<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:
    vector<pair<int, int>> getNeighbors(const vector<vector<int>>& grid, int row, int col) {
        vector<pair<int, int>> neighbors;

        if (row > 0) neighbors.emplace_back(row - 1, col);
        if (row < grid.size() - 1) neighbors.emplace_back(row + 1, col);
        if (col > 0) neighbors.emplace_back(row, col - 1);
        if (col < grid[0].size() - 1) neighbors.emplace_back(row, col + 1);

        return neighbors;
    }

    int orangesRotting(vector<vector<int>>& grid) {
        queue<pair<int, int>> traversal;

        for (int row = 0; row < grid.size(); row++) {
            for (int col = 0; col < grid[0].size(); col++) {
                if (grid[row][col] == 2) traversal.emplace(row, col);
            }
        }

        int counter = -1;

        while (!traversal.empty()) {
            int level_size = traversal.size();

            for (int i = 0; i < level_size; i++) {
                pair<int, int> front = traversal.front();
                traversal.pop();

                for (const pair<int, int>& neighbor : getNeighbors(grid, front.first, front.second)) {
                    if (grid[neighbor.first][neighbor.second] != 1) continue;
                    traversal.push(neighbor);
                    grid[neighbor.first][neighbor.second] = 2;
                }

            }

            counter++;
        }

        for (int row = 0; row < grid.size(); row++) {
            for (int col = 0; col < grid[0].size(); col++) {
                if (grid[row][col] == 1) return -1;
            }
        }

        return counter < 0 ? 0 : counter;
    }
};
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