What is the brute force approach and complexity for Course Schedule II?

Each round, take the first course whose prerequisites are all taken; stop when none is free with time complexity O(N · (N + P)) and space complexity O(N).

What is the optimal approach and complexity for Course Schedule II?

Count prerequisites once; queue the courses with none left; return [] if not all were taken with optimal time complexity O(N + P) and space complexity O(N + P).

How do you identify the pattern for Course Schedule II?

Look for key signals in the prompt: "return an order" indicates We need the sequence itself, not just yes or no, so the queue order is the answer.; "[a, b] means b before a" indicates A directed arrow b → a: the prerequisite points to the course that needs it.; "return an empty list if impossible" indicates A cycle means some courses are never taken, so the order is short..

Course Schedule II

Medium·LeetCode #210 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

There are numCourses courses, labelled 0 to numCourses − 1. Each pair [a, b] in prerequisites means you must finish course b before you can take course a.

Return an order in which you can take all the courses. If several orders work, return any one of them. If no order works (the prerequisites form a loop), return an empty list.

Input

numCourses, and prerequisites, a list of pairs [a, b].

Output

A list with every course once, in an order that respects every pair. Or an empty list when no such order exists.

Example 1

Input: numCourses = 2, prerequisites = [[1, 0]]

Output: [0, 1]

To take course 1 you must finish course 0 first, so 0 comes before 1.

Example 2

Input: numCourses = 4, prerequisites = [[1, 0], [2, 0], [3, 1], [3, 2]]

Output: [0, 1, 2, 3] (or [0, 2, 1, 3])

Course 0 needs nothing. Courses 1 and 2 both need only 0. Course 3 needs both 1 and 2, so it comes last.

Example 3

Input: numCourses = 2, prerequisites = [[0, 1], [1, 0]]

Output: []

Course 0 needs 1 and course 1 needs 0. Neither can be taken first, so there is no valid order.

Constraints

• 1 ≤ numCourses ≤ 2000
• 0 ≤ prerequisites.length ≤ numCourses · (numCourses − 1)
• prerequisites[i].length == 2
• 0 ≤ a, b < numCourses
• a ≠ b
• All the pairs [a, b] are distinct.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“return an order”

→ We need the sequence itself, not just yes or no, so the queue order is the answer.

“[a, b] means b before a”

→ A directed arrow b → a: the prerequisite points to the course that needs it.

“return an empty list if impossible”

→ A cycle means some courses are never taken, so the order is short.

Approach Complexity Ladder

Brute Force — rescan for a free courseO(N · (N + P))O(N)

Each round, take the first course whose prerequisites are all taken; stop when none is free

Optimal — Kahn's queue, empty answer if shortO(N + P)O(N + P)

Count prerequisites once; queue the courses with none left; return [] if not all were taken

Brute Force Intuition

Keep track of the taken courses. In each round, rebuild the list of blocked courses from all the pairs, take the first course that is not blocked, and repeat. If a round finds no free course while some are left, they wait on each other, so return an empty list.

Recognition rule for this sub-pattern:

Queue all in-degree-0 nodes; removing one lowers its neighbours' in-degree.

Remember this · Topological Sort

Trigger

Prerequisites, dependencies, build order, "can all courses be finished?", ordering letters from sorted words.

Think

Is this a directed graph where some things must happen before others?

Key question

Which nodes currently have in-degree 0?

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