diff --git a/Dynamic Programming/minDelSum.py b/Dynamic Programming/minDelSum.py
new file mode 100644
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+++ b/Dynamic Programming/minDelSum.py	
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+"""
+Problem: 712. Minimum ASCII Delete Sum for Two Strings
+
+Given two strings s1 and s2, return the lowest ASCII 
+sum of deleted characters to make two strings equal.
+
+Platform: LeetCode
+Link: https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/
+Performance: 85 % faster, 75 % less memory (relative)
+Easy to read
+"""
+
+class Solution:
+    def minimumDeleteSum(self, s1: str, s2: str) -> int:
+        '''
+        Problem statement: the lowest ASCII sum
+        of deleted elements to obtain LCS
+        = find the Highest ASCII sum of the LCS
+        And let's call it 'x'
+        Then our answer is sum of ascii chars of s1 and s2
+        minus 2 * x
+        '''
+        
+        # Redifined LCS to obtain highest sum ASCII LCS
+        def lcs(X , Y):
+            # find the length of the strings
+            m = len(X)
+            n = len(Y)
+
+            # declaring the array for storing the dp values
+            L = [[None]*(n+1) for i in range(m+1)]
+
+            """Following steps build L[m+1][n+1] in bottom up fashion
+            Note: L[i][j] contains maximum ascii sum of LCS of X[0..i-1]
+            and Y[0..j-1]"""
+            for i in range(m+1):
+                for j in range(n+1):
+                    
+                    # Base case
+                    if i == 0 or j == 0 :
+                        L[i][j] = 0
+                    
+                    # If two characters are matching then
+                    # The formed LCS can have heigst sum
+                    # Or We have previously obtained the 
+                    # Heigest sum of ascii characters
+                    # We need the max only
+                    elif X[i-1] == Y[j-1]:
+                        L[i][j] = max(L[i-1][j-1]+ord(X[i-1]), L[i-1][j] , L[i][j-1])
+                    
+                    # Else get the max without the character
+                    else:
+                        L[i][j] = max(L[i-1][j] , L[i][j-1])
+            return L[m][n]
+        
+        # Find the sum of all the ASCII values of 's1' chars
+        r1 = 0
+        for i in s1:
+            r1 += ord(i)
+        
+        # Find the sum of all the ASCII values of 's2' chars
+        r2 = 0
+        for i in s2:
+            r2 += ord(i)
+        
+        # Return the answer
+        return r1 + r2 - 2 * lcs(s1, s2)
+