<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:
    void dfs(int i, int j, vector<vector<int>> &arr, vector<vector<int>> &vis) {
       
        if (i < 0 || j < 0 || i >= arr.size() || j >= arr[0].size() || vis[i][j] || arr[i][j] == 0) return;

        
        vis[i][j] = 1;

        // Visit all four neighboring cells
        dfs(i + 1, j, arr, vis);
        dfs(i - 1, j, arr, vis);
        dfs(i, j + 1, arr, vis);
        dfs(i, j - 1, arr, vis);
    }

    int numEnclaves(vector<vector<int>>& arr) {
        int n = arr.size();
        int m = arr[0].size();

        // Visited matrix to track visited cells
        vector<vector<int>> vis(n, vector<int>(m, 0));

        // Traverse the boundary rows and columns and perform DFS on '1's
        for (int i = 0; i < n; i++) {
            // First and last columns
            if (arr[i][0] == 1 && !vis[i][0]) {
                dfs(i, 0, arr, vis);
            }
            if (arr[i][m-1] == 1 && !vis[i][m-1]) {
                dfs(i, m-1, arr, vis);
            }
        }

        for (int j = 0; j < m; j++) {
            // First and last rows
            if (arr[0][j] == 1 && !vis[0][j]) {
                dfs(0, j, arr, vis);
            }
            if (arr[n-1][j] == 1 && !vis[n-1][j]) {
                dfs(n-1, j, arr, vis);
            }
        }

        // Count the number of unvisited '1's (these are the enclaves)
        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (arr[i][j] == 1 && vis[i][j] == 0) {
                    ans++;
                }
            }
        }

        return ans;
    }
};

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