<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 n = grid.size();
        int m = grid[0].size();
        queue<pair<int, int>> q;

        // Push all boundary cells with value 1 into the queue and mark them as visited (by setting to 0)
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i == 0 || j == 0 || i == n - 1 || j == m - 1) {
                    if (grid[i][j] == 1) {
                        q.push({i, j});
                        grid[i][j] = 0; // Mark as visited
                    }
                }
            }
        }

        // Directions for BFS (up, down, left, right)
        int dx[4] = {1, 0, -1, 0};
        int dy[4] = {0, -1, 0, 1};

        // Perform BFS to eliminate all connected land (1's) from the boundaries
        while (!q.empty()) {
            int x = q.front().first;
            int y = q.front().second;
            q.pop();

            for (int i = 0; i < 4; i++) {
                int xi = x + dx[i];
                int yi = y + dy[i];
                if (xi >= 0 && xi < n && yi >= 0 && yi < m && grid[xi][yi] == 1) {
                    grid[xi][yi] = 0; // Mark as visited
                    q.push({xi, yi});
                }
            }
        }

        // Count the remaining 1's that are not connected to the boundary
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    count++;
                }
            }
        }

        return count;
    }
};

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