<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>

int visit(vector<vector<int>>& arr, vector<vector<bool>>& visited, int i,
          int j) {

    if (i < 0 || i >= arr.size())
        return 0;
    if (j < 0 || j >= arr[0].size())
        return 0;

    if (visited[i][j] == true)
        return 0;
    if (arr[i][j] == 0)
        return 0;

    visited[i][j] = true;

    int cnt = 1;

    cnt = cnt + visit(arr, visited, i + 1, j);
    cnt = cnt + visit(arr, visited, i - 1, j);
    cnt = cnt + visit(arr, visited, i, j + 1);
    cnt = cnt + visit(arr, visited, i, j - 1);

    return cnt;
}

class Solution {
public:
    int minDays(vector<vector<int>>& arr) {

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

        vector<bool> v(m, false);

        vector<pair<int, int>> prr;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (arr[i][j] == 1)
                    prr.push_back(make_pair(i, j));
            }
        }

        int island = 0;
        vector<vector<bool>> visited(n, v);

        // check if already there are more than one island
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {

                if (arr[i][j] == 1 && visited[i][j] == false) {
                    visit(arr, visited, i, j);
                    island++;
                }
            }
        }

        if (island > 1 || island == 0 ) return 0;

        for (auto p : prr) {

            int x = p.first;
            int y = p.second;

            arr[x][y] = 0;
            island = 0;

            vector<bool> v1(m, false);
            vector<vector<bool>> visited1(n, v1);

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {

                    if (arr[i][j] == 1 && visited1[i][j] == false) {
                        visit(arr, visited1, i, j);
                        island++;
                    }
                }
            }

            if (island > 1 || island == 0 )
                return 1;

            arr[x][y] = 1;
        }

        return 2;
    }
};
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