<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 {
    int timer;
    
    void dfs(int i, int par, unordered_map<int, vector<int>>& adj, 
             unordered_map<int, int>& low, unordered_map<int, int>& time,
             set<int>& pts, unordered_set<int>& vis) {
        vis.insert(i);
        time[i] = low[i] = timer++;
        int childs = 0;
        
        for (auto j : adj[i]) {
            if (j == par) continue; // Skip the parent node
            if (vis.find(j) == vis.end()) { // If the node is not visited
                childs++; // Increment child count
                dfs(j, i, adj, low, time, pts, vis);
                low[i] = min(low[i], low[j]);
                // If time[i] <= low[j] and it's not a root node, i is an articulation point
                if (time[i] <= low[j] && par != -1) {
                    pts.insert(i);
                }
            } else {
                low[i] = min(low[i], time[j]); // Update low[i] for back edges
            }
        }
        
        // If the current node is root and has more than one child, it's an articulation point
        if (par == -1 && childs > 1) {
            pts.insert(i);
        }
    }

    vector<int> articulationPoints(int n, unordered_map<int, vector<int>>& adj) {
        unordered_set<int> vis;
        timer = 0;
        unordered_map<int, int> low;
        unordered_map<int, int> time;
        set<int> pts;
        
        for (auto& a : adj) {
            int i = a.first;
            if (vis.find(i) == vis.end()) {
                dfs(i, -1, adj, low, time, pts, vis);
            }
        }
        
        vector<int> ans;
        for (auto i : pts) {
            ans.push_back(i);
        }
        if (ans.size() == 0) return {-1};
        
        return ans;
    }

    void dfsi(int i, int j, vector<vector<int>>& grid, vector<vector<int>>& vis) {
        vis[i][j] = 1;
        int n = grid.size();
        int m = grid[0].size();
        if (i - 1 >= 0 && grid[i - 1][j] == 1 && vis[i - 1][j] == 0) dfsi(i - 1, j, grid, vis);
        if (j - 1 >= 0 && grid[i][j - 1] == 1 && vis[i][j - 1] == 0) dfsi(i, j - 1, grid, vis);
        if (i + 1 < n && grid[i + 1][j] == 1 && vis[i + 1][j] == 0) dfsi(i + 1, j, grid, vis);
        if (j + 1 < m && grid[i][j + 1] == 1 && vis[i][j + 1] == 0) dfsi(i, j + 1, grid, vis);
    }

public:
    int minDays(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> visi(n, vector<int>(m, 0));
        int cnt1=0;
        for(int i=0;i<n;i++)for(int j=0;j<m;j++)if(grid[i][j]==1)cnt1++;

        if(cnt1==1 )return cnt1;

        int is = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1 && visi[i][j] == 0) {
                    dfsi(i, j, grid, visi);
                    is++;
                }
            }
        }
        if (is != 1) return 0;

        unordered_map<int, vector<int>> mp;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    if (i - 1 >= 0 && grid[i - 1][j] == 1) mp[i * m + j].push_back((i - 1) * m + j);
                    if (i + 1 < n && grid[i + 1][j] == 1) mp[i * m + j].push_back((i + 1) * m + j);
                    if (j - 1 >= 0 && grid[i][j - 1] == 1) mp[i * m + j].push_back(i * m + j - 1);
                    if (j + 1 < m && grid[i][j + 1] == 1) mp[i * m + j].push_back(i * m + j + 1);
                }
            }
        }

        vector<int> ans = articulationPoints(mp.size(), mp);
    
        if (ans.size() > 1 || ans[0] != -1) return 1;
        return 2;
    }
};

19980