<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>

"""
Idea:
- Split each cell into a three by three


b (blank)
- 0 0 0
- 0 0 0
- 0 0 0

/
- 0 0 1
- 0 1 0
- 1 0 0

\\
- 1 0 0
- 0 1 0
- 0 0 1
- 
"""

class Solution:
    def regionsBySlashes(self, grid: List[str]) -> int:
        res = 0
        n = len(grid)
        ZOOM = 5
        zoom_size = n * ZOOM
        newGrid = [[0 for _ in range(zoom_size)] for _ in range(zoom_size)]

        def dfs(r, c):
            nonlocal n
            if newGrid[r][c] == 0:
                newGrid[r][c] = 1
                for dr, dc in [[-1, 0], [1, 0], [0, 1], [0, -1]]:
                    newR = r + dr
                    newC = c + dc
                    if 0 <= newR < zoom_size and 0 <= newC < zoom_size and newGrid[newR][newC] == 0:
                        dfs(newR, newC)

        for i in range(n):
            for j in range(n):
                if grid[i][j] == "/":
                    for k in range(ZOOM):
                        newGrid[i * ZOOM + k][j * ZOOM + (ZOOM - 1 - k)] = 1
                elif grid[i][j] == "\\":
                    for k in range(ZOOM):
                        newGrid[i * ZOOM + k][j * ZOOM + k] = 1
            
        for i in range(zoom_size):
            for j in range(zoom_size):
                if newGrid[i][j] == 0:
                    res += 1
                    dfs(i, j)

        return res
        
19843