<p>Given an <code>m x n</code> 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 <code>grid[i][j]</code> can be:</p>

<ul>
	<li><code>1</code> which means go to the cell to the right. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j + 1]</code>)</li>
	<li><code>2</code> which means go to the cell to the left. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j - 1]</code>)</li>
	<li><code>3</code> which means go to the lower cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i + 1][j]</code>)</li>
	<li><code>4</code> which means go to the upper cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i - 1][j]</code>)</li>
</ul>

<p>Notice that there could be some signs on the cells of the grid that point outside the grid.</p>

<p>You will initially start at the upper left cell <code>(0, 0)</code>. A valid path in the grid is a path that starts from the upper left cell <code>(0, 0)</code> and ends at the bottom-right cell <code>(m - 1, n - 1)</code> following the signs on the grid. The valid path does not have to be the shortest.</p>

<p>You can modify the sign on a cell with <code>cost = 1</code>. You can modify the sign on a cell <strong>one time only</strong>.</p>

<p>Return <em>the minimum cost to make the grid have at least one valid path</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid1.png" style="width: 400px; height: 390px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --&gt; (0, 1) --&gt; (0, 2) --&gt; (0, 3) change the arrow to down with cost = 1 --&gt; (1, 3) --&gt; (1, 2) --&gt; (1, 1) --&gt; (1, 0) change the arrow to down with cost = 1 --&gt; (2, 0) --&gt; (2, 1) --&gt; (2, 2) --&gt; (2, 3) change the arrow to down with cost = 1 --&gt; (3, 3)
The total cost = 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid2.png" style="width: 350px; height: 341px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,3],[3,2,2],[1,1,4]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> You can follow the path from (0, 0) to (2, 2).
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid3.png" style="width: 200px; height: 192px;" />
<pre>
<strong>Input:</strong> grid = [[1,2],[4,3]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 4</code></li>
</ul>

max_int = 999999999
moves: Dict = {
        1: (0, 1),
        2: (0, -1),
        3: (1, 0),
        4: (-1, 0),
    }

class Solution:
    
    class Coordinate:
        i: int
        j: int
        cost: int
        visited: bool
        move_id: int

        def __repr__(self):
            return str((self.i, self.j, self.cost))

        def __init__(self, i: int, j: int, move_id: int, cost: int = max_int):
            self.i = i
            self.j = j
            self.cost = cost
            self.move_id = move_id
            self.visited = False

        def mark_visited(self):
            self.visited = True
        
        def is_valid_move(self, move_id: int, n: int, m: int) -> bool:
            move = moves[move_id]
            new_i = self.i + move[0]
            new_j = self.j + move[1]
            return new_i >= 0 and new_i < n and new_j >= 0 and new_j < m

        def move_cost(self, move_id: int) -> int:
            add_cost = (0 if move_id == self.move_id else 1)
            return self.cost + (0 if move_id == self.move_id else 1)

            

    def minCost(self, grid: List[List[int]]) -> int: 
        def insert_pq(queue: List, c: Solution.Coordinate) -> None:
            if c in queue:
                queue.remove(c)
            for i in range(len(queue)):
                if queue[i].cost <= c.cost:
                    queue.insert(i, c)
                    return
            queue.append(c)

        # We have to find the least expensive path between (0,0) and (n-1,m-1) 
        # Create data structures
        n, m = len(grid), len(grid[0])
        cells = [
            [Solution.Coordinate(i, j, grid[i][j]) for j in range(m)]
            for i in range(n)
        ]
        cells[0][0].cost = 0
        cells[0][0].visited = True
        # Create a queue with an initial element (0,0)
        queue = [cells[0][0]]
        
        # While queue not empty
        while queue:
        #   Pop coordinate c
            c = queue.pop()
        #   For all coordinate neighbours ngb:
            for move_id in moves.keys():

                if c.is_valid_move(move_id, n, m):
                    move = moves[move_id]
                    ngb = cells[c.i + move[0]][c.j + move[1]]
                    move_cost = c.move_cost(move_id)
                    if move_cost < ngb.cost:
                        ngb.cost = move_cost
                    if (ngb.i,ngb.j) != (n-1,m-1) and not ngb.visited:
                        insert_pq(queue, ngb)
                    c.visited = True 
        return cells[n-1][m-1].cost

479