<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:
    vector<pair<int, int>> directions = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

    void bfs(vector<vector<int>>& grid, int row, int col) {
        int m = grid.size();
        int n = grid[0].size();
        queue<pair<int, int>> q;
        q.push({row, col});
        grid[row][col] = 0; // Mark this cell as visited (change land to sea)

        while (!q.empty()) {
            auto [r, c] = q.front();
            q.pop();

            // Explore all 4 directions
            for (auto& [dr, dc] : directions) {
                int nr = r + dr;
                int nc = c + dc;

                // If the new position is within bounds and is a land cell
                if (nr >= 0 && nr < m && nc >= 0 && nc < n && grid[nr][nc] == 1) {
                    grid[nr][nc] = 0; // Mark it as visited
                    q.push({nr, nc});
                }
            }
        }
    }

    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();

        // Perform BFS/DFS on boundary rows and columns
        for (int i = 0; i < m; ++i) {
            if (grid[i][0] == 1) bfs(grid, i, 0);        // Left boundary
            if (grid[i][n - 1] == 1) bfs(grid, i, n - 1); // Right boundary
        }

        for (int j = 0; j < n; ++j) {
            if (grid[0][j] == 1) bfs(grid, 0, j);        // Top boundary
            if (grid[m - 1][j] == 1) bfs(grid, m - 1, j); // Bottom boundary
        }

        // Count remaining land cells (enclaves)
        int enclaveCount = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    enclaveCount++;
                }
            }
        }

        return enclaveCount;
    }
};

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