<p>An <code>n x n</code> grid is composed of <code>1 x 1</code> squares where each <code>1 x 1</code> square consists of a <code>&#39;/&#39;</code>, <code>&#39;\&#39;</code>, or blank space <code>&#39; &#39;</code>. These characters divide the square into contiguous regions.</p>

<p>Given the grid <code>grid</code> represented as a string array, return <em>the number of regions</em>.</p>

<p>Note that backslash characters are escaped, so a <code>&#39;\&#39;</code> is represented as <code>&#39;\\&#39;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/15/1.png" style="width: 200px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [&quot; /&quot;,&quot;/ &quot;]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/15/2.png" style="width: 200px; height: 198px;" />
<pre>
<strong>Input:</strong> grid = [&quot; /&quot;,&quot;  &quot;]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/15/4.png" style="width: 200px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [&quot;/\\&quot;,&quot;\\/&quot;]
<strong>Output:</strong> 5
<strong>Explanation: </strong>Recall that because \ characters are escaped, &quot;\\/&quot; refers to \/, and &quot;/\\&quot; refers to /\.
</pre>

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

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>&#39;/&#39;</code>, <code>&#39;\&#39;</code>, or <code>&#39; &#39;</code>.</li>
</ul>

class Solution:
    def regionsBySlashes(self, grid: List[str]) -> int:
        neighbors = {
            '/': {
                'U': {(-1, 0, 'D'), (0, 0, 'L')}, 
                'R': {(0,  1, 'L'), (0, 0, 'D')}, 
                'L': {(0, -1, 'R'), (0, 0, 'U')}, 
                'D': {(1,  0, 'U'), (0, 0, 'R')}
            },
            '\\': {
                'U': {(-1, 0, 'D'), (0, 0, 'R')}, 
                'R': {(0,  1, 'L'), (0, 0, 'U')}, 
                'L': {(0, -1, 'R'), (0, 0, 'D')}, 
                'D': {(1,  0, 'U'), (0, 0, 'L')}
            },
        }
        visited = set()
        def dfs(r, c, d):
            if r < 0 or c < 0 or r >= len(grid) or c >= len(grid[0]):
                return
            visited.add((r, c, d))
            val = grid[r][c]
            check = neighbors[val][d] if val != ' ' else neighbors['/'][d].union(neighbors['\\'][d])
            for dr, dc, nextd in check:
                if (r + dr, c + dc, nextd) not in visited:
                    dfs(r + dr, c + dc, nextd)
        ans = 0
        for r in range(len(grid)):
            for c in range(len(grid[0])):
                for d in 'URLD':
                    if (r, c, d) not in visited:
                        dfs(r, c, d)
                        ans += 1
        return ans
19951