<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:
    // Brute Force
    void bfs(int n, int m, vector<vector<int>> &grid, int i, int j, vector<vector<bool>> &visited){
        queue<int> pos;
        pos.push(i * m + j);
        visited[i][j] = true;

        while(!pos.empty()){
            int size = pos.size();
            while(size--){
                int node = pos.front();
                pos.pop();

                int row = node / m;
                int col = node % m;

                int xPos[4] = {1, -1, 0, 0};
                int yPos[4] = {0, 0, 1, -1};

                for(int k=0; k<4; k++){
                    int x = xPos[k] + row;
                    int y = yPos[k] + col;

                    if(x >= 0 && x < n && y >= 0 && y < m && !visited[x][y] && grid[x][y] == 1){
                        pos.push(x * m + y);
                        visited[x][y] = true;
                    }
                }
            }
        }
    }


    int findIsland(int n, int m, vector<vector<int>> &grid){

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

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

        return island;
    }

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

        int island = findIsland(n, m, grid);
        if(island != 1)   return 0;

        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(grid[i][j] == 1){
                    grid[i][j] = 0;

                    int island = findIsland(n, m, grid);
                    if(island != 1)   return 1;

                    grid[i][j] = 1;
                }       
            }
        }

        return 2;
    }
};
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