<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:
        m, n = len(grid), len(grid[0])
        new_grid = [[1 for _ in range(3 * n)] for _ in range(3 * m)]
        res = 0

        for i in range(m):
            for j in range(n):
                if grid[i][j] == '/':
                    new_grid[3*i][3*j+2] = 0
                    new_grid[3*i+1][3*j+1] = 0
                    new_grid[3*i+2][3*j] = 0
                if grid[i][j] == '\\':
                    new_grid[3*i][3*j] = 0
                    new_grid[3*i+1][3*j+1] = 0
                    new_grid[3*i+2][3*j+2] = 0
        
        def dfs(i, j):
            new_grid[i][j] = 0
            for x, y in [(1,0), (0,1), (-1,0), (0,-1)]:
                ii, jj = i+x, j+y
                if 0 <= ii < 3 * m and 0 <= jj < 3 * n and new_grid[ii][jj] == 1:
                    dfs(ii, jj)
        
        for i in range(3 * m):
            for j in range(3 * n):
                if new_grid[i][j] == 1:
                    res += 1
                    dfs(i, j)
        return res
17450