<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 numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        
        // Perform DFS from all boundary land cells
        for (int r = 0; r < m; ++r) {
            if (grid[r][0] == 1) dfs(r, 0, grid);
            if (grid[r][n - 1] == 1) dfs(r, n - 1, grid);
        }
        
        for (int c = 0; c < n; ++c) {
            if (grid[0][c] == 1) dfs(0, c, grid);
            if (grid[m - 1][c] == 1) dfs(m - 1, c, grid);
        }
        
        // Count remaining land cells that are enclaved
        int count = 0;
        for (int r = 0; r < m; ++r) {
            for (int c = 0; c < n; ++c) {
                if (grid[r][c] == 1) {
                    ++count;
                }
            }
        }
        
        return count;
    }

private:
    // Helper function for DFS
    void dfs(int r, int c, vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        
        // Directions for 4-directional movement
        vector<pair<int, int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        
        // Mark this cell as visited (escaped)
        grid[r][c] = 0;
        
        for (const auto& dir : directions) {
            int nr = r + dir.first;
            int nc = c + dir.second;
            
            if (nr >= 0 && nr < m && nc >= 0 && nc < n && grid[nr][nc] == 1) {
                dfs(nr, nc, grid);
            }
        }
    }
};

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