<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:
    void bfs(vector<vector<int>>& visited, int i, int j, vector<vector<char>>& grid, int n, int m){
        queue<pair<int,int>> q;
        q.push({i, j});
        while(!q.empty()){
            auto temp = q.front();
            q.pop();
            i = temp.first;
            j = temp.second;
            if(i+1<n && grid[i+1][j] == '1' && !visited[i+1][j]) {q.push({i+1, j});visited[i+1][j] = 1;}
            if(j+1<m && grid[i][j+1] == '1' && !visited[i][j+1]) {q.push({i, j+1});visited[i][j+1] = 1;}
            if(i-1>-1 && grid[i-1][j] == '1' && !visited[i-1][j]) {q.push({i-1, j});visited[i-1][j] = 1;}
            if(j-1>-1 && grid[i][j-1] == '1' && !visited[i][j-1]) {q.push({i, j-1});visited[i][j-1] = 1;}
        }
    }

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