<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>

import heapq
class Solution:
    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
        exit = []
        n,m = len(maze),len(maze[0])
        for i in range(n):
            for j in range(m):
                if i==0 or j==0 or i==n-1 or j ==m-1:
                    if maze[i][j]=='.' and [i,j] != entrance:
                        exit.append((i,j))
        
        dist = [[float('inf')]*m for _ in range(n)]
        dist[entrance[0]][entrance[1]] =0
        direction = [(1,0),(0,1),(-1,0),(0,-1)]
        q = [(0,entrance[0],entrance[1])]
        while q:
            wt, x,y= heapq.heappop(q)
            print(wt,x,y)
            if (x,y) in exit:
                return wt
            for dx,dy in direction:
                nx = x+dx
                ny = y+dy
                if 0<=nx<n and 0<=ny<m and maze[nx][ny]!='+':
                    if dist[nx][ny]>wt+1:
                        dist[nx][ny]=wt+1
                        heapq.heappush(q,(dist[nx][ny],nx,ny))
        return -1

17500