<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>

#include <ranges>

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

class Solution {
public:
    static constexpr int countSubIslands(const std::vector<std::vector<int>>& grid1,
                               const std::vector<std::vector<int>>& grid2) {
        const int ROWS = grid1.size() + 2;
        const int COLS = grid1[0].size() + 2;

        // We want to keep as little as possible in the cache, so we will make
        // the seen array contain all the information. bit 0 is if grid2 is set
        // and bit 1 is if grid1 is set.

        auto seen =
            std::make_unique_for_overwrite<unsigned char[]>(ROWS * COLS);
        for (int x = 0; x < COLS; x++) {
            seen[0 * COLS + x] = 0;
            seen[(ROWS - 1) * COLS + x] = 0;
        }
        for (int y = 1; y < ROWS - 1; y++) {
            seen[y * COLS + 0] = 0;
            seen[y * COLS + (COLS - 1)] = 0;
        }

        for (int i = 1; i < ROWS - 1; i++) {
            for (int j = 1; j < COLS - 1; j++) {
                seen[i * COLS + j] = grid1[i - 1][j - 1];
                seen[i * COLS + j] <<= 1;
                seen[i * COLS + j] |= grid2[i - 1][j - 1];
            }
        }

        const auto testAndClear = [&](const auto key) {
            if (seen[key] & 1) {
                seen[key] ^= 1;
                return true;
            } else {
                return false;
            }
        };

        auto Q = std::make_unique_for_overwrite<int[]>(COLS * ROWS);
        int Qidx = 0;

        const auto tryAdd = [&](const auto key) {
            if (testAndClear(key)) {
                Q[Qidx++] = key;
            }
        };

        int ans = 0;
        for (int i = 1; i < ROWS - 1; i++) {
            for (int j = 1; j < COLS - 1; j++) {
                const auto firstKey = i * COLS + j;
                if (!testAndClear(firstKey))
                    continue;

                bool goodSub = true;

                Q[Qidx++] = firstKey;

                while (0 < Qidx) {
                    const auto curKey = Q[--Qidx];

                    // If we don't got the second.
                    goodSub = goodSub && seen[curKey] & 2;

                    tryAdd(curKey - COLS); // Up
                    tryAdd(curKey + COLS); // Down
                    tryAdd(curKey - 1);    // Left
                    tryAdd(curKey + 1);    // Right
                }

                ans += goodSub;
            }
        }

        return ans;
    }
};
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