Sponsored
Sponsored
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
2
The C solution sets up a memo array for memoization to store calculated maximums for each house start index. It runs the helper function robHelper which recursively looks for the maximum stealing options, caching results for efficiency. It applies this to the scenarios that exclude the first and last house respectively.