213. House Robber II - Medium

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 a list of non-negative integers nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

 

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 = [0]
Output: 0

 

similar to house robber i, but has 2 cases here:
1. rob nums[0] (cannot rob nums[nums.length - 1]) -> get max1
2. rob nums[nums.length - 1] (cannot rob nums[0]) -> get max2
compare max1 and max2, the larger one is the result

time = O(n), space = O(1) optimal

class Solution {
    public int rob(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        if(nums.length == 1) {
            return nums[0];
        }
        return Math.max(rob(nums, 0, nums.length - 2), rob(nums, 1, nums.length - 1));
    }
    
    public int rob(int[] nums, int start, int end) {
        int prevRob = 0, prevNotRob = 0, rob = 0, notRob = 0;
        for(int i = start; i <= end; i++) {
            rob = prevNotRob + nums[i];
            notRob = Math.max(prevNotRob, prevRob);
            
            prevRob = rob;
            prevNotRob = notRob;
        }
        return Math.max(rob, notRob);
    }
}