<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) {
        if (grid.size() == 0) {
            return 0;
        }

        int fresh = 0;
        int row = grid.size();
        int col = grid[0].size();
        int time = 0;
        vector<int> dirX = { -1, 0, 1, 0 };
        vector<int> dirY = { 0, 1, 0, -1 };

        queue<int> qX;
        queue<int> qY;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (grid[i][j] == 1) {
                    fresh++;
                }
                else if (grid[i][j] == 2) {
                    qX.push(i);
                    qY.push(j);
                }
            }
        }
        if (fresh == 0) {
            return 0;
        }
        while (!qX.empty() && fresh != 0) {
            int size = qX.size();
            for (int i = 0; i < size; i++) {
                int x = qX.front();
                qX.pop();
                int y = qY.front();
                qY.pop();
                for (int d = 0; d < 4; d++) {
                    int X = x + dirX[d];
                    int Y = y + dirY[d];
                    if (X >= 0 && X < row && Y >= 0 && Y < col && grid[X][Y] == 1) {
                        grid[X][Y] = 2;
                        fresh--;
                        qX.push(X);
                        qY.push(Y);
                    }
                }
            }
            time++;
        }
        if (fresh != 0) {
            return -1;
        }
        return time;
    }
   
};
16600