
			}
		}
	}
	for _, q := range queries {
		ans = append(ans, f[q[0]][q[1]])
	}
	return
}

# Accepted solution for LeetCode #1462: Course Schedule IV
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]

// Accepted solution for LeetCode #1462: Course Schedule IV
struct Solution;
use std::collections::HashSet;

impl Solution {
    fn check_if_prerequisite(
        n: i32,
        prerequisites: Vec<Vec<i32>>,
        queries: Vec<Vec<i32>>,
    ) -> Vec<bool> {
        let n = n as usize;
        let mut graph: Vec<HashSet<usize>> = vec![HashSet::new(); n];
        for edge in prerequisites {
            let u = edge[0] as usize;
            let v = edge[1] as usize;
            graph[u].insert(v);
        }
        let mut closures: Vec<HashSet<usize>> = vec![HashSet::new(); n];
        for i in 0..n {
            let mut visited = vec![false; n];
            Self::dfs(i, &mut visited, &mut closures, &graph, i);
        }
        queries
            .into_iter()
            .map(|v| closures[v[0] as usize].contains(&(v[1] as usize)))
            .collect()
    }

    fn dfs(
        u: usize,
        visited: &mut Vec<bool>,
        closures: &mut Vec<HashSet<usize>>,
        graph: &[HashSet<usize>],
        start: usize,
    ) {
        if !visited[u] {
            visited[u] = true;
            for &v in &graph[u] {
                closures[start].insert(v);
                Self::dfs(v, visited, closures, graph, start);
            }
        }
    }
}

#[test]
fn test() {
    let n = 2;
    let prerequisites = vec_vec_i32![[1, 0]];
    let queries = vec_vec_i32![[0, 1], [1, 0]];
    let res = vec![false, true];
    assert_eq!(
        Solution::check_if_prerequisite(n, prerequisites, queries),
        res
    );
    let n = 2;
    let prerequisites = vec_vec_i32![];
    let queries = vec_vec_i32![[1, 0], [0, 1]];
    let res = vec![false, false];
    assert_eq!(
        Solution::check_if_prerequisite(n, prerequisites, queries),
        res
    );
    let n = 3;
    let prerequisites = vec_vec_i32![[1, 2], [1, 0], [2, 0]];
    let queries = vec_vec_i32![[1, 0], [1, 2]];
    let res = vec![true, true];
    assert_eq!(
        Solution::check_if_prerequisite(n, prerequisites, queries),
        res
    );
    let n = 3;
    let prerequisites = vec_vec_i32![[1, 0], [2, 0]];
    let queries = vec_vec_i32![[0, 1], [2, 0]];
    let res = vec![false, true];
    assert_eq!(
        Solution::check_if_prerequisite(n, prerequisites, queries),
        res
    );
    let n = 5;
    let prerequisites = vec_vec_i32![[0, 1], [1, 2], [2, 3], [3, 4]];
    let queries = vec_vec_i32![[0, 4], [4, 0], [1, 3], [3, 0]];
    let res = vec![true, false, true, false];
    assert_eq!(
        Solution::check_if_prerequisite(n, prerequisites, queries),
        res
    );
}

// Accepted solution for LeetCode #1462: Course Schedule IV
function checkIfPrerequisite(n: number, prerequisites: number[][], queries: number[][]): boolean[] {
    const f = Array.from({ length: n }, () => Array(n).fill(false));
    prerequisites.forEach(([a, b]) => (f[a][b] = true));
    for (let k = 0; k < n; ++k) {
        for (let i = 0; i < n; ++i) {
            for (let j = 0; j < n; ++j) {
                f[i][j] ||= f[i][k] && f[k][j];
            }
        }
    }
    return queries.map(([a, b]) => f[a][b]);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^3)

Space

O(n^2)

Approach Breakdown

REPEATED DFS

O(V × E) time

O(V) space

Repeatedly find a vertex with no incoming edges, remove it and its outgoing edges, and repeat. Finding the zero-in-degree vertex scans all V vertices, and we do this V times. Removing edges touches E edges total. Without an in-degree array, this gives O(V × E).

TOPOLOGICAL SORT

O(V + E) time

O(V + E) space

Build an adjacency list (O(V + E)), then either do Kahn's BFS (process each vertex once + each edge once) or DFS (visit each vertex once + each edge once). Both are O(V + E). Space includes the adjacency list (O(V + E)) plus the in-degree array or visited set (O(V)).

Shortcut: Process each vertex once + each edge once → O(V + E). Same as BFS/DFS on a graph.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.