SDE - 1

Given a matrix of dimension m*n where each cell in the matrix can have values 0, 1 or 2 which has the following meaning: 

0: Empty cell
1: Cells have fresh oranges
2: Cells have rotten oranges
Determine what is the minimum time required so that all the oranges become rotten. A rotten orange at index [i,j] can rot other fresh orange at indexes [i-1,j], [i+1,j], [i,j-1], [i,j+1] (up, down, left and right). If it is impossible to rot every orange then simply return -1.

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

You are given a 2-dimensional array/list having N rows and M columns, which is filled with ones(1) and zeroes(0). 1 signifies land, and 0 signifies water.

A cell is said to be connected to another cell, if one cell lies immediately next to the other cell, in any of the eight directions (two vertical, two horizontal, and four diagonals).

A group of connected cells having value 1 is called an island. Your task is to find the number of such islands present in the matrix.

You have to implement the pop function of Max Priority Queue and implement using a heap.

Functions :

a) push(int x) : 'x' has to be inserted in the priority queue. This has been implemented already

b) pop() : return the maximum element in the priority queue, if priority queue is empty then return '-1'.

Example:

We perform the following operations on an empty priority queue:

When operation push(5) is performed, we insert 1 in the priority queue.

When operation push(2) is performed, we insert 2 in the priority queue. 

When operation pop() is performed, we remove the maximum element from the priority queue and print which is 5.

When operation push(3) is performed, we insert 1 in the priority queue.

When operation pop() is performed, we remove the maximum element from the priority queue and print which is 3.

You are given a 2D matrix 'ARR' (containing either ‘0’ or ‘1’) of size 'N' x 'M', where each row is in sorted order.

Find the 0-based index of the first row with the maximum number of 1's.

Note :

If two rows have the same number of 1’s, return the row with a lower index.

If no row exists where at-least one '1' is present, return -1.

Example:

Input: ‘N’ = 3, 'M' = 3
'ARR' = 
[     [ 1,  1,  1 ],
      [ 0,  0,  1 ],
      [ 0,  0,  0 ]   ]

Output: 0

Explanation: The 0th row of the given matrix has the maximum number of ones.

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