Last updated on October 5th, 2024 at 05:51 pm

Here, We see Course Schedule 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

Course Schedule 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 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.

1. Course Schedule LeetCode Solution C++

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<int> adj[numCourses];
        vector<int> indegree(numCourses, 0);
        vector<int> ans;
        for(auto x: prerequisites){
            adj[x[0]].push_back(x[1]);
            indegree[x[1]]++;
        }
        queue<int> q;
        for(int i = 0; i < numCourses; i++){
            if(indegree[i] == 0){
                q.push(i);
            }
        }
        while(!q.empty()){
            auto t = q.front();
            ans.push_back(t);
            q.pop();
            for(auto x: adj[t]){
                indegree[x]--;
                if(indegree[x] == 0){
                    q.push(x);
                }
            }
        }
        return ans.size() == numCourses;
    }
};

2. Course Schedule LeetCode Solution Java

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        ArrayList[] graph = new ArrayList[numCourses];
        int[] degree = new int[numCourses];
        Queue queue = new LinkedList();
        int count=0;
        for(int i=0;i<numCourses;i++)
            graph[i] = new ArrayList();
        for(int i=0; i<prerequisites.length;i++){
            degree[prerequisites[i][1]]++;
            graph[prerequisites[i][0]].add(prerequisites[i][1]);
        }
        for(int i=0; i<degree.length;i++){
            if(degree[i] == 0){
                queue.add(i);
                count++;
            }
        }
        while(queue.size() != 0){
            int course = (int)queue.poll();
            for(int i=0; i<graph[course].size();i++){
                int pointer = (int)graph[course].get(i);
                degree[pointer]--;
                if(degree[pointer] == 0){
                    queue.add(pointer);
                    count++;
                }
            }
        }
        if(count == numCourses)
            return true;
        else    
            return false;
    }
}

3. Course Schedule Solution JavaScript

var canFinish = 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;
};

4. Course Schedule Solution Python

class Solution(object):
    def canFinish(self, numCourses, prerequisites):
        adj = [[] for _ in range(numCourses)]
        indegree = [0] * numCourses
        ans = []
        for pair in prerequisites:
            course = pair[0]
            prerequisite = pair[1]
            adj[prerequisite].append(course)
            indegree[course] += 1
        queue = deque()
        for i in range(numCourses):
            if indegree[i] == 0:
                queue.append(i)
        while queue:
            current = queue.popleft()
            ans.append(current)
            for next_course in adj[current]:
                indegree[next_course] -= 1
                if indegree[next_course] == 0:
                    queue.append(next_course)
        return len(ans) == numCourses