<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:
        # parse tiles one by one. if a slash is found, check
        # the existence of endpoints in a seen set
        # edges always return true
        # if the slash makes a connection (both points are in the set)
        # then add 1 to answer
        # store new points in set and continue
        n = len(grid)
        parent = {}
        rank = {}

        def find(a: Tuple[int]) -> Tuple[int]:
            if a not in parent:
                parent[a] = a
                return a
            pa = parent[a]
            if pa == a:
                return a
            ret = find(parent[a])
            parent[a] = ret
            return ret

        def union(a: Tuple[int], b: Tuple[int]) -> None:
            pa, pb = find(a), find(b)
            if pa == pb:
                return
            if pa not in rank:
                rank[pa] = 1
            if pb not in rank:
                rank[pb] = 1
            if rank[pa] < rank[pb]:
                pa, pb = pb, pa
            parent[pb] = pa
            rank[pb] = rank[pa]

        for r in range(n):
            for c in range(n):
                # missing empty space and union with surrounding
                # squares
                union((r, c, 0), (r - 1, c, 2))
                union((r, c, 3), (r, c - 1, 1))

                if grid[r][c] == '/':
                    union((r, c, 0), (r, c, 3))
                    union((r, c, 1), (r, c, 2))
                elif grid[r][c] == '\\':
                    union((r, c, 0), (r, c, 1))
                    union((r, c, 2), (r, c, 3))
                else:
                    union((r, c, 0), (r, c, 1))
                    union((r, c, 1), (r, c, 2))
                    union((r, c, 2), (r, c, 3))
        for k in parent.keys():
            find(k)

        #print(parent)
        return len(set(parent.values()))
19516