<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 bfs(self, maze, start, startList):
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        visited = set()

        while startList:
            row, col, length = startList.pop(0)
            visited.add((row, col))

            for rOffset, cOffset in directions:
                r, c = row + rOffset, col + cOffset
            
                if [r, c] == start:
                    return length + 1
                if r < 0 or r >= len(maze) or c < 0 or c >= len(maze[0]) or maze[r][c] == '+' or (r, c) in visited:
                    continue
                startList.append((r, c, length + 1))
                visited.add((r, c))
        return -1

    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
        rows = len(maze)
        cols = len(maze[0])
        startList = []

        for i in range(rows):
            for j in range(cols):
                if maze[i][j] == '.' and (i == 0 or i == rows - 1 or j == 0 or j == cols - 1) and [i, j] != entrance:
                    startList.append((i, j, 0))
        
        return self.bfs(maze, entrance, startList)
        
# class Solution:
#     def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
#         st = []
#         rows, cols = len(maze), len(maze[0])

#         for i in range(rows):
#             for j in range(cols):
#                 if maze[i][j] == "." and (i == 0 or i == rows - 1 or j == 0 or j == cols - 1) and [i, j] != entrance:
#                     st.append((i, j, 0))

#         return self.bfs(st, maze, entrance)

#     def bfs(self, st: List[tuple], maze: List[List[str]], start: List[int]) -> int:
#         directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
#         visited = set()

#         while st:
#             row, col, length = st.pop(0)
#             visited.add((row, col))

#             for r_offset, c_offset in directions:
#                 r, c = row + r_offset, col + c_offset

#                 if [r, c] == start:
#                     return length + 1
#                 if r < 0 or c < 0 or r >= len(maze) or c >= len(maze[0]) or maze[r][c] == "+" or (r, c) in visited:
#                     continue

#                 st.append((r, c, length + 1))
#                 visited.add((r, c))

#         return -1
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