<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:
        ROWS1, COLS1 = len(grid), len(grid[0])
        ROWS2, COLS2 = 3 * ROWS1, 3 * COLS1
        grid2 = [[0] * COLS2 for _ in range(ROWS2)]

        for r in range(ROWS1):
            for c in range(COLS1):
                r2, c2 = r * 3, c * 3
                if grid[r][c] == "/" :
                    grid2[r2][c2+2] = 1
                    grid2[r2+1][c2+1] = 1
                    grid2[r2+2][c2] = 1
                elif grid[r][c] == "\\":
                    grid2[r2][c2] = 1
                    grid2[r2+1][c2+1] = 1
                    grid2[r2+2][c2+2] = 1    

        def dfs(r, c, vist):
            if (r < 0 or c < 0 or r == ROWS2 or c == COLS2 or
                grid2[r][c] != 0 or (r, c) in visit):
                return
            visit.add((r, c))
            neighbors = [[r+1, c], [r, c+1], [r-1, c], [r, c-1]]

            for nr, nc in neighbors:
                dfs(nr, nc, visit)

        visit = set()
        res = 0

        for r in range(ROWS2):
            for c in range(COLS2):
                if grid2[r][c] == 0 and (r, c) not in visit:
                    dfs(r, c, visit)
                    res += 1
        return res                                    
22779