Done DP 2

Problem1.py

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+#Paint House
+
+#Time complexity = O(n*3) so basically On
+#Space complexity O1
+#for each new house check the minimium to cost with each of the colors, by choosing the minimum possible till previous house from the remaining colors
+
+class Solution:
+    def minCost(self, costs: List[List[int]]) -> int:
+        red = costs[0][0]
+        blue = costs[0][1]
+        green = costs[0][2]
+
+        for i in range(1, len(costs)):
+            oldRed = red
+            red = costs[i][0] + min(blue, green)
+
+            oldBlue = blue
+            blue = costs[i][1] + min(oldRed, green)
+
+            green = costs[i][2] + min(oldRed, oldBlue)
+
+        return min(red,min(green,blue))

Problem2.py

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+#Coin Change II
+
+#Time complexity = O(n+1*m+1) where n is length of coins and m is the amount
+#Space complexity O(m+1)
+
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        arr = [0]*(amount+1)
+        arr[0] = 1
+
+        for coin in coins:
+            for j in range(0,amount+1):
+                noChoose = arr[j]
+                choose = 0
+                if j >=coin:
+                    choose = arr[j-coin]
+                arr[j] = choose + noChoose
+        return arr[amount]