<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>

def get_ks(c):
    if c == " ":
        return [0]
    elif c == "/":
        return [1, 2]
    else:
        return [3, 4]


def indices_to_uf_id(i_row, j_col, k_type):
    return i_row * 1000 + j_col * 5  + k_type


class Solution:
    def regionsBySlashes(self, grid: List[str]) -> int:
        uf = {}

        def find(x):
            if x != uf[x]:
                uf[x] = find(uf[x])
            return uf[x]

        for i, row in enumerate(grid):
            for j, char in enumerate(row):
                for k in get_ks(char):
                    curr_id = indices_to_uf_id(i, j, k)
                    uf[curr_id] = curr_id

                    # chain top to here
                    if i > 0 and k in (0, 1, 4):
                        top_ids = [indices_to_uf_id(i - 1, j, k_t) for k_t in [0, 2, 3]]
                        top_ids = [top_id for top_id in top_ids if top_id in uf]
                        top_id = top_ids[0]

                        uf[find(top_id)] = curr_id

                    if j > 0 and k in (0, 1, 3):
                        left_ids = [indices_to_uf_id(i, j - 1, k_t) for k_t in [0, 2, 4]]
                        left_ids = [left_id for left_id in left_ids if left_id in uf]
                        left_id = left_ids[0]

                        uf[find(left_id)] = curr_id

        return len({find(x) for x in uf})
20495