SDE - Intern

You are given an 'N * N' matrix of integers where each row and each column is sorted in increasing order. You are given a target integer 'X'.

Find the position of 'X' in the matrix. If it exists then return the pair {i, j} where 'i' represents the row and 'j' represents the column of the array, otherwise return {-1,-1}

For example:

If the given matrix is:
[ [1, 2, 5],
  [3, 4, 9],
  [6, 7, 10]] 
We have to find the position of 4. We will return {1,1} since A[1][1] = 4.

Given two binary trees T and S, check whether tree S has exactly the same structure and node values with a subtree of T, i.e., check if tree S is a subtree of the tree T.

A subtree of a tree T is a tree S consisting of a node in T and all of its descendants in T. The subtree corresponding to the root node is the entire tree; the subtree corresponding to any other node is called a proper subtree.

You are given an integer 'N'. For a given 'N' x 'N' chessboard, find a way to place 'N' queens such that no queen can attack any other queen on the chessboard.

A queen can be killed when it lies in the same row, or same column, or the same diagonal of any of the other queens. You have to print all such configurations.

You have a binary tree of 'N' unique nodes and a Start node from where the tree will start to burn. Given that the Start node will always exist in the tree, your task is to print the time (in minutes) that it will take to burn the whole tree.

It is given that it takes 1 minute for the fire to travel from the burning node to its adjacent node and burn down the adjacent node.

For Example :

For the given binary tree: [1, 2, 3, -1, -1, 4, 5, -1, -1, -1, -1]
Start Node: 3

    1
   / \
  2   3
     / \
    4   5

Output: 2

Explanation :
In the zeroth minute, Node 3 will start to burn.

After one minute, Nodes (1, 4, 5) that are adjacent to 3 will burn completely.

After two minutes, the only remaining Node 2 will be burnt and there will be no nodes remaining in the binary tree. 

So, the whole tree will burn in 2 minutes.

You are given a 'Binary Tree'.

Return the bottom view of the binary tree.

Note :

1. A node will be in the bottom-view if it is the bottom-most node at its horizontal distance from the root. 

2. The horizontal distance of the root from itself is 0. The horizontal distance of the right child of the root node is 1 and the horizontal distance of the left child of the root node is -1. 

3. The horizontal distance of node 'n' from root = horizontal distance of its parent from root + 1, if node 'n' is the right child of its parent.

4. The horizontal distance of node 'n' from root = horizontal distance of its parent from the root - 1, if node 'n' is the left child of its parent.

5. If more than one node is at the same horizontal distance and is the bottom-most node for that horizontal distance, including the one which is more towards the right.

Example:

Input: Consider the given Binary Tree:

Output: 4 2 6 3 7

Explanation:
Below is the bottom view of the binary tree.

1 is the root node, so its horizontal distance = 0.
Since 2 lies to the left of 0, its horizontal distance = 0-1= -1
3 lies to the right of 0, its horizontal distance = 0+1 = 1
Similarly, horizontal distance of 4 = Horizontal distance of 2 - 1= -1-1=-2
Horizontal distance of 5 = Horizontal distance of 2 + 1=  -1+1 = 0
Horizontal distance of 6 = 1-1 =0
Horizontal distance of 7 = 1+1 = 2

The bottom-most node at a horizontal distance of -2 is 4.
The bottom-most node at a horizontal distance of -1 is 2.
The bottom-most node at a horizontal distance of 0 is 5 and 6. However, 6 is more towards the right, so 6 is included.
The bottom-most node at a horizontal distance of 1 is 3.
The bottom-most node at a horizontal distance of 2 is 7.

Hence, the bottom view would be 4 2 6 3 7