LeetCode #210 — MEDIUM

Course Schedule II

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 b_i first if you want to take course a_i.

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

Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].

Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]

Constraints:

1 <= numCourses <= 2000
0 <= prerequisites.length <= numCourses * (numCourses - 1)
prerequisites[i].length == 2
0 <= a_i, b_i < numCourses
a_i != b_i
All the pairs [a_i, b_i] are distinct.

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 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 the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

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]]

Example 2

4
[[1,0],[2,0],[3,1],[3,2]]

Example 3

1
[]

Core Insight

What unlocks the optimal approach

This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
<a href="https://www.youtube.com/watch?v=ozso3xxkVGU" target="_blank">Topological Sort via DFS</a> - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via <a href="http://en.wikipedia.org/wiki/Topological_sorting#Algorithms" target="_blank">BFS</a>.

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 #210: Course Schedule II
class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        List<Integer>[] g = new List[numCourses];
        Arrays.setAll(g, k -> new ArrayList<>());
        int[] indeg = new int[numCourses];
        for (var p : prerequisites) {
            int a = p[0], b = p[1];
            g[b].add(a);
            ++indeg[a];
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < numCourses; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }
        int[] ans = new int[numCourses];
        int cnt = 0;
        while (!q.isEmpty()) {
            int i = q.poll();
            ans[cnt++] = i;
            for (int j : g[i]) {
                if (--indeg[j] == 0) {
                    q.offer(j);
                }
            }
        }
        return cnt == numCourses ? ans : new int[0];
    }
}

// Accepted solution for LeetCode #210: Course Schedule II
func findOrder(numCourses int, prerequisites [][]int) []int {
	g := make([][]int, numCourses)
	indeg := make([]int, numCourses)
	for _, p := range prerequisites {
		a, b := p[0], p[1]
		g[b] = append(g[b], a)
		indeg[a]++
	}
	q := []int{}
	for i, x := range indeg {
		if x == 0 {
			q = append(q, i)
		}
	}
	ans := []int{}
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		ans = append(ans, i)
		for _, j := range g[i] {
			indeg[j]--
			if indeg[j] == 0 {
				q = append(q, j)
			}
		}
	}
	if len(ans) == numCourses {
		return ans
	}
	return []int{}
}

# Accepted solution for LeetCode #210: Course Schedule II
class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        g = defaultdict(list)
        indeg = [0] * numCourses
        for a, b in prerequisites:
            g[b].append(a)
            indeg[a] += 1
        ans = []
        q = deque(i for i, x in enumerate(indeg) if x == 0)
        while q:
            i = q.popleft()
            ans.append(i)
            for j in g[i]:
                indeg[j] -= 1
                if indeg[j] == 0:
                    q.append(j)
        return ans if len(ans) == numCourses else []

// Accepted solution for LeetCode #210: Course Schedule II
impl Solution {
    pub fn find_order(num_courses: i32, prerequisites: Vec<Vec<i32>>) -> Vec<i32> {
        let n = num_courses as usize;
        let mut adjacency = vec![vec![]; n];
        let mut entry = vec![0; n];
        // init
        for iter in prerequisites.iter() {
            let (a, b) = (iter[0], iter[1]);
            adjacency[b as usize].push(a);
            entry[a as usize] += 1;
        }
        // construct deque & reslut
        let mut deque = std::collections::VecDeque::new();
        for index in 0..n {
            if entry[index] == 0 {
                deque.push_back(index);
            }
        }
        let mut result = vec![];
        // bfs
        while !deque.is_empty() {
            let head = deque.pop_front().unwrap();
            result.push(head as i32);
            // update degree of entry
            for &out_entry in adjacency[head].iter() {
                entry[out_entry as usize] -= 1;
                if entry[out_entry as usize] == 0 {
                    deque.push_back(out_entry as usize);
                }
            }
        }
        if result.len() == n {
            result
        } else {
            vec![]
        }
    }
}

// Accepted solution for LeetCode #210: Course Schedule II
function findOrder(numCourses: number, prerequisites: number[][]): number[] {
    const g: number[][] = Array.from({ length: numCourses }, () => []);
    const indeg: number[] = new Array(numCourses).fill(0);
    for (const [a, b] of prerequisites) {
        g[b].push(a);
        indeg[a]++;
    }
    const q: number[] = [];
    for (let i = 0; i < numCourses; ++i) {
        if (indeg[i] === 0) {
            q.push(i);
        }
    }
    const ans: number[] = [];
    while (q.length) {
        const i = q.shift()!;
        ans.push(i);
        for (const j of g[i]) {
            if (--indeg[j] === 0) {
                q.push(j);
            }
        }
    }
    return ans.length === numCourses ? ans : [];
}

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(V + E)

Space

O(V + E)

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.