<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>


// BFS
// Time complexity - O(N*M*4)
// Space complexity- O(N*M + N*M)
typedef pair<int,int> pi;
class Solution {
    public:
    
    int x[4]={-1,0,+1,0};
    int y[4]={0,+1,0,-1};
    
    void bfs(vector<vector<int>>& grid,int i,int j,vector<vector<int>>& visited)
    {
        int n=grid.size();
        int m=grid[0].size();
        visited[i][j]=1;
        
        queue<pi> q;
        q.push({i,j});
        
        while(!q.empty())
        {
            i=q.front().first;
            j=q.front().second;
            q.pop();
            
            for(int k=0;k<4;k++)
            {
                int r=i+x[k];
                int c=j+y[k];
                
                if(r>=0 and r<n and c>=0 and c<m and !visited[r][c] and grid[r][c]==1)
                {
                    visited[r][c]=1;
                    q.push({r,c});
                }
            }
        }
    }
    int numEnclaves(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        
        vector<vector<int>> visited(n,vector<int>(m,0));
        
        // 1st row and last row
        for(int j=0;j<m;j++)
        {
            if(grid[0][j] and !visited[0][j])
            bfs(grid,0,j,visited);
            
            if(grid[n-1][j] and !visited[n-1][j])
            bfs(grid,n-1,j,visited);
        }
        
        // 1st col and last col
        for(int i=0;i<n;i++)
        {
            if(grid[i][0] and !visited[i][0])
            bfs(grid,i,0,visited);
            
            if(grid[i][m-1] and !visited[i][m-1])
            bfs(grid,i,m-1,visited);
        }
        
        int cnt=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(grid[i][j] and !visited[i][j])
                cnt++;
            }
        }
        return cnt;
    }
};
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