<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:
    const int dx[4] = {-1, 1, 0, 0};
    const int dy[4] = {0, 0, -1, 1};

    int minDays(vector<vector<int>>& grid) {
        if (numOfIslands(grid) != 1) {
            return 0;
        }        

        int n = grid.size();
        int m = grid[0].size();

        // 1 day
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (grid[i][j] == 1){
                    grid[i][j] = 0;
                    if (numOfIslands(grid) != 1) {
                        return 1;
                    }
                    grid[i][j] = 1;
                }
            }
        }


        return 2;
    }

    int numOfIslands(vector<vector<int>> grid) {
        int n = grid.size();
        int m = grid[0].size();

        vector<vector<bool>> visited(n, vector<bool> (m));

        auto bfs = [&](int su, int sv) -> void {
            queue<pair<int, int>> que;
            
            que.push({su, sv});
            visited[su][sv] = true;

            while (!que.empty()) {
                auto [u, v] = que.front();
                que.pop();
                
                for (int i = 0; i < 4; ++i) {
                    int x = u + dx[i];
                    int y = v + dy[i];

                    if (x < 0 || y < 0 || x == n || y == m) {
                        continue;
                    }

                    if (grid[x][y] == 0 || visited[x][y]) {
                        continue;
                    }

                    visited[x][y] = true;
                    que.push({x, y});
                }            
            }
        };

        int cntIslands = 0;
        for (int i = 0; i < n; ++i) { 
            for (int j = 0; j < m; ++j) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    bfs(i, j);
                    cntIslands++;
                }
            }
        }

        return cntIslands;

    }
};
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