<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>

#pragma GCC optimize("O3", "unroll-loops")

static auto speedup = [](){
    ios_base::sync_with_stdio(false);
    cout.tie(NULL);
    cin.tie(NULL);
    return NULL;
}();

constexpr int cyMask = 511 << 9, cyStep = 1 << 9;

class Solution {
public:
    int countSubIslands(vector<vector<int>>& g1, vector<vector<int>>& g2) {
        // support variables
        bool isValid;
        int res = 0, maxX = g1[0].size(), maxY = g1.size(), pos = 0,
        st[maxX-- * maxY--], curr;
        // parsing g2
        for (int y = 0, cx, cy, cyBit; y <= maxY; y++) {
            for (int x = 0; x <= maxX; x++) {
                // starting our hunt for new unexplored bits of land
                if (g2[y][x] == 1) {
                    isValid = true;
                    // dfs(x, y);
                    st[pos++] = x + (y << 9);
                    while (pos) {
                        // extracting the head of the stack
                        curr = st[--pos];
                        cx = curr & 511, cy = curr >> 9, cyBit = curr & cyMask;
                        // marking the current cell as visited
                        g2[cy][cx] = -1;
                        // checking the equivalent in g1 if valid or not
                        isValid &= g1[cy][cx];
                        // storing the adjacent cells:
                        // north
                        if (cy && g2[cy - 1][cx] == 1) st[pos++] = cx + cyBit - cyStep;
                        // west
                        if (cx && g2[cy][cx - 1] == 1) st[pos++] = cx - 1 + cyBit;
                        // south
                        if (cy < maxY && g2[cy + 1][cx] == 1) st[pos++] = cx + cyBit + cyStep;
                        // east
                        if (cx < maxX && g2[cy][cx + 1] == 1) st[pos++] = cx + 1 + cyBit;
                    }
                    res += isValid;
                }
            }
        }
        return res;
    }
};
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