213. House Robber II

1. Description

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.

2. Example

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

3. Constraints

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

4. Solutions

Dynamic Programming

n = nums.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int rob(const vector<int> &nums) {
        const int n = nums.size();

        if (n == 1) {
            return nums[0];
        } else if (n == 2) {
            return max(nums[0], nums[1]);
        }

        int rob_last = rob_range(nums, 0, n - 2);
        int rob_last = rob_range(nums, 1, n - 1);

        return max(rob_last, rob_last);
    }

private:
    int rob_range(const vector<int> &nums, int left, int right) {
        int prev_no = 0, prev_yes = 0;
        for (int i = left; i <= right; ++i) {
            int cur_no = max(prev_no, prev_yes);
            int cur_yes = prev_no + nums[i];

            prev_no = cur_no;
            prev_yes = cur_yes;
        }

        return max(prev_no, prev_yes);
    }
};