<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 n;
    int m;
    
    void dfs(int i, int j, vector<int> &visited, vector<vector<char>>& grid) {
        visited[i * m + j] = 1;
        if (j < m - 1 && grid[i][j + 1] == '1' && !visited[i * m + (j + 1)]) //right
            dfs(i, j + 1, visited, grid);
        
        if (i < n - 1 && grid[i + 1][j] == '1' && !visited[(i + 1) * m + j]) // down
            dfs(i + 1, j, visited, grid);
        
        if (j > 0 && grid[i][j - 1] == '1' && !visited[i * m + (j - 1)]) //left
            dfs(i, j - 1, visited, grid);
        
        if (i > 0 && grid[i - 1][j] == '1' && !visited[(i - 1) * m + j]) // up
            dfs(i - 1, j, visited, grid);
    }
    
    int numIslands(vector<vector<char>>& grid) {
        n = grid.size(), m = grid[0].size();
        int island = 0;
        
        vector<int> visited(n * m, 0);
        for (int i = 0; i< n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == '1' && !visited[i * m + j]){
                    dfs(i, j, visited, grid);
                    island++;
                }
            }
        }
        return island;
    }
};
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