MEDIUM NC#85 Graphs / BFS

994. Rotting Oranges

📖 Problem

You are given an m x n grid where each cell has one of three values: - 0 represents an empty cell - 1 represents a fresh orange - 2 represents 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.

🧠 Visual Learning Aid

1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases

JS/TS Refreshers

  • Array methods (`push`, `pop`, `shift`, `slice`)
  • Object/Map/Set usage patterns
  • Function parameter and return typing
  • Level-order traversal
  • Queue discipline
  • Shortest-step interpretation

Logical Thinking Concepts

  • Define invariants before coding
  • Check edge cases first (`[]`, single element, duplicates)
  • Estimate time/space before implementation
  • Apply BFS reasoning pattern

💡 Approach

  • Use BFS for level-by-level processing
  • Add all initially rotten oranges to queue
  • Process level by level, marking adjacent fresh oranges
  • Track minutes elapsed
  • Stop when no fresh oranges remain
  • Time: O(m * n), Space: O(m * n)

🛠️ Hints & Pitfalls

Hints

  • Use BFS for level-by-level processing
  • Add all initially rotten oranges to queue
  • Process level by level, marking adjacent fresh oranges

Common Pitfalls

  • Track minutes elapsed
  • Stop when no fresh oranges remain
  • Time: O(m * n), Space: O(m * n)

🧪 Test Cases

Hidden tests on submit: 1

Test Case 1
Not run
Input:
orangesRotting([[0,2]]);
Expected:
-1
Test Case 2
Not run
Input:
orangesRotting(grid: number[][]);
Expected:
Computed from hidden reference
Test Case 3
Not run
Input:
orangesRotting([ [2,1,1], [1,1,0], [1,0,2], [0,2,1] ]);
Expected:
Computed from hidden reference

📝 Code Editor

📚 Reference Solution

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