<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(int row, int col, vector<vector<int>> &grid, vector<vector<int>> &vis, vector<vector<int>> &ans, int drow[], int dcol[]){
        n = grid.size();
        m = grid[0].size();
        vis[row][col] = 1;
        ans[row][col] = 0;
        for(int i=0; i<4; i++){
            int nrow = row + drow[i];
            int ncol = col + dcol[i];

            if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && !vis[nrow][ncol] && grid[nrow][ncol]==1){
                dfs(nrow, ncol, grid, vis, ans, drow, dcol);
            }
        }
    }

    int n, m;
    int numEnclaves(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        int count = 0;

        vector<vector<int>>vis(n, vector<int>(m, 0));
        vector<vector<int>> ans = grid;
        
        int drow[] = {-1, 0, 1, 0};
        int dcol[] = {0, 1, 0, -1};

        for(int j=0; j<m; j++){
            if(!vis[0][j] && grid[0][j] == 1)
                dfs(0, j, grid, vis, ans, drow, dcol);
            if(!vis[n-1][j] && grid[n-1][j] == 1)
                dfs(n-1, j, grid, vis, ans, drow, dcol);
        }
        for(int i=0; i<n; i++){
            if(!vis[i][0] && grid[i][0] == 1)
                dfs(i, 0, grid, vis, ans, drow, dcol);
            if(!vis[i][m-1] && grid[i][m-1] == 1)
                dfs(i, m-1, grid, vis, ans, drow, dcol);
        }
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(!vis[i][j] && ans[i][j] == 1)
                count++;
            }
        }
        return count;
    }
};
//TC: O(N*M) SC: O(N*M)
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