Rotting Oranges

You are given a rectangular grid whose cells contain 0 for empty, 1 for a fresh orange, or 2 for a rotten orange.
For this exercise, assume every minute each fresh orange orthogonally adjacent to a rotten orange becomes rotten, and all changes in one minute happen simultaneously.
Return the minimum minutes until no fresh orange remains. Return -1 if that is impossible. Return 0 when no fresh orange exists initially.

Function
orangesRotting(grid: int[][]) → int

Examples
Example 1
grid = [[2,1,1],[1,1,0],[0,1,1]]
return = 4
The final fresh orange is reached by the fourth simultaneous breadth-first layer.
Example 2
grid = [[2,1,1],[0,1,1],[1,0,1]]
return = -1
The lower-left fresh orange is disconnected from every rotten orange.
Example 3
grid = [[0,2]]
return = 0
No fresh orange exists.

Constraints
1 <= grid.length <= 500.
1 <= grid[row].length <= 500.
Every row has the same length.
Each cell is 0, 1, or 2.

The trick is multi-source BFS. Scan the grid once, push every rotten cell into a queue, and count the fresh ones. Then process the queue level by level. Each level is one minute. When a fresh neighbor turns rotten, decrement the fresh counter and enqueue it. The pitfall is the minute count. If you increment per cell instead of per level, your answer is way too big. Also don't increment after the last level that adds nothing, or you're off by one. At the end, if fresh is still above zero, return -1. If fresh was zero at the start, return 0 before the loop. Mutate the grid in place to mark visited, so you skip a separate set. Time is O(rows times cols). If you freeze on the level-counting logic during the live OA, StealthCoder is the safety net that shows you the clean version.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.