House Robber LeetCode Solution

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

House Robber 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, 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.

• C++ Solution
• Java Solution
• JavaScript Solution
• Python Solution

House Robber Leetcode Solution 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;
    }
};
Code language: C++ (cpp)

House Robber Leetcode Solution 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;
    }
}
Code language: Java (java)

House Robber Leetcode Solution 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;
};
Code language: JavaScript (javascript)

House Robber Leetcode Solution 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)
Code language: Python (python)