Completed BFS-1

BinaryTreeLevelOrderTraversal.java

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+// Time Complexity : O(n)
+// Space Complexity : O(W) - width of the tree
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+// Your code here along with comments explaining your approach
+/*
+Have a queue where we initially add root and poll the elements level by level by taking snapshot of the
+size of the queue. We also make sure to add current node's left and right nodes if available. This way of
+using queue's FIFO principle to process the nodes and also taking snapshot approach helps us traverse
+elements in a level order way.
+ */
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    List<List<Integer>> result;
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        this.result = new ArrayList<>();
+        traverseListHelper(root);
+        return result;
+    }
+
+    private void traverseListHelper(TreeNode root) {
+        if(root == null)
+            return;
+        Queue<TreeNode> queue = new LinkedList<>();
+        queue.add(root);
+
+        while(!queue.isEmpty()) {
+            int size = queue.size();
+            List<Integer> nodeList = new ArrayList<>();
+            for(int i = 0 ; i < size ; i++) {
+                TreeNode temp = queue.poll();
+                nodeList.add(temp.val);
+
+                if(temp.left != null)
+                    queue.add(temp.left);
+                if(temp.right != null)
+                    queue.add(temp.right);
+            }
+            result.add(nodeList);
+        }
+    }
+}

CourseSchedule.java

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+// Time Complexity : O(V + E) / O(N + P)
+// Space Complexity : O(V + E) / O(N + P)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+// Your code here along with comments explaining your approach
+/*
+when we try to draw the dependencies of this problem, we can see its a graph and idea is to identify if
+there is a cycle present.If cycle is present, we cant finish all courses.So, firstly, we construct an adjacency
+map and have an indegree array to optimize the traversal.Then, we have a queue and add the independent courses
+first and then traverse through that course's dependent courses using adjacency map and add those courses to
+the queue if they become independent.At last, we check if our count has reached to the required number of courses or not.
+ */
+class Solution {
+    public boolean canFinish(int numCourses, int[][] prerequisites) {
+        int[] indegree = new int[numCourses];
+        Map<Integer, List<Integer>> map = new HashMap<>();
+
+        //arr[0] - dependent course, arr[1] - independent course
+        for(int[] arr : prerequisites) {
+            //increment dependent course values
+            indegree[arr[0]]++;
+            //map of independent to dependent courses
+            map.putIfAbsent(arr[1], new ArrayList<>());
+            map.get(arr[1]).add(arr[0]);
+        }
+
+        Queue<Integer> queue = new LinkedList<>();
+        int count = 0;
+
+        //add independent courses first to the queue (behaviour like root)
+        for(int i = 0 ; i < indegree.length ; i++) {
+            if(indegree[i] == 0) {
+                queue.add(i);
+                count++;
+            }
+        }
+
+        //no independent courses at all -> cycle
+        if(queue.isEmpty())
+            return false;
+        //all are independent courses
+        if(count == numCourses)
+            return true;
+
+        while(!queue.isEmpty()) {
+            int course = queue.poll();
+            List<Integer> dependentList = map.get(course);
+            if(dependentList != null) {
+                for(int dependent : dependentList) {
+                    indegree[dependent]--;
+                    if(indegree[dependent] == 0) {
+                        queue.add(dependent);
+                        count++;
+                    }
+                }
+            }
+        }
+        return count == numCourses ? true : false;
+    }
+}