19. Course Schedule

mediumAsked at Notion

Detect whether a directed dependency graph has a cycle — the same topological check Notion runs when resolving database rollup chains and formula dependencies to prevent circular reference crashes.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-26

Problem

There are numCourses courses labeled 0 to numCourses-1. You are given prerequisites, where prerequisites[i] = [a, b] means you must take course b before course a. Return true if you can finish all courses (i.e., no cycle exists in the dependency graph).

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 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 — cycle detected.

Approaches

1. DFS cycle detection (three-color)

Mark nodes as unvisited / in-progress / done. If DFS hits an in-progress node, a back edge (cycle) exists.

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 state = new Array(numCourses).fill(0); // 0=unvisited,1=visiting,2=done
  function dfs(node) {
    if (state[node] === 1) return false; // cycle
    if (state[node] === 2) return true;  // already verified
    state[node] = 1;
    for (const next of adj[node]) {
      if (!dfs(next)) return false;
    }
    state[node] = 2;
    return true;
  }
  for (let i = 0; i < numCourses; i++) {
    if (!dfs(i)) return false;
  }
  return true;
}

Tradeoff:

2. BFS topological sort (Kahn's algorithm)

Compute in-degrees; enqueue zero-in-degree nodes; process queue, decrement neighbor 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 next of adj[node]) {
      if (--inDegree[next] === 0) queue.push(next);
    }
  }
  return processed === numCourses;
}

Tradeoff:

Notion-specific tips

Notion database formulas can reference other property values — circular references cause infinite recalculation. Interviewers want you to name topological sort explicitly and explain Kahn's algorithm vs. DFS-coloring as alternative detection strategies. Bringing up the Notion formula context earns extra points.

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