16. Course Schedule

mediumAsked at Chegg

Detect if all courses can be finished given prerequisites — directly mirrors Chegg's curriculum graph where course dependencies must be cycle-free.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Problem

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

Constraints

1 <= numCourses <= 2000
0 <= prerequisites.length <= 5000
prerequisites[i].length == 2
No self-loops, no duplicate pairs

Examples

Example 1

Input

numCourses = 2, prerequisites = [[1,0]]

Output

true

Example 2

Input

numCourses = 2, prerequisites = [[1,0],[0,1]]

Output

false

Approaches

1. DFS cycle detection with visited states

Three-color DFS: unvisited (0), in-stack (1), done (2) — cycles show up when we revisit an in-stack node.

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);
  function dfs(node) {
    if (state[node] === 1) return false;
    if (state[node] === 2) return true;
    state[node] = 1;
    for (const nb of adj[node]) if (!dfs(nb)) return false;
    state[node] = 2;
    return true;
  }
  for (let i = 0; i < numCourses; i++) if (!dfs(i)) return false;
  return true;
}

Tradeoff:

2. Topological sort (Kahn's BFS)

Build in-degree counts, queue all zero-in-degree nodes, process iteratively reducing in-degrees — if all nodes are processed, no cycle exists.

Time: O(V+E)
Space: O(V+E)

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

Tradeoff:

Chegg-specific tips

Chegg's platform has literal course dependency graphs; mentioning Kahn's algorithm in the context of curriculum scheduling will resonate strongly with the interviewer.

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 Chegg interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →