House Robber: Dynamic Programming

dynamic_programming/house_robber.rb

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+# 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.
+
+#
+# Approach 1: Dynamic Programming
+#
+
+# Complexity Analysis
+#
+# Time Complexity: O(N) since we process at most N recursive calls, thanks to
+# caching, and during each of these calls, we make an O(1) computation which is
+# simply making two other recursive calls, finding their maximum, and populating
+# the cache based on that.
+#
+# Space Complexity: O(N) which is occupied by the cache and also by the recursion stack
+
+def rob(nums, i = nums.length - 1)
+  return 0 if i < 0
+
+  [rob(nums, i - 2) + nums[i], rob(nums, i - 1)].max
+end
+
+nums = [1, 2, 3, 1]
+puts rob(nums)
+# Output: 4
+
+nums = [2, 7, 9, 3, 1]
+puts rob(nums)
+# Output: 12