Done DP-3

Problem1.py

@@ -0,0 +1,25 @@
+#Delete and Earn
+
+#time complexity -> On
+# space complexity -> On
+import math
+class Solution:
+    def deleteAndEarn(self, nums: List[int]) -> int:
+        if len(nums) == 1:
+            return nums[0]
+        # We'll have to convert the given array to house robber pattern
+        # get the array of length 10^4 as that is the max number possible
+        numsArr = [0] * 10001
+        maxNum = 0
+        minNum = math.inf
+
+        for num in nums:
+            maxNum = max(maxNum,num)
+            minNum = min(minNum,num)
+            numsArr[num] +=num
+
+        #Now use houseroober method, i.e for each index try to maximise the sum
+        numsArr[minNum+1] = max(numsArr[minNum], numsArr[minNum+1])
+        for i in range(minNum+2,maxNum+1):
+            numsArr[i] = max(numsArr[i-1],(numsArr[i]+numsArr[i-2]))
+        return numsArr[maxNum]

Problem2.py

@@ -0,0 +1,28 @@
+#Minimum Falling Path Sum
+
+#Time complexity O(mxn)
+#Space complexity O(m) where m is the number of columns in the matrix
+#The idea is to calculate for each index starting from row 2 what's the minimum sum we can achieve in that index. to minimise the sum at each index the logic will be
+# sum of price at that index and the minimimum of the three/two from the top row. Dependinging if it's a middle index or at the edges of the matrix
+#This will give the minimum sum at each index that is possible. In the end we'll just find in all the index which has the minimimu sum and return that
+class Solution:
+    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
+        columns = len(matrix[0])
+        rows = len(matrix)
+        psblValuesAtEachColumn = [0]*columns
+        for j in range(0,columns):
+            psblValuesAtEachColumn[j] = matrix[0][j]
+
+        for i in range (1, rows):
+            previousTopVal = None
+            for j in range(0,columns):
+                currentTopVal = psblValuesAtEachColumn[j]
+                if previousTopVal == None:
+                    psblValuesAtEachColumn[j] = matrix[i][j] + min(psblValuesAtEachColumn[j],psblValuesAtEachColumn[j+1])
+                    previousTopVal = currentTopVal
+                elif j==columns-1:
+                    psblValuesAtEachColumn[j] = matrix[i][j] + min(previousTopVal, psblValuesAtEachColumn[j])
+                else:
+                    psblValuesAtEachColumn[j] = matrix[i][j] + min(previousTopVal, psblValuesAtEachColumn[j],psblValuesAtEachColumn[j+1])
+                    previousTopVal = currentTopVal
+        return min(psblValuesAtEachColumn)