<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 m = grid.size();
        int n = grid[0].size();
        vector<vector<int>> visited = grid;
        //making queue in which we will fill rotten oranges
        queue<pair<int, int>> q;
        int countFreshOrange = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (visited[i][j] == 2) {
                    q.push({i, j});
                }
                if (visited[i][j] == 1) {
                    countFreshOrange++;
                }
            }
        }
        //q.empty() means there is no rotten orange in the grid and countFreshOrange = 0 means we will rotten the freshoranges in 0 mins
        if (countFreshOrange == 0)
            return 0;
        if (q.empty())
            return -1;
        
        int minutes = -1;
        // we will cover four directions i.e. up, down, left, right
        vector<pair<int, int>> dirs = {{1, 0},{-1, 0},{0, -1},{0, 1}};
        while (!q.empty()) {
            int size = q.size();
            while (size--) {
                auto [x, y] = q.front();
                q.pop();
                for (auto [dx, dy] : dirs) {
                    int i = x + dx;
                    int j = y + dy;
                    if (i >= 0 && i < m && j >= 0 && j < n && visited[i][j] == 1) {
                        visited[i][j] = 2;
                        countFreshOrange--;
                        q.push({i, j});
                    }
                }
            }
            minutes++;
        }
        
        if (countFreshOrange == 0)
            return minutes;
        return -1;
    }
};
16500