# Java 2D Array

### Problem Statement :

```You are given a 6*6 2D array. An hourglass in an array is a portion shaped like this:

a b c
d
e f g
For example, if we create an hourglass using the number 1 within an array full of zeros, it may look like this:

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

Actually, there are many hourglasses in the array above. The three leftmost hourglasses are the following:

1 1 1     1 1 0     1 0 0
1           0             0
1 1 1     1 1 0     1 0 0
The sum of an hourglass is the sum of all the numbers within it. The sum for the hourglasses above are 7, 4, and 2, respectively.
In this problem you have to print the largest sum among all the hourglasses in the array.

Input Format

There will be exactly 6 lines, each containing 6 integers seperated by spaces. Each integer will be between -9 and 9 inclusive.

Output Format

Print the answer to this problem on a single line.```

### Solution :

```                            ```Solution in C :

import java.util.*;
import java.io.*;

class Solution{
public static void main(String []argh){
int[][] arr = new int[10][10];
Scanner sc = new Scanner(System.in);
for(int i=0;i<6;i++){
for(int j=0;j<6;j++){
arr[i][j]=sc.nextInt();

}
}
int maxi=-100000;
for(int i=0;i<6;i++){
for(int j=0;j<6;j++){
if(i<=3 && j<=3){
int sum=arr[i][j]+arr[i][j+1]+arr[i][j+2]+(arr[i+1][j+1])+arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2];
if(sum>maxi) maxi=sum;
}
}
}
System.out.println(maxi);
}
}```
```

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

## Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

## Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from