<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 dfs(vector<vector<int>>&b,vector<vector<bool>>&temp,int i,int j)
    {
        if(i<0 || j<0 || i>=b.size() || j>=b[0].size() || temp[i][j] || b[i][j] ==0)return;
        temp[i][j] = true;
        dfs(b,temp,i-1,j);
         dfs(b,temp,i+1,j);
          dfs(b,temp,i,j-1);
           dfs(b,temp,i,j+1);
        
    }
    int numEnclaves(vector<vector<int>>& board) {
        vector<vector<bool>>temp(board.size(),vector<bool>(board[0].size(),false));
        int n = board.size();
        int m = board[0].size();
        for(int i = 0;i<m;i++)
        {
            if(board[0][i] == 1)
            {
                dfs(board,temp,0,i);
            }
            if(board[n-1][i] == 1)
            {
                dfs(board,temp,n-1,i);
            }
        }
        for(int i = 1;i<n-1;i++)
        {
            if(board[i][0] == 1)
            {
                dfs(board,temp,i,0);
            }
             if(board[i][m-1] == 1)
            {
                dfs(board,temp,i,m-1);
            }
        }
        int count = 0;
        for(int i = 0;i<n;i++)
        {
            for(int j = 0;j<m;j++)
            {
                if(board[i][j]  == 1 && temp[i][j] == false)
                {
                    count++;
                }
            }
        }
        return count;
    }
};
44817