Added codes for 9 Feb

GeeksForGeeks/February/9-2-24/GFG.java

@@ -0,0 +1,175 @@
+//{ Driver Code Starts
+//Initial Template for Java
+
+
+
+
+//Initial Template for Java
+
+//Contributed by Sudarshan Sharma
+import java.util.LinkedList; 
+import java.util.Queue; 
+import java.io.*;
+import java.util.*;
+
+class Node{
+    int data;
+    Node left;
+    Node right;
+    Node(int data){
+        this.data = data;
+        left=null;
+        right=null;
+    }
+}
+
+
+class GfG {
+
+    static Node buildTree(String str){
+
+        if(str.length()==0 || str.charAt(0)=='N'){
+            return null;
+        }
+
+        String ip[] = str.split(" ");
+        // Create the root of the tree
+        Node root = new Node(Integer.parseInt(ip[0]));
+        // Push the root to the queue
+
+        Queue<Node> queue = new LinkedList<>(); 
+
+        queue.add(root);
+        // Starting from the second element
+
+        int i = 1;
+        while(queue.size()>0 && i < ip.length) {
+
+            // Get and remove the front of the queue
+            Node currNode = queue.peek();
+            queue.remove();
+
+            // Get the current node's value from the string
+            String currVal = ip[i];
+
+            // If the left child is not null
+            if(!currVal.equals("N")) {
+
+                // Create the left child for the current node
+                currNode.left = new Node(Integer.parseInt(currVal));
+                // Push it to the queue
+                queue.add(currNode.left);
+            }
+
+            // For the right child
+            i++;
+            if(i >= ip.length)
+                break;
+
+            currVal = ip[i];
+
+            // If the right child is not null
+            if(!currVal.equals("N")) {
+
+                // Create the right child for the current node
+                currNode.right = new Node(Integer.parseInt(currVal));
+
+                // Push it to the queue
+                queue.add(currNode.right);
+            }
+            i++;
+        }
+
+        return root;
+    }
+    void inOrder(Node node) {
+        if (node == null) {
+            return;
+        }
+
+        inOrder(node.left);
+        System.out.print(node.data + " ");
+
+        inOrder(node.right);
+    }
+
+	public static void main (String[] args) throws IOException{
+	        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+
+	        int t=Integer.parseInt(br.readLine());
+
+	        while(t-- > 0){
+	            String s = br.readLine();
+    	    	Node root = buildTree(s);
+			    Solution g=new Solution();
+                System.out.println(g.isSumProperty(root));
+
+	        }
+	}
+}
+
+
+// } Driver Code Ends
+
+
+//User function Template for Java
+
+
+/*Complete the function below
+Node is as follows:
+class Node{
+	int data;
+	Node left,right;
+	
+	Node(int key)
+	{
+	    data = key;
+	    left = right = null;
+	}
+}
+
+*/
+class Solution
+{
+    //Function to check whether all nodes of a tree have the value 
+    //equal to the sum of their child nodes.
+    public static int isSumProperty(Node root)
+    {
+        // add your code here
+        Queue<Node> q = new LinkedList<Node>();
+
+        q.offer(root);
+        q.offer(null);
+
+        while (!q.isEmpty()) 
+        {
+            Node curr = q.poll();
+
+            if (curr == null) 
+            {
+                if (!q.isEmpty())
+                    q.offer(null);
+                continue;
+            }
+
+            int sum = 0;
+
+            if (curr.left != null) 
+            {
+                q.offer(curr.left);
+                sum += curr.left.data;
+            }
+
+            if (curr.right != null) 
+            {
+                q.offer(curr.right);
+                sum += curr.right.data;
+            }
+
+            if (sum != curr.data && sum != 0)
+                return 0;
+        }
+
+        return 1;
+    }
+}

GeeksForGeeks/February/9-2-24/README.md

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+Time complexity - O(n)
+Space complexity - O(h)

LeetCode/February/9-2-24/README.md

@@ -0,0 +1,2 @@
+Time complexity - O(n^2)
+Space complexity - O(n)

LeetCode/February/9-2-24/Solution.java

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+class Solution 
+{
+    public List<Integer> largestDivisibleSubset(int[] nums) 
+    {
+        final int n = nums.length;
+        List<Integer> ans = new ArrayList<>();
+
+        int[] sizeEndsAt = new int[n];
+        int[] prevIndex = new int[n];
+        int maxSize = 0;
+        int index = -1;
+
+        Arrays.fill(sizeEndsAt, 1);
+        Arrays.fill(prevIndex, -1);
+        Arrays.sort(nums);
+
+        for (int i = 0; i < n; ++i) 
+        {
+            for (int j = i - 1; j >= 0; --j)
+                if (nums[i] % nums[j] == 0 && sizeEndsAt[i] < sizeEndsAt[j] + 1) 
+                {
+                    sizeEndsAt[i] = sizeEndsAt[j] + 1;
+                    prevIndex[i] = j;
+                }
+
+            if (maxSize < sizeEndsAt[i]) 
+            {
+                maxSize = sizeEndsAt[i];
+                index = i;
+            }
+        }
+
+        while (index != -1) 
+        {
+            ans.add(nums[index]);
+            index = prevIndex[index];
+        }
+
+        return ans;
+    }
+}