There are a total of numCourses courses you have to take, labeled from 0 to numCourses-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
Example 1:
Input: numCourses = 2, prerequisites = [[1,0]] Output: true Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: numCourses = 2, prerequisites = [[1,0],[0,1]] Output: false Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
Constraints:
- The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
- You may assume that there are no duplicate edges in the input prerequisites.
1 <= numCourses <= 10^5
class Solution {
public:
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
// Check cycle.
vector<vector<int>> graph = vector<vector<int>>(numCourses, vector<int>());
for (auto prereq : prerequisites) {
graph[prereq[0]].push_back(prereq[1]);
}
vector<int> graph_state = vector<int>(numCourses, 0);
for (int node = 0; node < numCourses; node++) {
if (graph_state[node] == 0 && cycle(graph, graph_state, node)) {
return false;
}
}
return true;
}
// DFS.
bool cycle(vector<vector<int>>& graph, vector<int>& state, int node) {
if (state[node] == 2) {
return false;
}
if (state[node] == 1) {
return true;
}
state[node] = 1;
for (auto neighbor : graph[node]) {
if (cycle(graph, state, neighbor)) {
return true;
}
}
state[node] = 2;
return false;
}
};
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