House Robber II

Problem

https://leetcode.com/problems/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.

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

Pattern

Array, Dynamic Programming

Approaches

Dynamic Programming

Code

def rob(nums: list[int]) -> int:
    N = len(nums)

    if len(nums) == 1:
        return nums[0]

    money = [0] * (N - 1)
    money1 = [0] * (N - 1)
    for i, num in enumerate(nums[:-1]):
        if i == 0:
            money[i] = num
        elif i == 1:
            money[i] = max(money[i - 1], num)
        else:
            money[i] = max(money[i - 1], num + money[i - 2])

    for i, num in enumerate(nums[1:]):
        if i == 0:
            money1[i] = num
        elif i == 1:
            money1[i] = max(money1[i - 1], num)
        else:
            money1[i] = max(money1[i - 1], num + money1[i - 2])

    return max(money[-1], money1[-1])

Test

>>> from house_robber_ii__dynamic_programming import rob
>>> rob([2, 3, 2])
3
>>> rob([1, 2, 3, 1])
4
>>> rob([1, 2, 3])
3

house_robber_ii__dynamic_programming.rob(nums: list[int]) → int