SDE - Intern

You have been given a Snake and Ladder Board with 'N' rows and 'N' columns with the numbers written from 1 to (N*N) starting from the bottom left of the board, and alternating direction each row.

For example

For a (6 x 6) board, the numbers are written as follows:

You start from square 1 of the board (which is always in the last row and first column). On each square say 'X', you can throw a dice which can have six outcomes and you have total control over the outcome of dice throw and you want to find out the minimum number of throws required to reach the last cell.
Some of the squares contain Snakes and Ladders, and these are possibilities of a throw at square 'X':
You choose a destination square 'S' with number 'X+1', 'X+2', 'X+3', 'X+4', 'X+5', or 'X+6', provided this number is <= N*N.
If 'S' has a snake or ladder, you move to the destination of that snake or ladder.  Otherwise, you move to S.
A board square on row 'i' and column 'j' has a "Snake or Ladder" if board[i][j] != -1. The destination of that snake or ladder is board[i][j].

Note :

You can only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do not continue moving - you have to ignore the snake or ladder present on that square.

For example, if the board is:
-1 1 -1
-1 -1 9
-1 4 -1
Let's say on the first move your destination square is 2  [at row 2, column 1], then you finish your first move at 4 [at row 1, column 2] because you do not continue moving to 9 [at row 0, column 0] by taking the ladder from 4.

A square can also have a Snake or Ladder which will end at the same cell.
For example, if the board is:
-1 3 -1
-1 5 -1
-1 -1 9
Here we can see Snake/Ladder on square 5 [at row 1, column 1] will end on the same square 5.

You are given a Doubly Linked List of ‘N’ positive integers. Your task is to delete a node at position ‘POS’ in the linked list.

Note:

Assume that the Indexing for the linked list starts from 0.

EXAMPLE:

Input: ‘N’ = 5, 'LIST' = [1, 1, 2, 3, 4, -1], ‘POS’ = 1.

Output: 1 < - > 2 < - > 3 < - > 4 

Here in the given list, we can see that the node at position 1 is deleted.

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