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https://github.com/TheAlgorithms/Ruby
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87 lines
2.3 KiB
Ruby
87 lines
2.3 KiB
Ruby
# You are a professional robber planning to rob houses along a street.
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# Each house has a certain amount of money stashed, the only constraint stopping you
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# from robbing each of them is that adjacent houses have security systems connected
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# and it will automatically contact the police if two adjacent houses
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# were broken into on the same night.
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#
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# Given an integer array nums representing the amount of money of each house,
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# return the maximum amount of money you can rob tonight without alerting the police.
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#
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# Example 1:
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#
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# Input: nums = [1,2,3,1]
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# Output: 4
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# Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
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# Total amount you can rob = 1 + 3 = 4.
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#
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# Example 2:
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#
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# Input: nums = [2,7,9,3,1]
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# Output: 12
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# Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
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# Total amount you can rob = 2 + 9 + 1 = 12.
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#
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# Approach 1: Dynamic Programming
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#
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# Complexity Analysis
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#
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# Time Complexity: O(N) since we process at most N recursive calls, thanks to
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# caching, and during each of these calls, we make an O(1) computation which is
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# simply making two other recursive calls, finding their maximum, and populating
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# the cache based on that.
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#
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# Space Complexity: O(N) which is occupied by the cache and also by the recursion stack
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def rob(nums, i = nums.length - 1)
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return 0 if i < 0
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[rob(nums, i - 2) + nums[i], rob(nums, i - 1)].max
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end
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nums = [1, 2, 3, 1]
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puts rob(nums)
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# Output: 4
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nums = [2, 7, 9, 3, 1]
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puts rob(nums)
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# Output: 12
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#
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# Approach 2: Optimized Dynamic Programming
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#
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# Time Complexity
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#
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# Time Complexity: O(N) since we have a loop from N−2 and we use the precalculated
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# values of our dynamic programming table to calculate the current value in the table
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# which is a constant time operation.
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#
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# Space Complexity: O(1) since we are not using a table to store our values.
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# Simply using two variables will suffice for our calculations.
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#
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def rob(nums)
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dp = Array.new(nums.size + 1)
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(nums.size + 1).times do |i|
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dp[i] = if i == 0
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0
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elsif i == 1
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nums[0]
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else
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[dp[i - 2] + nums[i - 1], dp[i - 1]].max
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end
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end
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dp[-1]
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end
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nums = [1, 2, 3, 1]
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puts rob(nums)
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# Output: 4
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nums = [2, 7, 9, 3, 1]
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puts rob(nums)
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# Output: 12
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