SDE - 1

You have been given a Binary Tree of distinct integers and two nodes ‘X’ and ‘Y’. You are supposed to return the LCA (Lowest Common Ancestor) of ‘X’ and ‘Y’.

The LCA of ‘X’ and ‘Y’ in the binary tree is the shared ancestor of ‘X’ and ‘Y’ that is located farthest from the root.

Note :

You may assume that given ‘X’ and ‘Y’ definitely exist in the given binary tree.

For example :

For the given binary tree

LCA of ‘X’ and ‘Y’ is highlighted in yellow colour.

You are given a starting position for a rat which is stuck in a maze at an initial point (0, 0) (the maze can be thought of as a 2-dimensional plane). The maze would be given in the form of a square matrix of order 'N' * 'N' where the cells with value 0 represent the maze’s blocked locations while value 1 is the open/available path that the rat can take to reach its destination. The rat's destination is at ('N' - 1, 'N' - 1). Your task is to find all the possible paths that the rat can take to reach from source to destination in the maze. The possible directions that it can take to move in the maze are 'U'(up) i.e. (x, y - 1) , 'D'(down) i.e. (x, y + 1) , 'L' (left) i.e. (x - 1, y), 'R' (right) i.e. (x + 1, y).

Note:

Here, sorted paths mean that the expected output should be in alphabetical order.

For Example:

Given a square matrix of size 4*4 (i.e. here 'N' = 4):
1 0 0 0
1 1 0 0
1 1 0 0
0 1 1 1 
Expected Output:
DDRDRR DRDDRR 
i.e. Path-1: DDRDRR and Path-2: DRDDRR

The rat can reach the destination at (3, 3) from (0, 0) by two paths, i.e. DRDDRR and DDRDRR when printed in sorted order, we get DDRDRR DRDDRR.

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.