<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 {
public:
    vector<vector<int>> a;

    void bfs(int i, int j, vector<vector<bool>>&vis){
        vis[i][j] = true;
        int n  = vis.size();

        queue<pair<int, int>>q;
        q.push({i, j});

        int dx[] = {1, -1, 0, 0};
        int dy[] = {0, 0, 1, -1};

        while(!q.empty()){
            int x = q.front().first;
            int y = q.front().second;
            q.pop();

            for(int i=0; i<4; i++){
                for(int j=0; j<4; j++){
                    int nx = x + dx[i];
                    int ny = y + dy[i];
                    if(nx >= n || ny >= n || nx < 0 || ny < 0 || vis[nx][ny] || a[nx][ny] == 1) continue;
                    vis[nx][ny] = true;
                    q.push({nx, ny});
                }
            }
        }
    }

    int regionsBySlashes(vector<string>& grid) {
        int n = grid.size();
        a.resize(3 * n, vector<int>(3 * n, 0));
        int i = 0, j = 0;

        for (string s : grid) {
            for (char c : s) {
                if (c == ' ') {
                    j += 3;
                    if (j >= 3 * n) {
                        j = 0;
                        i += 3;
                    }
                } else if (c == '/') {
                    if (i + 1 < 3 * n && j + 1 < 3 * n) {
                        a[i][j + 2] = 1;
                        a[i + 1][j + 1] = 1;
                        a[i + 2][j] = 1;
                    }
                    j += 3;
                    if (j >= 3 * n) {
                        j = 0;
                        i += 3;
                    }
                } else {
                    if (i + 1 < 3 * n && j + 1 < 3 * n) {
                        a[i][j] = 1;
                        a[i + 1][j + 1] = 1;
                        a[i + 2][j + 2] = 1;
                    }
                    j += 3;
                    if (j >= 3 * n) {
                        j = 0;
                        i += 3;
                    }
                }
            }
        }

        // for (const auto& row : a) {
        //     for (int val : row) cout << val << " ";
        //     cout << endl;
        // }

        n *= 3;

        vector<vector<bool>>vis(n, vector<bool>(n, false));
        int cnt = 0;

        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(vis[i][j] || a[i][j] == 1) continue;
                cout<<endl<<"NEW "<<i<<" "<<j<<endl;
                bfs(i, j, vis);
                cnt++;
            }
        }

        return cnt;
    }
};

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