<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>>& board) {
        int m = board.size() , n = board[0].size();
        vector<vector<int>>visited(m,vector<int>(n,0));
        queue<vector<int>>q;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if((i==0 || j==0 || i==m-1 || j==n-1) && board[i][j]==1){
                    visited[i][j]=1;
                    q.push({i,j});
                }
            }
        }
        while(!q.empty()){
            int row = q.front()[0];
            int col = q.front()[1];
            q.pop();
            vector<int>dr = {-1,0,1,0};
            vector<int>dc = {0,1,0,-1};
            for(int i=0;i<4;i++){
                int nr = row + dr[i];
                int nc = col + dc[i];
                if(0 <= nr && nr < m && 0 <= nc && nc < n && board[nr][nc]==1 && !visited[nr][nc]){
                    visited[nr][nc]=1;
                    q.push({nr,nc});
                }
            }
        }
        int cnt = 0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(!visited[i][j] && board[i][j]==1) cnt++;
            }
        }
        return cnt;
    }
};
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