198.House_Robber_(Medium).md

198. House Robber (Medium)

Date and Time: Jul 2, 2024, 11:55 (EST)

Link: https://leetcode.com/problems/house-robber/

Question:

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [1, 2, 3, 1]

Output: 4

Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: nums = [2, 7, 9, 3, 1]

Output: 12

Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). Total amount you can rob = 2 + 9 + 1 = 12.

Example 3:

Input: [1, 4, 2, 5]

Output: 9

Edge Case:

Input: [2, 1, 1, 2]

Output: 4

KeyPoints:

1. For each house in dp[i], we want to find if we should take the i-1th (previous) house or nums[i] + i-2 (current) house. So we should have 0 at the very beginning of dp[] and manually add dp[1] = nums[0], because the first house has no choice to choose, but any house after the first one have choice.

Example 1: nums = [1,2,3,1], dp = [0, 1, 2, 3, 1].

2. After the base case, we start from range(2, len(nums)+1), and follow the recurrence relation dp[i] = max(dp[i-1], dp[i-2] + nums[i]). Finally we just return the last element in dp[len(nums)], which will be the maximum you can rob.

Example 1: dp = [0, 1, 2, 4, 4], return dp[len(nums)] = 4.

Python Solution:

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

Time Complexity: $O(n)$
Space Complexity: $O(n)$