<p>You are given an <code>m x n</code> <code>grid</code> where each cell can have one of three values:</p>

<ul>
	<li><code>0</code> representing an empty cell,</li>
	<li><code>1</code> representing a fresh orange, or</li>
	<li><code>2</code> representing a rotten orange.</li>
</ul>

<p>Every minute, any fresh orange that is <strong>4-directionally adjacent</strong> to a rotten orange becomes rotten.</p>

<p>Return <em>the minimum number of minutes that must elapse until no cell has a fresh orange</em>. If <em>this is impossible, return</em> <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/16/oranges.png" style="width: 650px; height: 137px;" />
<pre>
<strong>Input:</strong> grid = [[2,1,1],[1,1,0],[0,1,1]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[2,1,1],[0,1,1],[1,0,1]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,2]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Since there are already no fresh oranges at minute 0, the answer is just 0.
</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;= 10</code></li>
	<li><code>grid[i][j]</code> is <code>0</code>, <code>1</code>, or <code>2</code>.</li>
</ul>

class Solution {
public:
    int orangesRotting(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        
        bool rotten = true; 
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(grid[i][j] == 1){
                        rotten = false;
                    }
            }
            cout << endl;
        }
        if(rotten){
            return 0;
        }

        for(int time = 2; time < n * m + 1; time++){
            cout << time << endl;
            for(int i = 0; i < n; i++){
                for(int j = 0; j < m; j++){
                    cout << grid[i][j] << " ";
                }
                cout << endl;
            }
            cout << endl;
            for(int i = 0; i < n; i++){
                for(int j = 0; j < m; j++){
                    if(grid[i][j] == time){
                        if(i > 0 && grid[i - 1][j] == 1){
                            grid[i - 1][j] = time + 1;
                        }
                        if(i < n - 1 && grid[i + 1][j] == 1){
                            grid[i + 1][j] = time + 1;
                        }
                        if(j > 0 && grid[i][j - 1] == 1){
                            grid[i][j - 1] = time + 1;
                        }
                        if(j < m - 1 && grid[i][j + 1] == 1){
                            grid[i][j + 1] = time + 1;
                        }
                    }
                }
            }
            bool fresh = false;
            for(int i = 0; i < n; i++){
                for(int j = 0; j < m; j++){
                    if(grid[i][j] == 1){
                        fresh = true;
                    }
                }
            }
            if(!fresh){
                return time - 1;
            }
        }
        return -1;
    }
};
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