4. House Robber II

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.

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

Constraints:

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

Approach

• Solve House robber for 2 ranges one without first and one without last element and compute the max the reason being that first and last cannot be together
• To solve single element edge case Max the above with the first element
• Most likely copy method excludes the end index
• Time: O(n) Space: O(1)

	class Solution {
    public int rob(int[] nums) {
        int max = Math.max(hr(Arrays.copyOfRange(nums,1,nums.length)),
            hr(Arrays.copyOfRange(nums,0,nums.length - 1))); 
        return Math.max(nums[0],max);
    }
 
    public int hr(int[] nums) {
        int l = 0, r = 0;
        for (int n: nums) {
            int t = Math.max(r,l+n);
            l = r;
            r = t;
        }
        return r;
    }
}