<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 UnionFind {
public:
    vector<int> parent;
    UnionFind(int n) : parent(n) {
        iota(parent.begin(), parent.end(), 0);
    }
    
    int find(int x) {
        if (x != parent[x])
            parent[x] = find(parent[x]);
        return parent[x];
    }
    
    void unite(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        if (rootX != rootY)
            parent[rootX] = rootY;
    }
};

class Solution {
public:
    int regionsBySlashes(vector<string>& grid) {
        int n = grid.size();
        UnionFind uf(4 * n * n);
        
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int index = 4 * (i * n + j);
                char c = grid[i][j];
                
                // Union the four parts of this cell
                if (c == '/') {
                    uf.unite(index + 0, index + 3); // top-right with bottom-left
                    uf.unite(index + 1, index + 2); // top-left with bottom-right
                } else if (c == '\\') {
                    uf.unite(index + 0, index + 1); // top-left with top-right
                    uf.unite(index + 2, index + 3); // bottom-left with bottom-right
                } else { // space
                    uf.unite(index + 0, index + 1);
                    uf.unite(index + 1, index + 2);
                    uf.unite(index + 2, index + 3);
                }
                
                // Union with neighboring cells
                if (i > 0) // union with the cell above
                    uf.unite(index + 0, (index - 4 * n) + 2);
                if (j > 0) // union with the cell on the left
                    uf.unite(index + 3, (index - 4) + 1);
            }
        }
        
        // Count distinct regions
        int regions = 0;
        for (int i = 0; i < 4 * n * n; ++i) {
            if (uf.find(i) == i)
                ++regions;
        }
        
        return regions;
    }
};

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