# Course Schedule II LeetCode Solution

Here, We see Course Schedule II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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 the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Example 1:
Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].

Example 2:
Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].

Example 3:
Input: numCourses = 1, prerequisites = []
Output: [0]

## Course Schedule II LeetCode SolutionC++

``````class Solution{
public:
bool kahnAlgo(vector<vector<int>> &adj, int n, vector<int> &indegree, vector<int> &ans)
{
queue<int> q;
for (int i = 0; i < n; i++)
{
if (indegree[i] == 0)
q.push(i);
}
int count = 0;
while (!q.empty()){
int curr = q.front();
q.pop();
indegree[a] -= 1;
if (indegree[a] == 0)
q.push(a);
}
ans.push_back(curr);
count++;
}
if (count != n)
return false;
return true;
}

vector<int> findOrder(int numCourses, vector<vector<int>> &prerequisites)
{
int n = prerequisites.size();
vector<int> indegree(numCourses, 0);
for (int i = 0; i < n; i++){
indegree[prerequisites[i][0]] += 1;
}
vector<int> ans;
return ans;
return {};
}
};```Code language: PHP (php)```

## Course Schedule II LeetCode SolutionJava

``````class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
int[] inDeg = new int[numCourses];
List<Integer>[] chl = new ArrayList[numCourses];
for (int i = 0; i < numCourses; i++) {
chl[i] = new ArrayList<Integer>();
}
int pre;
int cour;
for (int[] pair : prerequisites) {
pre = pair[1];
cour = pair[0];
inDeg[cour]++;
}
int[] res = new int[numCourses];
int k = 0;
for (int i = 0; i < numCourses; i++) {
if (inDeg[i] == 0) {
res[k++] = i;
}
}
if (k == 0) {
return new int[0];
}
int j = 0;
List<Integer> tmp;
while (k < numCourses) {
tmp = chl[res[j++]];
for (int id : tmp) {
if (--inDeg[id] == 0) {
res[k++] = id;
}
}
if (j == k) {
return new int[0];
}
}
return res;
}
}```Code language: PHP (php)```

## Course Schedule II SolutionJavaScript

``````var findOrder = function(numCourses, prerequisites) {
const order = [];
const queue = [];
const graph = new Map();
const indegree = Array(numCourses).fill(0);
for (const [e, v] of prerequisites) {
if (graph.has(v)) {
graph.get(v).push(e);
} else {
graph.set(v, [e]);
}
indegree[e]++;
}
for (let i = 0; i < indegree.length; i++) {
if (indegree[i] === 0) queue.push(i);
}
while (queue.length) {
const v = queue.shift();
if (graph.has(v)) {
for (const e of graph.get(v)) {
indegree[e]--;
if (indegree[e] === 0) queue.push(e);
}
}
order.push(v);
}
return numCourses === order.length ? order : [];
};```Code language: JavaScript (javascript)```

## Course Schedule II SolutionPython

``````class Solution(object):
def findOrder(self, numCourses, prerequisites):
self.graph = collections.defaultdict(list)
self.res = []
for pair in prerequisites:
self.graph[pair[0]].append(pair[1])
self.visited = [0 for x in xrange(numCourses)]
for x in xrange(numCourses):
if not self.DFS(x):
return []
return self.res

def DFS(self, node):
if self.visited[node] == -1:
return False
if self.visited[node] == 1:
return True
self.visited[node] = -1
for x in self.graph[node]:
if not self.DFS(x):
return False
self.visited[node] = 1
self.res.append(node)
return True``````
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