<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:
    void bfs(int i,int j,vector<vector<int>>& nums,vector<vector<int>>& visited,vector<vector<int>>& ways){
        int m=nums.size(),n=nums[0].size();

        queue<pair<int,int>> q;

        int x[]={-1,0,1,0};
        int y[]={0,1,0,-1};
        q.push({i,j});
        visited[i][j]=1;
        while(!q.empty()){
            int r=q.front().first;
            int c=q.front().second;
            q.pop();

            for(int i=0;i<4;i++){
                int nr=r+x[i],nc=c+y[i];
                if(nr>=0 and nr<m and nc>=0 and nc<n and nums[nr][nc]==1){
                    if(!visited[nr][nc]){
                        visited[nr][nc]=1;
                        q.push({nr,nc});
                    }
                    // ways[r][c]++;
                    ways[nr][nc]++;
                }
            }
        }
    }
    int count_islands(vector<vector<int>>& nums){
        int m=nums.size(),n=nums[0].size();

        vector<vector<int>> ways(m,vector<int>(n,0)),visited(m,vector<int>(n,0));
        int cnt=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                // int node_no=i*n+j;
                if(nums[i][j]==1 and !visited[i][j]){
                    cnt++;
                    bfs(i,j,nums,visited,ways);
                }
            }
        }

        return cnt;
    }
    int minDays(vector<vector<int>>& nums) {
        int m=nums.size(),n=nums[0].size();
        int initial_cnt=count_islands(nums);

        if(initial_cnt!=1){
            return 0;
        }

        //check for 1
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(nums[i][j]==1){
                    nums[i][j]=0;
                    int cnt= count_islands(nums);
                    nums[i][j]=1;
                    if(cnt!=1){
                        return 1;
                    }
                }
            }
        }

        return 2;
    }
};
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