<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 Solution {
public:
    void init(){
        for(int i = 0;i < mIdxList.size();++i){
            mIdxList[i] = i;
        }
    }
    int getRootIndex(int id){
        while(id != mIdxList[id]){
            id = mIdxList[id];
        }
        return id;
    }
    void connect(int id1,int id2){
        int root1 = getRootIndex(id1);
        int root2 = getRootIndex(id2);
        if(root1 == root2){
            return;
        }
        mIdxList[root2] = root1;
    }
    void connect(int i,int j,int i2,int j2){
        int id1 = i*mCol + j;
        int id2 = i2*mCol+j2;
        connect(id1,id2);
    }
    vector<int> mIdxList;
    int mRow;
    int mCol;
    bool valid(int i,int j,const vector<vector<int>>& grid){
        return i>=0 && j >=0 && i < mRow && j < mCol && grid[i][j];
    }
    int islandCount(const vector<vector<int>>& grid){
        set<int> idSet;
        vector<pair<int,int>> posList;
        for(int i = 0;i < mRow;++i){
            for(int j = 0;j < mCol;++j){
                if(grid[i][j] == 0){
                    continue;
                }
                posList.push_back({i,j});
                if(valid(i+1,j,grid)){
                    connect(i,j,i+1,j);
                }
                if(valid(i-1,j,grid)){
                    connect(i,j,i-1,j);
                }
                if(valid(i,j+1,grid)){
                    connect(i,j,i,j+1);
                }
                if(valid(i,j-1,grid)){
                    connect(i,j,i,j-1);
                }
            }
        }
        for(auto&pos:posList){
            int i = pos.first;
            int j = pos.second;
            int id= i*mCol+j;
            idSet.insert(getRootIndex(id));
        }
        return idSet.size();
    }
    int minDays(vector<vector<int>>& grid) {
        mRow = grid.size();
        mCol = grid[0].size();
        mIdxList.resize(mRow*mCol);
        init();

        if(islandCount(grid) != 1){
            return 0;
        }

        vector<pair<int,int>> posList;
        for(int i = 0;i < mRow;++i){
            for(int j = 0;j < mCol;++j){
                if(grid[i][j] == 1)
                    posList.push_back({i,j});
            }
        }
        for(auto& pos:posList){
            init();
            grid[pos.first][pos.second] = 0;
            if(islandCount(grid) != 1){
                return 1;
            }
            grid[pos.first][pos.second] = 1;
        }
        return 2;
    }
};
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