completed dp2

CoinChange2.py

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+# Time Complexity : O(m*n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : dp[i] is number of ways to make amount i.
+# dp[i] = dp[i - coin]
+
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        dp = [0] * (amount + 1)
+        dp[0] = 1
+
+        for coin in coins:
+            for a in range(coin, amount + 1):
+                dp[a] += dp[a-coin]
+
+        return dp[amount]
+

PaintHouse.py

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+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : We work backwards from the last house.
+# For each house, update the color costs based on the min of the other two colors from the next house.
+
+class Solution:
+    def minCost(self, costs: List[List[int]]) -> int:
+        num_houses = len(costs)
+        num_colors = len(costs[0])
+
+        costR = costs[num_houses-1][0]
+        costB = costs[num_houses-1][1]
+        costG = costs[num_houses-1][2]
+
+
+        for i in range(num_houses - 2, -1, -1):
+            tempR = costR
+            tempB = costB 
+
+            costR = costs[i][0] + min(costB, costG)
+            costB = costs[i][1] + min(tempR, costG)
+            costG = costs[i][2] + min(tempR, tempB)
+
+        return min(costR, costB, costG)       
+