<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();

        queue<pair<int,int>> q;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j] == 2){
                    q.push({i+1,j});
                    q.push({i,j+1});
                    q.push({i-1,j});
                    q.push({i,j-1});
                }
            }
        }

        int time=0;

        while(!q.empty()){
            int total = q.size();
            time++;

            while(total--){
                auto current = q.front();
                q.pop();
                int x = current.first;
                int y = current.second;

                if(x>=0 && y>=0 && x<n && y<m && grid[x][y] == 1){
                    q.push({x+1,y});
                    q.push({x,y+1});
                    q.push({x,y-1});
                    q.push({x-1,y});

                    grid[x][y]=2;
                }
            }
        }

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

        if(time == 0){
            return 0;
        }
        return time-1;
    }
};
16600