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
2int robLinear(int* nums, int len) {
3 if (len == 0) return 0;
4 if (len == 1) return nums[0];
5 int prev1 = 0, prev2 = 0;
6 for (int i = 0; i < len; i++) {
7 int temp = prev1;
8 prev1 = (nums[i] + prev2 > prev1) ? nums[i] + prev2 : prev1;
9 prev2 = temp;
10 }
11 return prev1;
12}
13
14int rob(int* nums, int numsSize) {
15 if (numsSize == 1) return nums[0];
16 return (robLinear(nums, numsSize - 1) > robLinear(nums + 1, numsSize - 1))
17 ? robLinear(nums, numsSize - 1)
18 : robLinear(nums + 1, numsSize - 1);
19}
20
This C solution defines a helper function robLinear
for the simpler linear version of the House Robber problem, then uses this to calculate the maximum rob amount in two scenarios: excluding the first and last house respectively. The final result is the maximum of these two computed 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
2var robHelper = function(nums, start, end, memo) {
3 if (start >= end) return 0;
4 if (memo[start] !== undefined) return memo[start];
5 memo[start] = Math.max(nums[start] + robHelper(nums, start + 2, end, memo), robHelper(nums, start + 1, end, memo));
6 return memo[start];
7};
8
9var rob = function(nums) {
10 if (nums.length === 1) return nums[0];
11 const memo1 = {};
12 const memo2 = {};
13 return Math.max(robHelper(nums, 0, nums.length - 1, memo1), robHelper(nums, 1, nums.length, memo2));
14};
15
This JavaScript solution applies memoization utilizing plain objects to store precomputed solutions. The function robHelper
determines the best regular house rob outcomes recursively while updating cache objects accordingly for each scenario considered.