When we have 3 houses it becomes a little less intuitive and this is where the dynamic programming comes in. Particularly the bottom up(iterative) process. If we are searching for the nth answer, we can solve a smaller subproblem for the ith answer to solve for n.

xxxxxxxxxx

let arr = new Array(len);

arr[0] = nums[0];

arr[1] = Math.max(nums[0], nums[1]);

for (let i = 2; i < len; i++) {

    arr[i] = Math.max(arr[i - 2] + nums[i], arr[i - 1]);

}

return arr[arr.length - 1];