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17. Course Schedule

mediumAsked at Autodesk

Decide if you can finish all courses given prerequisite pairs that form a directed graph.

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

Problem

There are numCourses labeled from 0 to numCourses-1. You are given prerequisite pairs [a, b] meaning you must take b before a. Return true if you can finish all courses, i.e., the directed prerequisite graph has no cycle.

Constraints

  • 1 <= numCourses <= 2000
  • 0 <= prerequisites.length <= 5000

Examples

Example 1

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

Example 2

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

Approaches

1. DFS color marking

Color nodes white/gray/black; gray reentry means a cycle.

Time
O(V+E)
Space
O(V+E)
function dfs(u){ if (color[u]===1) return false; if (color[u]===2) return true; color[u]=1; for (v of adj[u]) if (!dfs(v)) return false; color[u]=2; return true; }

Tradeoff:

2. Kahn's BFS topological sort

Compute indegrees, push zero-indegree nodes onto a queue, and pop while decrementing neighbors. If you pop all V nodes there's no cycle.

Time
O(V+E)
Space
O(V+E)
function canFinish(n, prereqs) {
  const adj = Array.from({length: n}, () => []);
  const indeg = new Array(n).fill(0);
  for (const [a, b] of prereqs) { adj[b].push(a); indeg[a]++; }
  const q = [];
  for (let i = 0; i < n; i++) if (indeg[i] === 0) q.push(i);
  let taken = 0;
  while (q.length) {
    const u = q.shift();
    taken++;
    for (const v of adj[u]) if (--indeg[v] === 0) q.push(v);
  }
  return taken === n;
}

Tradeoff:

Autodesk-specific tips

Topo-order reasoning maps directly onto Autodesk's dependency DAGs in parametric modeling (Fusion timeline, Revit family graph).

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Output

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