Course Schedule II LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Course Schedule II There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

  • For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].

Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]

Constraints:

  • 1 <= numCourses <= 2000
  • 0 <= prerequisites.length <= numCourses * (numCourses - 1)
  • prerequisites[i].length == 2
  • 0 <= ai, bi < numCourses
  • ai != bi
  • All the pairs [ai, bi] are distinct.

Course Schedule II Solutions

Time: O(∣V∣+∣E∣)
Space: O(∣V∣+∣E∣)

C++

class Solution {
 public:
  vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
    vector<int> ans;
    vector<vector<int>> graph(numCourses);
    vector<int> inDegree(numCourses);
    queue<int> q;

    // build graph
    for (const auto& p : prerequisites) {
      const int u = p[1];
      const int v = p[0];
      graph[u].push_back(v);
      ++inDegree[v];
    }

    // topology
    for (int i = 0; i < numCourses; ++i)
      if (inDegree[i] == 0)
        q.push(i);

    while (!q.empty()) {
      const int u = q.front();
      q.pop();
      ans.push_back(u);
      for (const int v : graph[u])
        if (--inDegree[v] == 0)
          q.push(v);
    }

    return ans.size() == numCourses ? ans : vector<int>();
  }
};

Java

 class Solution {
  public int[] findOrder(int numCourses, int[][] prerequisites) {
    List<Integer> ans = new ArrayList<>();
    List<Integer>[] graph = new List[numCourses];
    int[] inDegree = new int[numCourses];
    Queue<Integer> q = new ArrayDeque<>();

    for (int i = 0; i < numCourses; ++i)
      graph[i] = new ArrayList<>();

    // build graph
    for (int[] p : prerequisites) {
      final int u = p[1];
      final int v = p[0];
      graph[u].add(v);
      ++inDegree[v];
    }

    // topology
    for (int i = 0; i < numCourses; ++i)
      if (inDegree[i] == 0)
        q.offer(i);

    while (!q.isEmpty()) {
      final int u = q.poll();
      ans.add(u);
      for (final int v : graph[u])
        if (--inDegree[v] == 0)
          q.offer(v);
    }

    if (ans.size() == numCourses)
      return ans.stream().mapToInt(i -> i).toArray();
    return new int[] {};
  }
}

Python


class Solution:
  def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
    ans = []
    graph = [[] for _ in range(numCourses)]
    inDegree = [0] * numCourses
    q = deque()

    # build graph
    for v, u in prerequisites:
      graph[u].append(v)
      inDegree[v] += 1

    # topology
    for i, degree in enumerate(inDegree):
      if degree == 0:
        q.append(i)

    while q:
      u = q.popleft()
      ans.append(u)
      for v in graph[u]:
        inDegree[v] -= 1
        if inDegree[v] == 0:
          q.append(v)

    return ans if len(ans) == numCourses else []

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