<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>

inline const auto optimize = []() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  return 0;
}();


class Solution {
public:
    static int countSubIslands(const std::vector<std::vector<int>>& grid1, const std::vector<std::vector<int>>& grid2) {
        // Well this is a bit too easy, we just have to make sure that we explore islands and that
        // for each piece there is a piece in the other index.

        const int ROWS = grid1.size();
        const int COLS = grid1[0].size();

        std::vector<char> seen(ROWS*COLS, false);
        for (int i = 0; i < ROWS; i++) {
            for (int j = 0; j < COLS; j++) {
                seen[i*COLS + j] = !grid2[i][j];
            }
        }

        std::vector<std::pair<int,int>> Q;

        const auto tryAdd = [&](const int y, const int x) {
            if (y < 0 || ROWS <= y || x < 0 || COLS <= x)
                return;

            const int key = y*COLS + x;
            if (seen[key])
                return;
            
            seen[key] = true;
            Q.emplace_back(y, x);
        };

        int ans = 0;
        for (int i = 0; i < ROWS; i++) {
            for (int j = 0; j < COLS; j++) {
                if (seen[i*COLS + j]) {
                    continue;
                }
                seen[i*COLS + j] = true;
                
                // ok, so now then we are going to be doing a dfs here just. While we are doing that
                // we are going to be seeing if we are a sub

                bool goodSub = true;


                Q.clear();
                Q.emplace_back(i, j);

                while (!Q.empty()) {
                    const auto [y, x] = Q.back();
                    Q.pop_back();

                    if (goodSub && !grid1[y][x])
                        goodSub = false;

                    tryAdd(y-1, x);
                    tryAdd(y, x-1);
                    tryAdd(y, x+1);
                    tryAdd(y+1, x);
                }

                ans += goodSub;
            }
        }
        
        return ans;
    }
};
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