<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) {
        if (count_islands(grid) != 1) {
            return 0;
        }
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 0) {
                    continue;
                }
                grid[i][j] = 0;
                int islands_cnt = count_islands(grid);
                grid[i][j] = 1;
                if (islands_cnt != 1) {
                    return 1;
                }
            }
        }
        return 2;
    }

private:
    int count_islands(const vector<vector<int>>& grid) {
        vector<vector<int>> lookup(grid.size(), vector<int>(grid[0].size()));
        int islands_cnt = 0;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 0 || lookup[i][j]) {
                    continue;
                }
                ++islands_cnt;
                floodfill(grid, i, j, &lookup);
            }
        }
        return islands_cnt;
    }

    void floodfill(const vector<vector<int>>& grid, int i, int j, vector<vector<int>> *lookup) {
        static const vector<pair<int, int>> directions{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        vector<pair<int, int>> stk = {{i, j}};
        (*lookup)[i][j] = 1;
        while (!stk.empty()) {
            auto [i, j] = stk.back(); stk.pop_back();
            for (const auto& [di, dj] : directions) {
                const auto& [ni, nj] = pair(i + di, j + dj);
                if (!(0 <= ni && ni < grid.size() &&
                      0 <= nj && nj < grid[0].size() &&
                      grid[ni][nj] &&
                      !(*lookup)[ni][nj])) {
                    continue;
                }
                (*lookup)[ni][nj] = 1;
                stk.emplace_back(ni, nj);
            }
        }
    }
};

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