You are given an `m x n` grid representing kitchen cells. Each cell holds one of three values: - `0` — an empty cell, - `1` — a cell with a fresh fruit, - `2` — a cell with a spoiled fruit. Each minute, any fresh fruit that is **4-directionally adjacent** (up, down, left, right) to a spoiled fruit becomes spoiled. Return the **minimum number of minutes** that must elapse until no cell has a fresh fruit. If this is impossible (some fresh fruit can never be reached), return `-1`. **Approach (multi-source BFS):** Treat every initially-spoiled cell as a BFS source and push them all into a queue. Count the number of fresh fruits. Process the queue layer by layer; each completed layer corresponds to one elapsed minute. When a spoiled cell spoils a fresh neighbor, decrement the fresh count and enqueue the newly-spoiled cell. Stop when the queue is empty. If any fresh fruit remains, return `-1`; otherwise return the number of layers processed. **Correctness (high level):** BFS from all initial sources simultaneously expands the spoiled region one ring at a time, so a fruit is reached at exactly the minute equal to its shortest grid-distance to the nearest initial source. The number of completed expansion layers therefore equals the time at which the last reachable fruit spoils. Any fresh fruit never dequeued is in no connected component containing a source, so it can never spoil, giving `-1`. **Complexity:** Time `O(m·n)` — each cell is enqueued and processed at most once. Space `O(m·n)` for the queue in the worst case.