<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:
        graph = {}
        m, n = len(grid), len(grid[0])
        
        def addEdges(i, j):
            graph[(i, j)] = {}
            neighbors = [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]
            for each in neighbors:
                x, y = each
                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
        
        
        for i in range(m):
            for j in range(n):
                addEdges(i, j)
        
        def convert(x, y):
            return x * n + y
        
        def BFS(graph):
            q = deque()
            q.append([0, 0, 0])
            distances = [float(inf)] * (m * n)
            
            while q:
                cost, x, y = q.popleft()
                if (x, y) == (m - 1, n - 1):
                    return cost
                
                idx = convert(x, y)
                if distances[idx] < cost:
                    continue
                
                distances[idx] = cost
                for node, nxtCost in graph[(x, y)].items():
                    nxtIndex = convert(node[0], node[1])
                    if distances[nxtIndex] <= cost + nxtCost:
                        continue
                    
                    distances[nxtIndex] = cost + nxtCost
                    if nxtCost == 0:
                        q.appendleft([cost, node[0], node[1]])
                    else:
                        q.append([cost + 1, node[0], node[1]])
        
        def dijkstra(graph):
            distances = [float(inf)] * (m * n)
            myheap = [[0, 0, 0]]
            
            while myheap:
                cost, x, y = heappop(myheap)
                if (x, y) == (m - 1, n - 1):
                    return cost
                
                idx = convert(x, y)
                if distances[idx] < cost:
                    continue
                else:
                    distances[idx] = cost
                    for node, nxtCost in graph[(x, y)].items():
                        total = cost + nxtCost
                        nxtIndex = convert(node[0], node[1])
                        if distances[nxtIndex] <= total:
                            continue
                        else:
                            distances[nxtIndex] = total
                            heappush(myheap, [total, node[0], node[1]])
            
            return distances[-1]
        
        return BFS(graph)
23400