Rotting Oranges

https://leetcode.com/problems/rotting-oranges/?envType=problem-list-v2&envId=7p5x763

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.

For Rotting Oranges, the cleanest solution is multi-source BFS (queue). Every rotten orange (2) is a starting point at time 0. Each minute is one BFS “layer”, and rotting spreads to 4-directional neighbors.

Why BFS is the right tool

• The problem asks for the minimum minutes → shortest “time” to reach all fresh oranges.
• Rot spreads simultaneously from all rotten oranges each minute → multi-source BFS models this exactly.

What’s going wrong in your rough approach

• You’re trying to “walk” up/down/left/right with while loops from a single (r, c). But rotting happens from all rotten oranges at once, not one at a time.
• If you mutate the grid while scanning it in the same minute, you accidentally allow oranges to rot multiple steps in one minute (chain reaction) unless you carefully separate “current minute” vs “next minute”. BFS handles that separation naturally.
• Many syntax issues: while r in range(...) isn’t valid, r-- isn’t Python, if r = 0 should be ==, and boundary checks like r <= len(grid) should be r < len(grid).

Correct BFS algorithm

1. Count all fresh oranges (fresh).
2. Push all rotten orange positions into a queue with minute 0.