<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 {
    const vector<int> delta = {-1, 0, 1, 0};
    unordered_map<int, int> low, tin;
    int timer = 0;
    bool articulation_point = 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) {
                        articulation_point = true;
                    }
                    low[u] = min(low[u], low[v]);
                }
            }
        if (par == -1 && child_count > 1)
            articulation_point = true;
    }

public:
    int minDays(vector<vector<int>>& grid) {
        const int n = grid.size(), m = grid[0].size();
        const int V = n * m;
        unordered_map<int, vector<int>> adj;
        int vertex_count = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (grid[i][j]) {
                    int id = i * m + j;
                    adj[id] = {};
                    for (int k = 0; k < 4; k++) {
                        int r = i + delta[k], c = j + delta[(k + 1) % 4];
                        if (r < n && c < m && r >= 0 && c >= 0 && grid[r][c]) {
                            int id_adj = r * m + c;
                            adj[id].push_back(id_adj);
                        }
                    }
                }
        if (adj.size() <= 1)
            return adj.size();
        bool multiple_dfs = 0;
        for (auto& it : adj)
            if (!low.count(it.first)) {
                if (multiple_dfs)
                    return 0;
                dfs(adj, it.first, -1);
                multiple_dfs = 1;
            }
        if (articulation_point)
            return 1;
        return 2;
    }
};
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