<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 {
    void DFS(int i, int j, int n, int m, vector <vector <int>>& grid) {
        if(i < 0 or j < 0 or i >= n or j >= m or grid[i][j] == 0) {
            return ;
        }
        grid[i][j] = 0;
        DFS(i+1, j, n, m, grid);
        DFS(i, j+1, n, m, grid);
        DFS(i, j-1, n, m, grid);
        DFS(i-1, j, n, m, grid);
    }
    int countNumberOfIslands(int first, int second, int n, int m, vector <vector <int>> grid) {
        grid[first][second] = 0;
        // cout << first << "-" << second << ";";
        int cnt = 0;
         for(int i = 0; i<n; i++) {
            for(int j = 0; j<m; j++) {
                if(grid[i][j] == 1) {
                    DFS(i, j, n, m, grid);
                    cnt++;
                }
            }
         }
         return cnt;
    }
public:
    int minDays(vector<vector<int>>& grid) {
        // check number of islands.
        // If more than one islands, then return 0;
        // If exactly one island is available, make that disconnected in minimum number of moves.
        // How to disconnect ?
        int n = grid.size(), m = grid[0].size(), count = 0;
        auto tempGrid = grid;
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<m; j++) {
                if(tempGrid[i][j] == 1) {
                    count++;
                    DFS(i, j, n, m, tempGrid);
                }
            }
        }
        if(count > 1 or count == 0) return 0;
        // Check if we can do it in 1 day
        tempGrid = grid;
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<m; j++) {
                if(tempGrid[i][j] == 1) {
                    if(countNumberOfIslands(i, j, n, m, tempGrid) > 1 or countNumberOfIslands(i, j, n, m, tempGrid) == 0) return 1;
                }
            }
        }
        return 2;
    }
};
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