Leetcode

Course Schedule II

May 26, 2022 by aAshish

Approach 1: DFS

  • Time:O(|V| + |E|), where |V| = \texttt{numCourses} and |E| = |\texttt{prerequisites}|
  • Space:O(|V| + |E|), where |V| = \texttt{numCourses} and |E| = |\texttt{prerequisites}|

C++

 
enum class State { INIT, VISITING, VISITED };

class Solution {
 public:
  vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
    vector<int> ans;
    vector<vector<int>> graph(numCourses);
    vector<State> state(numCourses);

    for (const auto& p : prerequisites)
      graph[p[1]].push_back(p[0]);

    for (int i = 0; i < numCourses; ++i)
      if (hasCycle(graph, i, state, ans))
        return {};

    reverse(begin(ans), end(ans));
    return ans;
  }

 private:
  bool hasCycle(const vector<vector<int>>& graph, int u, vector<State>& state,
                vector<int>& ans) {
    if (state[u] == State::VISITING)
      return true;
    if (state[u] == State::VISITED)
      return false;

    state[u] = State::VISITING;
    for (const int v : graph[u])
      if (hasCycle(graph, v, state, ans))
        return true;
    state[u] = State::VISITED;
    ans.push_back(u);

    return false;
  }
};

JAVA

 
enum State { INIT, VISITING, VISITED }

class Solution {
  public int[] findOrder(int numCourses, int[][] prerequisites) {
    Deque<Integer> ans = new ArrayDeque<>();
    List<Integer>[] graph = new List[numCourses];
    State[] state = new State[numCourses];

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

    for (int[] p : prerequisites)
      graph[p[1]].add(p[0]);

    for (int i = 0; i < numCourses; ++i)
      if (hasCycle(graph, i, state, ans))
        return new int[] {};

    return ans.stream().mapToInt(i -> i).toArray();
  }

  private boolean hasCycle(List<Integer>[] graph, int u, State[] state, Deque<Integer> ans) {
    if (state[u] == State.VISITING)
      return true;
    if (state[u] == State.VISITED)
      return false;

    state[u] = State.VISITING;
    for (final int v : graph[u])
      if (hasCycle(graph, v, state, ans))
        return true;
    state[u] = State.VISITED;
    ans.addFirst(u);

    return false;
  }
}

Python

 
from enum import Enum


class State(Enum):
  INIT = 0
  VISITING = 1
  VISITED = 2


class Solution:
  def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
    ans = []
    graph = [[] for _ in range(numCourses)]
    state = [State.INIT] * numCourses

    for v, u in prerequisites:
      graph[u].append(v)

    def hasCycle(u: int) -> bool:
      if state[u] == State.VISITING:
        return True
      if state[u] == State.VISITED:
        return False

      state[u] = State.VISITING
      if any(hasCycle(v) for v in graph[u]):
        return True
      state[u] = State.VISITED
      ans.append(u)

      return False

    if any(hasCycle(i) for i in range(numCourses)):
      return []

    return ans[::-1]

Approach 2: Topology

  • 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];

    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
    Queue<Integer> q = IntStream.range(0, numCourses)
                           .filter(i -> inDegree[i] == 0)
                           .boxed()
                           .collect(Collectors.toCollection(ArrayDeque::new));

    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 []