<p>You are given an <code>m x n</code> matrix <code>maze</code> (<strong>0-indexed</strong>) with empty cells (represented as <code>&#39;.&#39;</code>) and walls (represented as <code>&#39;+&#39;</code>). You are also given the <code>entrance</code> of the maze, where <code>entrance = [entrance<sub>row</sub>, entrance<sub>col</sub>]</code> denotes the row and column of the cell you are initially standing at.</p>

<p>In one step, you can move one cell <strong>up</strong>, <strong>down</strong>, <strong>left</strong>, or <strong>right</strong>. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the <strong>nearest exit</strong> from the <code>entrance</code>. An <strong>exit</strong> is defined as an <strong>empty cell</strong> that is at the <strong>border</strong> of the <code>maze</code>. The <code>entrance</code> <strong>does not count</strong> as an exit.</p>

<p>Return <em>the <strong>number of steps</strong> in the shortest path from the </em><code>entrance</code><em> to the nearest exit, or </em><code>-1</code><em> if no such path exists</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/nearest1-grid.jpg" style="width: 333px; height: 253px;" />
<pre>
<strong>Input:</strong> maze = [[&quot;+&quot;,&quot;+&quot;,&quot;.&quot;,&quot;+&quot;],[&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;+&quot;],[&quot;+&quot;,&quot;+&quot;,&quot;+&quot;,&quot;.&quot;]], entrance = [1,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There are 3 exits in this maze at [1,0], [0,2], and [2,3].
Initially, you are at the entrance cell [1,2].
- You can reach [1,0] by moving 2 steps left.
- You can reach [0,2] by moving 1 step up.
It is impossible to reach [2,3] from the entrance.
Thus, the nearest exit is [0,2], which is 1 step away.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/nearesr2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> maze = [[&quot;+&quot;,&quot;+&quot;,&quot;+&quot;],[&quot;.&quot;,&quot;.&quot;,&quot;.&quot;],[&quot;+&quot;,&quot;+&quot;,&quot;+&quot;]], entrance = [1,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> There is 1 exit in this maze at [1,2].
[1,0] does not count as an exit since it is the entrance cell.
Initially, you are at the entrance cell [1,0].
- You can reach [1,2] by moving 2 steps right.
Thus, the nearest exit is [1,2], which is 2 steps away.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/nearest3-grid.jpg" style="width: 173px; height: 93px;" />
<pre>
<strong>Input:</strong> maze = [[&quot;.&quot;,&quot;+&quot;]], entrance = [0,0]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are no exits in this maze.
</pre>

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

<ul>
	<li><code>maze.length == m</code></li>
	<li><code>maze[i].length == n</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>maze[i][j]</code> is either <code>&#39;.&#39;</code> or <code>&#39;+&#39;</code>.</li>
	<li><code>entrance.length == 2</code></li>
	<li><code>0 &lt;= entrance<sub>row</sub> &lt; m</code></li>
	<li><code>0 &lt;= entrance<sub>col</sub> &lt; n</code></li>
	<li><code>entrance</code> will always be an empty cell.</li>
</ul>

class Solution:
    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
        graph = [] 
        m, n = len(maze), len(maze[0])
        visited = [ [False]*n  for i in range(m)]


        for i in range(m):
            row = []
            for j in range(n):
                if maze[i][j] == "+":
                    row.append(float("inf"))
                elif i==0 or j==0 or i==m-1 or j==n-1:
                    row.append(0)
                else:
                    row.append(-1) # no value yet
            graph.append(row)
        graph[entrance[0]][entrance[1]] = -1
        
        res = float("inf")
        q = deque([ ( 0,(entrance[0],entrance[1]) )])
        while q:
            d, (i,j) = q.popleft()

            if i<0 or j<0 or i==m or j==n:
                continue # don't visit this node its not valid
            
            if visited[i][j]:
                continue # you already visited this node
            if graph[i][j] == 0:
                res = min(d,res)
                continue # 
            if graph[i][j] == float("inf"):
                continue

            visited[i][j]=True
            q.append( (d+1 , (i-1 ,j ) ) )
            q.append( (d+1 , (i+1 ,j ) ) )
            q.append( (d+1 , (i ,j -1) ) )
            q.append( (d+1 , (i ,j +1) ) )

        # res = visit(entrance[0],entrance[1],graph,visited)
        # print(graph)
        if res == float("inf"): return -1
        return res

def visit(i,j,graph,visited):
    m,n = len(graph),len(graph[0])
    if i<0 or j<0 or i==m or j==n:
        return float("inf")
    if visited[i][j]:
        return float("inf")

    # if graph[i][j]!=-1:
    #     return graph[i][j] # if 0 or float(inf)
    ## only visit what is not visited
    visited[i][j]=True
    minDist = float("inf") ## calculat what is the smallest dist from neighbor
    minDist = min(minDist,visit(i-1,j,graph,visited))
    minDist = min(minDist,visit(i+1,j,graph,visited)) 
    minDist = min(minDist,visit(i,j+1,graph,visited))
    minDist = min(minDist,visit(i,j-1,graph,visited))
    graph[i][j] = 1+minDist #min(1+minDist, graph[i][j] if graph[i][j] != -1 else float("inf"))
    visited[i][j] = False
    return 1+minDist
    
        


        
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