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

        vector<vector<bool>>visited(rows, vector<bool>(cols, false));

        queue<pair<int, int>>q;     // row and col indices

        for (int i = 0; i<cols; i++){
            if (grid[0][i]==1){
                q.push({0, i});
                visited[0][i] = true;
            }
            if (grid[rows-1][i]==1){
                q.push({rows-1, i});
                visited[rows-1][i] = true;
            }
        }
        for (int i = 0; i<rows; i++){
            if (grid[i][0]==1){
                q.push({i, 0});
                visited[i][0] = true;
            }
            if (grid[i][cols-1]==1){
                q.push({i, cols-1});
                visited[i][cols-1] = true;
            }
        }

        vector<vector<int>>dirs = {{0,1}, {1,0}, {-1,0}, {0,-1}};

        while(!q.empty()){
            auto curr = q.front();
            int currRow = curr.first;
            int currCol = curr.second;

            q.pop();

            for (auto dir : dirs){
                int newRow = currRow + dir[0];
                int newCol = currCol + dir[1];

                if (newRow >=0 && newRow < rows && newCol >=0 && newCol < cols && grid[newRow][newCol]==1 && !visited[newRow][newCol]){
                    visited[newRow][newCol] = true;
                    q.push({newRow, newCol});
                }
            }
        }

        int ans = 0;
        
        for (int i = 0; i<rows; i++){
            for (int j = 0; j<cols; j++){
                if (grid[i][j] == 1 && !visited[i][j]){
                    ans++;
                }
            }
        }
        return ans;
    }
};
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