<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) {
        // want to run dfs from the first one
        vector<pair<int,int>> dir={{0,-1},{0,1},{1,0},{-1,0}};
        const int m=grid.size();
        const int n=grid[0].size();
        int ans=0;
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[0].size();j++){
                if (grid[i][j]){
                    int seen=0;
                    stack<pair<int,int>> q;
                    q.emplace(i,j);
                    grid[i][j]=0;
                    bool reach_wall=0;


                    while (!q.empty()){
                        auto [a,b]=q.top();
                        q.pop();
                        seen++;

                        for(auto [c,d]:dir){
                            int x=a+c;
                            int y=b+d;

                            if (x<0 || x>=m || y<0 || y>=n){
                                // then 
                                reach_wall|=1;
                                continue;
                            }

                            if (grid[x][y]){
                                q.emplace(x,y);
                                grid[x][y]=0;
                            }

                        }

                        
                    }

                    if (!reach_wall)
                        ans+=seen;

                }
            }
        }
        return ans;
    }
};
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