# 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

```.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}```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;
}
}
```Code language: Java (java)```

## 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);
}
}
```Code language: Java (java)```
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