<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:
    int dx[4] = {-1,0,1,0};
    int dy[4] = {0,-1,0,1};
    unordered_map<int,int> low,tin;
    int timer = 0;
    bool articulationPoint = false;

    void dfs(unordered_map<int,vector<int>> &adj,int u,int par) {
        low[u] = tin[u] = ++timer;
        int child_count = 0;

        for(int v : adj[u]) {
            if(v != par) {
                if(low.count(v)) {
                    low[u] = min(low[u],tin[v]);
                }
                else {
                    child_count++;
                    dfs(adj,v,u);
                    if(low[v] >= tin[u] && par != -1) articulationPoint = true;
                    low[u] = min(low[u],low[v]);
                }
            }
        }

        if(par == -1 && child_count > 1) articulationPoint = true;
    }
    int minDays(vector<vector<int>>& grid) {
        const int m = grid.size(),n = grid[0].size();
        const int v = m*n;

        unordered_map<int,vector<int>> adj;
        int vertex_count = 0;

        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {
                if(grid[i][j]) {
                    int id = i*n + j;
                    adj[id] = {};

                    for(int k=0;k<4;k++) {
                        int x = i+dx[k],y = j+dy[k];
                        if(x >= 0 && x < m && y>=0 && y<n && grid[x][y]) {
                            int id_adj = x*n + y;
                            adj[id].push_back(id_adj);
                        }
                    }
                }
            }
        }

        if(adj.size() <= 1) return adj.size();
        bool multiple_dfs = false;
        for(auto &it : adj) {
            if(!low.count(it.first)) {
                if(multiple_dfs) return 0;
            }
            dfs(adj,it.first,-1);
            multiple_dfs = true;
        }

        if(articulationPoint) return 1;
        return 2;
    }
};
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