## 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:3Explanation: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:4Explanation: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

**Constraints:**

`1 <= nums.length <= 100`

`0 <= nums[i] <= 1000`

## The solution in Java code

```
class Solution {
public int rob(int[] nums) {
if (nums.length ==1) return nums[0];
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:4Explanation: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:

```
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);
}
}
```