<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:
    pair<int, int> next[4] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    void dfs(int i, int j, vector<vector<int>>& vis, vector<vector<int>>& a) {
        vis[i][j] = 1;
        for (int k = 0; k < 4; k++) {
            int newi = i + next[k].first;
            int newj = j + next[k].second;
            if (newi >= 0 && newi < a.size() && newj >= 0 &&
                newj < a[0].size() && !vis[newi][newj] && a[newi][newj] == 1) {
                dfs(newi, newj, vis, a);
            }
        }
    }
    int caller(vector<vector<int>>& a) {
        int n = a.size();
        int m = a[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));
        int c = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) {
                if (!vis[i][j] && a[i][j] == 1) {
                    c++;
                    dfs(i, j, vis, a);
                }
            }
        return c;
    }
    int minDays(vector<vector<int>>& a) {
        int n = a.size();
        int m = a[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));
        int c = 0;
        int s = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) {
                if (!vis[i][j] && a[i][j] == 1) {
                    c++;
                    dfs(i, j, vis, a);
                }
                if (a[i][j] == 1)
                    s++;
            }
        if (c != 1)
            return 0;
        if (s <= 2)
            return s;
        int ans = 2;
        for (int i = 0; i < a.size(); i++)
            for (int j = 0; j < m; j++) {
                if (a[i][j] == 1) {
                    int c = 0;
                    if (i > 0 && a[i - 1][j])
                        c++;
                    if (i < n - 1 && a[i + 1][j])
                        c++;
                    if (j > 0 && a[i][j - 1])
                        c++;
                    if (j < m - 1 && a[i][j + 1])
                        c++;
                    ans = min(ans, c);
                }
            }
        if (ans == 1)
            return 1;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) {
                if(a[i][j]==1)
                {
                    a[i][j]=0;
                    if(caller(a)!=1)
                    return 1;
                    a[i][j]=1;
                }
            }
        return ans;
    }
};
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