<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 {
private:
    void bfs(vector<vector<int>>& grid, int i, int j){
        queue<pair<int,int>>q;
        q.push({i, j});
        grid[i][j]=0;

        vector<vector<int>> coor={{0,1}, {1,0}, {-1,0}, {0,-1}};
        int m = grid.size();
        int n = grid[0].size();

        while(!q.empty()){
            auto t=q.front();
            q.pop();
            for(auto it: coor){
                int row = it[0] + t.first, col = it[1] + t.second;
                if(row>=0 && col>=0 && row<m && col<n){
                    if (grid[row][col]==1) {
                        q.push({row, col});
                        grid[row][col]=0;
                        // vis[row][col]=1;
                    }
                }
            }
        }
        
    }
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int ans = 0;
        // vector<vector<int>> vis(m, vector<int>(n, 0));

        for(int i=0;i<n;i++){
            if(grid[0][i]==1) bfs(grid, 0, i);
            if(grid[m-1][i]==1) bfs(grid, m-1, i);
        }
        for(int i=1;i<m-1;i++){
            if(grid[i][0]==1) bfs(grid, i, 0);
            if(grid[i][n-1]==1) bfs(grid, i, n-1);
        }

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    ans++;
                }
            }
        }
        return ans;
        
    }
};
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