<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();
        int m = grid[0].size();

        queue<pair<int,int>> q;
        for (int i = 0; i < m; i++)
        {
            if (grid[0][i] == 1)
            {
                q.push({0,i});
                grid[0][i] = 2;
            }
            if (grid[n-1][i] == 1)
            {
                q.push({n-1,i});
                grid[n-1][i] = 2;
            }
        }
        for (int i = 0; i < n; i++)
        {
            if (grid[i][0] == 1)
            {
                q.push({i,0});
                grid[i][0] = 2;
            }
            if (grid[i][m-1] == 1)
            {
                q.push({i,m-1});
                grid[i][m-1] = 2;
            }
        }

        vector<vector<int>> directions = {{-1,0},{1,0},{0,1},{0,-1}};
        while (!q.empty())
        {
            auto node = q.front();
            int row = node.first;
            int col = node.second;
            q.pop();

            for (auto dir: directions)
            {
                int newRow = row + dir[0];
                int newCol = col + dir[1];

                if (newRow < n && newRow >= 0 && newCol < m && newCol >= 0 &&
                grid[newRow][newCol] == 1)
                {
                    q.push({newRow,newCol});
                    grid[newRow][newCol] = 2;
                }
            }


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