House Robber II

For the linear House Robber problem, the decision at a house is binary: skip it and keep the best answer through the previous house, or take it and add its value to the best answer two houses back. Therefore dp[i] = max(dp[i - 1], nums[i] + dp[i - 2]). Keep only the last two states for O(1) extra space, and handle empty and one-house inputs before the recurrence. For House Robber II, the first and last houses are adjacent, so no valid solution can take both. Solve two linear ranges, excluding the first and excluding the last, then take the maximum. Do not use the linear recurrence over the whole circle. Test one and two houses, equal values, all zero values, and cases where the endpoints tempt an invalid choice. Time is O(n) and space is O(1) beyond the input.