<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 x, int y, int n, int m, vector<vector<int>> &mat, vector<vector<int>> &boundary){

        boundary[x][y] = 1;

        // 4 neighbours
        vector<int> dx = {-1, 0, +1, 0};
        vector<int> dy = {0, +1, 0, -1};

        for(int i=0;i<4;i++){
            int newRow = x + dx[i];    
            int newCol = y + dy[i];   

            if(newRow >= 0 && newRow < n && newCol >= 0 && newCol < m && mat[newRow][newCol] == 1 && boundary[newRow][newCol] == 0){
                dfs(newRow, newCol, n, m, mat, boundary);               
            }
        }
    }
    int numEnclaves(vector<vector<int>>& mat) {
        
        int n = mat.size();
        int m = mat[0].size();
        
        vector<vector<int>> boundary(n, vector<int> (m, 0));           // or visited to store elements which are adjacent to boundary

        // Traverse matrix boundary wise
        for(int j=0;j<m;j++){
            if(mat[0][j] == 1 && boundary[0][j] == 0){       // top horizontal
                dfs(0, j, n, m, mat, boundary);
            }

            if(mat[n-1][j] == 1 && boundary[n-1][j] == 0){  // bottom horizontal
                dfs(n-1, j, n, m, mat, boundary);
            }         
        }
        
        for(int i=0;i<n;i++){
            if(mat[i][0] == 1 && boundary[i][0] == 0){       // left vertical
                dfs(i, 0, n, m, mat, boundary);
            }

            if(mat[i][m-1] == 1 && boundary[i][m-1] == 0){     // right vertical
                dfs(i, m-1, n, m, mat, boundary);
            }
        }

        // Now count enclaves
        int cnt = 0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mat[i][j] == 1 && boundary[i][j] == 0){       // means 1's which are not adjacent to boundary
                    cnt++;
                }
            }
        }

        return cnt;       
    }
};
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