<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:
        # use union find, simplified without ranking
        # take each char in string as sqaure and split into 4
        # sections by 2 diagonal lines
        # if / then union (south and east) and (north and west)
        # vice versa for \
        # if cell has adj cells, union bottom and right sides 
        # respectively
        parents = {}
        def find(x):
            if x not in parents:
                parents[x] = x
            while parents[x] != x:
                x = parents[x]
            return x

        def union(x, y):
            px, py = find(x), find(y)
            parents[px] = py

        n = len(grid)
        for i in range(n):
            for j in range(n):
                if grid[i][j] == "/":
                    union((i, j, 'n'), (i, j, 'w'))
                    union((i, j, 'e'), (i, j, 's'))
                elif grid[i][j] == "\\":
                    union((i, j, 'n'), (i, j, 'e'))
                    union((i, j, 's'), (i, j, 'w'))
                else:
                    union((i, j, 'n'), (i, j, 'w'))
                    union((i, j, 'w'), (i, j, 's'))
                    union((i, j, 's'), (i, j, 'e'))
                
                if j > 0:
                    union((i, j, 'w'), (i, j - 1, 'e'))
                if i > 0:
                    union((i, j, 'n'), (i - 1, j, 's'))

        return len(set(map(find, parents)))
16906