Done Dp-1

coin-change.py

@@ -0,0 +1,26 @@
+''' 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
+
+   Your code here along with comments explaining your approach
+
+   Approach : calculating min no of coins required at each amount
+'''
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        if amount == 0:
+            return 0
+        dp = [[amount+1 for _ in range(amount+1)] for _ in range(len(coins)+1)]
+        for i in range(len(coins)+1):
+            dp[i][0] = 0 
+        for i in range(1,len(coins)+1):
+            for j in range(1,(amount+1)):
+                if coins[i-1] > j:
+                    dp[i][j] = dp[i-1][j]
+                else:
+                    dp[i][j] = min(dp[i-1][j], dp[i][j-coins[i-1]] + 1)
+        if dp[i][j] == (amount+1):
+            return -1
+        else:
+            return dp[i][j]

house-robber.py

@@ -0,0 +1,19 @@
+''' Time Complexity : O(n) 
+    Space Complexity : O(n) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Your code here along with comments explaining your approach
+'''
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        n = len(nums)
+        if n == 1:
+            return nums[0]
+        if n == 2:
+            return max(nums)
+        dp = [0] * (n)
+        for i in range(n):
+            dp[i] = max(dp[i-1],dp[i-2]+nums[i])
+        return dp[n-1]