<p>You are given an <code>m x n</code> binary matrix <code>grid</code>, where <code>0</code> represents a sea cell and <code>1</code> represents a land cell.</p>

<p>A <strong>move</strong> consists of walking from one land cell to another adjacent (<strong>4-directionally</strong>) land cell or walking off the boundary of the <code>grid</code>.</p>

<p>Return <em>the number of land cells in</em> <code>grid</code> <em>for which we cannot walk off the boundary of the grid in any number of <strong>moves</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves1.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves2.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All 1s are either on the boundary or can reach the boundary.
</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;= 500</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    void bfs(vector<vector<int>>& grid,int i,int j,vector<vector<bool>>&visited)
    {
        if(i<0 || j<0 || i>=grid.size() || j>=grid[0].size())
        return ;
        grid[i][j]=0;
        visited[i][j]=true;
        if(i>0 && !visited[i-1][j] && grid[i-1][j]==1)
        bfs(grid,i-1,j,visited);
        if(j>0 && !visited[i][j-1] && grid[i][j-1]==1)
        bfs(grid,i,j-1,visited);
        if(i<grid.size()-1 && !visited[i+1][j] && grid[i+1][j]==1)
        bfs(grid,i+1,j,visited);
        if(j<grid[0].size()-1&& !visited[i][j+1] && grid[i][j+1]==1)
        bfs(grid,i,j+1,visited);


    }
    int numEnclaves(vector<vector<int>>& grid) {

        vector<vector<bool>>visited(grid.size(),vector<bool>(grid[0].size(),false));
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if((i==0 || j==0 ||i==grid.size()-1 || j==grid[0].size()-1) && grid[i][j]==1 && !visited[i][j]) 
                bfs(grid,i,j,visited);
            }
        }
        int c=0;
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if(grid[i][j]==1)
                c++;
            }
        }
        return c;
    }
};
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