<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 {
    vector<int> parent;
    vector<int> rank;
    int count;
    
public:
    int regionsBySlashes(vector<string>& grid) {
        int n = grid.size();
        int dots = n + 1;
        parent.resize(dots * dots);
        rank.resize(dots * dots, 1);
        count = 0;

        // Initialize Union-Find structure
        for (int i = 0; i < parent.size(); ++i) {
            parent[i] = i;
        }

        // Connect boundaries to the top-left corner (0,0)
        for (int i = 0; i < dots; ++i) {
            for (int j = 0; j < dots; ++j) {
                if (i == 0 || j == 0 || i == n || j == n) {
                    int cell = i * dots + j;
                    unionSets(0, cell);
                }
            }
        }

        // Process each cell and connect regions based on slashes
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '\\') {
                    int cell1 = i * dots + j;
                    int cell2 = (i + 1) * dots + (j + 1);
                    unionSets(cell1, cell2);
                } else if (grid[i][j] == '/') {
                    int cell1 = (i + 1) * dots + j;
                    int cell2 = i * dots + (j + 1);
                    unionSets(cell1, cell2);
                }
            }
        }

        return count;
    }
    
private:
    void unionSets(int a, int b) {
        int parentA = find(a);
        int parentB = find(b);
        if (parentA == parentB) {
            count++;
        } else {
            if (rank[parentA] > rank[parentB]) {
                parent[parentB] = parentA;
            } else if (rank[parentA] < rank[parentB]) {
                parent[parentA] = parentB;
            } else {
                parent[parentB] = parentA;
                rank[parentA]++;
            }
        }
    }
    
    int find(int a) {
        if (parent[a] == a) {
            return a;
        }
        return parent[a] = find(parent[a]);
    }
};
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