<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 dfs(int i, int j, vector<vector<int>>& grid, vector<vector<bool>> &visited)
    {        
        int m = grid.size();
        int n = grid[0].size();
        vector<pair<int, int>> dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        int num = 0;
        stack<pair<int, int>> pq;
        pq.push({i, j});
        bool flag = true;
        while (!pq.empty())
        {
            pair<int, int> vec = pq.top();
            int x1 = vec.first;
            int y1 = vec.second;
            if (x1 == 0 || y1 == 0 || x1 == (m - 1) || y1 == (n - 1))
            {
                flag = false;
                // break;
            }
            pq.pop();
            if (visited[x1][y1])
            {
                continue;
            }
            visited[x1][y1] = true;
            num++;
            for (auto& dir : dirs)
            {
                int x = x1 + dir.first;
                int y = y1 + dir.second;
                if (x >= 0 && y >= 0 && x < m && y < n)
                {
                    if (!visited[x][y] && grid[x][y] == 1)
                    {
                        pq.push({x, y});
                    }
                }
            }
        }
        if (flag == false)
        {
            num = 0;
        }
        return num;
    }

    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int num = 0;
        int max_num = 0;
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        for (int i = 0; i <= m - 1; i++)
        for (int j = 0; j <= n - 1; j++)
        {
            if (grid[i][j] == 1 && !visited[i][j])
            {
                num = dfs(i, j, grid, visited);
                max_num += num;
            }
        }
        return max_num;
    }
};
53672