<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:

    vector<pair<int,int>> dir;

    void dfs(int i, int j, vector<vector<int>>& grid, vector<vector<bool>>& vis, int& n, int& m) {
        vis[i][j] = true;

        for(int k=0; k<4; k++) {
            int x = i + dir[k].first, y = j + dir[k].second;
            if(x >= 0 and y >= 0 and x < n and y < m and grid[x][y] == 1 and vis[x][y] == false) {
                dfs(x,y,grid,vis,n,m);
            }
        }
    }

    int t = 0; // timer
    void dfs2(int i, int j, vector<vector<bool>>& vis, vector<vector<pair<int,int>>>& parent, vector<vector<int>>& firstTime, vector<vector<int>>& lowestTime, vector<vector<int>>& grid, bool& articulation, int& n, int& m) {
        t++;
        vis[i][j] = true;
        firstTime[i][j] = t;
        lowestTime[i][j] = t;

        int subgraphCnt = 0;

        for(int k=0; k<4; k++) {
            int x = i + dir[k].first, y = j + dir[k].second;
            if(x >= 0 and y >= 0 and x < n and y < m and grid[x][y] == 1) {

                if(vis[x][y] == false) {
                    parent[x][y] = {i,j};
                    subgraphCnt++;
                    dfs2(x, y, vis, parent, firstTime, lowestTime, grid, articulation, n, m);

                    // update lowest time
                    lowestTime[i][j] = min(lowestTime[i][j], lowestTime[x][y]);

                    if(parent[i][j].first == -1 and parent[i][j].second == -1 and subgraphCnt > 1) { // condition 1
                        
                        articulation = true;
                    }

                    if(parent[i][j].first != -1 and parent[i][j].second != -1 and firstTime[i][j] <= lowestTime[x][y]) { // condition 2
                        articulation = true;
                    }
                }
                
                if(x != parent[i][j].first or y != parent[i][j].second) { // YAHA OR LAGEGA
                    lowestTime[i][j] = min(lowestTime[i][j], firstTime[x][y]);
                }
                
            }
            
        }
    }

    int minDays(vector<vector<int>>& grid) {
        
        dir.resize(4);
        dir[0] = {1,0};
        dir[1] = {-1,0};
        dir[2] = {0,1};
        dir[3] = {0,-1};

        int n = grid.size(), m = grid[0].size();
        vector<vector<bool>> vis(n, vector<bool>(m, false));

        int components = 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(grid[i][j] == 1 and vis[i][j] == false) {
                    components++;
                    dfs(i,j,grid,vis,n,m);
                }
            }
        }

        if(components != 1) {
            // either no islands or already disconnected
            return 0;
        }
        
        // components == 1. and size of component == 1 then answer 1
        if(cnt1 == 1) {
            return 1;
        }

        // components == 1 and If there is an articulation point then ans = 1
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                vis[i][j] = false;
            }
        }


        vector<vector<pair<int,int>>> parent(n, vector<pair<int,int>>(m, {-1,-1}));
        vector<vector<int>> firstTime(n, vector<int>(m));
        vector<vector<int>> lowestTime(n, vector<int>(m));
        bool articulation = false; // as we need to find atleast 1 AP

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(grid[i][j] == 1 and vis[i][j] == false) {
                    dfs2(i,j,vis,parent,firstTime, lowestTime, grid, articulation, n, m);
                }
            }
        }

        if(articulation == true) {
            return 1;
        }

        // any graph can be disconnected by removing 2 nodes
        return 2;
    }
};
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