<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 row, int col,vector<vector<int>>& v, vector<vector<int>>& vis, int delRow[], int delCol[]) {
        vis[row][col]=1;
        v[row][col]=0;
        int n=v.size();
        int m=v[0].size();
        for(int i=0;i<4;i++) {
            int nRow=row+delRow[i];
            int nCol=col+delCol[i];
            if(nRow>=0 && nRow<n && nCol>=0 && nCol<m && vis[nRow][nCol]==0 && v[nRow][nCol]==1) {
                dfs(nRow,nCol,v,vis,delRow,delCol);
            }
        }
    }
    int numEnclaves(vector<vector<int>>& grid) {
        // Approach: First see surrounded regions problem. What we'll do is that whichever 1's that are connected to boundary directly or indirectly, we'll convert that to 0. And at last, we'll count the number if 1's.
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        vector<vector<int>> v = grid;
        int delRow[]={-1,0,+1,0};
        int delCol[]={0,+1,0,-1};
        // Traversing boundaries of the matrix
        for(int i=0;i<m;i++) { // 1st row and last row
            if(v[0][i]==1 && vis[0][i]==0) dfs(0,i,v,vis,delRow,delCol);
            if(v[n-1][i]==1 && vis[n-1][i]==0) dfs(n-1,i,v,vis,delRow,delCol);
        }
        for(int i=0;i<n;i++) {  // 1st col and last col
            if(v[i][0]==1 && vis[i][0]==0) dfs(i,0,v,vis,delRow,delCol);
            if(v[i][m-1]==1 && vis[i][m-1]==0) dfs(i,m-1,v,vis,delRow,delCol);
        }
        int c=0;
        for(auto it:v) {
            c+=count(it.begin(),it.end(),1);   // using inbuilt count method
        }
        return c;
    }
};
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