House Robber

Define helper(i) as the max money obtainable using only houses 0..i. At house i, either skip it (helper(i-1)) or rob it and add to the best from two houses back (helper(i-2) + nums[i]).

class Solution:
    def rob(self, nums: list[int]) -> int:
        def helper(i: int) -> int:
            if i < 0:
                return 0
            return max(helper(i - 1), helper(i - 2) + nums[i])
        return helper(len(nums) - 1)

Each call spawns two more, so the call tree doubles with depth → O(2ⁿ) time, even though only n distinct values of i ever occur (huge redundant recomputation).

Cache helper(i) so each of the n distinct subproblems is solved exactly once.

class Solution:
    def rob(self, nums: list[int]) -> int:
        memo = {}
        def helper(i: int) -> int:
            if i < 0:
                return 0
            if i in memo:
                return memo[i]
            memo[i] = max(helper(i - 1), helper(i - 2) + nums[i])
            return memo[i]
        return helper(len(nums) - 1)

n subproblems, O(1) work each beyond memoized recursive calls → O(n) time, O(n) space for the memo and call stack.

Build dp[i] = max money robbable from houses 0..i, filled left to right: dp[i] = max(dp[i-1], dp[i-2] + nums[i]).

class Solution:
    def rob(self, nums: list[int]) -> int:
        n = len(nums)
        if n == 1:
            return nums[0]
        dp = [0] * n
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])
        for i in range(2, n):
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
        return dp[-1]

Watch how each new house only ever reads dp[i-1] and dp[i-2] as it decides whether to rob or to skip: nums = [1, 2, 3, 1]:

One pass, constant work per house → O(n) time. Since dp[i] only ever needs the previous two entries, the array can be collapsed to two rolling variables for O(1) space, but the full array is shown here for clarity.