This approach involves splitting the problem into two scenarios: robbing from house 0 to house n-2, and from house 1 to house n-1. The aim is to apply the linear version of the House Robber problem to each scenario separately and then take the maximum of the two resulting summed values.
Time Complexity: O(n).
Space Complexity: O(1).
1
2class Solution {
3public:
4 int robLinear(vector<int>& nums, int start, int end) {
5 int prev1 = 0, prev2 = 0;
6 for (int i = start; i < end; i++) {
7 int temp = prev1;
8 prev1 = max(nums[i] + prev2, prev1);
9 prev2 = temp;
10 }
11 return prev1;
12 }
13 int rob(vector<int>& nums) {
14 int n = nums.size();
15 if (n == 1) return nums[0];
16 return max(robLinear(nums, 0, n - 1), robLinear(nums, 1, n));
17 }
18};
19This C++ solution uses a helper function robLinear for computing the linear house rob solution and applies it twice: once excluding the first house and once excluding the last house. It returns the maximum of these two values.
This approach implements a memoized version of the dynamic programming solution to avoid recomputing values. We explore two scenarios of the circle, taking into account the cycles explicitly by caching intermediate results.
Time Complexity: O(n).
Space Complexity: O(n) for the memoization array.
1
2class Solution:
3 def robHelper(self, nums, start, end, memo):
4 if start >= end:
5 return 0
6 if memo[start] >= 0:
7 return memo[start]
8 memo[start] = max(nums[start] + self.robHelper(nums, start + 2, end, memo), self.robHelper(nums, start + 1, end, memo))
9 return memo[start]
10
11 def rob(self, nums):
12 if len(nums) == 1:
13 return nums[0]
14 memo1 = [-1] * len(nums)
15 memo2 = [-1] * len(nums)
16 return max(self.robHelper(nums, 0, len(nums) - 1, memo1), self.robHelper(nums, 1, len(nums), memo2))
17This Python solution makes use of a memo array to store intermediate maximums to prevent redundant calculations. The robHelper function recursively determines possible rob outcomes and updates the cache, providing two results depending on whether the first house is considered or not.