diff --git a/207. Course Schedule.cpp b/207. Course Schedule.cpp
new file mode 100644
index 0000000..bd1fe4a
--- /dev/null
+++ b/207. Course Schedule.cpp	
@@ -0,0 +1,68 @@
+// ---------------- Approach ----------------
+//
+// This is a **cycle detection in a directed graph** problem.
+// - Each course is a node.
+// - A prerequisite pair [a, b] represents a directed edge b → a.
+// - If there is a cycle in this graph, it means courses are interdependent 
+//   and cannot all be finished.
+//
+// We use DFS with two arrays:
+// - visited[node]: to mark nodes that have been fully processed.
+// - pathVisit[node]: to mark nodes in the current DFS recursion stack.
+//
+// If during DFS we encounter a node that is already in the current path (pathVisit),
+// it means we found a cycle → return false.
+
+
+// ---------------- Intuition ----------------
+//
+// - Think of prerequisites as a graph of dependencies.
+// - If a cycle exists, some course depends on itself indirectly → impossible to finish.
+// - If no cycle exists, we can always complete courses in a valid topological order.
+
+
+// ---------------- Solution in Code ----------------
+
+class Solution {
+public:
+    bool dfs(vector<vector<int>> &adj, vector<int> &visited, vector<int> &pathVisit, int node) {
+        visited[node] = 1;
+        pathVisit[node] = 1;
+
+        for (int neighbor : adj[node]) {
+            if (!visited[neighbor]) {
+                if (dfs(adj, visited, pathVisit, neighbor)) {
+                    return true; // cycle detected
+                }
+            }
+            else if (pathVisit[neighbor]) {
+                return true; // back edge → cycle
+            }
+        }
+
+        pathVisit[node] = 0; // backtrack
+        return false;
+    }
+
+    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
+        // Build adjacency list from prerequisites
+        vector<vector<int>> adj(numCourses);
+        for (auto &edge : prerequisites) {
+            int a = edge[0], b = edge[1];
+            adj[b].push_back(a);
+        }
+
+        vector<int> visited(numCourses, 0);
+        vector<int> pathVisit(numCourses, 0);
+
+        // Run DFS from each unvisited node
+        for (int i = 0; i < numCourses; i++) {
+            if (!visited[i]) {
+                if (dfs(adj, visited, pathVisit, i)) {
+                    return false; // cycle found
+                }
+            }
+        }
+        return true;
+    }
+};