Firstly, let us establish the recurrence relationship since that will be used for both recursive and iterative using dp. The recurrence relationship looks like:

T(n,i) = Max of ( a[i]+T(n,i+2) , a[i+1] + T(n,i+3)).

The recursive code based on this is:

```
private static int recursive(int[] nums, int[] dp, int i)
{
if (i >= nums.length)
{
return 0;
}
if (nums.length == 1)
{
return nums[0];
}
if (nums.length == 2)
{
return Math.max(nums[0], nums[1]);
}
if (dp[i + 2] == -1)
{
dp[i + 2] = recursive(nums, dp, i + 2);
}
if (dp[i + 3] == -1)
{
dp[i + 3] = recursive(nums, dp, i + 3);
}
int even = nums[i] + dp[i + 2];
int odd = (i + 1 < nums.length ? nums[i + 1] : 0) + dp[i + 3];
int res = Math.max(even, odd);
return res;
}
```

The time complexity for this looks to be linear with memoization but I will let someone more knowledgeable confirm that.

Turning that into DP is fairly easy. This solution is O(n).

```
public int rob(int[] nums) {
if(nums.length==0) {
return 0;
}
int[] dp = new int[nums.length + 3];
if (nums.length == 1)
{
return nums[0];
}
if (nums.length == 2)
{
return Math.max(nums[0], nums[1]);
}
dp[0] = nums[0];
dp[1] = Math.max(nums[0], nums[1]);
int max = 0;
for (int i = 2; i < nums.length; i++)
{
dp[i] = Math.max(nums[i] + dp[i - 2],
dp[i - 1]);
max = Math.max(max, dp[i]);
}
return max;
}
```