<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(int i, int j, vector<vector<int>>& vis,
             vector<vector<int>>& board) {
        int n = board.size();
        int m = board[0].size();
        vis[i][j] = 1;

        vector<int> r1 = {-1, 0, 1, 0};
        vector<int> c1 = {0, 1, 0, -1};

        for (int k = 0; k < 4; k++) {
            int nrow = i + r1[k];
            int ncol = j + c1[k];

            if (nrow >= 0 && nrow < n && ncol >= 0 && ncol < m &&
                !vis[nrow][ncol] && board[nrow][ncol] == 1) {
                dfs(nrow, ncol, vis, board);
            }
        }
    }
    int numEnclaves(vector<vector<int>>& board) {
        int n = board.size();
        int m = board[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));

        // mark 1 and last coloumn
        for (int j = 0; j < m; j++) {
            if (!vis[0][j] && board[0][j] == 1) {
                dfs(0, j, vis, board);
            }
            if (!vis[n - 1][j] && board[n - 1][j] == 1) {
                dfs(n - 1, j, vis, board);
            }
        }
        // mark 1 and last row
        for (int i = 0; i < n; i++) {
            if (!vis[i][0] && board[i][0] == 1) {
                dfs(i, 0, vis, board);
            }
            if (!vis[i][m - 1] && board[i][m - 1] == 1) {
                dfs(i, m - 1, vis, board);
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cout << vis[i][j] << " ";
            }
            cout << endl;
        }

        int cnt = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!vis[i][j] && board[i][j] == 1) {
                    cnt++;
                }
            }
        }
        return cnt;
    }
};
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