<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:
    int minDays(vector<vector<int>>& grid) {
           ios_base::sync_with_stdio(false);
        cin.tie(nullptr); cout.tie(nullptr);
        int rows = grid.size();
        int cols = grid[0].size();
        int dx[4] = {0, 0, 1, -1};
        int dy[4] = {1, -1, 0, 0};
        int initialIslandCount = countIslands(grid, rows, cols, dx, dy);
        if (initialIslandCount != 1) {
            return 0;
        }
        for (int row = 0; row < rows; row++) {
            for (int col = 0; col < cols; col++) {
                if (grid[row][col] == 0) continue;  


                grid[row][col] = 0;
                int newIslandCount = countIslands(grid, rows, cols, dx, dy);


                if (newIslandCount != 1) return 1;


                grid[row][col] = 1;
            }
        }

        return 2;
    }

private:
    int countIslands(vector<vector<int>>& grid, int rows, int cols, int dx[], int dy[]) {
           ios_base::sync_with_stdio(false);
        cin.tie(nullptr); cout.tie(nullptr);
        vector<vector<int>> visited(rows, vector<int>(cols, 0));
     
        int islandCount = 0;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    islandCount++;
                    queue<pair<int, int>> q1;
                    q1.push({i, j});

                    while (!q1.empty()) {
                        auto it = q1.front();
                        q1.pop();
                        int x = it.first;
                        int y = it.second;

                        if (visited[x][y]) continue;
                        visited[x][y] = 1;

                        for (int k = 0; k < 4; k++) {
                            int nx = x + dx[k];
                            int ny = y + dy[k];

                            if (nx >= 0 && ny >= 0 && nx < rows && ny < cols &&
                                grid[nx][ny] == 1 && !visited[nx][ny]) {
                                q1.push({nx, ny});
                            }
                        }
                    }
                }
            }
        }
        return islandCount;
    }
};

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