House Robber II
Problem Description
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 <= 1000 <= nums[i] <= 1000
Solve it on LeetCode
Approaches
1. Recursive Solution with Memoization
Intuition:
In this problem, houses are arranged in a circle. This means the first house is a neighbor of the last house. To avoid robbing adjacent houses, the problem can be broken down into two subproblems:
- Rob houses from index
0ton-2(excluding the last house). - Rob houses from index
1ton-1(excluding the first house).
Use a helper function to recursively calculate the maximum money that can be robbed, while using memoization to avoid redundant calculations.
Code:
- Time Complexity: O(n), where n is the length of the array because each house is calculated once.
- Space Complexity: O(n), due to the space used for the memoization array.
2. Dynamic Programming with Optimized Space
Intuition:
This approach uses the same two subproblems as the previous method. However, we optimize the space complexity by storing only the last two results (as we only need these to calculate the current house's decision).
Code:
- Time Complexity: O(n), linear traversal of array.
- Space Complexity: O(1), space usage is constant due to only using fixed variables for calculations.