add solution: house-robber

Author: dusunax

house-robber/dusunax.py

@@ -0,0 +1,44 @@
+'''
+# Leetcode 198. House Robber
+
+use **dynamic programming** to solve this problem. (bottom-up approach) 🧩
+
+choose bottom-up approach for less space complexity.
+
+## DP relation
+
+```
+dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+```
+
+- **dp[i - 1]:** skip and take the value from the previous house
+- **dp[i - 2]:** rob the current house, add its value to the maximum money from two houses before
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the list just once to calculate the maximum money. = O(n)
+
+### SC is O(n):
+- using a list to store the maximum money at each house. = O(n)
+
+'''
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) == 1:
+            return nums[0]
+
+        dp = [0] * len(nums)
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+
+        for i in range(2, len(nums)):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+            
+        return dp[-1]