diff --git a/coin-change1.py b/coin-change1.py
new file mode 100644
index 00000000..308970f4
--- /dev/null
+++ b/coin-change1.py
@@ -0,0 +1,33 @@
+# Take a 1D array DP upto the length of amount+1 and assign it to infinity. Assign the first element to 0.
+# For each coin in coins, and each amount from 1 to amount, if the coins are lesser than or qual to the amount,
+# then update the index with the minimun of the current index or the element at the i-denomination index +1.
+# At last, return -1 if the element is greater than the amount else return the element at the last column.
+
+# Time complexity: O(n*m) n= number of coins, m = amount
+# Space complexity: O(m), m = amount
+
+
+
+
+class Solution(object):
+    def coinChange(self, coins, amount):
+        """
+        :type coins: List[int]
+        :type amount: int
+        :rtype: int
+        """
+        rows = len(coins)+1
+        cols = amount+1
+        dp=[None for i in range(cols)]
+        dp[0]=0
+        for i in range(1,cols):
+            dp[i]=float('inf')
+       
+        for i,c in enumerate(coins):
+            for j in range(1,cols):
+                if c<=j:
+                    dp[j]=min(dp[j],dp[j-c]+1)
+        if dp[cols-1] >amount:
+            return -1
+        else:
+            return dp[cols-1]
diff --git a/house-robber.py b/house-robber.py
new file mode 100644
index 00000000..e843643a
--- /dev/null
+++ b/house-robber.py
@@ -0,0 +1,31 @@
+# Take 2 variables, prev2 and prev where prev2 is the first elememt and prev is the max between 
+# the first and the second element for each remaining elements in nums, check the max between
+#  prev2+current element and prev elememt, the max is the current element. Prev2 becomes prev1 
+#  and prev1 becomes current, updat the new current
+
+#  Time complexity: O(n)
+#  Space complexity: O(1)
+
+
+
+
+class Solution(object):
+    def rob(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        l=len(nums)
+        if l==1:
+            return nums[0]
+        if l==2:
+            return max(nums[0],nums[1])
+        prev2=nums[0]
+        prev1=max(nums[1],nums[0])
+        for i in range(2,l):
+            current = max(prev1,prev2+nums[i])
+            prev2=prev1
+            prev1=current
+        return current
+
+        
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