Course Schedule LeetCode Solution – Queslers

Problem – Course Schedule

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] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

  • 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 <= ai, bi < numCourses
  • All the pairs prerequisites[i] are unique.

Course Schedule LeetCode Solution in Java

    public boolean canFinish(int n, int[][] prerequisites) {
        ArrayList<Integer>[] G = new ArrayList[n];
        int[] degree = new int[n];
        ArrayList<Integer> bfs = new ArrayList();
        for (int i = 0; i < n; ++i) G[i] = new ArrayList<Integer>();
        for (int[] e : prerequisites) {
            G[e[1]].add(e[0]);
            degree[e[0]]++;
        }
        for (int i = 0; i < n; ++i) if (degree[i] == 0) bfs.add(i);
        for (int i = 0; i < bfs.size(); ++i)
            for (int j: G[bfs.get(i)])
                if (--degree[j] == 0) bfs.add(j);
        return bfs.size() == n;
    }

Course Schedule LeetCode Solution in C++

    bool canFinish(int n, vector<vector<int>>& prerequisites) {
        vector<vector<int>> G(n);
        vector<int> degree(n, 0), bfs;
        for (auto& e : prerequisites)
            G[e[1]].push_back(e[0]), degree[e[0]]++;
        for (int i = 0; i < n; ++i) if (!degree[i]) bfs.push_back(i);
        for (int i = 0; i < bfs.size(); ++i)
            for (int j: G[bfs[i]])
                if (--degree[j] == 0) bfs.push_back(j);
        return bfs.size() == n;
    }

Course Schedule LeetCode Solution in Python

    def canFinish(self, n, prerequisites):
        G = [[] for i in range(n)]
        degree = [0] * n
        for j, i in prerequisites:
            G[i].append(j)
            degree[j] += 1
        bfs = [i for i in range(n) if degree[i] == 0]
        for i in bfs:
            for j in G[i]:
                degree[j] -= 1
                if degree[j] == 0:
                    bfs.append(j)
        return len(bfs) == n
Course Schedule LeetCode Solution Review:

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