# 213. House Robber II

## Medium

***

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***.

**Example 1:**

```
Input: nums = [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
```

**Example 2:**

```
Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
```

**Example 3:**

```
Input: nums = [1,2,3]
Output: 3
```

**Constraints:**

* `1 <= nums.length <= 100`
* `0 <= nums[i] <= 1000`

```python
class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0
        elif len(nums) == 1:
            return nums[0]
        elif len(nums) == 2:
            return max(nums[0], nums[1])
        total = len(nums)
        return max(self.robHouse1(nums[:total-1]), self.robHouse1(nums[1:total]))
        
    # Here I'm using my PB 198: House Robber Solution
    def robHouse1(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]
        elif len(nums) == 2:
            return max(nums[0], nums[1])
        length = len(nums)
        arr = [0]*len(nums)
        arr[0] = nums[0]
        arr[1] = nums[1]
        arr[2] = nums[0] + nums[2]
        for index in range(3, len(nums)):
            arr[index] = max(arr[index-3], arr[index-2]) + nums[index]
        return max(arr[length-1], arr[length-2])
        
```