<p>You are given two <code>m x n</code> binary matrices <code>grid1</code> and <code>grid2</code> containing only <code>0</code>&#39;s (representing water) and <code>1</code>&#39;s (representing land). An <strong>island</strong> is a group of <code>1</code>&#39;s connected <strong>4-directionally</strong> (horizontal or vertical). Any cells outside of the grid are considered water cells.</p>

<p>An island in <code>grid2</code> is considered a <strong>sub-island </strong>if there is an island in <code>grid1</code> that contains <strong>all</strong> the cells that make up <strong>this</strong> island in <code>grid2</code>.</p>

<p>Return the <em><strong>number</strong> of islands in </em><code>grid2</code> <em>that are considered <strong>sub-islands</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/10/test1.png" style="width: 493px; height: 205px;" />
<pre>
<strong>Input:</strong> grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
<strong>Output:</strong> 3
<strong>Explanation: </strong>In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/03/testcasex2.png" style="width: 491px; height: 201px;" />
<pre>
<strong>Input:</strong> grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
<strong>Output:</strong> 2 
<strong>Explanation: </strong>In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid1.length == grid2.length</code></li>
	<li><code>n == grid1[i].length == grid2[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>grid1[i][j]</code> and <code>grid2[i][j]</code> are either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
        int m = grid1.size();
        int n = grid1[0].size();
        
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        
        // in function call, where the return is bool and the parameters are of int and int
        function<bool(int, int)> dfs = [&](int i, int j) {
            // if grid2[i][j] is 0 then you need to return true and not false
            if (i < 0 || i >= m || j < 0 || j >= n || visited[i][j] || grid2[i][j] == 0)
                return true;
            
            visited[i][j] = true;
            
            // this is the main condition
            bool isSubIsland = grid1[i][j] == 1; // Check if it overlaps with grid1
            isSubIsland &= dfs(i+1, j);
            isSubIsland &= dfs(i-1, j);
            isSubIsland &= dfs(i, j+1);
            isSubIsland &= dfs(i, j-1);


            // if everything returns true from all the 4 directions
            return isSubIsland;
        };
        
        int count = 0;

        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid2[i][j] == 1 && !visited[i][j]) {
                    if (dfs(i, j)) {
                        ++count;
                    }
                }
            }
        }
        
        return count;
    }
};


// on the right grid the cell is considered if there is a corresponding land cell on the left cell

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