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
function numEnclaves(grid: number[][]): number {
    const [m, n] = [grid.length, grid[0].length];
    const dirs: number[] = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number) => {
        grid[i][j] = 0;
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y <= n && grid[x][y] === 1) {
                dfs(x, y);
            }
        }
    };
    for (let j = 0; j < n; ++j) {
        for (let i of [0, m - 1]) {
            if (grid[i][j] === 1) {
                dfs(i, j);
            }
        }
    }
    for (let i = 0; i < m; ++i) {
        for (let j of [0, n - 1]) {
            if (grid[i][j] === 1) {
                dfs(i, j);
            }
        }
    }
    return grid.flat().reduce((acc, cur) => acc + cur, 0);
}
