<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:
    int dfs(int x, int y, vector<vector<int>>& board, vector<vector<bool>> &visited, bool &has_boundry)
    {
        int m = board.size();
        int n = board[0].size();
        int dx[4] = {1,-1,0,0};
        int dy[4] = {0,0,1,-1};
        visited[x][y] = 1;
        if(x==0 || x==m-1 || y==0 || y==n-1){
            has_boundry = true;
        }
        int cnt=1;
        for(int i=0; i<4; i++)
        {
            int nx = x+dx[i];
            int ny = y+dy[i];
            if(nx >= 0 && nx < m && ny >=0 && ny < n && visited[nx][ny] == 0 && board[nx][ny] == 1)
            {
                cnt+=dfs(nx, ny, board, visited, has_boundry);
            }
        } 
        return cnt;  
    }

    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int ans = 0;
        vector<vector<bool>> visited(m,vector<bool>(n,0));
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j] == 1 && !visited[i][j])
                {
                    bool has_boundry = false;
                    int count = dfs(i, j, grid, visited, has_boundry);
                    if(!has_boundry)
                    {
                        ans += count;
                    }
                }
    
            }
        }
        return ans;
        
    }
};
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