<p>Given a 2D&nbsp;<code>grid</code> consists of <code>0s</code> (land)&nbsp;and <code>1s</code> (water).&nbsp; An <em>island</em> is a maximal 4-directionally connected group of <code><font face="monospace">0</font>s</code> and a <em>closed island</em>&nbsp;is an island <strong>totally</strong>&nbsp;(all left, top, right, bottom) surrounded by <code>1s.</code></p>

<p>Return the number of <em>closed islands</em>.</p>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png" style="width: 240px; height: 120px;" /></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
Islands in gray are closed because they are completely surrounded by water (group of 1s).</pre>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png" style="width: 160px; height: 80px;" /></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1],
&nbsp;              [1,0,0,0,0,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,1,0,1,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li><code>1 &lt;= grid.length, grid[0].length &lt;= 100</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;=1</code></li>
</ul>

class Solution {
public:
    int closedIsland(vector<vector<int>>& grid) {
        int dx[4]={1,0,-1,0};
        int dy[4]={0,1,0,-1};
        vector<vector<int>> visit(grid.size(), vector<int>(grid[0].size(),0));
        int n = grid.size();int m= grid[0].size();
        if(n==1 || m==1)return 0;
        function<void(int, int)> bfs =[&](int i, int j){
                visit[i][j]=1;
                queue<int> q;
                q.push(i);q.push(j);
                while(!q.empty()){
                    int x = q.front();q.pop();
                    int y = q.front();q.pop();
                    for(int k{};k<4;++k){
                        int a = x+ dx[k];
                        int b = y+ dy[k];
                        if(0<=a && a<n && 0<=b && b<m && grid[a][b]==0 &&visit[a][b]==0 ){
                            visit[a][b]=1;
                            q.push(a);q.push(b);
                        }
                    }
                }
        };
        for(int i{};i<n;i+=n-1)
        for(int j{};j<m;++j)
            if(grid[i][j]==0&& visit[i][j]==0)bfs(i,j);

        for(int i{};i<n;++i)
        for(int j{};j<m;j+=m-1)
            if(grid[i][j]==0 && visit[i][j]==0)bfs(i,j);
    
        int count =0;
        for(int i{};i<n;++i)
        for(int j{};j<m;++j)if(grid[i][j]==0 && visit[i][j]==0){
            bfs(i,j);
            ++count;}
        return count;
    }
};
18141