Merge pull request #178 from TheAlgorithms/house-robber

House Robber: Dynamic Programming approach

DIRECTORY.md

@@ -86,6 +86,7 @@
   * [Coin Change](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/coin_change.rb)
   * [Count Sorted Vowel Strings](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/count_sorted_vowel_strings.rb)
   * [Fibonacci](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/fibonacci.rb)
+  * [House Robber](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/house_robber.rb)
   * [Pascal Triangle Ii](https://github.com/TheAlgorithms/Ruby/blob/master/dynamic_programming/pascal_triangle_ii.rb)
 
 ## Maths

dynamic_programming/house_robber.rb

@@ -0,0 +1,87 @@
+# 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
+
+#
+# Approach 2: Optimized Dynamic Programming
+#
+
+# Time Complexity
+#
+# Time Complexity: O(N) since we have a loop from N−2 and we use the precalculated
+# values of our dynamic programming table to calculate the current value in the table
+# which is a constant time operation.
+#
+# Space Complexity: O(1) since we are not using a table to store our values.
+# Simply using two variables will suffice for our calculations.
+#
+
+def rob(nums)
+  dp = Array.new(nums.size + 1)
+
+  (nums.size + 1).times do |i|
+    dp[i] = if i == 0
+              0
+            elsif i == 1
+              nums[0]
+            else
+              [dp[i - 2] + nums[i - 1], dp[i - 1]].max
+            end
+  end
+
+  dp[-1]
+end
+
+nums = [1, 2, 3, 1]
+puts rob(nums)
+# Output: 4
+
+nums = [2, 7, 9, 3, 1]
+puts rob(nums)
+# Output: 12