<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:
        L = len(grid)
        R = C= 3*L
        g3 = [[0] *R for _ in range(C)]
        for i in range(L):
            for j in range(L):
                v = grid[i][j]
                if v == ' ':
                    for x in range(3):
                        for y in range(3):
                            g3[i*3+x][j*3+y] = 0
                elif v == '/':
                    for x in range(3):
                        for y in range(3):
                            if x+y == 2:
                                g3[i*3+x][j*3+y] = 1
                            else:
                                g3[i*3+x][j*3+y] = 0
                elif v == '\\': # one char, len == 1
                    for x in range(3):
                        for y in range(3):
                            if x == y:
                                g3[i*3+x][j*3+y] = 1
                            else:
                                g3[i*3+x][j*3+y] = 0
        parent = [r*C+c for r in range(R) for c in range(C)]

        def find(x):
            if parent[x] != x:
                parent[x] = find(parent[x])
            return parent[x]
        
        def union(x,y):
            parent[find(x)] = find(y)

        res = R*C
        for i in range(R):
            for j in range(C):
                if g3[i][j] != 0:
                    res -= 1
                    continue
                for di, dj in (-1,0), (1,0), (0,-1), (0,1):
                    ni = i+di
                    nj = j+dj
                    if ni<0 or ni>R-1 or nj<0 or nj>C-1 or g3[ni][nj] != 0:
                        continue
                    r1 = find(i*C+j)
                    r2 = find(ni*C+nj)
                    if r1 != r2:
                        union(i*C+j, ni*C+nj)
                        res -= 1
        return res



        




        
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