LeetCode 213: House Robber II

Meta Data

#

Difficulty: medium First Attempt: 2026-04-25 Source: Day 11 learning note

Study Context

#

This article is rebuilt from the exact LeetCode section in the learning note. I kept the note’s repair points, comparison points, and common mistakes, while removing unrelated non-LeetCode material from the same day.

Learning Note Extract

#

Problem 2 - LC 213 House Robber II Timed Re-Solve

#

• Status: Good enough.
• Pattern: Circular array -> split into two linear robber problems.

Key Constraint

#

first house and last house are adjacent

So a valid answer must be one of:

exclude last house
exclude first house

Interview-Ready Explanation

#

Because the houses are arranged in a circle, the first and last houses are adjacent, so I cannot rob both. I split the problem into two linear House Robber I cases: rob nums[:-1] or rob nums[1:], then take the maximum of those two answers.

Organized Notes

#

The circular constraint is the whole problem. Once house 0 and house n - 1 are adjacent, a single linear robber pass can accidentally choose both. Splitting into nums[:-1] and nums[1:] removes that conflict: every valid optimal answer either excludes the last house or excludes the first house. The helper is exactly LC 198 with two rolling variables, and the n == 1 case must be handled before slicing.

Clean Solution

#

The note above captures the reasoning and the mistakes to avoid. The implementation below is the version I would submit.

from typing import List

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]

        def rob_line(arr):
            prev2 = prev1 = 0
            for x in arr:
                prev2, prev1 = prev1, max(prev1, prev2 + x)
            return prev1

        return max(rob_line(nums[:-1]), rob_line(nums[1:]))

Complexity

#

Time O(n), Space O(1).

Mistakes To Watch

#

• Running LC 198 directly on a circle.
• Forgetting n=1.

Final Interview Explanation

#

I would first call out the circular edge between the first and last house. Because an optimal solution cannot include both, I solve two linear robber subproblems, excluding one end each time, and return the larger result. The single-house case must be handled separately.