## The challenge

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 = 
Output: 0```

Constraints:

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

## The solution in Java code

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 `````` ``````class Solution { public int rob(int[] nums) { if (nums.length ==1) return nums; return Math.max(rob(nums, 0, nums.length-2), rob(nums, 1, nums.length-1)); } private int rob(int[]nums, int start, int end) { int a = 0, b = 0; for (int i = start; i <= end; i++) { int temp = b; if (nums[i] + a > b) b = nums[i] + a; a = temp; } return b; } } ``````

## An Alternate View at the Acceptance Criteria

If we take the first example and say that:

```Input: nums = [2,3,2]
Output: 4
Explanation: You can rob house 1 (money = 2) and rob house 3 (money = 2), because they are NOT adjacent houses. As they have house 2 (money = 3) inbetween them.```

If we take this new viewpoint, then we can write the following code to resolve the problem:

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 `````` ``````class Solution { public int rob(int[] nums) { // start from 0, then skip until end, build a total int total1 = 0; // start from 1, then skip until end, build a total int total2 = 0; for (int i=0; i<=nums.length; i++) { int val = (i==nums.length) ? 0 : nums[i]; if (i%2==0) total1 += val; else total2 += val; } // compare the totals and give the higher one return Math.max(total1, total2); } } ``````