<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:
        row1,col1 = len(grid),len(grid[0])
        row2,col2 = 3*row1,3*col1
        grid2 = [[0]*col2 for _ in range(row2)]     # convert normal grid into 3 times

        for i in range(row1):
            for j in range(col1):
                r2,c2 = i*3,j*3
                if grid[i][j]=='\\':
                    grid2[r2][c2]=1
                    grid2[r2+1][c2+1]=1
                    grid2[r2+2][c2+2]=1
                elif grid[i][j]=='/':
                    grid2[r2][c2+2]=1
                    grid2[r2+1][c2+1]=1
                    grid2[r2+2][c2]=1
        
        def dfs(r,c,visist):
            if(r<0 or c<0 or r==row2 or c==col2
            or grid2[r][c]!=0 or (r,c) in visit):
                return 
            visit.add((r,c))
            neighbour = [[r+1,c],[r,c+1],[r-1,c],[r,c-1]]
            for nr,nc in neighbour:
                dfs(nr,nc,visit)

        visit=set()
        res=0
        for r in range(row2):
            for c in range(col2):
                if grid2[r][c]==0 and (r,c) not in visit:
                    dfs(r,c,visit)
                    res+=1
        return res

22888