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Course Schedule⚓︎

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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.

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:

  • 1 <= numCourses <= 2000
  • 0 <= prerequisites.length <= 5000
  • prerequisites[i].length == 2
  • 0 <= a_i, b_i < numCourses
  • All the pairs prerequisites[i] are unique.

Solution⚓︎

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class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> edges(numCourses);
        vector<int> inDegrees(numCourses);

        for (const auto& info : prerequisites) {
            edges[info[1]].push_back(info[0]);
            inDegrees[info[0]]++;
        }

        queue<int> q;
        for (int i = 0; i < numCourses; i++) {
            if (inDegrees[i] == 0)
                q.push(i);
        }

        int count = 0;
        while (!q.empty()) {
            int current = q.front();
            q.pop();
            count++;
            for (int v : edges[current]) {
                if (--inDegrees[v] == 0)
                    q.push(v);
            }
        }

        return count == numCourses;
    }
};
  • Time complexity: \(O(E+V)\);
  • Space complexity: \(O(E+V)\).