SDE - 1

The lowest common ancestor (LCA) is a concept in graph theory and computer science.

Let ‘T’ be a rooted tree with ‘N’ nodes. The lowest common ancestor is defined between two nodes, ‘u’ and ‘v’, as the lowest node in ‘T’ that has both ‘u’ and ‘v’ as descendants (where we allow a node to be a descendant of itself). - Wikipedia

For the given tree, The LCA of nodes 5 and 8 will be node 2, as node 2 is the first node that lies in the path from node 5 to root node 1 and from node 8 to root node 1.

Path from node 5 to root node looks like 5 → 2 → 1.

Path from node 8 to root node looks like 8 → 6 → 2 → 1.

Since 2 is the first node that lies in both paths. Hence LCA will be 2.

Given any two nodes ‘u’ and ‘v’, find the LCA for the two nodes in the given Tree ‘T’.

Note: For each test case, the tree is rooted at node 1.

Given a binary number as an integer N, convert it into decimal and print.

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.