<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 ("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx2,tune=native")
struct UnionFind {
    vector<int> parent, rank;
    vector<char> isSubIsland;
    explicit UnionFind(int n) : parent(vector<int>(n)), rank(vector<int>(n, 1)), isSubIsland(vector<char>(n, true)) {
        iota(parent.begin(), parent.end(), 0);
    }
    int find(int node) {
        if(node == parent[node]) return node;
        return parent[node] = find(parent[node]);
    }
    bool unionize(int node1, int node2) {
        int p1 = find(node1);
        int p2 = find(node2);
        if(p1 == p2) return false; // Same parent
// Addon for UnionFind 
        isSubIsland[p1] = isSubIsland[p2] = (isSubIsland[p1] && isSubIsland[p2]);
        if(rank[p1] >= rank[p2]) {
            // Append p2 under p1
            parent[p2] = p1;
            rank[p1] += rank[p2];
        }else {
            parent[p1] = p2;
            rank[p2] += rank[p1];
        }
        return true;
    }
// Helper function for this problem
    int reduceWhenUnion(int node1, int node2) {
        int p1 = find(node1);
        int p2 = find(node2);
        // When unioning 0,0 -> no need to do anything whether p1 == p2 or not
        // When unioning 1,0/0,1 -> they get connected so that one becomes a 0 so reduce 1
        // When unioning 1,1 -> reduce 1 only IF they are not connected (p1 != p2)
        //                      because if they are connected, its the same island so no need to do anything
        //                      if they aren't connected, then 2 sub-islands are being merged so the result is 1 sub-island. (reduce 1)
        return p1 != p2 && (isSubIsland[p1] || isSubIsland[p2]);
    }
};
// Non-hyper optimized solution
class Solution {
public:
    static UnionFind uf;
    int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
        int M = (int) grid1.size(), N = (int) grid1[0].size();
        int ans = 0;
        for(int i = 0; i < M; i++) {
            for(int j = 0; j < N; j++) {
                if(grid2[i][j]) {
                    int key = i*N + j;
                    // As traversing the grid, since we are reusing 500*500 UnionFind,
                    // reset the parent, rank and isSubIsland
                    uf.parent[key] = key; uf.rank[key] = 1;
                    uf.isSubIsland[key] = (grid1[i][j] == 1);
                    if(uf.isSubIsland[key]) ans++;
                    if(i > 0 && grid2[i-1][j]) {
                        int topKey = key-N;
                        ans -= uf.reduceWhenUnion(key, topKey);
                        uf.unionize(key, topKey);
                    }
                    if(j > 0 && grid2[i][j-1]) {
                        int leftKey = key-1;
                        ans -= uf.reduceWhenUnion(key, leftKey);
                        uf.unionize(key, leftKey);
                    }
                }
            }
        }
        return ans;
    }
};
UnionFind Solution::uf(250'000);
// Hyper-optimized IO solution
auto y123 = ([]() -> bool {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string s;
    UnionFind& uf = Solution::uf;
    ofstream out("user.out");
    bool grid1[500][500], grid2[500][500];
    while(getline(cin, s)) {
        int idx = 2, M = 0, N = 0, ans = 0, SL = (int) s.length();
        while(idx < SL) {
            N = 0;
            while(s[idx] == '0' || s[idx] == '1') {
                grid1[M][N++] = (s[idx] == '1');
                idx += 2;
            }
            M++; idx += 2;
        }
        getline(cin, s);
        idx = 2; SL = (int) s.length();
        for(int i = 0; i < M; i++,idx+=2) {
            for(int j = 0; j < N; j++,idx+=2) {
                if(grid2[i][j] = (s[idx] == '1')) {
                    int key = i*N + j;
                    uf.parent[key] = key;
                    uf.rank[key] = 1;
                    if((uf.isSubIsland[key] = grid1[i][j] == 1)) ans++;
                    if(i > 0 && grid2[i-1][j]) {
                        int topKey = key-N;
                        ans -= uf.reduceWhenUnion(key, topKey);
                        uf.unionize(key, topKey);
                    }
                    if(j > 0 && grid2[i][j-1]) {
                        int leftKey = key-1;
                        ans -= uf.reduceWhenUnion(key, leftKey);
                        uf.unionize(key, leftKey);
                    }
                }
            }
        }
        out << ans << '\n';
    }
    out.flush();
    exit(0);
    return true;
})();
13302