<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:
        
        r = len(grid)
        c = len(grid[0])
        
        cell_map = {1: (0, 1), 2: (0, -1), 3: (1, 0), 4: (-1, 0)}
        
        """
        hash_set = set()
        
        dp = {}
        
        def min_value(i, j): 
            
            if i == r - 1 and j == c - 1:
                return 0
            
            ans_value = sys.maxsize
            
            if i < 0 or i >= r or j < 0 or j >= c:
                return ans_value
            
            if (i, j) in hash_set:
                return ans_value
            
            if dp.get((i, j)):
                return dp[(i, j)]
            
            hash_set.add((i, j))
            for value in cell_map:
                row = i + cell_map[value][0]
                col = j + cell_map[value][1]

                if row < 0 or row >= r or col < 0 or col >= c:
                    continue
                val = min_value(row, col)
                if value != grid[i][j] and val != sys.maxsize:
                    val += 1
                ans_value = min(ans_value, val)
            hash_set.remove((i, j))  
            
            dp[(i, j)] = ans_value  
            
            return ans_value
        # min_value(0, 0)
        # return dp[(0, 0)]
        """
    
        
#         class Custom:
#             def __init__(self, i, j, cost):
#                 self.i = i
#                 self.j = j
#                 self.cost = cost
                
#             def __lt__(self, cust):
#                 self.cost < cust.cost
                
        def bfs():
                
            visited = {}
            queue = [(0, 0, 0)]
            ans = sys.maxsize
            while queue:
                cost, i, j = heapq.heappop(queue)
               
                if i == r - 1 and j == c - 1:
                    return cost
                if visited.get((i, j)):
                    continue
                visited[(i, j)] = True    
                for value in cell_map:
                    row = i + cell_map[value][0]
                    col = j + cell_map[value][1]
                    if row < 0 or row >= r or col < 0 or col >= c or visited.get((row, col)):
                        continue
                    heapq.heappush(queue, (cost + (1 if grid[i][j] != value else 0), row, col))
            
            return -1
        
        return bfs()

                    
                                    

        
                
            
            
                    
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