<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>>& nums) {
        int n=nums.size(),m=nums[0].size();
        int ans=0;

        int delrow[]={0,0,-1,1};
        int delcol[]={-1,1,0,0};

        queue<pair<int,int>> q;

        for(int i=0;i<n;i++)
        {
            if(nums[i][0])
            {
                nums[i][0]=0;
                q.push({i,0});
            }
            if(nums[i][m-1])
            {
                nums[i][m-1]=0;
                q.push({i,m-1});
            }
        }

        for(int j=1;j<m-1;j++)
        {
            if(nums[0][j])
            {
                nums[0][j]=0;
                q.push({0,j});
            }
            if(nums[n-1][j])
            {
                nums[n-1][j]=0;
                q.push({n-1,j});
            }
        }

        while(!q.empty())
        {
            int x=q.front().first,y=q.front().second;

            q.pop();

            for(int i=0;i<4;i++)
            {
                int r=delrow[i]+x,c=delcol[i]+y;
                if(r<n&&c<m&&r>=0&&c>=0)
                {
                    if(nums[r][c])
                    {
                        nums[r][c]=0;
                        q.push({r,c});
                    }
                }
            }
        }

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(nums[i][j])
                ans++;
            }
        }

        return ans;
    }
};
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