<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 counter = 0;
    bool isVisited[501][501] = {};
    bool isBoundary = false;
    int numEnclaves(vector<vector<int>>& grid) {
        for(int i = 0; i <= 500; i++)
            for(int j = 0; j <= 500; j++)
                isVisited[i][j] = false;
        int ans = 0;
        for(int i = 0; i < grid.size(); i++)
            for(int j = 0; j < grid[i].size(); j++) {
                if(isVisited[i][j] || grid[i][j] == 0)
                    continue;
                counter = 0;
                isBoundary = false;
                dfs(i, j, grid);
                if(!isBoundary)
                    ans += counter;
            }
        return ans;
    }
    void dfs(int y, int x, vector<vector<int>>& grid) {
        counter++;
        isVisited[y][x] = true;
        if(y == 0 || y == grid.size()-1 || x == 0 || x == grid[0].size()-1)
            isBoundary = true;
        int adj[4][2], adjSize = 0;
        if(y-1 > -1 && grid[y-1][x] == 1) {
            adj[adjSize][0] = y-1;
            adj[adjSize++][1] = x;
        }
        if(y+1 < grid.size() && grid[y+1][x] == 1) {
            adj[adjSize][0] = y+1;
            adj[adjSize++][1] = x;
        }
        if(x-1 > -1 && grid[y][x-1] == 1) {
            adj[adjSize][0] = y;
            adj[adjSize++][1] = x-1;
        }
        if(x+1 < grid[0].size() && grid[y][x+1] == 1) {
            adj[adjSize][0] = y;
            adj[adjSize++][1] = x+1;
        }
        for(int i = 0; i < adjSize; i++)
            if(!isVisited[adj[i][0]][adj[i][1]])
                dfs(adj[i][0], adj[i][1], grid);
    }
};
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