207. Course Schedule
1. Description
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] = [$a_i$, $b_i$] indicates that you must take course $b_i$ first if you want to take course $a_i$.
- For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.
Return true if you can finish all courses. Otherwise, return false.
2. Example
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.
3. Constraints
- 1 <= numCourses <= 2000
- 0 <= prerequisites.length <= 5000
- prerequisites[i].length == 2
- 0 <= $a_i$, $b_i$ < numCourses
- All the pairs prerequisites[i] are unique.
4. Solutions
Breadth-First Search
m is the number of nodes, n is the number of edges
Time complexity: O(m + n)
Space complexity: O(m + n)
class Solution {
public:
bool canFinish(int n, const vector<vector<int>> &prerequisites) {
vector<int> requires_count(n, 0);
vector<vector<int>> enable_courses(n, vector<int>{});
for (const auto &p : prerequisites) {
enable_courses[p[1]].push_back(p[0]);
++requires_count[p[0]];
}
queue<int> courses;
for (int i = 0; i < n; ++i) {
if (requires_count[i] == 0) {
courses.push(i);
}
}
int finished_count = 0;
while (!courses.empty()) {
int course = courses.front();
courses.pop();
++finished_count;
for (int e : enable_courses[course]) {
--requires_count[e];
if (requires_count[e] == 0) {
courses.push(e);
}
}
}
return finished_count == n;
}
};