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
2public class Solution {
3 private int RobLinear(int[] nums, int start, int end) {
4 int prev1 = 0, prev2 = 0;
5 for (int i = start; i < end; i++) {
6 int temp = prev1;
7 prev1 = Math.Max(nums[i] + prev2, prev1);
8 prev2 = temp;
9 }
10 return prev1;
11 }
12
13 public int Rob(int[] nums) {
14 if (nums.Length == 1) return nums[0];
15 return Math.Max(RobLinear(nums, 0, nums.Length - 1), RobLinear(nums, 1, nums.Length));
16 }
17}
18This C# solution defines RobLinear for the linear version of the problem and solves it for two cases: one excluding the first and another excluding the last house. The maximum of these two outcomes is returned.
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
2public class Solution {
3 private Dictionary<int, int> memo = new Dictionary<int, int>();
4 private int RobHelper(int[] nums, int start, int end) {
5 if (start >= end) return 0;
6 if (memo.ContainsKey(start)) return memo[start];
7 int result = Math.Max(nums[start] + RobHelper(nums, start + 2, end), RobHelper(nums, start + 1, end));
8 memo[start] = result;
9 return result;
10 }
11
12 public int Rob(int[] nums) {
13 if (nums.Length == 1) return nums[0];
14 memo.Clear();
15 int case1 = RobHelper(nums, 0, nums.Length - 1);
16 memo.Clear();
17 int case2 = RobHelper(nums, 1, nums.Length);
18 return Math.Max(case1, case2);
19 }
20}
21This C# solution employs a Dictionary for managing memoization to store and retrieve calculated subproblems conveniently. The method RobHelper determines the maximum taking turns at each index, from which two scenarios for circle robbing are evaluated for the final yield optimum.