Course Schedule Problem
Using Graph Topological Sort

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Course Schedule 2 - Task Ordering Using Topological Sort

Topic Contents

Problem Examples Solution Algorithm Code Time Space

Problem Statement

You are given n tasks labeled from 0 to n - 1 and an array of prerequisite pairs. Each pair [a, b] indicates that task b must be completed before task a.

Return any valid order in which the tasks can be completed. If no such order exists (due to a cycle), return an empty array.

Examples❯

n	Prerequisites	Valid Output	Description
2	[[1, 0]]	[0, 1]	Task 1 depends on 0, so 0 must come first
4	[[1, 0], [2, 0], [3, 1], [3, 2]]	[0, 1, 2, 3]	Task 3 depends on both 1 and 2, which in turn depend on 0
1	[]	[0]	Single task with no prerequisites
3	[[0, 1], [1, 2], [2, 0]]	[]	Cycle detected, impossible to complete all tasks
5	[]	[0, 1, 2, 3, 4]	No dependencies, any order is valid

Solution❯

Understanding the Problem

The Course Schedule problem involves scheduling a set of courses given a list of prerequisite pairs. Each pair indicates that one course must be completed before another. The challenge is to determine if it's possible to complete all the courses, and if so, in what order.

Representing Courses as a Graph

We treat each course as a node in a directed graph. If course A is a prerequisite for course B, we create a directed edge from A to B. This graph structure helps us track dependencies.

Detecting Cycles Using Topological Sort

To determine a valid order, we use topological sorting, which only works on Directed Acyclic Graphs (DAGs). If a cycle exists (i.e., circular dependency), it’s impossible to complete all courses.

Steps Involved

Build an adjacency list to represent the graph.
Track the in-degree (number of prerequisites) for each course.
Add courses with zero in-degree to a queue—they can be taken immediately.
Process the queue: for each course, reduce the in-degree of dependent courses. If a dependent course’s in-degree becomes zero, add it to the queue.

Valid Scenarios

If we can process all courses (i.e., the result list has all courses), then a valid order exists and we return it.

Invalid Scenarios

If we cannot process all courses—because a cycle prevents progress—then we return an empty array.

Example

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: Both are valid topological orders satisfying all prerequisites.

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: []
Explanation: A cycle exists (0 → 1 → 0), so no valid order is possible.

Algorithm Steps

Create an adjacency list for the graph.
Create an array to track in-degree for each task.
Populate the adjacency list and in-degree array based on prerequisites.
Initialize a queue with all tasks having in-degree 0.
While the queue is not empty:

Remove a task from the queue and add it to the result list.
For each neighbor of that task, reduce its in-degree by 1.
If any neighbor’s in-degree becomes 0, add it to the queue.

After processing, if the result contains all tasks, return it.
Otherwise, return an empty array (cycle detected).

Code

JavaScript

function findOrder(n, prerequisites) {
  const adj = Array.from({ length: n }, () => []);
  const inDegree = new Array(n).fill(0);

  for (const [a, b] of prerequisites) {
    adj[b].push(a);
    inDegree[a]++;
  }

  const queue = [], result = [];
  for (let i = 0; i < n; i++) {
    if (inDegree[i] === 0) queue.push(i);
  }

  while (queue.length > 0) {
    const node = queue.shift();
    result.push(node);
    for (const neighbor of adj[node]) {
      inDegree[neighbor]--;
      if (inDegree[neighbor] === 0) queue.push(neighbor);
    }
  }

  return result.length === n ? result : [];
}

console.log("Order for 2 tasks:", findOrder(2, [[1, 0]]));
console.log("Order for 4 tasks:", findOrder(4, [[1, 0], [2, 0], [3, 1], [3, 2]]));

Time Complexity

Case	Time Complexity	Explanation
Best Case	O(V + E)	In all cases, we must visit every node and edge to build the graph and perform topological sorting.
Average Case	O(V + E)	Each course and each prerequisite relation must be examined exactly once.
Worst Case	O(V + E)	Even with complex dependencies, we still process every node and edge once for sorting and cycle detection.

Space Complexity

O(V + E)

Explanation: We use an adjacency list (O(E)) and an in-degree array and queue (O(V)) to store and process the graph.

⬅ Previous Topic

Cycle Detection in Directed Graph using BFS

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Course Schedule 2 - Task Ordering Using Topological Sort

Overview of Techniques14
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Sorting6
❯

Arrays32
❯

Arrays - Binary Search28
❯

Strings18
❯

Matrix3
❯

Matrix - Binary Search5
❯

Binary Trees36
❯

Binary Search Trees14
❯

Graphs46
❯

Tries1
❯

Course Schedule Problem
Using Graph Topological Sort

Problem Statement

Examples❯

Solution❯

Understanding the Problem

Representing Courses as a Graph

Detecting Cycles Using Topological Sort

Steps Involved

Valid Scenarios

Invalid Scenarios

Example

Algorithm Steps

Code

Time Complexity

Space Complexity

Support ProgramGuru.org❯

Course Schedule ProblemUsing Graph Topological Sort

Problem Statement

Examples❯

Solution❯

Understanding the Problem

Representing Courses as a Graph

Detecting Cycles Using Topological Sort

Steps Involved

Valid Scenarios

Invalid Scenarios

Example

Algorithm Steps

Code

Time Complexity

Space Complexity

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Course Schedule Problem
Using Graph Topological Sort