<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(), n = grid[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        queue<pair<int, int>> bound;
        for(int i=0; i<n; i++) {
            if(grid[0][i]) {
                bound.push({0, i});
                visited[0][i] = true;
            }
            if(grid[m-1][i]) {
                bound.push({m-1, i});
                visited[m-1][i] = true;
            }
        }
        for(int i=1; i<m-1; i++) {
            if(grid[i][0]) {
                bound.push({i, 0});
                visited[i][0] = true;
            }
            if(grid[i][n-1]) {
                bound.push({i, n-1});
                visited[i][n-1] = true;
            }
        }
        vector<pair<int, int>> dels = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        while(!bound.empty()) {
            int x = bound.front().first, y = bound.front().second;
            bound.pop();
            for(auto del: dels) {
                int i = x + del.first, j = y + del.second;
                if(i >= 0 && i < m && j >= 0 && j < n && grid[i][j] == 1 && !visited[i][j]) {
                    bound.push({i, j});
                    visited[i][j] = true;
                }
            }
        }
        int count = 0;
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(grid[i][j] && !visited[i][j]) count++;
            }
        }
        
        return count;
    }
};
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