Topological Sort and BFS

Author: Arjiit

BFS/courseSchedule.java

@@ -0,0 +1,45 @@
+class Solution {
+    public boolean canFinish(int numCourses, int[][] prerequisites) {
+        /*
+        This is a dependency type question because inorder to complete a course, we need to complete it's 
+        pre-requistes, hence we can apply topological sort here.
+        We can initialize an array with index representing the courses and the value representing the number
+        of pre-requistes to complete that particular course essentially representing a graph with nodes 
+        directing inside it as the pre-requistes. We then apply BFS and if there are no-prerequistes left
+        we add it to queue. In the end, we check if the inDegree array is empty or not. If it is not empty, 
+        that means it course cannot be finished.
+        T.C. - O(V+E), Space - O(V)
+        */
+        
+        int[] inDegree = new int[numCourses];
+        
+        for (int i=0; i< prerequisites.length;i++) {
+            inDegree[prerequisites[i][0]]++;
+        }
+        
+        Queue<Integer> q = new LinkedList<>();
+        for (int j=0; j<inDegree.length; j++) {
+            if (inDegree[j] == 0) {
+                q.offer(j);
+            }
+        }
+        while(!q.isEmpty()) {
+            int out = q.poll();
+            for (int i=0; i<prerequisites.length; i++) {
+                if (prerequisites[i][1] == out) {
+                    inDegree[prerequisites[i][0]]--;
+                    if (inDegree[prerequisites[i][0]] == 0) {
+                        q.offer(prerequisites[i][0]);
+                    }
+                }
+            }
+            
+        }
+        for (int i=0; i<inDegree.length; i++) {
+            if (inDegree[i] != 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}