Done DP-2

coin-change-2.py

@@ -0,0 +1,42 @@
+#-------------Solution 1 : 2D Matrix--------
+''' Time Complexity : O(m*n)
+    Space Complexity : O(m*n)
+
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+'''    
+
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        n = len(coins)
+        dp = [[0 for _ in range(amount+1)] for _ in range(n+1)]
+        dp[0][0] = 1
+        for i in range(1,n+1):
+            for j in range(amount+1):
+                coin = coins[i-1]
+                if coin > j:
+                    dp[i][j] = dp[i-1][j]
+                else:
+                    dp[i][j] = dp[i-1][j] + dp[i][j-coin]
+        return dp[n][amount]
+
+
+#-------------Solution 2 : 1D Array--------
+''' Time Complexity : O(m*n)
+    Space Complexity : O(m)
+
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+'''    
+
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        n = len(coins)
+        dp = [0 for _ in range(amount+1)]
+        dp[0] = 1
+        for i in range(0,n):
+            for j in range(amount+1):
+                coin = coins[i]
+                if coin <= j:
+                    dp[j] = dp[j] + dp[j-coin]
+        return dp[amount]

paint-house.py

@@ -0,0 +1,24 @@
+#-------------Solution 1 : 2D Matrix--------
+''' Time Complexity : O(m*n)
+    Space Complexity : O(m*n)
+
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+'''    
+class Solution:
+    def minCost(self, costs: List[List[int]]) -> int:
+        rows= len(costs)
+        cols = len(costs[0])
+        dp = [[0 for _ in range(cols)] for _ in range(rows)]
+        for i in range(rows):
+            for j in range(cols):
+                if i == 0 :
+                    dp[i][j] = costs[i][j]
+                else:
+                    if j == 0:
+                        dp[i][j] = min(costs[i][j]+dp[i-1][1], costs[i][j]+dp[i-1][2])
+                    if j == 1:
+                        dp[i][j] = min(costs[i][j]+dp[i-1][0], costs[i][j]+dp[i-1][2])
+                    if j == 2:
+                        dp[i][j] = min(costs[i][j]+dp[i-1][0], costs[i][j]+dp[i-1][1])
+        return min(dp[rows-1][0],dp[rows-1][1],dp[rows-1][2])