diff --git a/problem-1.py b/problem-1.py
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index 00000000..5b5d0709
--- /dev/null
+++ b/problem-1.py
@@ -0,0 +1,26 @@
+# LEETCODE PROBLEM 740 DELETE AND EARN
+# TIME COMPLEXITY: O(N+ max(N)) where N denotes the number of elements in an array
+# SPACE COMPLEXITY:O(max(n))
+# Any problem you faced while coding this: I am having difficulty in optimizing this code to a better solution
+
+
+class Solution(object):
+    def deleteAndEarn(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        max_val=0
+        for i in nums:
+            max_val=max(max_val,i)
+        arr=[0]*(max_val+1)
+        for i in nums:
+            arr[i]+=i
+        prev=arr[0]
+        curr=max(arr[0],arr[1])
+
+        for i in range(2,max_val+1):
+            temp=curr
+            curr=max(curr,arr[i]+prev)
+            prev=temp
+        return curr
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diff --git a/problem-2.py b/problem-2.py
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index 00000000..c18b003d
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+++ b/problem-2.py
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+# LEETCODE PROBLEM 931. MINIMUM FALLING PATH SUM
+# TIME COMPLEXITY: O(N*N) where N denotes the number of cols/rows present
+# SPACE COMPLEXITY: O(N) 
+# Any problem you faced while coding this: I had some hiccup while writing the 1D array code mainly on how the optimization is looking and 
+# while thinking of the logic I was pretty clear on it but it took me some time while getting it converted to code
+
+class Solution(object):
+    def minFallingPathSum(self, matrix):
+        """
+        :type matrix: List[List[int]]
+        :rtype: int
+        """
+        #Approach 1
+        # m=len(matrix)
+        # n=len(matrix[0])
+        # dp=[[0]*n for _ in range(m)]
+        # for i in range(n):
+        #     dp[0][i]=matrix[0][i]
+        # for i in range(1,m):
+        #     for j in range(n):
+        #         if j==0:
+        #             dp[i][j]=matrix[i][j]+min(dp[i-1][j],dp[i-1][j+1])
+        #         elif j==n-1:
+        #             dp[i][j]=matrix[i][j]+min(dp[i-1][j],dp[i-1][j-1])
+        #         else:
+        #             dp[i][j]=matrix[i][j]+min(dp[i-1][j],min(dp[i-1][j+1],dp[i-1][j-1]))
+        # return min(dp[m-1])
+
+        #m=len(matrix)
+
+        #Approach 2
+        n=len(matrix)
+        dp=list(matrix[0])
+        for i in range(1,n):
+            new_dp=[0]*n
+            for j in range(n):
+                if j==0:
+                    new_dp[j]=matrix[i][j]+min(dp[j],dp[j+1])
+                elif j==n-1:
+                    new_dp[j]=matrix[i][j]+min(dp[j],dp[j-1])
+                else:
+                    new_dp[j]=matrix[i][j]+min(dp[j],min(dp[j+1],dp[j-1]))
+            dp=new_dp
+        return min(dp)
+