House Robber II

Problem Description

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night. Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Key Insights

• The problem is a variation of the "House Robber" problem where the houses are arranged in a circle.
• We cannot rob both the first and last house due to their adjacency.
• The solution involves considering two separate cases:
  1. Rob houses from the first to the second-to-last (excluding the last house).
  2. Rob houses from the second to the last (excluding the first house).
• The final answer will be the maximum amount obtained from the two cases.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

To solve the problem, we can use dynamic programming. The idea is to use two helper functions that compute the maximum amount of money that can be robbed from a linear arrangement of houses. We will then call these functions twice:

1. Once excluding the last house.
2. Once excluding the first house. We will return the maximum of the two results.

The dynamic programming approach involves maintaining a running total of the maximum amounts that can be robbed up to each house, ensuring that no two adjacent houses are robbed.

Code Solutions