🧩 LeetCode 207 — Course Schedule

🧠 Intuition

The problem is about determining if all courses can be completed given their prerequisites.
Think of each course as a node in a directed graph, and each prerequisite pair [a, b] as a directed edge from b → a — meaning b must be completed before a.
If there’s a cycle, it means a course depends on itself (directly or indirectly), so it’s impossible to complete all courses.
Hence, the task reduces to detecting if the directed graph has a cycle.

🚀 Approach

We use Kahn’s Algorithm (Topological Sorting using BFS) to determine if all nodes can be visited without a cycle.

1. Build the Graph

   • Create an adjacency list adj for the courses.
   • Maintain an inDegree array to store how many prerequisites each course has.

2. Initialize Queue

   • Add all courses with inDegree == 0 (no prerequisites) into a queue.

3. Process Courses

   • While the queue isn’t empty:
     • Pop a course x from the queue and increment a count of completed courses.
     • For each dependent course in adj[x], decrement its inDegree.
     • If any course’s inDegree becomes 0, push it into the queue.

4. Check for Cycles

   • If count == V (all courses processed), return true.
   • Otherwise, a cycle exists — return false.

💻 Code Solution (C++)

class Solution {
public:
    bool canFinish(int V, vector<vector<int>>& prerequisites) {
        vector<vector<int>> adj(V);
        vector<int> inDeg(V);
        
        for (int i = 0; i < prerequisites.size(); i++) {
            int a = prerequisites[i][0];
            int b = prerequisites[i][1];
            adj[b].push_back(a);
            inDeg[a]++;
        }

        queue<int> q;
        for (int i = 0; i < V; i++) {
            if (inDeg[i] == 0)
                q.push(i);
        }

        int count = 0;
        while (!q.empty()) {
            int x = q.front();
            q.pop();
            count++;

            for (int i = 0; i < adj[x].size(); i++) {
                inDeg[adj[x][i]]--;
                if (inDeg[adj[x][i]] == 0)
                    q.push(adj[x][i]);
            }
        }

        return count == V;
    }
};

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