213. House Robber II

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system 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.

class Solution:
    def rob(self, nums: List[int]) -> int:
        """
        - if we rob the first house, we can't rob the last house
        - run two dps, one where we always skip the first house
          one where we always take from the first house
        - max of the last house from the skipping first house dp
          and the second last house from the taking first house dp
        """
        if len(nums) == 1:
            return nums[0]
        if len(nums) == 2:
            return max(nums[0], nums[1])
 
        take_dp = [nums[0], nums[0]]
        for i in range(2, len(nums)-1):
            take_dp.append(max(take_dp[i-1], nums[i]+take_dp[i-2]))
 
        skip_dp = [0, nums[1]]
        for i in range(2, len(nums)):
            skip_dp.append(max(skip_dp[i-1], nums[i]+skip_dp[i-2]))
 
        return max(take_dp[-1], skip_dp[-1])

• consider two cases where we either take the first house or not.
  • if we take the first house, we can’t consider the last house, and vice versa.

table without id file.inlinks as Backlinks
where file.name = this.file.name

Related.

References.

Categories:: dynamic-programming, array