MEDIUM NC#102 Blind #23 Dynamic Programming 1D
213. House Robber II
📖 Problem
Houses are arranged in a circle (first and last are adjacent). Return the maximum amount of money you can rob without alerting the police.
🧠 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
- •Array DP table updates
- •State transition thinking
- •Base case initialization
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Dynamic Programming reasoning pattern
- •Apply Recursion reasoning pattern
💡 Approach
- → Two cases: rob houses 0 to n-2, or rob houses 1 to n-1
- → Use same logic as House Robber I for each case
- → Return max of both cases
- → Time: O(n), Space: O(1)
🧭 Prerequisites
🛠️ Hints & Pitfalls
Hints
- •Two cases: rob houses 0 to n-2, or rob houses 1 to n-1
- •Use same logic as House Robber I for each case
- •Return max of both cases
Common Pitfalls
- •Time: O(n), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
rob2([2,3,2]); Expected:
3 Test Case 2
Not run Input:
rob2([1,2,3,1]); Expected:
4 Test Case 3
Not run Input:
rob2([1,2,3]); Expected:
3 📝 Code Editor
📤 Output