Intuition

You can finish all courses exactly when the prerequisite graph has no cycle. Topological sort (Kahn’s algorithm) succeeds iff every node can be ordered, i.e. no cycle.

Approach

Build an adjacency list and in-degree count (edge b -> a).
Queue all zero in-degree courses; repeatedly remove one and decrement its neighbours.
If the number removed equals n, there is no cycle.

Complexity

Time — O(V+E): each node and edge processed once.
Space — O(V+E): adjacency list and queue.

Full Solution — all languages

class Solution:
    def canFinish(self, n, edges):
        adj=[[] for _ in range(n)]
        indeg=[0]*n
        for a,b in edges:
            adj[b].append(a); indeg[a]+=1
        from collections import deque
        q=deque(i for i in range(n) if indeg[i]==0)
        seen=0
        while q:
            u=q.popleft(); seen+=1
            for v in adj[u]:
                indeg[v]-=1
                if indeg[v]==0: q.append(v)
        return seen==n

import java.util.*;

class Solution {
    public boolean canFinish(int n, int[][] edges) {
        List<List<Integer>> adj=new ArrayList<>(); for(int i=0;i<n;i++)adj.add(new ArrayList<>()); int[] indeg=new int[n]; for(int[] e:edges){adj.get(e[1]).add(e[0]);indeg[e[0]]++;} Deque<Integer> q=new ArrayDeque<>(); for(int i=0;i<n;i++)if(indeg[i]==0)q.add(i); int seen=0; while(!q.isEmpty()){int u=q.poll();seen++;for(int v:adj.get(u))if(--indeg[v]==0)q.add(v);} return seen==n;
    }
}

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    bool canFinish(int n, vector<vector<int>>& edges) {
        vector<vector<int>> adj(n); vector<int> indeg(n,0); for(auto& e:edges){adj[e[1]].push_back(e[0]);indeg[e[0]]++;} queue<int> q; for(int i=0;i<n;i++)if(indeg[i]==0)q.push(i); int seen=0; while(!q.empty()){int u=q.front();q.pop();seen++;for(int v:adj[u])if(--indeg[v]==0)q.push(v);} return seen==n;
    }
};

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {boolean}
 */
var canFinish = function(n, edges) {
    const adj=Array.from({length:n},()=>[]),indeg=new Array(n).fill(0); for(const e of edges){adj[e[1]].push(e[0]);indeg[e[0]]++;} const q=[]; for(let i=0;i<n;i++)if(indeg[i]===0)q.push(i); let seen=0; while(q.length){const u=q.shift();seen++;for(const v of adj[u]){if(--indeg[v]===0)q.push(v);}} return seen===n;
};

Course Schedule

Example 1

Example 2

Constraints