<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 dfs(vector<vector<int>>&grid,int i ,int j,vector<pair<int,int> >&dir){
        for(int k=0;k<dir.size();k++){
            if(i+dir[k].first>=0 && i+dir[k].first<=grid.size()-1 && j+dir[k].second>=0 && j+dir[k].second<=grid[i].size()-1 && grid[i+dir[k].first][j+dir[k].second]==1){
                //visited[i+dir[k].first][j+dir[k].second]=1;
                grid[i+dir[k].first][j+dir[k].second]=0;
                dfs(grid,i+dir[k].first,j+dir[k].second,dir);
            }
        }
        //return 0;
    }
    int numEnclaves(vector<vector<int>>& grid) {
        queue<pair<int,int> > q;
        //int cost=0;
        vector<pair<int,int>> dir = {{0,1},{0,-1},{1,0},{-1,0}};
        //vector<vector<int>> visited(grid.size(),vector<int>(grid[0].size(),0));

        int onec=0;
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[i].size();j++){
                //if(grid[i][j]==1) onec++;
                if((i==0 || i==grid.size()-1) && grid[i][j]==1){
                    q.push({i,j});
                }
                if((j==0 || j==grid[i].size()-1)&&(i>0 && i<grid.size()-1)&&grid[i][j]==1){
                    q.push({i,j});
                }
            }
        }
        while(!q.empty()){
            pair<int,int> t = q.front();
            q.pop();
            if(grid[t.first][t.second]){
                // visited[t.first][t.second]=1;
                grid[t.first][t.second]=0;
                dfs(grid,t.first,t.second,dir);
            }
            
        }
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[i].size();j++){
                if(grid[i][j]==1) onec++;
            }
        }
        return onec;

    }
};
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