Given a m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:
1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])
2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])
3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])
4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])
Notice that there could be some invalid signs on the cells of the grid which points outside the grid.
You will initially start at the upper left cell (0,0). A valid path in the grid is a path which starts from the upper left cell (0,0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path doesn't have to be the shortest.
You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.
Return the minimum cost to make the grid have at least one valid path.
Example 1:
Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
Output: 3
Explanation: You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)
The total cost = 3.
Example 2:
Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
Output: 0
Explanation: You can follow the path from (0, 0) to (2, 2).
Example 3:
Input: grid = [[1,2],[4,3]]
Output: 1
Example 4:
Input: grid = [[2,2,2],[2,2,2]]
Output: 3
Example 5:
Input: grid = [[4]]
Output: 0
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n <= 100
from collections import deque

class Solution:
    def minCost(self, grid: List[List[int]]) -> int:
        right, left, down, up = (0, 1), (0, -1), (1, 0), (-1, 0)
        
        direction_map = {
            1: right,
            2: left,
            3: down,
            4: up
        }
        
        directions = [right, left, down, up]
        visited = set()
        
        def in_bounds(i, j):
            return 0 <= i < len(grid) and 0 <= j < len(grid[i])
        
        def dfs(i, j):                                           
            # not in bounds
            if not in_bounds(i, j) or (i, j) in visited:
                return []
            
            visited.add((i, j))

            sign = grid[i][j]
            direction = direction_map[sign]
            next_i, next_j = i + direction[0], j + direction[1]
            return [(i, j)] + dfs(next_i, next_j)
                
                    
        reachable = dfs(0, 0)
        curr_cost = 0
        while reachable:
            next_reachable = []
            for (i, j) in reachable:
                if i == len(grid) - 1 and j == len(grid[i]) - 1:
                    return curr_cost
                
                for d in directions:
                    next_reachable += dfs(i + d[0], j + d[1])
            reachable = next_reachable
            curr_cost += 1
                    
        return -1
                
                    

from collections import deque

class Solution:
    def minCost(self, grid: List[List[int]]) -> int:
        right, left, down, up = (0, 1), (0, -1), (1, 0), (-1, 0)
        
        direction_map = {
            1: right,
            2: left,
            3: down,
            4: up
        }
        
        directions = [right, left, down, up]
        visited = set()
        
        def in_bounds(i, j):
            return 0 <= i < len(grid) and 0 <= j < len(grid[i])
        
        def dfs(i, j):                                           
            # not in bounds
            if not in_bounds(i, j) or (i, j) in visited:
                return []
            
            visited.add((i, j))

            sign = grid[i][j]
            direction = direction_map[sign]
            next_i, next_j = i + direction[0], j + direction[1]
            return [(i, j)] + dfs(next_i, next_j)
                
                    
        reachable = dfs(0, 0)
        curr_cost = 0
        while reachable:
            next_reachable = []
            for (i, j) in reachable:
                if i == len(grid) - 1 and j == len(grid[i]) - 1:
                    return curr_cost
                
                for d in directions:
                    next_reachable += dfs(i + d[0], j + d[1])
            reachable = next_reachable
            curr_cost += 1
                    
        return -1
                
                    
