# LeetCode - House Robber II

## Problem statement

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.

Problem statement taken from: https://leetcode.com/problems/house-robber-ii/

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
``````

### Explanation

This problem is an extension of our old blog post House Robber. The tricky part of the problem is the houses are arranged in a circle. The first house is the neighbor of the last one, which means one of the houses cannot be robbed.

We need to run two cases here:

• Select the first house and ignore the last one. We calculate the max amount that can be looted from nums[0] to nums[len - 2].
• Select the last house and ignore the first one. We calculate the max amount that can be looted from nums[1] to nums[len - 1].

Let's check the algorithm here:

``````// rob(nums) method
- set n = nums.size();

// for empty array.
- if n == 0
- return 0

// for an array of size 1
- if n == 1
- return nums[0]

- return max(robHelper(nums, 0, n - 2), robHelper(nums, 1, n - 1))

// robHelper(nums, l, r)
- set include = 0
exclude = 0
tmp = 0

- loop for i = l; i <= r; i++
- set tmp = max(include, exclude)

- update include = exclude + nums[i]

- set exclude = tmp

- return max(include, exclude)
``````

Let's check our solutions in C++, Golang, and Javascript.

#### C++ solution

``````class Solution {
public:
int robHelper(vector<int>& nums, int l, int r) {
int include = 0, exclude = 0, tmp;

for(int i = l ; i <= r ; i++) {
tmp = max(include , exclude);
include = exclude + nums[i];
exclude = tmp;
}

return max(include , exclude);
}

int rob(vector<int>& nums) {
int n = nums.size();

if(n == 0) {
return 0;
}

if(n == 1) {
return nums[0];
}

return max(robHelper(nums, 0, n - 2) , robHelper(nums, 1, n - 1));
}
};``````

#### Golang solution

``````func max(a, b int) int {
if a > b {
return a
}

return b
}

func robHelper(nums []int, l, r int) int {
include, exclude, tmp := 0, 0, 0

for i := l; i <= r; i++ {
tmp = max(include, exclude)
include = exclude + nums[i]
exclude = tmp
}

return max(include, exclude)
}

func rob(nums []int) int {
n := len(nums)

if n == 0 {
return 0
}

if n == 1 {
return nums[0]
}

return max(robHelper(nums, 0, n - 2), robHelper(nums, 1, n - 1))
}``````

#### Javascript solution

``````var robHelper = function(nums, l, r) {
let include = 0, exclude = 0, tmp;

for(let i = l; i <= r; i++) {
tmp = Math.max(include, exclude);
include = exclude + nums[i];
exclude = tmp;
}

return Math.max(include, exclude);
}

var rob = function(nums) {
let n = nums.length;

if(n == 0) {
return 0;
}

if(n == 1) {
return nums[0];
}

return Math.max(robHelper(nums, 0, n - 2), robHelper(nums, 1, n - 1));
};``````

Let's dry run our algorithm for a given input.

``````Input: nums = [1, 3, 2, 6, 8, 4]

// rob(vector<int>& nums)
Step 1: n = nums.size()
= 6

Step 2: n == 0
6 == 0
false

Step 3: n == 1
6 == 1
false

Step 4: return max(robHelper(nums, 0, n - 2), robHelper(nums, 1, n - 1))
return max(robHelper(nums, 0, 4), robHelper(nums, 1, 5))

First call for robHelper(nums, 0, 4)
Step 5: include = 0
exclude = 0
tmp

Step 6: loop for i = l; i <= r
i = 0
0 <= 4
true

tmp = max(include , exclude)
= max(0, 0)
= 0

include = exclude + nums[i]
= 0 + nums[0]
= 0 + 1
= 1

exclude = tmp
= 0

i++
i = 1

Step 7: loop for i <= r
1 <= 4
true

tmp = max(include , exclude)
= max(1, 0)
= 1

include = exclude + nums[i]
= 0 + nums[1]
= 0 + 3
= 3

exclude = tmp
= 1

i++
i = 2

Step 8: loop for i <= r
2 <= 4
true

tmp = max(include , exclude)
= max(3, 1)
= 3

include = exclude + nums[i]
= 1 + nums[2]
= 1 + 2
= 3

exclude = tmp
= 3

i++
i = 3

Step 9: loop for i <= r
3 <= 4
true

tmp = max(include , exclude)
= max(3, 3)
= 3

include = exclude + nums[i]
= 3 + nums[3]
= 3 + 6
= 9

exclude = tmp
= 3

i++
i = 4

Step 10: loop for i <= r
4 <= 4
true

tmp = max(include , exclude)
= max(9, 3)
= 9

include = exclude + nums[i]
= 3 + nums[4]
= 3 + 8
= 11

exclude = tmp
= 9

i++
i = 5

Step 11: loop for i <= r
5 <= 4
false

Step 12: return max(include, exclude)
max(11, 9)

We return 11

// we come back to step 4 and compute robHelper(nums, 1, 5)

Step 13: return max(11, robHelper(nums, 1, 5))

call for robHelper(nums, 1, 5)
Step 14: include = 0
exclude = 0
tmp

Step 15: loop for i = l; i <= r
i = 1
1 <= 5
true

tmp = max(include , exclude)
= max(0, 0)
= 0

include = exclude + nums[i]
= 0 + nums[1]
= 0 + 3
= 3

exclude = tmp
= 0

i++
i = 2

Step 16: loop for i <= r
i = 2
2 <= 5
true

tmp = max(include , exclude)
= max(3, 0)
= 3

include = exclude + nums[i]
= 0 + nums[2]
= 0 + 2
= 2

exclude = tmp
= 3

i++
i = 3

Step 17: loop for i <= r
i = 3
3 <= 5
true

tmp = max(include , exclude)
= max(2, 3)
= 3

include = exclude + nums[i]
= 3 + nums[3]
= 3 + 6
= 9

exclude = tmp
= 3

i++
i = 4

Step 18: loop for i <= r
i = 4
4 <= 5
true

tmp = max(include , exclude)
= max(9, 3)
= 9

include = exclude + nums[i]
= 3 + nums[4]
= 3 + 8
= 11

exclude = tmp
= 9

i++
i = 5

Step 19: loop for i <= r
i = 5
5 <= 5
true

tmp = max(include , exclude)
= max(11, 9)
= 11

include = exclude + nums[i]
= 9 + nums[5]
= 9 + 4
= 13

exclude = tmp
= 11

i++
i = 6

Step 20: loop for i <= r
i = 6
6 <= 5
false