house_robber_ii 

 1from functools import cache
 2
 3
 4# @leet start
 5class Solution:
 6    def rob(self, nums: list[int]) -> int:
 7        """
 8        This question is like house robber, but you're robbing a circular
 9        array instead of a line.
10
11        This is the same as house robber 1, but you can pass in a boolean that
12        allows you to see if you've robbed the house before.
13        """
14
15        @cache
16        def dp(i, robbed):
17            if i >= len(nums) or (i == len(nums) - 1 and robbed):
18                return 0
19            rob_curr = nums[i] + dp(i + 2, robbed if i != 0 else True)
20            rob_next = dp(i + 1, robbed)
21            return max(rob_curr, rob_next)
22
23        return dp(0, False)
24
25
26def test():
27    assert 2 + 2 == 4
class Solution: 
 6class Solution:
 7    def rob(self, nums: list[int]) -> int:
 8        """
 9        This question is like house robber, but you're robbing a circular
10        array instead of a line.
11
12        This is the same as house robber 1, but you can pass in a boolean that
13        allows you to see if you've robbed the house before.
14        """
15
16        @cache
17        def dp(i, robbed):
18            if i >= len(nums) or (i == len(nums) - 1 and robbed):
19                return 0
20            rob_curr = nums[i] + dp(i + 2, robbed if i != 0 else True)
21            rob_next = dp(i + 1, robbed)
22            return max(rob_curr, rob_next)
23
24        return dp(0, False)
def rob(self, nums: list[int]) -> int: 
 7    def rob(self, nums: list[int]) -> int:
 8        """
 9        This question is like house robber, but you're robbing a circular
10        array instead of a line.
11
12        This is the same as house robber 1, but you can pass in a boolean that
13        allows you to see if you've robbed the house before.
14        """
15
16        @cache
17        def dp(i, robbed):
18            if i >= len(nums) or (i == len(nums) - 1 and robbed):
19                return 0
20            rob_curr = nums[i] + dp(i + 2, robbed if i != 0 else True)
21            rob_next = dp(i + 1, robbed)
22            return max(rob_curr, rob_next)
23
24        return dp(0, False)

This question is like house robber, but you're robbing a circular array instead of a line.

This is the same as house robber 1, but you can pass in a boolean that allows you to see if you've robbed the house before.

def test(): 
27def test():
28    assert 2 + 2 == 4