Skip to main content

19. Course Schedule

mediumAsked at Glassdoor

Glassdoor models skill prerequisites and career-path dependencies as directed graphs — cycle detection via topological sort is the pattern they reach for when verifying those graphs are actually traversable.

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] means you must take course bi before ai. Return true if you can finish all courses, false if a cycle makes it impossible.

Constraints

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

Examples

Example 1

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

Explanation: Take course 0 first, then course 1. No cycle.

Example 2

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

Explanation: Course 0 requires 1 and course 1 requires 0 — a cycle makes completion impossible.

Approaches

1. DFS cycle detection

Build an adjacency list and run DFS, tracking nodes in the current recursion stack. A back edge (visiting a gray node) indicates 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);

  // 0 = unvisited, 1 = in stack, 2 = done
  const state = new Array(numCourses).fill(0);

  function hasCycle(node) {
    if (state[node] === 1) return true;
    if (state[node] === 2) return false;
    state[node] = 1;
    for (const nb of adj[node]) {
      if (hasCycle(nb)) return true;
    }
    state[node] = 2;
    return false;
  }

  for (let i = 0; i < numCourses; i++) {
    if (hasCycle(i)) return false;
  }
  return true;
}

Tradeoff:

2. BFS Kahn's algorithm (topological sort)

Count in-degrees. Repeatedly enqueue nodes with in-degree 0 and decrement neighbors. If the total processed equals numCourses, 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 > 0) {
    const node = queue.shift();
    processed++;
    for (const nb of adj[node]) {
      inDegree[nb]--;
      if (inDegree[nb] === 0) queue.push(nb);
    }
  }
  return processed === numCourses;
}

Tradeoff:

Glassdoor-specific tips

Glassdoor's graph-heavy data model (companies → reviews → roles → skills) means engineers regularly reason about dependency chains. Interviewers want to hear you name the two approaches — DFS with coloring vs. Kahn's BFS — and know when each is cleaner. For interview settings, Kahn's iterative version is often easier to debug out loud. Verify your in-degree counting on a small example before claiming correctness.

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

Practice these live with InterviewChamp.AI

Drill Course Schedule and other Glassdoor interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →