# 198. House Robber

## The Problem

Link to original problem on Leetcode.

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.

Examples

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.

Constraints
• 1 <= nums.length <= 100
• 0 <= nums[i] <= 400

## My Solution

A leetcode problem with a story! This is quite rare, and a real treat!1 This is a dynamic programming problem, and as such we can solve it either recursively or iteratively.

### Recursive Solution

To solve this problem recursively, we should think about what sub-problem we're trying to solve. Whenever I'm at a house at index, I have to decide: am I better off robbing this house, or waiting until I get to the next house? If I rob this house, I get the value of its possessions nums[index], plus the value of all the houses I robbed before if the last house I robbed was the house from two before it at nums[index - 2]. If I skip this house, I get the value of all the houses I robbed before this house ending at the house immediately before it at nums[index - 1]. If index starts at the last value of nums, then we can write a recursive function to figure out the sum of the best sequence of houses by working back from index. I can work backward until I'm out of houses to compare—that is to say, the index is less than 0. That's our base case.

// This function is modified to accept a second argument// from how it's scaffolded in leetcode: index. We set a// default index value of nums.length - 1 to signify we// are starting at the last value of nums.function rob(nums: number[], index = nums.length - 1): number {	if (index < 0) {		// Indices less than zero are the base case; we are		// out of houses to rob! Non-existant houses have		// a value of zero.		return 0;	} else {		// If we're not yet out of houses to rob, we return the		// maximum of this house plus the house two before, or		// the house one before. We recursively call the function		// to get the sum of other house's values.		return Math.max(			rob(nums, index - 2) + nums[index],			rob(nums, index - 1)		);	}};

This function is correct, but it's not performant. Because we call the function again twice for each comparison, our time complexity is $O(2^{n})$! If you try running it in leetcode, it will time out. We need to improve it with memoization.

### Recursive Solution + Memoization

To improve on the previous solution, we add memoization to remember values we've previously computed. This dramatically speeds up our solution and gives us a time complexity of $O(n)$. Here, we pass around a Map with the index as key and the result of our comparisons as the value.

// We've once again modified the function. Now it also takes// a JavaScript Map to use for our memoization, which we// initialize as a new empty Map.function rob(nums: number[], index = nums.length - 1, memo = new Map<number, number>()): number {	if (index < 0) {		return 0;	} else if (memo.has(index)) {		// Instead of always computing non-base cases, we'll first		// check to see if they exist in our memo Map and return		// that if we find it.		return memo.get(index);	} else {		const result = Math.max(			rob(nums, index - 2, memo) + nums[index],			rob(nums, index - 1, memo)		);		// Instead of returning the result of comparisons right		// away, we first add it to the memo Map so we can find		// it later if we need it.		memo.set(index, result);		return result;	}};

### Iterative Solution

Our recursive solution was top-down. That is to say, we started at the farthest house and worked backward. For our iterative solution, we'll instead go bottom-up. We'll calculate answers to sub-problems near the beginning in order to answer more sub-problems as we go along. The same sub-problem logic still holds: for each house we're at, we want to know if we're better off taking the sum of this house and all the houses previously robbed as of two house before, or just the sum of all house robbed previously as of one house before.

We could use an array to remember the optimal possible result for any given index in nums, but we don't have to. All we really need are two variables to remember the sums for one house back and two houses back. With each iteration across nums, we can just update those two variables. So not only do we get time complexity of $O(n)$, but we also get space complexity of $O(1)$!.

function rob(nums: number[]): number {	if (nums.length === 0) return 0;	let oneHouseBack = 0, twoHousesBack = 0, temp = oneHouseBack;	for (let thisHouse of nums) {		// We're about to reassign the value of oneHouseBack,		// so we want to store the current value somewhere.		// We put it in this temp variable, because we're going		// to update twoHousesBack to be the current value of		// oneHouseBack, and oneHouseBack to the max of our		// comparison.		temp = oneHouseBack;		oneHouseBack = Math.max(twoHousesBack + thisHouse, oneHouseBack)		twoHousesBack = temp;	}	return oneHouseBack};

1. If you like tough data structures and algorithms problems with a story, check out Advent of Code for more and better puzzles. ↩︎