Last updated on January 20th, 2025 at 11:01 pm

Here, we see a House Robber LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Dynamic Programming

Companies

Airbnb, LinkedIn

Level of Question

Medium

House Robber LeetCode Solution

1. Problem Statement

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems 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 = [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 2:
Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

2. Coding Pattern Used in Solution

The coding pattern used in the provided code is Dynamic Programming (DP). Specifically, it is a “Fibonacci-style DP” or “House Robber Problem DP”. This pattern is used to solve problems where the solution to the current state depends on the solutions to previous states, and the goal is to optimize a decision-making process.

3. Code Implementation in Different Languages

3.1 House Robber C++

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size(), pre = 0, cur = 0;
        for (int i = 0; i < n; i++) {
            int temp = max(pre + nums[i], cur);
            pre = cur;
            cur = temp;
        }
        return cur;
    }
};

3.2 House Robber Java

class Solution {
    public int rob(int[] nums) {
        int pre = 0, cur = 0;
        for (int num : nums) {
            final int temp = Integer.max(pre + num, cur);
            pre = cur;
            cur = temp;
        }
        return cur;
    }
}

3.3 House Robber JavaScript

var rob = function(nums) {
    if (!nums.length) return 0;
    if (nums.length === 1) return nums[0];
    if (nums.length === 2) return Math.max(nums[0], nums[1]);
    
    let maxAtTwoBefore = nums[0];
    let maxAtOneBefore = Math.max(nums[0], nums[1]);
    
    for (let i = 2; i < nums.length; i++) {
        const maxAtCurrent = Math.max(nums[i] + maxAtTwoBefore, maxAtOneBefore);
        
        maxAtTwoBefore = maxAtOneBefore;
        maxAtOneBefore = maxAtCurrent;
    }
    return maxAtOneBefore;
};

3.4 House Robber Python

class Solution(object):
    def rob(self, 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)

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n)	O(1)
Java	O(n)	O(1)
JavaScript	O(n)	O(1)
Python	O(n)	O(1)

• The code is an efficient implementation of the Dynamic Programming pattern for the “House Robber Problem.”
• It optimizes space by using only two variables instead of a full DP array.
• The time complexity is linear, and the space complexity is constant, making it optimal for this problem.