<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:
        n = len(grid)
        parent = {}
        
        # Initialize the union-find parent for each triangle
        def find(x):
            if parent[x] != x:
                parent[x] = find(parent[x])
            return parent[x]
        
        def union(x, y):
            rootX = find(x)
            rootY = find(y)
            if rootX != rootY:
                parent[rootX] = rootY
        
        # Each cell is represented by 4 triangles: 0 (top-left), 1 (top-right), 2 (bottom-right), 3 (bottom-left)
        def get_index(i, j, k):
            return (i * n + j) * 4 + k
        
        # Initialize all triangles
        for i in range(n):
            for j in range(n):
                for k in range(4):
                    parent[get_index(i, j, k)] = get_index(i, j, k)
                        
        for i in range(n):
            for j in range(n):
                if grid[i][j] == '/':
                    union(get_index(i, j, 0), get_index(i, j, 3))
                    union(get_index(i, j, 1), get_index(i, j, 2))
                elif grid[i][j] == '\\':
                    union(get_index(i, j, 0), get_index(i, j, 1))
                    union(get_index(i, j, 2), get_index(i, j, 3))
                else:
                    union(get_index(i, j, 0), get_index(i, j, 1))
                    union(get_index(i, j, 1), get_index(i, j, 2))
                    union(get_index(i, j, 2), get_index(i, j, 3))
                
                # Union with right and bottom cells if possible
                if j + 1 < n:
                    union(get_index(i, j, 1), get_index(i, j + 1, 3))
                if i + 1 < n:
                    union(get_index(i, j, 2), get_index(i + 1, j, 0))
        
        # Count the number of distinct roots
        return sum(find(x) == x for x in parent)
24736