diff --git a/coin-change-II.py b/coin-change-II.py
new file mode 100644
index 00000000..d6cbb3c3
--- /dev/null
+++ b/coin-change-II.py
@@ -0,0 +1,32 @@
+# Time Complexity : O(n^2)
+# Space Complexity : O(n^2)
+# Did this code successfully run on Leetcode : yes was able to solve the question
+# Any problem you faced while coding this : No
+
+class Solution:
+
+    def change(self, amount, coins):
+        n = len(coins)
+        m = amount
+
+        dp = [0 for each in range(1 + m)]
+        dp[0] = 1
+
+        for i in range(1, n + 1):
+            for j in range(1, m + 1):
+                if coins[i - 1] <= j:
+                    dp[j] = dp[j] + dp[j - coins[i - 1]]
+        
+        return dp[m]
+
+
+s = Solution()
+amount = 5
+coins = [1,2,5]
+print(s.change(amount, coins))
+amount = 3
+coins = [2]
+print(s.change(amount, coins))
+amount = 10
+coins = [10]
+print(s.change(amount, coins))
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diff --git a/paint-house.py b/paint-house.py
new file mode 100644
index 00000000..ade22871
--- /dev/null
+++ b/paint-house.py
@@ -0,0 +1,43 @@
+# Time Complexity : O(n^2)
+# Space Complexity : O(n^2)
+# Did this code successfully run on Leetcode : yes was able to solve the question
+# Any problem you faced while coding this : No
+
+class Solution:
+    # This solution takes the DP route, here we first check for the base case if the house is present to paint.
+    # second we put the minmum cost as -inf but this can be -1 as cost cannot be in -ve.
+    # third we iterate through the paint or color. 
+    # fourth we check if the current color is same as the previous color as we recursively pass the previously used color.
+    # fifth we will make the dp call with the next house and provide the prev color and also add the current color of current house.
+    # here we also take the minimum of the cost
+    # in end we send the min cost.
+    # the time complexity for this will be: O(n)
+    # the space complexity for this function will be: O(n)
+    def minCost(self, costs):
+        def dp(i, prev):
+            # base case
+            if i >= len(costs):
+                return 0
+            min_cost = float('inf')
+            for c in range(3):
+                if prev != c:
+                    min_cost = min(min_cost, dp(i + 1, c) + costs[i][c])
+            return min_cost
+        return dp(0, -1)
+    
+    # here we use the same costs matrices to compute the minimum cost for the houses to be painted
+    # Time complexity will be O(n)
+    # Space complexity will be O(1)
+    def minCost(self, costs):
+        for i in range(1, len(costs)):
+            costs[i][0] += min(costs[i-1][1], costs[i-1][2])
+            costs[i][1] += min(costs[i-1][0], costs[i-1][2])
+            costs[i][2] += min(costs[i-1][1], costs[i-1][1])
+        return min(costs[-1])
+
+s = Solution()
+costs = [[17,2,17],[16,16,5],[14,3,19]]
+print(s.minCost(costs))
+costs = [[7,6,2]]
+print(s.minCost(costs))
+        
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