<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) {
        if(grid.size() == 0)return 0;
        vector<vector<int>>dirs = {{0,1},{0,-1},{1,0},{-1,0}};
        int m  = grid.size();
        //cols = rows ? grid[0].size() : 0;
        int n = grid[0].size();
        int count = 0;
        queue<pair<int,int>>q;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j] == '1'){
                    count++;
                    q.push({i,j});
                    while(!q.empty()){
                        auto pos = q.front();
                        q.pop();
                        int u = pos.first;
                        int v = pos.second;
                        if(grid[u][v] == '1'){
                            grid[u][v] = '0';
                            for(auto dir : dirs){
                                int a = u + dir[0];
                                int b = v + dir[1];
                                if(a >= 0 && b >= 0 && a < m && b <n)
                                    q.push({a,b});
                            }
                        }
                    }
                }
            }
        }
        return count;
    }
    
};
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