909.-snakes-and-ladders.md

909. Snakes and Ladders

• Medium

• You are given an n x n integer matrix board where the cells are labeled from 1 to n2 in a Boustrophedon style starting from the bottom left of the board (i.e. board[n - 1][0]) and alternating direction each row.

  You start on square 1 of the board. In each move, starting from square curr, do the following:

  • Choose a destination square next with a label in the range [curr + 1, min(curr + 6, n2)].
    • This choice simulates the result of a standard 6-sided die roll: i.e., there are always at most 6 destinations, regardless of the size of the board.
  • If next has a snake or ladder, you must move to the destination of that snake or ladder. Otherwise, you move to next.
  • The game ends when you reach the square n2.

  A board square on row r and column c has a snake or ladder if board[r][c] != -1. The destination of that snake or ladder is board[r][c]. Squares 1 and n2 do not have a snake or ladder.

  Note that you 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 follow the subsequent snake or ladder.

  • For example, suppose the board is [[-1,4],[-1,3]], and on the first move, your destination square is 2. You follow the ladder to square 3, but do not follow the subsequent ladder to 4.

  Return the least number of moves required to reach the square n2. If it is not possible to reach the square, return -1.

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/09/23/snakes.png)

```
Input: board = [[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,35,-1,-1,13,-1],[-1,-1,-1,-1,-1,-1],[-1,15,-1,-1,-1,-1]]
Output: 4
Explanation: 
In the beginning, you start at square 1 (at row 5, column 0).
You decide to move to square 2 and must take the ladder to square 15.
You then decide to move to square 17 and must take the snake to square 13.
You then decide to move to square 14 and must take the ladder to square 35.
You then decide to move to square 36, ending the game.
This is the lowest possible number of moves to reach the last square, so return 4.
```

Analysis

The first thing that comes to my mind is DP. Each position on the board comes either from rolling the dice 1~6 or directly from a ladder or snake. So here is my first version of the code.

####This is wrong

class Solution:
    def snakesAndLadders(self, board: List[List[int]]) -> int:
        n=len(board)
        self.short={}
        self.seen={}
        counter=0
        for i in reversed(range(n)):
            if (n-i) %2==1:
                for j in range(n):
                    counter+=1
                    if board[i][j]!=-1:
                        self.short[counter]=board[i][j]
            if (n-i)%2==0:
                for j in reversed(range(n)):
                    counter+=1
                    if board[i][j]!=-1:
                        self.short[counter]=board[i][j]
        return self.helper(n**2)            
                    
        
    def helper(self,position):
        if position in self.seen:
            return self.seen[position]+1
        if position<8:
            return 1
        mini=min(self.helper(position-1),self.helper(position-2),self.helper(position-3),self.helper(position-4),self.helper(position-5),self.helper(position-6))+1
        if position in self.short:
            print(position,self.short)
            mini=min(mini,self.helper(self.short[position]))
        self.seen[position]=mini
        return mini

However when I started to run it I realized that "snake" is causing a lot of problem for me and it will create a endless circle since it is pointing from the higher values to the lower values.

class Solution:
    def snakesAndLadders(self, board):
        n = len(board)
        need = {1: 0}
        bfs = [1]
        for x in bfs:
            refuse_steps = False
            for i in range(min(n**2, x + 6), x, -1):
                a, b = (i - 1) // n, (i - 1) % n
                nxt = board[~a][b if a % 2 == 0 else ~b]
                if nxt > 0: i = nxt
                if i == n * n: return need[x] + 1
                if nxt == -1 and refuse_steps: continue
                if nxt == -1: refuse_steps = True
                if i not in need:
                    need[i] = need[x] + 1
                    bfs.append(i)
        return -1