House Robber II
House Robber, but the houses stand in a circle, so the first and last house are neighbours.
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Lesson notes
Try it before you watch
Restate the problem in your own words, list the edge cases, and sketch a solution with its running time. Then play the video and compare.
Reveal the key idea
The first and last house cannot both be robbed, so solve the linear House Robber problem twice (without the first house and without the last) and take the better result.
Pattern: Dynamic programming. Define a state, write the recurrence between states, and compute each state only once.
Complexity
| Measure | Bound |
|---|---|
| Time | O(n) |
| Extra space | O(1) |
Walkthroughs often start from a simpler approach first; aim to reach these bounds. New to Big-O? Read understanding algorithmic complexity.