LeetCode Solutions
207. Course Schedule
Time: $O(|V| + |E|)$, where $|V| = \texttt{numCourses}$ and $|E| = |\texttt{prerequisites}|$ Space: $O(|V| + |E|)$, where $|V| = \texttt{numCourses}$ and $|E| = |\texttt{prerequisites}|$
enum class State { kInit, kVisiting, kVisited };
class Solution {
public:
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
vector<vector<int>> graph(numCourses);
vector<State> state(numCourses);
for (const vector<int>& p : prerequisites)
graph[p[1]].push_back(p[0]);
for (int i = 0; i < numCourses; ++i)
if (hasCycle(graph, i, state))
return false;
return true;
}
private:
bool hasCycle(const vector<vector<int>>& graph, int u, vector<State>& state) {
if (state[u] == State::kVisiting)
return true;
if (state[u] == State::kVisited)
return false;
state[u] = State::kVisiting;
for (const int v : graph[u])
if (hasCycle(graph, v, state))
return true;
state[u] = State::kVisited;
return false;
}
};
enum State { kInit, kVisiting, kVisited }
class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
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))
return false;
return true;
}
private boolean hasCycle(List<Integer>[] graph, int u, State[] state) {
if (state[u] == State.kVisiting)
return true;
if (state[u] == State.kVisited)
return false;
state[u] = State.kVisiting;
for (final int v : graph[u])
if (hasCycle(graph, v, state))
return true;
state[u] = State.kVisited;
return false;
}
}
from enum import Enum
class State(Enum):
kInit = 0
kVisiting = 1
kVisited = 2
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
graph = [[] for _ in range(numCourses)]
state = [State.kInit] * numCourses
for a, b in prerequisites:
graph[b].append(a)
def hasCycle(u: int) -> bool:
if state[u] == State.kVisiting:
return True
if state[u] == State.kVisited:
return False
state[u] = State.kVisiting
if any(hasCycle(v) for v in graph[u]):
return True
state[u] = State.kVisited
return False
return not any(hasCycle(i) for i in range(numCourses))