<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>

class Solution:
    def minCost(self, grid: List[List[int]]) -> int:
        m,n = len(grid), len(grid[0])

        graph = defaultdict(dict)
        directions = [(-1, 0), (0, -1), (1, 0), (0, 1)]

        def buildGraph():
            for i in range(m):
                for j in range(n):
                    graph[(i,j)] = {}
                    neighbors = [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]

                    for x, y in neighbors:
                        if x < 0 or y < 0 or x > m-1 or y > n-1:
                            continue
                        else:
                            graph[(i,j)][(x, y)] = 1
                    if grid[i][j] == 1:
                        if j != n-1:
                            graph[(i,j)][(i, j+1)] = 0
                    elif grid[i][j] == 2:
                        if j != 0:
                            graph[(i,j)][(i, j-1)] = 0
                    elif grid[i][j] == 3:
                        if i < m-1:
                            graph[(i,j)][(i+1, j)] = 0
                    else:
                        if i != 0:
                            graph[(i,j)][(i-1, j)] = 0


        buildGraph()
        dq = deque()
        dq.append([0,0,0])
        dist = {}
        for i in range(m):
            for j in range(n):
                dist[(i, j)] = float("inf")

        while dq:
            i,j, cost = dq.popleft()
            if i == m-1 and j == n-1:
                return cost

            if dist[(i, j)] < cost:
                continue

            dist[(i, j)] = cost
            for node, nextCost in graph[(i, j)].items():
                x, y = node[0], node[1]
                print(node, nextCost)
                if dist[(x,y)] <= cost + nextCost:
                    continue
                dist[(x, y)] = cost + nextCost
                if nextCost == 0:
                    dq.appendleft([x, y, cost])
                else:
                    dq.append([x, y, cost + 1])
543