SDE - 2

Given a sequence of ‘N’ space-separated non-negative integers A[1],A[2],A[3],......A[i]…...A[n]. Where each number of the sequence represents the height of the line drawn at point 'i'. Hence on the cartesian plane, each line is drawn from coordinate ('i',0) to coordinate ('i', 'A[i]'), here ‘i’ ranges from 1 to ‘N’. Find two lines, which, together with the x-axis forms a container, such that the container contains the most area of water.

Note :

1. You can not slant the container i.e. the height of the water is equal to the minimum height of the two lines which define the container.

2. Do not print anything, you just need to return the area of the container with maximum water.

Example

For the above Diagram, the first red marked line is formed between coordinates (2,0) and (2,10), and the second red-marked line is formed between coordinates (5,0) and (5,9). The area of water contained between these two lines is (height* width) = (5-2)* 9 = 27, which is the maximum area contained between any two lines present on the plane. So in this case, we will return 3* 9=27.

Given a ‘N’ * ’M’ maze with obstacles, count and return the number of unique paths to reach the right-bottom cell from the top-left cell. A cell in the given maze has a value '-1' if it is a blockage or dead-end, else 0. From a given cell, we are allowed to move to cells (i+1, j) and (i, j+1) only. Since the answer can be large, print it modulo 10^9 + 7.

For Example :

Consider the maze below :
0 0 0 
0 -1 0 
0 0 0

There are two ways to reach the bottom left corner - 

(1, 1) -> (1, 2) -> (1, 3) -> (2, 3) -> (3, 3)
(1, 1) -> (2, 1) -> (3, 1) -> (3, 2) -> (3, 3)

Hence the answer for the above test case is 2.

Given a binary tree with N number of nodes, check if that input tree is Partial BST (Binary Search Tree) or not. If yes, return true, return false otherwise.

A binary search tree (BST) is said to be a Partial BST if it follows the following properties.

• The left subtree of a node contains only nodes with data less than and equal to the node’s data.
• The right subtree of a node contains only nodes with data greater than and equal to the node’s data.
• Both the left and right subtrees must also be partial binary search trees.

Example:

Input:

Answer:

Level 1: 

All the nodes in the left subtree of 4 (2, 1, 3) are smaller 
than 4, all the nodes in the right subtree of the 4 (5) are 
larger than 4.

Level 2 :

For node 2:
All the nodes in the left subtree of 2 (1) are smaller than 
2, all the nodes in the right subtree of the 2 (3) are larger than 2.
For node 5:
The left and right subtree for node 5 is empty.

Level 3:

For node 1:
The left and right subtree for node 1 are empty.
For node 3:
The left and right subtree for node 3 are empty.
Because all the nodes follow the property of a Partial binary 
search tree, the above tree is a Partial binary search tree.

You are given a linked list of 'n' nodes and an integer 'k', where 'k' is less than or equal to 'n'.

Your task is to reverse the order of each group of 'k' consecutive nodes, if 'n' is not divisible by 'k', then the last group of nodes should remain unchanged.

For example, if the linked list is 1->2->3->4->5, and 'k' is 3, we have to reverse the first three elements, and leave the last two elements unchanged. Thus, the final linked list being 3->2->1->4->5.

Implement a function that performs this reversal, and returns the head of the modified linked list.

Example:

Input: 'list' = [1, 2, 3, 4], 'k' = 2

Output: 2 1 4 3

Explanation:
We have to reverse the given list 'k' at a time, which is 2 in this case. So we reverse the first 2 elements then the next 2 elements, giving us 2->1->4->3.

Note:

All the node values will be distinct.

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