<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>

#include <vector>
#include <algorithm>
#include <numeric>

using namespace std;

class Solution { 
    int time; // For keeping track of discovery times
    bool hasArticulationPoint;

    void dfs(vector<vector<int>>& grid, int x, int y) {
        int n = grid.size(), m = grid[0].size();
        vector<int> dx = {1, -1, 0, 0};
        vector<int> dy = {0, 0, 1, -1};
        if (x < 0 || y < 0 || x >= n || y >= m || grid[x][y] == 0) return;
        grid[x][y] = 0;
        for (int d = 0; d < 4; d++) dfs(grid, x + dx[d], y + dy[d]);
    }

    bool checkConnected(vector<vector<int>> grid) {
        int components = 0;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j]) {
                    components++;
                    if (components > 1) return false; // More than one connected component
                    dfs(grid, i, j);
                }
            }
        }
        return components == 1;
    }

    void dfs2(vector<vector<int>>& grid, vector<vector<int>>& disc, vector<vector<int>>& low, vector<vector<int>>& parent, int x, int y) {
        disc[x][y] = low[x][y] = ++time;
        int children = 0;
        int n = grid.size(), m = grid[0].size();
        vector<int> dx = {1, -1, 0, 0};
        vector<int> dy = {0, 0, 1, -1};

        for (int d = 0; d < 4; d++) {
            int newX = x + dx[d], newY = y + dy[d];
            if (newX < 0 || newY < 0 || newX >= n || newY >= m || grid[newX][newY] == 0) continue;
            if (newX == parent[x][y] && newY == parent[x][y]) continue; // Skip the parent node

            if (disc[newX][newY] == -1) { // If not visited
                parent[newX][newY] = x * m + y; // Encode the parent cell as a single integer
                children++;
                dfs2(grid, disc, low, parent, newX, newY);

                low[x][y] = min(low[x][y], low[newX][newY]);

                // Check if the current node is an articulation point
                if (parent[x][y] == -1 && children > 1) {
                    hasArticulationPoint = true; // Root node with more than one child
                }
                if (parent[x][y] != -1 && low[newX][newY] >= disc[x][y]) {
                    hasArticulationPoint = true; // Non-root node condition
                }
            } else {
                low[x][y] = min(low[x][y], disc[newX][newY]); // Update low-link value
            }
        }
    }

    bool hasArticulationPointInGraph(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector<int>> disc(n, vector<int>(m, -1));
        vector<vector<int>> low(n, vector<int>(m, -1));
        vector<vector<int>> parent(n, vector<int>(m, -1));
        time = 0;
        hasArticulationPoint = false;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] && disc[i][j] == -1) {
                    dfs2(grid, disc, low, parent, i, j);
                    if (hasArticulationPoint) return true;
                }
            }
        }
        return hasArticulationPoint;
    }

public:
    int minDays(vector<vector<int>>& grid) { 
        if (!checkConnected(grid)) return 0;
        int cnt = 0;
        for (auto &curr : grid) cnt += accumulate(curr.begin(), curr.end(), 0);
        if (cnt <= 2) return cnt;

        if (hasArticulationPointInGraph(grid)) return 1;

        // If no articulation point, return 2 because at least two days are needed.
        return 2;
    }
};

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