> For the complete documentation index, see [llms.txt](https://interviews.woojiahao.com/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://interviews.woojiahao.com/problems-guide/dynamic-programming-roadmap/linear-sequence/house-robber.md).

# House Robber

## Transitions

1. Robber chooses to rob the current house
2. Robber does not rob the current house

## Top-down

The naive top-down approach would be trying all possibilities given the transitions:

```python
def house_robber(houses):
    def rob(i):
        if i >= len(houses):
            return 0
        return max(rob(i + 1), rob(i + 2) + houses[i])
    return rob(0)
```

If house `i` isn't robbed, then we can move on to house `i + 1` to try again.

If house `i` is robbed, then we can only start robbing from house `i + 2` onwards.

This generates the following recurrence relation:

$$
dp(i) = \begin{cases}
0, i >= |houses|\\
\max(dp(i+1), dp(i+2)+houses\[i])
\end{cases}
$$

## Deriving bottom-up

We will apply the trick of "re-framing the problem" to derive a bottom-up solution. Given a prefix of $$i-1$$ houses, can the optimal answer for house `i` be derived?

To rob house `i`, we must do so when we had not robbed the previous house, thus, we must have robbed at most house `i - 2`.

If house `i` is not going to be robbed, then the most amount of money we can rob from house `0` to `i` is the same as if we only looked out houses `0` to `i - 1`.

As a result, we can form a new recurrence relation:

$$
dp(i) = \begin{cases}
houses\[0], i =0\\
\max(houses\[0], houses\[1]), i = 1\\
\max(dp(i-1), dp(i-2)+houses\[i])
\end{cases}
$$

The base cases arise from the following observations:

1. If you only have 1 house, the optimal choice is always to rob it
2. If you have 2 houses, you can either rob the first but not the second or rob the second but not the first so we pick the maximum of the two

## Bottom-up

We will apply the $$n$$ state caching optimization where $$n = 2$$.

```python
def house_robber(houses):
    if len(houses) == 1: return houses[0]
    h1, h2 = houses[0], max(houses[0], houses[1])
    for i in range(2, len(houses)):
        h1, h2 = h2, max(h2, h1 + houses[i])
    return h2
```

Note that `h1` corresponds to `dp(i-2)` and `h2` corresponds to `dp(i-1)`.


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter, and the optional `goal` query parameter:

```
GET https://interviews.woojiahao.com/problems-guide/dynamic-programming-roadmap/linear-sequence/house-robber.md?ask=<question>&goal=<endgoal>
```

`ask` is the immediate question: it should be specific, self-contained, and written in natural language.
`goal` is optional and describes the broader end goal you are ultimately trying to accomplish on behalf of the user. GitBook uses it to tailor the answer towards what is most useful for that goal.

The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.