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24. Course Schedule

mediumAsked at Box

Detect a cycle in a directed prerequisite graph — Box solves the same problem when validating that enterprise workflow automation rules and folder-permission inheritance chains contain no circular dependencies.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

There are numCourses courses labeled 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi before course ai. Return true if you can finish all courses, or false if there is a cycle making it impossible.

Constraints

  • 1 <= numCourses <= 2000
  • 0 <= prerequisites.length <= 5000
  • prerequisites[i].length == 2
  • 0 <= ai, bi < numCourses
  • All prerequisite pairs are unique

Examples

Example 1

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

Explanation: Take course 0 first, then course 1.

Example 2

Input
numCourses = 2, prerequisites = [[1,0],[0,1]]
Output
false

Explanation: Courses 0 and 1 depend on each other — cycle detected.

Approaches

1. Brute force — DFS with recursion stack

Build adjacency list, then DFS from each unvisited node tracking a 'in current path' set. A back-edge into the current path signals a cycle.

Time
O(V + E)
Space
O(V + E)
function canFinish(numCourses, prerequisites) {
  const adj = Array.from({length: numCourses}, () => []);
  for (const [a, b] of prerequisites) adj[b].push(a);
  const UNVISITED = 0, VISITING = 1, VISITED = 2;
  const state = new Array(numCourses).fill(UNVISITED);
  function dfs(node) {
    if (state[node] === VISITING) return false; // cycle
    if (state[node] === VISITED) return true;
    state[node] = VISITING;
    for (const nb of adj[node]) {
      if (!dfs(nb)) return false;
    }
    state[node] = VISITED;
    return true;
  }
  for (let i = 0; i < numCourses; i++) {
    if (!dfs(i)) return false;
  }
  return true;
}

Tradeoff:

2. Optimal — Kahn's algorithm (topological sort BFS)

Compute in-degrees; enqueue nodes with in-degree 0; each dequeue reduces neighbors' in-degrees. If all nodes are processed, no cycle exists.

Time
O(V + E)
Space
O(V + E)
function canFinish(numCourses, prerequisites) {
  const adj = Array.from({length: numCourses}, () => []);
  const inDegree = new Array(numCourses).fill(0);
  for (const [a, b] of prerequisites) {
    adj[b].push(a);
    inDegree[a]++;
  }
  const queue = [];
  for (let i = 0; i < numCourses; i++) {
    if (inDegree[i] === 0) queue.push(i);
  }
  let processed = 0;
  while (queue.length) {
    const node = queue.shift();
    processed++;
    for (const nb of adj[node]) {
      inDegree[nb]--;
      if (inDegree[nb] === 0) queue.push(nb);
    }
  }
  return processed === numCourses;
}

Tradeoff:

Box-specific tips

Box rates both DFS and Kahn's approaches equally — what separates candidates is the ability to articulate the real-world analogue. Mentioning that Box's permission-propagation engine must validate that group-inheritance graphs are acyclic before applying changes to millions of files shows you understand why the algorithm matters beyond the whiteboard.

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