# Rotting Oranges Problem

## Description

LeetCode Problem 994.

You are given an m x n grid where each cell can have one of three values:

• 0 representing an empty cell,
• 1 representing a fresh orange, or
• 2 representing a rotten orange.

Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1.

Example 1:

``````1
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Input: grid = [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
``````

Example 2:

``````1
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Input: grid = [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
``````

Example 3:

``````1
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Input: grid = [[0,2]]
Output: 0
Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
``````

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 10
• grid[i][j] is 0, 1, or 2.

## Sample C++ Code

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class Solution {
public:
int orangesRotting(vector<vector<int>>& grid) {
queue<pair<int,int>> bfs;
int count_zeroes = 0;
int rows = grid.size();
int cols = grid[0].size();
for (int i = 0; i < grid.size(); i++) {
for (int j = 0; j < grid[0].size(); j++) {
if (grid[i][j] == 2) {
bfs.push({i, j});
}
else if (grid[i][j] == 0) {
++count_zeroes;
}
}
}
int oranges = (rows * cols) - count_zeroes;
int num_el = bfs.size();
int minutes = 0;
while (num_el < oranges && !bfs.empty()) {
++minutes;
int n = bfs.size();
for (int i = 0; i < n; i++) {
int r = bfs.front().first;
int c = bfs.front().second;
bfs.pop();
if (r > 0 && grid[r - 1][c] == 1) {
bfs.push({r - 1, c});
grid[r - 1][c] = 2;
}
if (r < rows - 1 && grid[r + 1][c] == 1) {
bfs.push({r + 1, c});
grid[r + 1][c] = 2;
}
if (c > 0 && grid[r][c-1] == 1) {
bfs.push({r, c - 1});
grid[r][c - 1] = 2;
}
if (c < cols - 1 && grid[r][c + 1] == 1) {
bfs.push({r, c + 1});
grid[r][c + 1] = 2;
}
}
num_el += bfs.size();
if (num_el == oranges)
return minutes;
}
if (num_el < oranges)
return -1;
return minutes;
}
};
``````