<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 bfs(vector<vector<int>>& grid, int x, int y) {
    int rowSize = grid.size();
    int colSize = grid[0].size();
    queue<pair<int, int>> q;
    q.push({x, y});
    grid[x][y] = -1; // Mark as visited

    vector<vector<int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    while (!q.empty()) {
        auto [curX, curY] = q.front();
        q.pop();
        for (auto& dire : directions) {
            int newX = curX + dire[0];
            int newY = curY + dire[1];
            if (newX >= 0 && newX < rowSize && newY >= 0 && newY < colSize && grid[newX][newY] == 1) {
                grid[newX][newY] = -1; // Mark as visited
                q.push({newX, newY});
            }
        }
    }
}
    int numEnclaves(vector<vector<int>>& grid) {
        int rowSize = grid.size();
    int colSize = grid[0].size();

    // Perform BFS for boundary land cells
    for (int i = 0; i < rowSize; ++i) {
        for (int j = 0; j < colSize; ++j) {
            if ((i == 0 || i == rowSize - 1 || j == 0 || j == colSize - 1) && grid[i][j] == 1) {
                bfs(grid, i, j);
            }
        }
    }

    // Count the remaining land cells
    int count = 0;
    for (int i = 1; i < rowSize - 1; ++i) {
        for (int j = 1; j < colSize - 1; ++j) {
            if (grid[i][j] == 1) {
                ++count;
            }
        }
    }
    return count;
    }
};
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