<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 dfs(int i, int j, vector<vector<int>>& grid) {
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == 0) {
            return;
        }

        // Mark the cell as visited
        grid[i][j] = 0;

        dfs(i + 1, j, grid);
        dfs(i - 1, j, grid);
        dfs(i, j + 1, grid);
        dfs(i, j - 1, grid);
    }

    int countIslands(vector<vector<int>>& grid) {
        int count = 0;
        vector<vector<int>> temp = grid;  // Copy the grid

        for (int i = 0; i < temp.size(); i++) {
            for (int j = 0; j < temp[0].size(); j++) {
                if (temp[i][j] == 1) {
                    dfs(i, j, temp);
                    count++;
                }
            }
        }
        return count;
    }

    int minDays(vector<vector<int>>& grid) {
        // Step 1: Check initial number of islands
        vector<vector<int>> temp = grid;
        int numOfIslands =countIslands(temp);
        if (numOfIslands == 0) return 0;
        if ( numOfIslands >= 2) return 0;  // Already more than 1 island

        // Step 2: Try changing each land cell (1) to water (0) and check if it splits into multiple islands
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == 1) {
                    // Temporarily change the cell to water
                    temp = grid;  // Reset the grid before each modification
                    temp[i][j] = 0;

                    // Check if the modified grid results in more than 1 island or 0 islands
                    int islands = countIslands(temp);
                    if (islands >= 2 || islands == 0) {
                        return 1;  // One cell change results in disconnection
                    }
                }
            }
        }

        // Step 3: If no single cell change disconnects the grid, return 2
        return 2;  // Max 2 days required
    }
};

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