House robber II

We can calculate the house robber using the same logic as the problem House robber, but we can't take the last element as it will create a circle. So, we can actually take 2 sets of numbers, one will be from start to 1 item before last, and another set will be the from second item till last item. Then we will return the maximum of these 2 set as result.

class Solution:
    def rob(self, nums: List[int]) -> int:
        def countSum(i, nums, memo):
            if i in memo:
                return memo[i]
            if i >= len(nums):
                return 0

            include = nums[i] + countSum(i+2, nums, memo)
            exclude = countSum(i+1, nums, memo)

            memo[i] = max(include, exclude)
            return memo[i]

        if len(nums) == 0:
            return 0
        elif len(nums) == 1:
            return nums[0]
        else:
            return max(countSum(0, nums[1:],{}), countSum(0, nums[:-1], {}))

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