Last updated on October 5th, 2024 at 09:08 pm

Here, We see House Robber II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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House Robber II LeetCode Solution

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.

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

1. House Robber II LeetCode Solution C++

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size(); 
        if (n < 2) return n ? nums[0] : 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;
    }
};

2. House Robber II LeetCode Solution Java

public class Solution {
    public int rob(int[] nums) {
        if (nums.length == 0)
            return 0;
        if (nums.length < 2)
            return nums[0];
        
        int[] startFromFirstHouse = new int[nums.length + 1];
        int[] startFromSecondHouse = new int[nums.length + 1];
        
        startFromFirstHouse[0]  = 0;
        startFromFirstHouse[1]  = nums[0];
        startFromSecondHouse[0] = 0;
        startFromSecondHouse[1] = 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]);
    }
}

3. House Robber II Solution JavaScript

var rob = function(nums) {
    if (nums.length < 2) {
        return nums[0] || 0;
    }
    const memo1 = [nums[0]];
    const memo2 = [0, nums[1]];
    for (let i=1; i<nums.length - 1; i++) {
        memo1[i] = Math.max(nums[i] + (memo1[i - 2] || 0), memo1[i - 1]);
    }
    for (let i=2; i<nums.length; i++) {
        memo2[i] = Math.max(nums[i] + memo2[i - 2], memo2[i - 1]);
    }
    return Math.max(memo1.pop(), memo2.pop());
};

4. House Robber II Solution Python

class Solution(object):
    def rob(self, nums):
        def simple_rob(nums):
            rob, not_rob = 0, 0
            for num in nums:
                rob, not_rob = not_rob + num, max(rob, not_rob)
            return max(rob, not_rob)
        if not nums:
            return 0
        elif len(nums) == 1:
            return nums[0]
        else:
            return max(simple_rob(nums[1:]), simple_rob(nums[:-1]))