<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 x, int y, vector<vector<int>>& grid) {
        if( x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] == 0) {
            return;
        }
        grid[x][y] = 0;
        vector<int> dirs = {0, 1, 0, -1, 0};
        for(int i = 0; i < 4; i++) {
            int nx = x + dirs[i];
            int ny = y + dirs[i+1];
            dfs(nx, ny, grid);
        }
    }
    int numEnclaves(vector<vector<int>>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();
        for(int i = 0; i < rows; i++) {
            if(grid[i][0] == 1) dfs(i, 0, grid);
            if(grid[i][cols - 1] == 1) dfs(i, cols-1, grid);
        }
        for(int i = 0; i < cols; i++) {
            if(grid[0][i] == 1) dfs(0, i, grid);
            if(grid[rows-1][i] == 1) dfs(rows-1, i, grid);
        }
        int count = 0;
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(grid[i][j] == 1) count++;
            }
        }
        return count;
    }
};
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