From e4d0519687bebebd1da078e5d1c1e3f95d420510 Mon Sep 17 00:00:00 2001
From: Anjali <anjalisugandhapai@gmail.com>
Date: Mon, 20 Oct 2025 01:38:40 +0530
Subject: [PATCH] course schedule problem

---
 4_graph_algorithms/Course_Schedule.cpp | 97 ++++++++++++++++++++++++++
 1 file changed, 97 insertions(+)
 create mode 100644 4_graph_algorithms/Course_Schedule.cpp

diff --git a/4_graph_algorithms/Course_Schedule.cpp b/4_graph_algorithms/Course_Schedule.cpp
new file mode 100644
index 0000000..12e292b
--- /dev/null
+++ b/4_graph_algorithms/Course_Schedule.cpp
@@ -0,0 +1,97 @@
+#include <bits/stdc++.h>
+using namespace std;
+/*
+Course Schedule / Topological Sorting Problem
+
+Prerequisite:
+Graph theory, Topological sorting, Kahn’s Algorithm (BFS-based approach).
+
+Main Idea:
+We are given a set of courses and prerequisite relationships (a → b), meaning
+you must complete course 'a' before course 'b'. This forms a Directed Graph.
+The task is to determine a valid order of completing all courses — a Topological Order.
+
+Solution:
+1. Represent courses as graph nodes (1...n) and prerequisites as directed edges.
+2. Compute indegree[] for each node (number of prerequisites).
+3. Push all nodes with indegree = 0 (no prerequisites) into a queue.
+4. Repeatedly remove a node from the queue:
+   - Add it to the result order.
+   - Reduce the indegree of all its neighbors by 1.
+   - If a neighbor’s indegree becomes 0, push it into the queue.
+5. If all nodes are processed → we found a valid order.
+   If not → there exists a cycle, and it's IMPOSSIBLE to finish all courses.
+
+Why It Works:
+Kahn’s Algorithm ensures that each course is taken only when all its prerequisites are done.
+If a cycle exists (mutual dependencies), some nodes will never reach indegree 0.
+
+Complexity:
+Time  : O(n + m)   — we visit each node and edge once.
+Memory: O(n + m)   — for adjacency list and indegree array.
+
+Alternative:
+A DFS-based topological sort can also be used with recursion and a stack,
+but Kahn’s Algorithm (BFS) is iterative and cycle-detection-friendly.
+*/
+
+int32_t main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, m;
+    cin >> n >> m;
+
+    // Adjacency list representation of the graph
+    // adj[a] will store all nodes b such that there is an edge a -> b
+    vector<vector<int>> adj(n + 1);
+
+    // indegree[i] = number of prerequisites (incoming edges) for course i
+    vector<int> indegree(n + 1, 0);
+
+    // Reading m prerequisite relations
+    for (int i = 0; i < m; i++) {
+        int a, b;
+        cin >> a >> b;
+        adj[a].push_back(b);  // a must come before b
+        indegree[b]++;        // increase indegree of b
+    }
+
+    // Queue to store all courses with no prerequisites (indegree = 0)
+    queue<int> q;
+
+    for (int i = 1; i <= n; i++) {
+        if (indegree[i] == 0)
+            q.push(i); // can be taken first
+    }
+
+    // This will store the valid topological order
+    vector<int> order;
+
+    // Process nodes with no incoming edges
+    while (!q.empty()) {
+        int u = q.front();
+        q.pop();
+        order.push_back(u);
+
+        // Reduce indegree of all neighbors (courses depending on u)
+        for (int v : adj[u]) {
+            indegree[v]--;
+            // If indegree becomes 0, it can now be taken
+            if (indegree[v] == 0)
+                q.push(v);
+        }
+    }
+
+    // If we couldn't include all courses, there’s a cycle → impossible to finish all
+    if ((int)order.size() != n) {
+        cout << "IMPOSSIBLE\n";
+    } else {
+        // Print valid topological order
+        for (int x : order)
+            cout << x << " ";
+        cout << "\n";
+    }
+
+    return 0;
+}