<p>You are given an <code>m x n</code> binary grid <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water. An <strong>island</strong> is a maximal <strong>4-directionally</strong> (horizontal or vertical) connected group of <code>1</code>&#39;s.</p>

<p>The grid is said to be <strong>connected</strong> if we have <strong>exactly one island</strong>, otherwise is said <strong>disconnected</strong>.</p>

<p>In one day, we are allowed to change <strong>any </strong>single land cell <code>(1)</code> into a water cell <code>(0)</code>.</p>

<p>Return <em>the minimum number of days to disconnect the grid</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land1.jpg" style="width: 500px; height: 169px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]

<strong>Output:</strong> 2
<strong>Explanation:</strong> We need at least 2 days to get a disconnected grid.
Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land2.jpg" style="width: 404px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [[1,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Grid of full water is also disconnected ([[1,1]] -&gt; [[0,0]]), 0 islands.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class UnionFind {
public:
    vector<int> rank, root, size;
    int cnt;
    
    UnionFind(int n, int cnt) : rank(n), root(n), size(n, 1), cnt(cnt) {
        for(int i = 0; i < n; i++) 
            root[i] = i;
    }

    int find(int x) {
        if(x == root[x])
            return x;
        return root[x] = find(root[x]);
    }

    void unionSet(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        if(rootX != rootY) {
            if(rank[rootX] > rank[rootY]) 
                root[rootY] = rootX;
            else if(rank[rootX] < rank[rootY])
                root[rootX] = rootY;
            else {
                root[rootY] = rootX;
                rank[rootX]++;
            }
            int newSize = size[rootX] + size[rootY];
            size[rootX] = size[rootY] = newSize;
            cnt--;
        }
    }
};

class Solution {
public:
    void unionCells(vector<vector<int>> &grid, UnionFind &uf) {
       int m = grid.size(), n = grid[0].size();

        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++) 
                if(grid[i][j]) {
                    if(i + 1 < m && grid[i + 1][j])
                        uf.unionSet(i * n + j, (i + 1) * n + j);

                    if(j + 1 < n && grid[i][j + 1])
                        uf.unionSet(i * n + j, i * n + (j + 1));
                }        
    }

    int minDays(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int land = 0;
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                land += grid[i][j];

        UnionFind uf(m * n, land);
        unionCells(grid, uf);

        if(uf.cnt != 1)
            return 0;

        // delete 1 cell
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                if(grid[i][j]) {
                    grid[i][j] = 0;
                    
                    UnionFind cur(m * n, land - 1);
                    unionCells(grid, cur);
                    if(cur.cnt != 1)
                        return 1;
                    
                    grid[i][j] = 1;
                }
        
        return 2;
    }
};
/*
minimum number of adjacent land cells can only be 0~2
=> answer must be 0~2
=> check whether answer is 0 or 1, otherwise it's 2
*/
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