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# 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) {
degree[e]++;
}
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].push_back(e), degree[e]++;
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 =  * 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``````
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