213 House Robber II - Medium

Problem:

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.

Example 1:

Input: nums = [2,3,2] Output: 3 Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.

Example 2:

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 3:

Input: nums = [1,2,3] Output: 3

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 1000

Problem Analysis:

similar to 198-house-robber198 House Robber - MediumProblem 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 th, with 1 additional constraint that the list of houses are arranged in a circle
base case is when the length is 1
then we have to perform dp on 2 subarrays. 1 without the first element and 1 without the last element.
then we just have to return the max of these 3 values

Solutions:

class Solution:
    def rob(self, nums: List[int]) -> int:
        # base case: length 1, return the first element
        # then do dp on 2 subarray; 1 without the first element, 1 without the last element
        return max(nums[0], self.dynamic_programming(nums[1:]), self.dynamic_programming(nums[:-1]))

    def dynamic_programming(self, nums: List[int]) -> int:
        rob1, rob2 = 0, 0
        
        for n in nums:
            temp = max(rob1 + n, rob2)
            rob1 = rob2
            rob2 = temp
        return rob2

Walter Teng.

213 House Robber II - Medium

Problem:

Problem Analysis:

Solutions:

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