Last updated on October 5th, 2024 at 05:26 pm

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.

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Course Schedule II 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]

1. Course Schedule II LeetCode Solution C++

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();
            for (auto a : adj[curr]){
                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<vector<int>> adj(numCourses);
        vector<int> indegree(numCourses, 0);
        for (int i = 0; i < n; i++){
            adj[prerequisites[i][1]].push_back(prerequisites[i][0]);
            indegree[prerequisites[i][0]] += 1;
        }
        vector<int> ans;
        if (kahnAlgo(adj, numCourses, indegree, ans))
            return ans;
        return {};
    }
};

2. Course Schedule II LeetCode Solution Java

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];
            chl[pre].add(cour);
            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;
    }
}

3. Course Schedule II Solution JavaScript

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 : [];
};

4. Course Schedule II Solution Python

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