<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 {
private:
    int timer = 0;

    void dfs(int node, int parent, unordered_map<int, vector<int>>& graph, unordered_map<int, int>& low, unordered_map<int, int>& disc, vector<int>& articulations) {
        disc[node] = low[node] = ++timer;
        int childCount = 0;
        for (int neighbor : graph[node]) {
            if (disc.count(neighbor) == 0) { // if neighbor is not visited
                childCount++;
                dfs(neighbor, node, graph, low, disc, articulations);
                low[node] = min(low[node], low[neighbor]);

                if ((parent == -1 && childCount > 1) || (parent != -1 && low[neighbor] >= disc[node])) {
                    articulations.push_back(node);
                }
            } else if (neighbor != parent) { // ensure we do not update low[node] with parent
                low[node] = min(low[node], disc[neighbor]);
            }
        }
    }

    vector<int> findArticulationPoints(int root, unordered_map<int, vector<int>>& graph) {
        vector<int> articulations;
        unordered_map<int, int> low, disc;
        dfs(root, -1, graph, low, disc, articulations);
        return articulations;
    }

    int countIslands(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<bool>> visited(n, vector<bool>(m, false));
        int islandCount = 0;
        
        function<void(int, int)> dfs = [&](int i, int j) {
            visited[i][j] = true;
            pair<int, int> directions[] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
            for (auto dir : directions) {
                int x = i + dir.first;
                int y = j + dir.second;
                if (x >= 0 && x < n && y >= 0 && y < m && grid[x][y] == 1 && !visited[x][y]) {
                    dfs(x, y);
                }
            }
        };

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    ++islandCount;
                    dfs(i, j);
                }
            }
        }

        return islandCount;
    }

public:
    int minDays(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        
        // Step 1: Check if the grid is already disconnected (multiple islands)
        if (countIslands(grid) != 1) {
            return 0;
        }

        // Step 2: Convert grid into a graph
        unordered_map<int, vector<int>> graph;
        int root = -1;
        pair<int, int> directions[] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    int u = i * m + j;

                    if (root == -1)
                        root = u;

                    for (auto dir : directions) {
                        int x = i + dir.first;
                        int y = j + dir.second;
                        if (x >= 0 && x < n && y >= 0 && y < m && grid[x][y] == 1) {
                            int v = x * m + y;
                            graph[u].push_back(v);
                            graph[v].push_back(u);
                        }
                    }
                }
            }
        }

        // Step 3: Find articulation points
        vector<int> articulations = findArticulationPoints(root, graph);   

        // Special case: If there's only one land cell, it takes 1 day to disconnect.
        if (graph.size() == 1) {
            return 1;
        }

        // If there are no articulation points, it will take 2 days to disconnect
        if (articulations.empty()) {
            return 2;
        }

        // Otherwise, it can be done in 1 day
        return 1;
    }
};

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