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

        auto dfs = [&](int i_, int j_){

            queue<pair<int, int>>q;
            q.push({i_, j_});
            grid[i_][j_] = 0;

            while(!q.empty()){

                auto [i__, j__] = q.front();
                q.pop();

                for( auto [itx, ity] : v ){

                    int i = i__ + itx, j = j__ + ity;

                    if(i > -1 && i < n && j > -1 && j < m && grid[i][j]){

                        grid[i][j] = 0;
                        q.push({i, j});

                    }
                }
            }
        };

        for(int k = 0; k < n; k++){

            if(grid[k][0]) dfs(k, 0);
            if(grid[k][m - 1]) dfs(k, m - 1);
        
        }

        for(int k = 0; k < m; k++){

            if(grid[0][k]) dfs(0, k);
            if(grid[n - 1][k]) dfs(n - 1, k);

        }

        int ans = 0;

        for(auto iit : grid)
            for(auto it : iit)
                if(it) ans++;
        
        return ans;

    }
};
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