206.Course Schedule.cpp

206.Course Schedule.cpp

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+// 🧠 Approach
+
+// This problem can be modeled as a Directed Graph Cycle Detection problem.
+// Each course is represented as a node, and a directed edge b → a means you must take course b before course a.
+
+// To determine if all courses can be finished, we must ensure no cycle exists in this graph.
+// We use Kahn’s Algorithm (Topological Sorting using BFS) to detect cycles.
+
+// 🚀 Algorithm Steps
+
+// Build the Graph:
+
+// - Create an adjacency list adj where adj[b] contains all courses that depend on b.
+// - Maintain an array inDeg to store the number of prerequisites (incoming edges) for each course.
+
+// Find Starting Nodes:
+
+// - Push all nodes with inDeg[i] == 0 (no prerequisites) into a queue — these can be taken first.
+// - Topological Sorting using BFS:
+// - While the queue isn’t empty:
+// - Pop a node x (a course that can be completed now).
+// - Increment a counter c for completed courses.
+// - For each neighbor adj[x][i], reduce its inDeg by 1 (as its prerequisite x is now done).
+// - If any neighbor’s inDeg becomes 0, push it into the queue.
+
+// Check Completion:
+
+// - If all courses are processed (c == V), return true.
+// - Otherwise, there’s a cycle → some courses depend on each other, so return false.
+
+
+// Solution in CPP:
+
+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]++;
+        }
+        vector<int> vis(V);
+        queue<int> q;
+
+        for (int i = 0; i < V; i++) {
+            if (inDeg[i] == 0)
+                q.push(i);
+        }
+        int c = 0;
+        while (!q.empty()) {
+            int x = q.front();
+            q.pop();
+            c++;
+
+            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 c == V;
+    }
+};