<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 {


vector<vector<int>> mt;
vector<vector<bool>> vis;
int n, m;

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

bool valid(int i, int j) {
	return i >= 0 && i < n && j >= 0 && j < m && !vis[i][j] && mt[i][j] == 1;
}

pair<int, bool> dfs(int i, int j) {
	int cnt = 1;
	bool boundary = i == 0 || j == 0 || i == n - 1 || j == m - 1;
	vis[i][j] = 1; // mark current cell as visited 
	for (vector<int> &di: d) {
		int i1 = i + di[0], j1 = j + di[1];
		if (valid(i1, j1)) { // unvisited valid cell
			pair<int, bool> p = dfs(i1, j1);
			cnt += p.first;
			boundary = boundary || p.second;
		}
	}	
	return {cnt, boundary};
}

public:
    int numEnclaves(vector<vector<int>>& mt1) {
		mt = mt1;
		n = mt.size(), m = mt[0].size();
		vis.resize(n, vector<bool>(m, false));
		int ans = 0;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				if (mt[i][j] == 1 && !vis[i][j]) {
					pair<int, bool> p = dfs(i, j);
					if (!p.second) {
						ans += p.first;
					}
				}
			}	
		}		     
		return ans;   
    }
};

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