<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 row = grid.size();
        int col = grid[0].size();
        vector<vector<int>> directions  = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        queue<pair<int, int>> oranges;
        int freshOnes = 0;
        int minMints = -1;

        for (int i = 0; i < row; ++i)
        {
            for (int j = 0; j < col; ++j)
            {
                if (grid[i][j] == 1) 
                {
                    freshOnes++;
                }
                else if (grid[i][j] == 2)
                {
                    oranges.push({i, j});
                }
            }
        }

        oranges.push({-1, -1});
        while(!oranges.empty())
        {
            pair<int, int> loc = oranges.front();
            oranges.pop();
            int x = loc.first;
            int y = loc.second;

             if (x == -1)
            {
                minMints++;
                if (!oranges.empty())
                {
                    oranges.push({-1, -1});
                }
            }
            else 
            {
                for (auto dir : directions)
                {
                    int nextX = x + dir[0];
                    int nextY = y + dir[1];
                    if (nextX >= 0 && nextX < row && nextY >=0 && nextY < col)
                    {
                        
                        if (grid[nextX][nextY] == 1) 
                        {
                            grid[nextX][nextY] = 2;
                            freshOnes--;
                            oranges.push({nextX, nextY});
                        }
                    }
                }
            }

            
            
        }
        return (freshOnes ==0) ? minMints : -1;
    }
};
17800