How to Solve the House Robber II Challenge in Java

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 = [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: 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:

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);
     }
}
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