<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>

from typing import List

class Solution:
    def regionsBySlashes(self, grid: List[str]) -> int:
        # Create a new grid 3 times the size of the original grid in both dimensions
        n = len(grid)
        newGrid = [[1 for _ in range(n * 3)] for _ in range(n * 3)]
    
        # Map each '/' and '\' to a 3x3 sub-grid
        for i in range(n):
            for j in range(len(grid[i])):
                if grid[i][j] == '/':
                    newGrid[i * 3][j * 3 + 2] = 0
                    newGrid[i * 3 + 1][j * 3 + 1] = 0
                    newGrid[i * 3 + 2][j * 3] = 0
                elif grid[i][j] == '\\':
                    newGrid[i * 3][j * 3] = 0
                    newGrid[i * 3 + 1][j * 3 + 1] = 0
                    newGrid[i * 3 + 2][j * 3 + 2] = 0
        
        # Directions for DFS traversal: up, left, down, right
        directions = [(-1, 0), (0, -1), (1, 0), (0, 1)]

        # Helper function to perform DFS and mark all connected cells as visited
        def dfs(x, y):
            newGrid[x][y] = 0  # Mark the current cell as visited
            for dx, dy in directions:
                newX, newY = x + dx, y + dy
                # Check if the next cell is within bounds and unvisited
                if 0 <= newX < len(newGrid) and 0 <= newY < len(newGrid[0]) and newGrid[newX][newY] == 1:
                    dfs(newX, newY)

        # Count the number of distinct regions
        regions = 0
        for i in range(len(newGrid)):
            for j in range(len(newGrid[i])):
                if newGrid[i][j] == 1:  # Found a new region
                    regions += 1
                    dfs(i, j)  # Perform DFS to mark the entire region
        
        return regions

17994