<p>Given an <code>m x n</code> 2D binary grid <code>grid</code> which represents a map of <code>&#39;1&#39;</code>s (land) and <code>&#39;0&#39;</code>s (water), return <em>the number of islands</em>.</p>

<p>An <strong>island</strong> is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;]
]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;]
]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>grid[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();

        vector<int> v(m*n, 0);
        for (int i = 0; i < v.size(); i++) {
            v[i] = i;
        }
        auto find = [&](int p) -> int {
            while (p != v[p]) {
                p = v[p];
            }
            return p;
        };

        int rnd = 0;
        auto doUnion = [&](int x, int y) {
            if (rnd%2 == 0) {
                v[find(x)] = find(y);
            } else {
                v[find(y)] = find(x);
            }
            rnd++;
        };

        #define POS_MAP(x, y) (x*n + y)

        vector<pair<int, int>> dirs = {
            {-1, 0}, {1, 0}, {0, 1}, {0, -1}
        };
        for (int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if (grid[i][j] == '0') {
                    continue;
                }    
    
                for (auto& dir: dirs) {
                    int x = i + dir.first;
                    int y = j + dir.second;
                    if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1') {
                        doUnion(POS_MAP(i, j), POS_MAP(x, y));
                    }
                }
            }
        }

        int cnt = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1' && find(POS_MAP(i, j)) == POS_MAP(i, j)) {
                    cnt ++;
                }
            }
        }
        return cnt;
    }
};
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