House Robber


  1. Robber chooses to rob the current house

  2. Robber does not rob the current house


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

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)={0,i>=āˆ£housesāˆ£maxā”(dp(i+1),dp(i+2)+houses[i])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 iāˆ’1i-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)={houses[0],i=0maxā”(houses[0],houses[1]),i=1maxā”(dp(iāˆ’1),dp(iāˆ’2)+houses[i])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


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

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).

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