Python solution with detailed explanation

  • 1


    House Robber

    • At each house, we have the option to either rob the house or not rob the house. This is the single most important insight we shall use while we develop our algorithm. We use this insight in both memoization and dynamic programming approach. In the former we move forward and in the latter we move backwards.


    • Sketch the recursion tree & go top down. Decide at last node.
    • rob(i) = max(rob_this_house, do_not_rob_this_house)
    • rob(i) = max(nums[i]+rob(i+2), rob(i+1))
    • rob(i) = 0 when we have i >= N - Base Base
    class Solution(object):
        def helper(self, i, nums, cache):
            N = len(nums)
            if i >= N:
                return 0
            if i in cache:
                return cache[i]
            cache[i] = max(nums[i]+self.helper(i+2, nums, cache), self.helper(i+1, nums, cache))
            return cache[i]
        def rob(self, nums):
            :type nums: List[int]
            :rtype: int
            return self.helper(0, nums, {})

    Table Filling Dynamic Programming

    • Bottom up.
    • rob(i) - Maximum money we can make by robbing houses from [0,1,2..i]
    • rob(i) = max(nums[i] + rob(i-2), rob(i-1)) : rob house(i) or not
    • rob(i) = 0 for i = -1 or -2
    • Move backwards. Assume we have solved the problem till house i-1. Then at house i ask the question - should we rob house i or not?
    • We only need constant space in this case.
    class Solution(object):
        def rob(self, nums):
            result = prev_prev = prev = 0
            for idx in range(len(nums)):
                result = max(prev_prev+nums[idx], prev)
                prev_prev = prev
                prev = result
            return result

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