LeetCode: House Robber

Posted on January 10, 2018July 26, 2020 by braindenny

House Robber

Similar Problems:

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 system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: dp + O(1) space

## https://code.dennyzhang.com/house-robber
## Basic Ideas: dynamic programming
##
## Similar to Kadane's algorithm.
## Track current's best candidate, which indicates not taking current item
##
## Complexity: Time O(n), Space O(1)
class Solution:
    def rob(self, nums: List[int]) -> int:
        n = len(nums)
        pp, res = 0, 0
        for i in range(n):
            tmp = res
            # don't take vs take
            res = max(res, pp+nums[i])
            pp = tmp # without current item
        return res

Solution: dp + O(1) space

## https://code.dennyzhang.com/house-robber
## Basic Ideas: dynamic programming
##
## Complexity: Time O(n), Space O(1)
class Solution:
    def rob(self, nums: List[int]) -> int:
        n = len(nums)
        if n==0: return 0
        if n==1: return nums[0]
        if n==2: return max(nums[0], nums[1])
        pp, p, v = nums[0], max(nums[1], nums[0]), 0
        res = 0
        for i in range(2, n):
            v = max(p, pp+nums[i])
            res = max(res, pp, p, v)
            pp, p = p, v
        return res

Solution: dp + O(n) space

## https://code.dennyzhang.com/house-robber
class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        ## Idea: Recursive way will timeout
        ##       DP: robs[i] the max profit so far.
        ##       How does DP formula work?
        ## Complexity:
        length = len(nums)
        if length == 0:
            return 0
        if length == 1:
            return nums[0]
        if length == 2:
            return max(nums[0], nums[1])

        robs = [None]*length
        robs[0] = nums[0]
        robs[1] = max(nums[0], nums[1])
        robs[2] = max(nums[0]+nums[2], nums[1])

        for i in range(3, length):
            robs[i] = max(robs[i-3]+nums[i-1], robs[i-2]+nums[i])
        return robs[-1]

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