DP 1 done

Problem1.py

@@ -0,0 +1,23 @@
+#Coin Change
+
+import math
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        prevArr = [math.inf] * (amount+1)
+        length = len(prevArr)
+        for coin in coins:
+            for i in range(0,length):
+                if i == 0:
+                    prevArr[i] = 0
+                    continue
+                if coin > i:
+                    #noChose, do nothing, already should have min in that index on array, if there is a possible path
+                    continue
+                else:
+                    #compare between choose and no choose option, in case the possible ways will be 1(for choosing) + min possible ways for achieving remaining value
+                    minPossible = min(prevArr[i], (1+prevArr[i-coin]))
+                    prevArr[i] = minPossible
+        if prevArr[length-1] == math.inf:
+            return -1
+        else:
+            return prevArr[length-1]

Problem2.py

@@ -0,0 +1,24 @@
+#House Robber
+
+#keep on calculating the max possible sum till current house by comparing 2 options, if we choose current then add that with the max till the house
+# 2 before and comapre it with just no choose category, in which you are just comparing max till prev house. For current house the max will become the
+# max of these 2 choices
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        totalHouses = len(nums)
+        if totalHouses == 1:
+            return nums[0]
+
+        if totalHouses == 2:
+            return max(nums[0], nums[1])
+
+        # we just need max possible count for last 2 houses
+        prevHouseMaxCount = max(nums[0], nums[1])
+        prevPrev = nums[0]
+
+        for i in range(2,totalHouses):
+            choseCurrentMaxCount = nums[i] + prevPrev
+            prevPrev = prevHouseMaxCount
+            prevHouseMaxCount = max(choseCurrentMaxCount, prevHouseMaxCount)
+
+        return prevHouseMaxCount