304 North Cardinal St.
Dorchester Center, MA 02124

# House Robber II LeetCode Solution

## Problem – House Robber II LeetCode Solution

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`

## House Robber II LeetCode Solution in C++

``````class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if (n < 2) return n ? nums : 0;
return max(robber(nums, 0, n - 2), robber(nums, 1, n - 1));
}
private:
int robber(vector<int>& nums, int l, int r) {
int pre = 0, cur = 0;
for (int i = l; i <= r; i++) {
int temp = max(pre + nums[i], cur);
pre = cur;
cur = temp;
}
return cur;
}
};
``````

## House Robber II LeetCode Solution in Python

``````class Solution:
def rob(self, nums):
def rob_helper(nums):
dp1, dp2 = 0, 0
for num in nums:
dp1, dp2 = dp2, max(dp1 + num, dp2)
return dp2

return max(nums + rob_helper(nums[2:-1]), rob_helper(nums[1:]))
``````

## House Robber II LeetCode Solution in Java

``````public class Solution {
public int rob(int[] nums) {
if (nums.length == 0)
return 0;
if (nums.length < 2)
return nums;

int[] startFromFirstHouse = new int[nums.length + 1];
int[] startFromSecondHouse = new int[nums.length + 1];

startFromFirstHouse  = 0;
startFromFirstHouse  = nums;
startFromSecondHouse = 0;
startFromSecondHouse = 0;

for (int i = 2; i <= nums.length; i++) {
startFromFirstHouse[i] = Math.max(startFromFirstHouse[i - 1], startFromFirstHouse[i - 2] + nums[i-1]);
startFromSecondHouse[i] = Math.max(startFromSecondHouse[i - 1], startFromSecondHouse[i - 2] + nums[i-1]);
}

return Math.max(startFromFirstHouse[nums.length - 1], startFromSecondHouse[nums.length]);
}
}
``````
##### House Robber II LeetCode Solution Review:

In our experience, we suggest you solve this House Robber II LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the House Robber II LeetCode Solution

Find on Leetcode

##### Conclusion:

I hope this House Robber II LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions