Number of Enclaves
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1020.Number%20of%20Enclaves/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1020.Number%20of%20Enclaves/images/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>
Depth-First Search; Breadth-First Search; Union Find; Array; Matrix
class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        const int dirs[5] = {-1, 0, 1, 0, -1};
        function<void(int, int)> dfs = [&](int i, int j) {
            grid[i][j] = 0;
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                    dfs(x, y);
                }
            }
        };
        for (int j = 0; j < n; ++j) {
            for (int i : {0, m - 1}) {
                if (grid[i][j] == 1) {
                    dfs(i, j);
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j : {0, n - 1}) {
                if (grid[i][j] == 1) {
                    dfs(i, j);
                }
            }
        }
        int ans = 0;
        for (const auto& row : grid) {
            ans += accumulate(row.begin(), row.end(), 0);
        }
        return ans;
    }
};
