Contents

# Leetcode 198. House robber

## Description

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 system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

### Example 1

``````Input: [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: [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.
``````

## Dumb solution

Initially, I came up with a pretty straitforward (and dumb) solution for the task. Just stupid checks of all variants of robbery, here is the code:

``````class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
} else if (nums.length == 1) {
return nums[0];
} else if (nums.length == 2) {
return Math.max(nums[0], nums[1]);
} else if (nums.length == 3) {
return Math.max(nums[0]+nums[2], nums[1]);
}
int[] a = new int[nums.length];
a[0] = nums[0];
a[1] = nums[1];
a[2] = nums[0] + nums[2];
int maxRob = Math.max(Math.max(a[0],a[1]), a[2]);
for (int i=3; i<nums.length; i++) {
a[i] = Math.max(nums[i]+a[i-2], nums[i]+a[i-3]);
if (a[i] > maxRob) {
maxRob = a[i];
}
}
return maxRob;
}
}
``````

## Better way

But more smart solution can be as easy as following (and it has `O(1)` space complexity, as well as `O(n)` time complexity):

``````class Solution {
public int rob(int[] nums) {
int a = 0, b = 0;
for (int num : nums){
int tmp = Math.max(num+a, b);
a = b;
b = tmp;
}
return Math.max(a, b);
}
}
``````

Let’s explain what goes on here. Variable `a` stores prev previous value, `b` contains previous value. And `b` is updated only if new value is bigger than its previous value. This is the key to this solution.