<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 n = grid.size();
        int m = grid[0].size();

        vector<vector<int>> vis(n, vector<int>(m, 0));
        queue<pair<int,int>> q;

        for (int i=0 ; i<n ; i++) {
            for (int j=0 ; j<m ; j++) {
                if (grid[i][j]) {
                    if (i+1 >= n || i-1 < 0 || j+1 >= m || j-1 < 0) {
                        vis[i][j] = 1;
                        q.push({i, j});
                    }
                }
            }
        }

        int delRow[] = {-1, 0, +1, 0};
        int delCol[] = {0, +1, 0, -1};

        while (!q.empty()) {
            int row = q.front().first;
            int col = q.front().second;

            q.pop();

            for (int i=0 ; i<4 ; i++) {
                int nRow = row + delRow[i];
                int nCol = col + delCol[i];

                if (nRow >= 0 && nRow < n && nCol >= 0 && nCol < m) {
                    if (grid[nRow][nCol] == 1 && !vis[nRow][nCol]) {
                        vis[nRow][nCol] = 1;
                        q.push({nRow, nCol});
                    }
                }
            }
        }

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

        return lands;
    }
};
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