<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 numEnclaves(vector<vector<int>>& grid) {
        int dirs[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        int m = grid.size(), n = grid.at(0).size();
        auto isIn = [&](int r, int c) -> bool {return r >= 0 && r < m && c > 0 && c < n;};
        std::stack<std::tuple<int, int>> stack;
        auto enque = [&](int r, int c) {if (grid[r][c] == 1) {stack.push({r, c});}};
        for (int i = 0; i < m; ++i) {
            enque(i, 0);
            enque(i, n - 1);
        }
        for (int j = 1; j < n - 1; ++j) {
            enque(0, j);
            enque(m - 1, j);
        }
        while (!stack.empty()) {
            auto [r, c] = stack.top();
            stack.pop();
            if (grid[r][c] == 0) {
                continue;
            }
            grid[r][c] = 0;
            for (int d = sizeof(dirs) / sizeof(dirs[0]) - 1; d >= 0; --d) {
                int i = r + dirs[d][0], j = c + dirs[d][1];
                if (isIn(i, j) && grid[i][j] == 1) {
                    stack.push({i, j});
                }
            }
        }
        int result = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                result += grid[i][j];
            }
        }
        return result;
    }
};
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